Thermodynamics - Physics at Oregon State University
Thermodynamics - Physics at Oregon State University
Thermodynamics - Physics at Oregon State University
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<strong>Thermodynamics</strong><br />
Henri J.F. Jansen<br />
Department of <strong>Physics</strong><br />
<strong>Oregon</strong> St<strong>at</strong>e <strong>University</strong><br />
August 19, 2010
Contents<br />
PART I. <strong>Thermodynamics</strong> Fundamentals 1<br />
1 Basic <strong>Thermodynamics</strong>. 3<br />
1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4<br />
1.2 Some definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . 7<br />
1.3 Zeroth Law of <strong>Thermodynamics</strong>. . . . . . . . . . . . . . . . . . 12<br />
1.4 First law: Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . 13<br />
1.5 Second law: Entropy. . . . . . . . . . . . . . . . . . . . . . . . . 18<br />
1.6 Third law of thermodynamics. . . . . . . . . . . . . . . . . . . . . 31<br />
1.7 Ideal gas and temper<strong>at</strong>ure. . . . . . . . . . . . . . . . . . . . . . 32<br />
1.8 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 36<br />
1.9 Problems for chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . 38<br />
2 Thermodynamic potentials and 49<br />
2.1 Internal energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . 51<br />
2.2 Free energies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57<br />
2.3 Euler and Gibbs-Duhem rel<strong>at</strong>ions. . . . . . . . . . . . . . . . . . 61<br />
2.4 Maxwell rel<strong>at</strong>ions. . . . . . . . . . . . . . . . . . . . . . . . . . . 64<br />
2.5 Response functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 65<br />
2.6 Rel<strong>at</strong>ions between partial deriv<strong>at</strong>ives. . . . . . . . . . . . . . . . . 68<br />
2.7 Conditions for equilibrium. . . . . . . . . . . . . . . . . . . . . . 72<br />
2.8 Stability requirements on other free energies. . . . . . . . . . . . 78<br />
2.9 A magnetic puzzle. . . . . . . . . . . . . . . . . . . . . . . . . . . 80<br />
2.10 Role of fluctu<strong>at</strong>ions. . . . . . . . . . . . . . . . . . . . . . . . . . 85<br />
2.11 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 94<br />
2.12 Problems for chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . 96<br />
3 Phase transitions. 107<br />
3.1 Phase diagrams. . . . . . . . . . . . . . . . . . . . . . . . . . . 108<br />
3.2 Clausius-Clapeyron rel<strong>at</strong>ion. . . . . . . . . . . . . . . . . . . . . . 116<br />
3.3 Multi-phase boundaries. . . . . . . . . . . . . . . . . . . . . . . . 121<br />
3.4 Binary phase diagram. . . . . . . . . . . . . . . . . . . . . . . . . 123<br />
3.5 Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e. . . . . . . . . . . . . . . . . . . 126<br />
3.6 Spinodal decomposition. . . . . . . . . . . . . . . . . . . . . . . . 138<br />
III
IV CONTENTS<br />
3.7 Generaliz<strong>at</strong>ions of the van der Waals equ<strong>at</strong>ion. . . . . . . . . . . 142<br />
3.8 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 143<br />
3.9 Problems for chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . 144<br />
PART II. <strong>Thermodynamics</strong> Advanced Topics 152<br />
4 Landau-Ginzburg theory. 153<br />
4.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154<br />
4.2 Order parameters. . . . . . . . . . . . . . . . . . . . . . . . . . . 156<br />
4.3 Landau theory of phase transitions. . . . . . . . . . . . . . . . . . 162<br />
4.4 Case one: a second order phase transition. . . . . . . . . . . . . . 164<br />
4.5 Order of a phase transition. . . . . . . . . . . . . . . . . . . . . . 169<br />
4.6 Second case: first order transition. . . . . . . . . . . . . . . . . . 170<br />
4.7 Fluctu<strong>at</strong>ions and Ginzburg-Landau theory. . . . . . . . . . . . . 176<br />
4.8 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 183<br />
4.9 Problems for chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . 184<br />
5 Critical exponents. 193<br />
5.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194<br />
5.2 Mean field theory. . . . . . . . . . . . . . . . . . . . . . . . . . . 200<br />
5.3 Model free energy near a critical point. . . . . . . . . . . . . . . . 208<br />
5.4 Consequences of scaling. . . . . . . . . . . . . . . . . . . . . . . . 211<br />
5.5 Scaling of the pair correl<strong>at</strong>ion function. . . . . . . . . . . . . . . 217<br />
5.6 Hyper-scaling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218<br />
5.7 Validity of Ginzburg-Landau theory. . . . . . . . . . . . . . . . . 219<br />
5.8 Scaling of transport properties. . . . . . . . . . . . . . . . . . . . 221<br />
5.9 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 225<br />
5.10 Problems for chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . 226<br />
6 Transport in <strong>Thermodynamics</strong>. 229<br />
6.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230<br />
6.2 Some thermo-electric phenomena. . . . . . . . . . . . . . . . . . . 233<br />
6.3 Non-equilibrium <strong>Thermodynamics</strong>. . . . . . . . . . . . . . . . . . 237<br />
6.4 Transport equ<strong>at</strong>ions. . . . . . . . . . . . . . . . . . . . . . . . . . 240<br />
6.5 Macroscopic versus microscopic. . . . . . . . . . . . . . . . . . . . 245<br />
6.6 Thermo-electric effects. . . . . . . . . . . . . . . . . . . . . . . . 252<br />
6.7 Extra questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 262<br />
6.8 Problems for chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . 262<br />
7 Correl<strong>at</strong>ion Functions. 265<br />
7.1 Description of correl<strong>at</strong>ed fluctu<strong>at</strong>ions. . . . . . . . . . . . . . . . 265<br />
7.2 M<strong>at</strong>hem<strong>at</strong>ical functions for correl<strong>at</strong>ions. . . . . . . . . . . . . . . 270<br />
7.3 Energy consider<strong>at</strong>ions. . . . . . . . . . . . . . . . . . . . . . . . . 275<br />
PART III. Additional M<strong>at</strong>erial 279
CONTENTS V<br />
A Questions submitted by students. 281<br />
A.1 Questions for chapter 1. . . . . . . . . . . . . . . . . . . . . . . 281<br />
A.2 Questions for chapter 2. . . . . . . . . . . . . . . . . . . . . . . 284<br />
A.3 Questions for chapter 3. . . . . . . . . . . . . . . . . . . . . . . 287<br />
A.4 Questions for chapter 4. . . . . . . . . . . . . . . . . . . . . . . 290<br />
A.5 Questions for chapter 5. . . . . . . . . . . . . . . . . . . . . . . 292<br />
B Summaries submitted by students. 295<br />
B.1 Summaries for chapter 1. . . . . . . . . . . . . . . . . . . . . . 295<br />
B.2 Summaries for chapter 2. . . . . . . . . . . . . . . . . . . . . . 297<br />
B.3 Summaries for chapter 3. . . . . . . . . . . . . . . . . . . . . . 299<br />
B.4 Summaries for chapter 4. . . . . . . . . . . . . . . . . . . . . . 300<br />
B.5 Summaries for chapter 5. . . . . . . . . . . . . . . . . . . . . . 302<br />
C Solutions to selected problems. 305<br />
C.1 Solutions for chapter 1. . . . . . . . . . . . . . . . . . . . . . . 305<br />
C.2 Solutions for chapter 2. . . . . . . . . . . . . . . . . . . . . . . 321<br />
C.3 Solutions for chapter 3. . . . . . . . . . . . . . . . . . . . . . . 343<br />
C.4 Solutions for chapter 4. . . . . . . . . . . . . . . . . . . . . . . 353
VI CONTENTS
List of Figures<br />
1.1 Carnot cycle in PV diagram . . . . . . . . . . . . . . . . . . . 20<br />
1.2 Schem<strong>at</strong>ics of a Carnot engine. . . . . . . . . . . . . . . . . . 21<br />
1.3 Two engines feeding eachother. . . . . . . . . . . . . . . . . . 22<br />
1.4 Two Carnot engines in series. . . . . . . . . . . . . . . . . . . 25<br />
2.1 Container with piston as internal divider. . . . . . . . . . . . 53<br />
2.2 Container where the internal degree of freedom becomes external<br />
and hence can do work. . . . . . . . . . . . . . . . . . . 54<br />
3.1 Model phase diagram for a simple model system. . . . . . . . 111<br />
3.2 Phase diagram for solid Ce. . . . . . . . . . . . . . . . . . . . . 112<br />
3.3 Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure,<br />
wrong picture. . . . . . . . . . . . . . . . . . . . . . . . . 113<br />
3.4 Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure,<br />
correct picture. . . . . . . . . . . . . . . . . . . . . . . . 114<br />
3.5 Volume versus pressure <strong>at</strong> the phase transition. . . . . . . . . 114<br />
3.6 Model phase diagram for a simple model system in V-T space.115<br />
3.7 Model phase diagram for a simple model system in p-V space.116<br />
3.8 Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure<br />
for both phases. . . . . . . . . . . . . . . . . . . . . . . . 118<br />
3.9 Typical binary phase diagram with regions L=liquid, A(B)=B<br />
dissolved in A, and B(A)=A dissolved in B. . . . . . . . . . . 124<br />
3.10 Solidific<strong>at</strong>ion in a reversible process. . . . . . . . . . . . . . . 125<br />
3.11 Typical binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, with regions L=liquid, A(B)=B dissolved in A, B(A)=A<br />
dissolved in B, and AB= AB with either A or B dissolved. . . 126<br />
3.12 Typical binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, where the intermedi<strong>at</strong>e region is too small to discern<br />
from a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127<br />
3.13 Impossible binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, where the intermedi<strong>at</strong>e region is too small to discern<br />
from a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127<br />
3.14 Graphical solution of equ<strong>at</strong>ion 3.24. . . . . . . . . . . . . . . . 131<br />
3.15 p-V curves in the van der Waals model. . . . . . . . . . . . . . 133<br />
VII
VIII LIST OF FIGURES<br />
3.16 p-V curves in the van der Waals model with neg<strong>at</strong>ive values<br />
of the pressure. . . . . . . . . . . . . . . . . . . . . . . . . . . . 133<br />
3.17 p-V curve in the van der Waals model with areas corresponding<br />
to energies. . . . . . . . . . . . . . . . . . . . . . . . . . . . 135<br />
3.18 Unstable and meta-stable regions in the van der Waals p-V<br />
diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138<br />
3.19 Energy versus volume showing th<strong>at</strong> decomposition lowers the<br />
energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141<br />
4.1 He<strong>at</strong> capacity across the phase transition in the van der Waals<br />
model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155<br />
4.2 He<strong>at</strong> capacity across the phase transition in an experiment. . 155<br />
4.3 Continuity of phase transition around critical point in p-T<br />
plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158<br />
4.4 Continuity of phase around singular point. . . . . . . . . . . . 159<br />
4.5 Continuity of phase transition around critical point in H-T<br />
plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160<br />
4.6 Magnetiz<strong>at</strong>ion versus temper<strong>at</strong>ure. . . . . . . . . . . . . . . . . 165<br />
4.7 Forms of the Helmholtz free energy. . . . . . . . . . . . . . . . 166<br />
4.8 Entropy near the critical temper<strong>at</strong>ure. . . . . . . . . . . . . . 167<br />
4.9 Specific he<strong>at</strong> near the critical temper<strong>at</strong>ure. . . . . . . . . . . . 168<br />
4.10 Magnetic susceptibility near the critical temper<strong>at</strong>ure. . . . . 169<br />
4.11 Three possible forms of the Helmholtz free energy in case 2. 171<br />
4.12 Values for m corresponding to a minimum in the free energy. 172<br />
4.13 Magnetiz<strong>at</strong>ion as a function of temper<strong>at</strong>ure. . . . . . . . . . . 173<br />
4.14 Hysteresis loop. . . . . . . . . . . . . . . . . . . . . . . . . . . . 174<br />
4.15 Critical behavior in first order phase transition. . . . . . . . . 176<br />
6.1 Essential geometry of a thermocouple. . . . . . . . . . . . . . . 234<br />
C.1 m versus T-H . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363<br />
C.2 Figure 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
INTRODUCTION IX<br />
Introduction.<br />
<strong>Thermodynamics</strong>??? Why? Wh<strong>at</strong>? How? When? Where? Many questions<br />
to ask, so we will start with the first one. A frequent argument against studying<br />
thermodynamics is th<strong>at</strong> we do not have to do this, since everything follows from<br />
st<strong>at</strong>istical mechanics. In principle, this is, of course, true. The argument, however,<br />
assumes th<strong>at</strong> we know the exact description of a system on the microscopic<br />
scale, and th<strong>at</strong> we can calcul<strong>at</strong>e the partition function. In practice, we can only<br />
calcul<strong>at</strong>e the partition function for a few simple cases, and in all other cases we<br />
need to make serious approxim<strong>at</strong>ions. This is where thermodynamics plays an<br />
invaluable role. In thermodynamics we derive basic equ<strong>at</strong>ions th<strong>at</strong> all systems<br />
have to obey, and we derive these equ<strong>at</strong>ions from a few basic principles. In<br />
this sense thermodynamics is a meta-theory, a theory of theories, very similar<br />
to wh<strong>at</strong> we see in a study of non-linear dynamics. <strong>Thermodynamics</strong> gives us<br />
a framework for the results derived in st<strong>at</strong>istical mechanics, and allows us to<br />
check if approxim<strong>at</strong>ions made in st<strong>at</strong>istical mechanical calcul<strong>at</strong>ions viol<strong>at</strong>e some<br />
of these basic results. For example, if the calcul<strong>at</strong>ed he<strong>at</strong> capacity in st<strong>at</strong>istical<br />
mechanics is neg<strong>at</strong>ive, we know we have a problem!<br />
There are some semantic issues with the words thermodynamics and st<strong>at</strong>istical<br />
mechanics. In the English speaking part of the world thermodynamics<br />
is often seen as a subset of the field of st<strong>at</strong>istical mechanics. In the German<br />
world it is often seen as an a different field from st<strong>at</strong>istical mechanics. I take<br />
the l<strong>at</strong>ter view. <strong>Thermodynamics</strong> is the field of physics describing thermal effects<br />
in m<strong>at</strong>ter in a manner which is independent of the microscopic details of<br />
the m<strong>at</strong>erial. St<strong>at</strong>istical mechanics starts <strong>at</strong> a microscopic model and derives<br />
conclusions for the macroscopic world, based on these microscopic details. In<br />
this course we discuss thermodynamics, we present equ<strong>at</strong>ions and conclusions<br />
which are independent of the microscopic details.<br />
<strong>Thermodynamics</strong> also gives us a language for the description of experimental<br />
results. It defines observable quantities, especially in the form of response<br />
functions. It gives the definitions of critical exponents and transport properties.<br />
It allows analyzing experimental d<strong>at</strong>a in the framework of simple models, like<br />
equ<strong>at</strong>ions of st<strong>at</strong>e. It provides a framework to organize experimental d<strong>at</strong>a. To<br />
say th<strong>at</strong> we do not need this is quite arrogant, and assumes th<strong>at</strong> if you cannot<br />
follow the (often very complic<strong>at</strong>ed) deriv<strong>at</strong>ions in st<strong>at</strong>istical mechanics you<br />
might as well give up. <strong>Thermodynamics</strong> is the meeting ground of experimenters<br />
and theorists. It gives the common language needed to connect experimental<br />
d<strong>at</strong>a and theoretical results.<br />
Classical mechanics has its limits of validity, and we need rel<strong>at</strong>ivity and/or<br />
quantum mechanics to extend the domain of this theory. <strong>Thermodynamics</strong> and<br />
st<strong>at</strong>istical mechanics do not have such a rel<strong>at</strong>ion, though, contrary to wh<strong>at</strong> people<br />
claim who believe th<strong>at</strong> we do not need thermodynamics. A prime example<br />
is the concept of entropy. Entropy is defined as a measurable quantity in ther-
X INTRODUCTION<br />
modynamics, and the definition relies both on the thermodynamic limit (a large<br />
system) and the existence of reservoirs (an even larger outside). We can also<br />
define entropy in st<strong>at</strong>istical mechanics, but purists will only call this an entropy<br />
analogue. It is a good one, though, and it reproduces many of the well known<br />
results. The st<strong>at</strong>istical mechanical definition of entropy can also be applied to<br />
very small systems, and to the whole universe. But problems arise if we now<br />
also want to apply the second law of thermodynamics in these cases. Small<br />
system obey equ<strong>at</strong>ions which are symmetric under time reversal, which contradicts<br />
the second law. W<strong>at</strong>ch out for Maxwell’s demons! On the large scale, the<br />
entropy of the universe is probably increasing (it is a very large system, and<br />
by definition isol<strong>at</strong>ed). But if the universe is in a well defined quantum st<strong>at</strong>e,<br />
the entropy is and remains zero! These are very interesting questions, but for<br />
a different course. Confusing paradoxes arise easily if one does not appreci<strong>at</strong>e<br />
th<strong>at</strong> thermodynamics is really a meta-theory, and when one applies concepts<br />
under wrong conditions.<br />
Another interesting question is the following. Do large systems obey the<br />
same equ<strong>at</strong>ions as small systems? Are there some new ingredients we need when<br />
we describe a large system? Can we simply scale up the microscopic models to<br />
arrive <strong>at</strong> the large scale, as is done in renormaliz<strong>at</strong>ion group theory? How does<br />
the arrow of time creep into the description when we go from the microscopic<br />
time reversible world to the macroscopic second law of thermodynamics? How<br />
do the large scale phenomena emerge from a microscopic description, and why<br />
do microscopic details become unimportant or remain observable? All good<br />
questions, but also for another course. Here we simply look <strong>at</strong> thermodynamics.<br />
And wh<strong>at</strong> if you disagree with wh<strong>at</strong> was said above? Keep reading nevertheless,<br />
because thermodynamics is also fun. Well, <strong>at</strong> least for me it is......<br />
The m<strong>at</strong>erial in front of you is not a textbook, nor is it an <strong>at</strong>tempt <strong>at</strong> a<br />
future textbook. There are many excellent textbooks on thermodynamics, and<br />
it is not very useful to add a new textbook of lower quality. Also, when you<br />
write a textbook you have to dot all the t-s and cross all the i-s, or something<br />
like th<strong>at</strong>. You get it, I am too lazy for th<strong>at</strong>. This set of notes is meant to be<br />
a tool to help you study the topic of thermodynamics. I have over the years<br />
collected the topics I found relevant, and working through these notes will give<br />
you a good basic understanding of wh<strong>at</strong> is needed in general. If any important<br />
topic is missing, I would like to know so I can add it. If you find a topic too far<br />
out, so be it. All mistakes in these notes are mine. If something is quite useful,<br />
it is stolen from somewhere else.<br />
You can simply take these notes and read them. After doing so, you will<br />
<strong>at</strong> least have seen the basic concepts, and be able to recognize them in the<br />
liter<strong>at</strong>ure. But a much better approach is to read these notes and use them as a<br />
start for further study. This could mean going to the library and looking up these<br />
topics in a number of books on thermodynamics. Nothing helps understanding<br />
more than seeing different descriptions of the same m<strong>at</strong>erial. If there is one skill<br />
th<strong>at</strong> is currently missing among many students, it is the capability of really<br />
using a library! Also, I do not want to give you examples of wh<strong>at</strong> I consider<br />
good textbooks. You should go find out. My opinion would only be a single
INTRODUCTION XI<br />
biased opinion anyhow.<br />
These notes started when I summarized discussions in class. In the current<br />
form, I have presented them as reading m<strong>at</strong>erial, to start class discussions.<br />
<strong>Thermodynamics</strong> can be taught easily in a non-lecture approach, and I am<br />
working on including more questions which could be discussed in class (they are<br />
especially lacking in the l<strong>at</strong>er parts). Although students feel uneasy with this<br />
approach, having a fear th<strong>at</strong> they miss something important, they should realize<br />
th<strong>at</strong> the purpose of these notes is to make sure th<strong>at</strong> all important m<strong>at</strong>erial is<br />
in front of them. Class discussions, of course, have to be guided. Sometimes a<br />
discussion goes in the wrong direction. This is fine for a while, but than the<br />
instructor should help bring it back to the correct p<strong>at</strong>h. Of course, the analysis<br />
of why the discussion took a wrong turn is extremely valuable, because one<br />
learns most often from one’s mistakes (<strong>at</strong> least, one should). To be honest,<br />
finding the right balance for each new group remains a challenge.<br />
The m<strong>at</strong>erial in these notes is sufficient for a quarter or a semester course.<br />
In a semester course one simply adds expansions to selected topics. Also, the<br />
m<strong>at</strong>erial should be accessible for seniors and first year gradu<strong>at</strong>e students. The<br />
m<strong>at</strong>hem<strong>at</strong>ics involved is not too hard, but calculus with many partial deriv<strong>at</strong>ives<br />
is always a bit confusing for everybody, and functional deriv<strong>at</strong>ives also need a bit<br />
of review. It is assumed th<strong>at</strong> basic m<strong>at</strong>erial covered in the introductory physics<br />
sequence is known, hence students should have some idea about temper<strong>at</strong>ure<br />
and entropy. Apart from th<strong>at</strong>, visit the library and discover some lower level<br />
texts on thermodynamics. Again, there are many good ones. And, if these<br />
textbooks are more than ten years old, do not discard them, because they are<br />
still as relevant as they were before. On the other hand, if you use the web<br />
as a source of inform<strong>at</strong>ion, be aware th<strong>at</strong> there are many web-sites posted by<br />
well-meaning individuals, which are full of wrong inform<strong>at</strong>ion. Nevertheless,<br />
browsing the web is a good exercise, since nothing is more important than to<br />
learn to recognize which inform<strong>at</strong>ion is incorrect!<br />
Problem solving is very important in physics, and in order to obtain a working<br />
knowledge of thermodynamics it is important to be able to do problems.<br />
Many problems are included, most of them with solutions. It is good to start<br />
problems in class, and to have a discussion of the general approach th<strong>at</strong> needs<br />
to be taken. When solving problems, for most people it is very beneficial to<br />
work in groups, and th<strong>at</strong> is encouraged. When you try to solve a problem and<br />
you get stuck, do not look <strong>at</strong> the solution! Go to other textbooks and try to<br />
find m<strong>at</strong>erial th<strong>at</strong> pertains to your problem. When you believe th<strong>at</strong> you have<br />
found the solution, then it is time to compare with the solution in the notes,<br />
and then you can check if the solution in the notes is correct.<br />
In many cases, solving problems in thermodynamics always follows the same<br />
general p<strong>at</strong>h. First you identify the independent st<strong>at</strong>e variables. If an experiment<br />
is performed <strong>at</strong> constant temper<strong>at</strong>ure, temper<strong>at</strong>ure is an independent<br />
st<strong>at</strong>e variable because it is controlled. Control means th<strong>at</strong> either we can set it<br />
<strong>at</strong> a certain value, or th<strong>at</strong> we can prevent changes in the variable. For example,<br />
if we discuss a gas in a closed container, the volume of the gas is an independent<br />
st<strong>at</strong>e variable, since the presence of the container makes it impossible for the
XII INTRODUCTION<br />
gas to expand or contract. Pressure, on the other hand, is not an independent<br />
st<strong>at</strong>e variable in this example, since we have no means of controlling it. Second,<br />
based on our determin<strong>at</strong>ion of independent st<strong>at</strong>e variables, we select the correct<br />
thermodynamic potential to work with. Finally, we calcul<strong>at</strong>e the response<br />
functions using this potential, and find rel<strong>at</strong>ions between these functions. Or<br />
we use these response functions to construct equ<strong>at</strong>ions of st<strong>at</strong>e using measured<br />
d<strong>at</strong>a. And so on.<br />
Problem solving is, however, only a part of learning. Another part is to ask<br />
questions. Why do I think this m<strong>at</strong>erial is introduced <strong>at</strong> this point? Wh<strong>at</strong> is<br />
the relevance? How does it build on the previous m<strong>at</strong>erial? Sometimes these<br />
questions are subjective, because wh<strong>at</strong> is obvious for one person can be obscure<br />
for another. The detailed order of topics might work well for one person but not<br />
for another. Consequently, it is also important to ask questions about your own<br />
learning. How did I understand the m<strong>at</strong>erial? Which steps did I make? Which<br />
steps were in the notes, and which were not? How did I fill in the gaps? In<br />
summary, one could say th<strong>at</strong> problem solving improves technical skills, which<br />
leads to a better prepar<strong>at</strong>ion to apply the knowledge. Asking questions improves<br />
conceptual knowledge, and leads to a better understanding how to approach new<br />
situ<strong>at</strong>ions. Asking questions about learning improves the learning process itself<br />
and will facilit<strong>at</strong>e future learning, and also to the limits of the current subject.<br />
Work in progress is adding more questions in the main text. There are<br />
more in the beginning than in the end, a common phenomenon. As part of<br />
their assignments, I asked students in the beginning which questions they would<br />
introduce. These questions are collected in an appendix. So, one should not<br />
take these questions as questions from the students (although quite a few are),<br />
but also as questions th<strong>at</strong> the students think are good to ask! In addition, I<br />
asked students to give a summary of each chapter. These responses are also<br />
given in an appendix.<br />
I provided this m<strong>at</strong>erial in the appendices, because I think it is useful in<br />
two different ways. If you are a student studying thermodynamics, it is good<br />
to know wh<strong>at</strong> others <strong>at</strong> your level in the educ<strong>at</strong>ional system think. If you are<br />
struggling with a concept, it is reassuring to see th<strong>at</strong> others are too, and to<br />
see with wh<strong>at</strong> kind of questions they came up to find a way out. In a similar<br />
manner, it it helpful to see wh<strong>at</strong> others picked out as the most important part<br />
of each chapter. By providing the summaries I do not say th<strong>at</strong> I agree with<br />
them (in fact, sometimes I do not)(Dutchmen rarely agree anyhow), but it gives<br />
a standard for wh<strong>at</strong> others picked out as important. And on the other hand,<br />
if you are teaching this course, seeing wh<strong>at</strong> students perceive to be the most<br />
important part of the content is extremely helpful.<br />
Finally, a thanks to all students who took my classes. Your input has been<br />
essential, your questions have lead to a better understanding of the m<strong>at</strong>erial,<br />
and your research interests made me include a number of topics in these notes<br />
which otherwise would have been left out.<br />
History of these notes:<br />
1991 Original notes for first three chapters written using the program EXP.
INTRODUCTION XIII<br />
1992 Extra notes developed for chapters four and five.<br />
2001 Notes on chapters four and five expanded.<br />
2002 Notes converted to L ATEX, significantly upd<strong>at</strong>ed, and chapter six added.<br />
2003 Notes corrected, minor additions to first five chapters, some additions to<br />
chapter six.<br />
2006 Correction of errors. Upd<strong>at</strong>ed section on fluctu<strong>at</strong>ions. Added m<strong>at</strong>erial on<br />
correl<strong>at</strong>ion functions in chapter seven, but this far from complete.<br />
2008 Upd<strong>at</strong>ed the m<strong>at</strong>erial on correl<strong>at</strong>ion functions and included the two equ<strong>at</strong>ions<br />
of st<strong>at</strong>e rel<strong>at</strong>ed to pair correl<strong>at</strong>ion functions.<br />
2010 Corrected errors and made small upd<strong>at</strong>es.
PART I<br />
<strong>Thermodynamics</strong> Fundamentals<br />
1
Chapter 1<br />
Basic <strong>Thermodynamics</strong>.<br />
The goal of this chapter is to introduce the basic concepts th<strong>at</strong> we use in thermodynamics.<br />
One might think th<strong>at</strong> science has only exact definitions, but th<strong>at</strong><br />
is certainly not true. In the history of any scientific topic one always starts<br />
with language. How do we describe things? Which words do we use and wh<strong>at</strong><br />
do they mean? We need to get to some common understanding of wh<strong>at</strong> terms<br />
mean, before we can make them equivalent to some m<strong>at</strong>hem<strong>at</strong>ical description.<br />
This seems r<strong>at</strong>her vague, but since our n<strong>at</strong>ural way of communic<strong>at</strong>ion is based<br />
on language, it is the only way we can proceed.<br />
We all have some idea wh<strong>at</strong> volume means. But it takes some discussion to<br />
discover th<strong>at</strong> our ideas about volume are all similar. We are able to arrive <strong>at</strong><br />
some definitions we can agree on. Similarly, we all have a good intuition wh<strong>at</strong><br />
temper<strong>at</strong>ure is. More importantly, we can agree how we measure volume and<br />
temper<strong>at</strong>ure. We can take an arbitrary object and put it in w<strong>at</strong>er. The rise of<br />
the w<strong>at</strong>er level will tell us wh<strong>at</strong> the volume of the object is. We can put an<br />
object in contact with a mercury thermometer, and read of the temper<strong>at</strong>ure.<br />
We have used these procedures very often, since they are reproducible and give<br />
us the same result for the same object. Actually, not the same, but in the same<br />
Gaussian distribution. We can do error analysis.<br />
In this chapter we describe the basic terms used in thermodynamics. All<br />
these terms are descriptions of wh<strong>at</strong> we can see. We also make connections with<br />
the ideas of energy and work. The words use to do so are all familiar, and we<br />
build on the vocabulary from a typical introductory physics course. We make<br />
m<strong>at</strong>hem<strong>at</strong>ical connections between our newly defined quantities, and postul<strong>at</strong>e<br />
four laws th<strong>at</strong> hold for all systems. These laws are independent of the n<strong>at</strong>ure of<br />
a system. The m<strong>at</strong>hem<strong>at</strong>ical formul<strong>at</strong>ion by necessity uses many variables, and<br />
we n<strong>at</strong>urally connect with partial deriv<strong>at</strong>ives and multi-variable calculus.<br />
And then there is this quantity called temper<strong>at</strong>ure. We all have a good<br />
”feeling” for it, and standard measurements use physical phenomena like thermal<br />
expansion to measure it. We need to be more precise, however, and define<br />
temper<strong>at</strong>ure in a complic<strong>at</strong>ed manner based on the efficiency of Carnot engines.<br />
At the end of the chapter we show th<strong>at</strong> our definition is equivalent to<br />
3
4 CHAPTER 1. BASIC THERMODYNAMICS.<br />
the definition of temper<strong>at</strong>ure measured by an ideal gas thermometer. Once we<br />
have made th<strong>at</strong> connection, we know th<strong>at</strong> our definition of temper<strong>at</strong>ure is the<br />
same as the common one, since all thermometers are calibr<strong>at</strong>ed against ideal<br />
gas thermometers.<br />
There are two reasons for us to define temper<strong>at</strong>ure in this complic<strong>at</strong>ed manner.<br />
First of all, it is a definition th<strong>at</strong> uses energy in stead of a thermal m<strong>at</strong>erials<br />
property as a basis. Second, it allows us to define an even more illustrious quantity,<br />
named entropy. This new quantity allows us to define thermal equilibrium<br />
in m<strong>at</strong>hem<strong>at</strong>ical terms as a maximum of a function. The principle of maximum<br />
entropy is the corner stone for all th<strong>at</strong> is discussed in the chapter th<strong>at</strong> follow.<br />
1.1 Introduction.<br />
Simple beginnings.<br />
Wh<strong>at</strong> st<strong>at</strong>e am I in?<br />
In the mechanical world of the 19th century, physics was very easy. All you<br />
needed to know were the forces acting on particles. After th<strong>at</strong> it was simply F =<br />
ma. Although this formalism is not hard, actual calcul<strong>at</strong>ions are only feasible<br />
when the system under consider<strong>at</strong>ion contains a few particles. For example,<br />
the motion of solid objects can be described this way if they are considered to<br />
be point particles. In this context, we have all played with Lagrangians and<br />
Hamiltonians. Liquids and gases, however, are harder to deal with, and are<br />
often described in a continuum approxim<strong>at</strong>ion. Everywhere in space one defines<br />
a mass density and a velocity field. The continuity and Newton’s equ<strong>at</strong>ions then<br />
lead to the time evolution of the flow in the liquid.<br />
Asking the right questions.<br />
The big difference between a solid and a liquid is complexity. In first approxim<strong>at</strong>ion<br />
a solid can be described by six coordin<strong>at</strong>es (center of mass, orient<strong>at</strong>ion),<br />
while a liquid needs a mass density field which is essentially an infinite set of<br />
coordin<strong>at</strong>es. The calcul<strong>at</strong>ion of the motion of a solid is rel<strong>at</strong>ively easy, especially<br />
if one uses conserv<strong>at</strong>ion laws for energy, momentum, and angular momentum.<br />
The calcul<strong>at</strong>ion of the flow of liquids is still hard, even today, and is often done<br />
on computers using finite difference or finite element methods. In the 19th century,<br />
only the simplest flow p<strong>at</strong>terns could be analyzed. In many cases these<br />
flow p<strong>at</strong>terns are only details of the overall behavior of a liquid. Very often one<br />
is interested in more general quantities describing liquids and gases. In the 19th<br />
century many important questions have been raised in connection with liquids
1.1. INTRODUCTION. 5<br />
and gases, in the context of steam engines. How efficient can a steam engine<br />
be? Which temper<strong>at</strong>ure should it oper<strong>at</strong>e <strong>at</strong>? Hence the problem is wh<strong>at</strong> we<br />
need to know about liquids and gases to be able to answer these fundamental<br />
questions.<br />
Divide and conquer.<br />
In thermodynamics we consider macroscopic systems, or systems with a large<br />
number of degrees of freedom. Liquids and gases certainly belong to this class of<br />
systems, even if one does not believe in an <strong>at</strong>omic model! The only requirement<br />
is th<strong>at</strong> the system needs a description in terms of density and velocity fields.<br />
Solids can be described this way. In this case the mass density field is given<br />
by a constant plus a small vari<strong>at</strong>ion. The time evolution of these devi<strong>at</strong>ions<br />
from equilibrium shows oscill<strong>at</strong>ory p<strong>at</strong>terns, as expected. The big difference<br />
between a solid and a fluid is th<strong>at</strong> the devi<strong>at</strong>ions from average in a solid are<br />
small and can be ignored in first approxim<strong>at</strong>ion.<br />
No details, please.<br />
In a thermodynamic theory we are never interested in the detailed functional<br />
form of the density as a function of position, but only in macroscopic or global<br />
averages. Typical quantities of interest are the volume, magnetic moment, and<br />
electric dipole moment of a system. These macroscopic quantities, which can<br />
be measured, are called thermodynamic or st<strong>at</strong>e variables. They uniquely determine<br />
the thermodynamic st<strong>at</strong>e of a system.<br />
St<strong>at</strong>e of a system.<br />
The definition of the st<strong>at</strong>e of a system is in terms of oper<strong>at</strong>ions. Wh<strong>at</strong><br />
are the possible quantities which we can measure? In other words, how do we<br />
assign numbers to the fields describing a m<strong>at</strong>erial? Are there any problems?<br />
For example, one might think th<strong>at</strong> it is easy to measure the volume of a liquid<br />
in a beaker. But how does this work close to the critical point where the index<br />
of refraction of the liquid and vapor become the same? How does this work if<br />
there is no gravity? In thermodynamics we simply assume th<strong>at</strong> we are able to<br />
measure some basic quantities like volume.<br />
Another question is how many st<strong>at</strong>e variables do we need to describe a<br />
system. For example we prepare a certain amount of w<strong>at</strong>er and pour it in a<br />
beaker, which we cap. The next morning we find th<strong>at</strong> the w<strong>at</strong>er is frozen. It<br />
is obvious th<strong>at</strong> the w<strong>at</strong>er is not in the same st<strong>at</strong>e, and th<strong>at</strong> the inform<strong>at</strong>ion we<br />
had about the system was insufficient to uniquely define the st<strong>at</strong>e of the system.<br />
In this case we omitted temper<strong>at</strong>ure.<br />
It can be more complic<strong>at</strong>ed, though. Suppose we have prepared many iron<br />
bars always <strong>at</strong> the same shape, volume, mass, and temper<strong>at</strong>ure. They all look
6 CHAPTER 1. BASIC THERMODYNAMICS.<br />
the same. But then we notice th<strong>at</strong> some bars affect a compass needle while others<br />
do not. Hence not all bars are the same and we need additional inform<strong>at</strong>ion to<br />
uniquely define the st<strong>at</strong>e of the bar. Using the compass needle we can measure<br />
the magnetic field and hence we find the magnetic moment of the bar.<br />
In a next step we use all bars with a given magnetic moment. We apply a<br />
force on these bars and see how much they expand, from which we calcul<strong>at</strong>e the<br />
elastic constants of the bar. We find th<strong>at</strong> different bars have different values of<br />
the elastic constants. Wh<strong>at</strong> are we missing in terms of st<strong>at</strong>e variables? This is<br />
certainly not obvious. We need inform<strong>at</strong>ion about defects in the structure, or<br />
inform<strong>at</strong>ion about the mass density field beyond the average value.<br />
Is it changing?<br />
Was I in a different st<strong>at</strong>e before?<br />
At this point it is important to realize th<strong>at</strong> a measured value of a st<strong>at</strong>e<br />
variable is always a time-average. The pressure in an ideal gas fluctu<strong>at</strong>es, but<br />
the time scale of the fluctu<strong>at</strong>ions is much smaller than the time scale of the<br />
externally applied pressure changes. Hence these short time fluctu<strong>at</strong>ions can be<br />
ignored and are averaged out in a measurement. This does imply a warning,<br />
though. If the fluctu<strong>at</strong>ions in the st<strong>at</strong>e variables are on a time scale comparable<br />
with the dur<strong>at</strong>ion of the experiment a standard thermodynamic description<br />
is useless. If they are on a very long time scale, however, we can use our<br />
thermodynamic description again. In th<strong>at</strong> case the motion is so slow and we<br />
can use linear approxim<strong>at</strong>ions for all changes.<br />
How do we change?<br />
The values of the st<strong>at</strong>e variables for a given system can be modified by applying<br />
forces. An increase in pressure will decrease the volume, a change in<br />
magnetic induction will alter the magnetic moment. The pressure in a gas in<br />
a container is in many cases equal to the pressure th<strong>at</strong> this container exerts on<br />
the gas in order to keep it within the volume of the container. It is possible to<br />
use this pressure to describe the st<strong>at</strong>e of the system and hence pressure (and<br />
magnetic induction) are also st<strong>at</strong>e variables. One basic question in thermodynamics<br />
is how these st<strong>at</strong>e variables change when external forces are applied. In<br />
a more general way, if a specific st<strong>at</strong>e variable is changed by external means,<br />
how do the other st<strong>at</strong>e variables respond?<br />
Number of variables, again.
1.2. SOME DEFINITIONS. 7<br />
The number of st<strong>at</strong>e variables we need to describe the st<strong>at</strong>e of a system depends<br />
on the n<strong>at</strong>ure of th<strong>at</strong> system. We expand somewh<strong>at</strong> more on the previous<br />
discussion. An ideal gas, for example, is in general completely characterized by<br />
its volume, pressure, and temper<strong>at</strong>ure. It is always possible to add more st<strong>at</strong>e<br />
variables to this list. Perhaps one decides to measure the magnetic moment of<br />
an ideal gas too. Obviously, th<strong>at</strong> changes our knowledge of the st<strong>at</strong>e of the ideal<br />
gas. If the value of this additional st<strong>at</strong>e variable is always the same, no m<strong>at</strong>ter<br />
wh<strong>at</strong> we do in our experiment, then this variable is not essential. But one can<br />
always design experiments in which this st<strong>at</strong>e variable becomes essential. The<br />
magnetic moment is usually measured by applying a very small magnetic induction<br />
to the system. This external field should be zero for all purposes. If it<br />
is not, then we have to add the magnetic moment to our list of st<strong>at</strong>e variables.<br />
It is also possible th<strong>at</strong> one is not aware of additional essential st<strong>at</strong>e variables.<br />
Experiments will often indic<strong>at</strong>e th<strong>at</strong> more variables are needed. An example is<br />
an experiment in which we measure the properties of a piece of iron as a function<br />
of volume, pressure, and temper<strong>at</strong>ure. At a temper<strong>at</strong>ure of about 770 ◦ C some<br />
abnormal behavior is observed. As it turns out, iron is magnetic below this<br />
temper<strong>at</strong>ure and in order to describe the st<strong>at</strong>e of an iron sample one has to<br />
include the magnetic moment in the list of essential st<strong>at</strong>e variables. An ideal<br />
gas in a closed container is a simple system, but if the gas is allowed to escape<br />
via a valve, the number of particles in this gas also becomes an essential st<strong>at</strong>e<br />
variable needed to describe the st<strong>at</strong>e of the system inside the container.<br />
Are measured values always sp<strong>at</strong>ial averages?<br />
Are there further classific<strong>at</strong>ions of st<strong>at</strong>es or processes?<br />
1.2 Some definitions.<br />
Two types of processes.<br />
If one takes a block of wood, and splits it into two pieces, one has performed<br />
a simple action. On the level of thermodynamic variables one writes something<br />
like V = V1 + V2 for the volumes and similar equ<strong>at</strong>ions for other st<strong>at</strong>e variables.<br />
The detailed n<strong>at</strong>ure of this process is, however, not accessible in this language.<br />
In addition, if we put the two pieces back together again, they do in general not<br />
stay together. The process was irreversible. In general, in thermodynamics one<br />
only studies the reversible behavior of macroscopic systems. An example would
8 CHAPTER 1. BASIC THERMODYNAMICS.<br />
be the study of the liquid to vapor transition. M<strong>at</strong>erial is slowly transported<br />
from one phase to another and can go back if the causes are reversed. The<br />
st<strong>at</strong>e variables one needs to consider in this case are the pressure, temper<strong>at</strong>ure,<br />
volume, interface area (because of surface tension), and perhaps others in more<br />
complic<strong>at</strong>ed situ<strong>at</strong>ions.<br />
When there is NO change.<br />
Obviously, all macroscopic systems change as a function of time. Most of<br />
these changes, however, are on a microscopic level and are not of interest. We<br />
are not able to measure them directly. Therefore, in thermodynamics one defines<br />
a steady st<strong>at</strong>e when all thermodynamic variables are independent of time. A<br />
resistor connected to a constant voltage is in a steady st<strong>at</strong>e. The current through<br />
the resistor is constant and although there is a flow of charge, there are no net<br />
changes in the resistor. The same amount of charge comes in as goes out.<br />
Thermodynamic equilibrium describes a more restricted situ<strong>at</strong>ion. A system<br />
is in thermodynamic equilibrium if it is in a steady st<strong>at</strong>e and if there are no<br />
net macroscopic currents (of energy, particles, etc) over macroscopic distances.<br />
There is some ambiguity in this definition, connected to the scale and magnitude<br />
of the currents. A vapor-liquid interface like the ocean, with large waves, is<br />
clearly not in equilibrium. But how small do the waves have to be in order th<strong>at</strong><br />
we can say th<strong>at</strong> the system is in equilibrium? If we discuss the thermal balance<br />
between oceans and <strong>at</strong>mosphere, are waves important? Also, the macroscopic<br />
currents might be very small. Glass, for example, is not in thermal equilibrium<br />
according to a strict definition, but the changes are very slow with a time scale<br />
of hundreds of years. Hence even if we cannot measure macroscopic currents,<br />
they might be there. We will in general ignore these situ<strong>at</strong>ions, since they tend<br />
not to be of interest on the time scale of the experiments!<br />
St<strong>at</strong>e functions.<br />
Wh<strong>at</strong> do you think about hysteresis loops in magnets?<br />
Once we understand the n<strong>at</strong>ure of thermal equilibrium, we can generalize<br />
the concept of st<strong>at</strong>e variables. A st<strong>at</strong>e function is any quantity which in thermodynamic<br />
equilibrium only depends on the values of the st<strong>at</strong>e variables and not<br />
on the history (or future?) of the sample. A simple st<strong>at</strong>e function would be the<br />
product of the pressure and volume. This product has a physical interpret<strong>at</strong>ion,<br />
but cannot be measured directly.<br />
Two types of variables.
1.2. SOME DEFINITIONS. 9<br />
Thermodynamic variables come in two varieties. If one takes a system in<br />
equilibrium the volume of the left half is only half the total volume (surprise)<br />
but the pressure in the left half is equal to the pressure of the total system.<br />
There are only two possibilities. Either a st<strong>at</strong>e variable scales linearly with the<br />
size of the system, or is independent of the size of the system. In other words,<br />
if we consider two systems in thermal equilibrium, made of identical m<strong>at</strong>erial,<br />
one with volume V1 and one with volume V2, a st<strong>at</strong>e variable X either obeys<br />
X1<br />
V1<br />
= X2<br />
V2 or X1 = X2. In the first case the st<strong>at</strong>e variable is called extensive and<br />
in the second case it is called intensive. Extensive st<strong>at</strong>e variables correspond to<br />
generalized displacements. For the volume this is easy to understand; increasing<br />
volume means displacing outside m<strong>at</strong>erial and doing work on it in the process.<br />
Intensive st<strong>at</strong>e variables correspond to generalized forces. The pressure is the<br />
force needed to change the volume. For each extensive st<strong>at</strong>e variable there is a<br />
corresponding intensive st<strong>at</strong>e variable and vice-versa.<br />
Extensive st<strong>at</strong>e variables correspond to quantities which can be determined,<br />
measured, or prescribed directly. The volume of a gas can be found by measuring<br />
the size of the container, the amount of m<strong>at</strong>erial can be measured using a<br />
balance. Intensive st<strong>at</strong>e variables are measured by making contact with something<br />
else. We measure temper<strong>at</strong>ure by using a thermometer, and pressure using<br />
a manometer. Such measurements require equilibrium between the sample and<br />
measuring device.<br />
Note th<strong>at</strong> this distinction limits where and how we can apply thermodynamics.<br />
The gravit<strong>at</strong>ional energy of a large system is not proportional to the amount<br />
of m<strong>at</strong>erial, but to the amount of m<strong>at</strong>erial to the five-thirds power. If the force<br />
of gravity is the dominant force internally in our system we need other theories.<br />
Electrical forces are also of a long range, but because we have both positive<br />
and neg<strong>at</strong>ive charges, they are screened. Hence in m<strong>at</strong>erials physics we normally<br />
have no fundamental problems with applying thermodynamics.<br />
Thermodynamic limit.<br />
At this point we are able to define wh<strong>at</strong> we mean by a large system. R<strong>at</strong>ios<br />
of an extensive st<strong>at</strong>e variable and the volume, like X<br />
V , are often called densities.<br />
It is customary to write these densities in lower case, x = X<br />
V . If the volume is too<br />
small, x will depend on the volume. Since X is extensive, this is not supposed to<br />
be the case. In order to get rid of the effects of a finite volume (surface effects!)<br />
one has to take the limit V → ∞. This is called the thermodynamic limit.<br />
All our m<strong>at</strong>hem<strong>at</strong>ical formulas are strictly speaking only correct in this limit.<br />
In practice, this means th<strong>at</strong> the volume has to be large enough in order th<strong>at</strong><br />
changes in the volume do not change the densities anymore. It is always possible<br />
to write x(V ) = x∞ +αV −1 +O(V −2 ). The magnitude of α decides which value<br />
of the volume is large enough.<br />
<strong>Physics</strong> determines the rel<strong>at</strong>ion between st<strong>at</strong>e variables.
10 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Why are all these definitions important? So far we have not discussed any<br />
physics. If all the st<strong>at</strong>e variables would be independent we could stop right here.<br />
Fortun<strong>at</strong>ely, they are not. Some st<strong>at</strong>e variables are rel<strong>at</strong>ed by equ<strong>at</strong>ions of st<strong>at</strong>e<br />
and these equ<strong>at</strong>ions contain the physics of the system. It is important to note<br />
th<strong>at</strong> these equ<strong>at</strong>ions of st<strong>at</strong>e only rel<strong>at</strong>e the values of the st<strong>at</strong>e variables when<br />
the system is in thermal equilibrium, in the thermodynamic limit! If a system<br />
is not in equilibrium, any combin<strong>at</strong>ion of st<strong>at</strong>e variables is possible. It is even<br />
possible to construct non-equilibrium systems in which the actual definition or<br />
measurement of certain st<strong>at</strong>e variables is not useful or becomes ambiguous.<br />
Simple examples of equ<strong>at</strong>ions of st<strong>at</strong>e are the ideal gas law pV = NRT and<br />
Curie’s law M = CNH<br />
T . The first equ<strong>at</strong>ion rel<strong>at</strong>es the product of the pressure<br />
p and volume V of an ideal gas to the number of moles of gas N and the<br />
temper<strong>at</strong>ure T. The constant of proportionality, R, is the molar gas constant,<br />
which is the product of Boltzmann’s constant kB and Avogadro’s number NA.<br />
The second equ<strong>at</strong>ion rel<strong>at</strong>es the magnetic moment M to the number of moles<br />
of <strong>at</strong>oms N, the magnetic field H, and the temper<strong>at</strong>ure T. The constant of proportionality<br />
is Curie’s constant C. Note th<strong>at</strong> in thermodynamics the preferred<br />
way of measuring the amount of m<strong>at</strong>erial is in terms of moles, which again can<br />
be defined independent of a molecular model. Note, too, th<strong>at</strong> in electricity and<br />
magnetism we always use the magnetiz<strong>at</strong>ion density in the Maxwell equ<strong>at</strong>ions,<br />
but th<strong>at</strong> in thermodynamics we define the total magnetiz<strong>at</strong>ion as the relevant<br />
quantity. This makes M an extensive quantity.<br />
Equ<strong>at</strong>ions of st<strong>at</strong>e are rel<strong>at</strong>ed to st<strong>at</strong>e functions. For any system we can<br />
define the st<strong>at</strong>e function Fst<strong>at</strong>e = pV − NRT . It will take on all kinds of values.<br />
We then define the special class of systems for which Fst<strong>at</strong>e ≡ 0, identical to<br />
zero, as an ideal gas. The right hand side then leads to an equ<strong>at</strong>ion of st<strong>at</strong>e,<br />
which can be used to calcul<strong>at</strong>e one of the basic st<strong>at</strong>e variables if others are<br />
known. The practical applic<strong>at</strong>ion of this idea is to look for systems for which<br />
Fst<strong>at</strong>e is small, with small defined in an appropri<strong>at</strong>e context. In th<strong>at</strong> case we can<br />
use the class with zero st<strong>at</strong>e function as a first approxim<strong>at</strong>ion of the real system.<br />
In many cases the ideal gas approxim<strong>at</strong>ion is a good start for a description of a<br />
real gas, and is a start for system<strong>at</strong>ic improvements of the description!<br />
How do we get equ<strong>at</strong>ions of st<strong>at</strong>e?<br />
Equ<strong>at</strong>ions of st<strong>at</strong>e have two origins. One can completely ignore the microscopic<br />
n<strong>at</strong>ure of m<strong>at</strong>ter and simply postul<strong>at</strong>e some rel<strong>at</strong>ion. One then uses the<br />
laws of thermodynamics to derive functional forms for specific st<strong>at</strong>e variables as<br />
a function of the others, and compares the predicted results with experiment.<br />
The ideal gas law has this origin. This procedure is exactly wh<strong>at</strong> is done in thermodynamics.<br />
One does not need a model for the detailed n<strong>at</strong>ure of the systems,<br />
but derives general conclusions based on the average macroscopic behavior of a<br />
system in the thermodynamic limit.<br />
In order to derive equ<strong>at</strong>ions of st<strong>at</strong>e, however, one has to consider the microscopic<br />
aspects of a system. Our present belief is th<strong>at</strong> all systems consist of <strong>at</strong>oms.
1.2. SOME DEFINITIONS. 11<br />
If we know the forces between the <strong>at</strong>oms, the theory of st<strong>at</strong>istical mechanics will<br />
tell us how to derive equ<strong>at</strong>ions of st<strong>at</strong>e. There is again a choice here. It is possible<br />
to postul<strong>at</strong>e the forces. The equ<strong>at</strong>ions of st<strong>at</strong>e could then be derived from<br />
molecular dynamics calcul<strong>at</strong>ions, for example. The other route derives these<br />
effective forces from the laws of quantum mechanics and the structure of the<br />
<strong>at</strong>oms in terms of electrons and nuclei. The interactions between the particles<br />
in the <strong>at</strong>oms are simple Coulomb interactions in most cases. These Coulomb<br />
interactions follow from yet a deeper theory, quantum electro-dynamics, and are<br />
only a first approxim<strong>at</strong>ion. These corrections are almost always unimportant in<br />
the study of m<strong>at</strong>erials and only show up <strong>at</strong> higher energies in nuclear physics<br />
experiments.<br />
Why do we need equ<strong>at</strong>ions of st<strong>at</strong>e?<br />
Equ<strong>at</strong>ions of st<strong>at</strong>es can be used to classify m<strong>at</strong>erials. They can be used<br />
to derive <strong>at</strong>omic properties of m<strong>at</strong>erials. For example, <strong>at</strong> low densities a gas<br />
of helium <strong>at</strong>oms and a gas of methane <strong>at</strong>oms both follow the ideal gas law.<br />
This indic<strong>at</strong>es th<strong>at</strong> in this limit the internal structure of the molecules does not<br />
affect the motion of the molecules! In both cases they seem to behave like point<br />
particles. L<strong>at</strong>er we will see th<strong>at</strong> other quantities are different. For example, the<br />
internal energy certainly is larger for methane where rot<strong>at</strong>ions and transl<strong>at</strong>ions<br />
play a role.<br />
Classific<strong>at</strong>ion of changes of st<strong>at</strong>e.<br />
Since a st<strong>at</strong>ic universe is not very interesting, one has to consider changes<br />
in the st<strong>at</strong>e variables. In a thermodynamic transform<strong>at</strong>ion or process a system<br />
changes one or more of its st<strong>at</strong>e variables. A spontaneous process takes place<br />
without any change in the externally imposed constraints. The word constraint<br />
in this context means an external description of the st<strong>at</strong>e variables for the system.<br />
For example, we can keep the volume of a gas the same, as well as the<br />
temper<strong>at</strong>ure and the amount of gas. Or if the temper<strong>at</strong>ure of the gas is higher<br />
than the temper<strong>at</strong>ure of the surroundings, we allow the gas to cool down. In<br />
an adiab<strong>at</strong>ic process no he<strong>at</strong> is exchanged between the system and the environment.<br />
A process is called isothermal if the temper<strong>at</strong>ure of the system remains<br />
the same, isobaric if the pressure does not change, and isochoric if the mass<br />
density (the number of moles of particles divided by the volume) is constant. If<br />
the change in the system is infinitesimally slow, the process is quasist<strong>at</strong>ic .<br />
Reversible process.<br />
The most important class of processes are those in which the system starts<br />
in equilibrium, the process is quasist<strong>at</strong>ic, and all the intermedi<strong>at</strong>e st<strong>at</strong>es and the<br />
final st<strong>at</strong>e are in equilibrium. These processes are called reversible. The process
12 CHAPTER 1. BASIC THERMODYNAMICS.<br />
can be described by a continuous p<strong>at</strong>h in the space of the st<strong>at</strong>e variables, and<br />
this p<strong>at</strong>h is restricted to the surfaces determined by the equ<strong>at</strong>ions of st<strong>at</strong>e for<br />
the system. By inverting all external forces, the p<strong>at</strong>h in the space of the st<strong>at</strong>e<br />
functions will be reversed, which prompted the name for this type of process.<br />
Reversible processes are important because they can be described m<strong>at</strong>hem<strong>at</strong>ically<br />
via the equ<strong>at</strong>ions of st<strong>at</strong>e. This property is lost for an irreversible process<br />
between two equilibrium st<strong>at</strong>es, where we only have a useful m<strong>at</strong>hem<strong>at</strong>ical description<br />
of the initial and final st<strong>at</strong>e. As we will see l<strong>at</strong>er, the second law of<br />
thermodynamics makes another distinction between reversible and irreversible<br />
processes.<br />
How does a process become irreversible?<br />
An irreversible process is either a process which happens too fast or which is<br />
discontinuous. The sudden opening of a valve is an example of the last case. The<br />
system starts out in equilibrium with volume Vi and ends in equilibrium with<br />
a larger volume Vf . For the intermedi<strong>at</strong>e st<strong>at</strong>es the volume is not well defined,<br />
though. Such a process takes us outside of the space of st<strong>at</strong>e variables we<br />
consider. It can still be described in the phase space of all system variables, and<br />
m<strong>at</strong>hem<strong>at</strong>ically it is possible to define the volume, but details of this definition<br />
will play a role in the description of the process. Another type of irreversible<br />
process is the same expansion from Vi to Vf in a controlled way. The volume<br />
is well-defined everywhere in the process, but the system is not in equilibrium<br />
in the intermedi<strong>at</strong>e st<strong>at</strong>es. The process is going too fast. In an ideal gas this<br />
would mean, for example, pV ̸= NRT for the intermedi<strong>at</strong>e stages.<br />
Are there general principles connecting the values of st<strong>at</strong>e variables, valid for<br />
all systems?<br />
1.3 Zeroth Law of <strong>Thermodynamics</strong>.<br />
General rel<strong>at</strong>ions.<br />
An equ<strong>at</strong>ion of st<strong>at</strong>e specifies a rel<strong>at</strong>ion between st<strong>at</strong>e variables which holds<br />
for a certain class of systems. It represents the physics particular to th<strong>at</strong> system.<br />
There are, however, a few rel<strong>at</strong>ions th<strong>at</strong> hold for all systems, independent of the<br />
n<strong>at</strong>ure of the system. Following an old tradition, these rel<strong>at</strong>ions are called the<br />
laws of thermodynamics. There are four of them, numbered 0 through 3. The<br />
middle two are the most important, and they have been paraphrased in the<br />
following way. Law one tells you th<strong>at</strong> in the game of thermodynamics you<br />
cannot win. The second law makes it even worse, you cannot break even.
1.4. FIRST LAW: ENERGY. 13<br />
Law zero.<br />
The zeroth law is rel<strong>at</strong>ively trivial. It discusses systems in equilibrium.<br />
Two systems are in thermal equilibrium if they are in contact and the total<br />
system, encompassing the two systems as subsystems, is in equilibrium. In other<br />
words, two systems in contact are in equilibrium if the individual systems are in<br />
equilibrium and there are no net macroscopic currents between the systems. The<br />
zeroth law st<strong>at</strong>es th<strong>at</strong> if equilibrium system A is in contact and in equilibrium<br />
with systems B and C (not necessarily <strong>at</strong> the same time, but A does not change),<br />
then systems B and C are also in equilibrium with each other. If B and C are not<br />
in contact, it would mean th<strong>at</strong> if we bring them in contact no net macroscopic<br />
currents will flow.<br />
Significance of law zero.<br />
The importance of this law is th<strong>at</strong> it enables to define universal standards<br />
for temper<strong>at</strong>ure, pressure, etc. If two different systems cause the same reading<br />
on the same thermometer, they have the same temper<strong>at</strong>ure. A temper<strong>at</strong>ure<br />
scale on a new thermometer can be set by comparing it with systems of known<br />
temper<strong>at</strong>ure. Therefore, the first law is essential, without it we would not be<br />
able to give a meaningful analysis of any experiment. We postul<strong>at</strong>e it a a law,<br />
because we have not seen any exceptions. We postul<strong>at</strong>e it as a law, because we<br />
absolutely need it. We cannot prove it to be true. It cannot be derived from<br />
st<strong>at</strong>istical mechanics, because in th<strong>at</strong> theory it is also a basic or fundamental<br />
assumption. But if one rejects it completely, one throws away a few hundred<br />
years of successful science. But wh<strong>at</strong> if the situ<strong>at</strong>ion is similar to Newton’s<br />
F = ma, where Einstein showed the limits of validity? Th<strong>at</strong> scenario is certainly<br />
possible, but we have not yet needed it, or seen any reason for its need. Also,<br />
it is completely unclear wh<strong>at</strong> kind of theory should be used to replace all wh<strong>at</strong><br />
we will explore in these notes.<br />
Are there any consequences for the sizes of the systems?<br />
1.4 First law: Energy.<br />
He<strong>at</strong> is energy flow.<br />
The first law of thermodynamics st<strong>at</strong>es th<strong>at</strong> energy is conserved. The change<br />
in internal energy U of a system is equal to the amount of he<strong>at</strong> energy added to<br />
the system minus the amount of work done by the system. It implies th<strong>at</strong> he<strong>at</strong>
14 CHAPTER 1. BASIC THERMODYNAMICS.<br />
is a form of energy. Technically, he<strong>at</strong> describes the flow of energy, but we are<br />
very sloppy in our use of words here. The formal st<strong>at</strong>ement of the first law is<br />
dU = ¯ dQ − ¯ dW (1.1)<br />
The amount of he<strong>at</strong> added to the system is ¯ dQ and the amount of work done<br />
by the system is ¯ dW . The m<strong>at</strong>hem<strong>at</strong>ical formul<strong>at</strong>ion of the first law also shows<br />
an important characteristic of thermodynamics. It is often possible to define<br />
thermodynamic rel<strong>at</strong>ions only via changes in the thermodynamic quantities.<br />
Note th<strong>at</strong> we define the work term as work done on the outside world. It<br />
represents a loss of energy of the system. This is the standard definition, and<br />
represents the fact th<strong>at</strong> in the original analysis one was interested in supplying<br />
he<strong>at</strong> to an engine, which then did work. Some books, however, try to be consistent,<br />
and write the work term as work done on the system. In th<strong>at</strong> case there<br />
is no minus sign in equ<strong>at</strong>ion 1.1. Although th<strong>at</strong> is somewh<strong>at</strong> ne<strong>at</strong>er, it causes<br />
too much confusion with standard texts. It is always much too easy to lose a<br />
minus sign.<br />
The first law again is a st<strong>at</strong>ement th<strong>at</strong> has always been observed to be<br />
true. In addition, the first law does follow directly in a st<strong>at</strong>istical mechanical<br />
tre<strong>at</strong>ment. We have no reason to doubt the validity of the first law, and we<br />
discard any proposals of engines th<strong>at</strong> cre<strong>at</strong>e energy out of nothing. But again,<br />
there is no absolute proof of its validity. And, again as well, if one discards<br />
the first law, all the remainder of these notes will be invalid as well. A theory<br />
without the first law is very difficult to imagine.<br />
The internal energy is a st<strong>at</strong>e variable.<br />
The internal energy U is a st<strong>at</strong>e variable and an infinitesimal change in<br />
internal energy is an exact differential. Since U is a st<strong>at</strong>e variable, the value of<br />
any integral ∫ dU depends only on the values of U <strong>at</strong> the endpoints of the p<strong>at</strong>h<br />
in the space of st<strong>at</strong>e variables, and not on the specific p<strong>at</strong>h between the endpoints.<br />
The internal energy U has to be a st<strong>at</strong>e variable, or else we could devise<br />
a process in which a system goes through a cycle and returns to its original st<strong>at</strong>e<br />
while loosing or gaining energy. For example, this could mean th<strong>at</strong> a burning<br />
piece of coal today would produce less he<strong>at</strong> than tomorrow. If the internal<br />
energy would not be a st<strong>at</strong>e variable, we would have sources of free energy.<br />
Exact differentials.<br />
The concept of exact differentials is very important, and hence we will illustr<strong>at</strong>e<br />
it by using some examples. Assume the function f is a st<strong>at</strong>e function of<br />
the st<strong>at</strong>e variables x and y only, f(x,y). For small changes we can write<br />
df =<br />
( )<br />
∂f<br />
dx +<br />
∂x y<br />
( )<br />
∂f<br />
dy (1.2)<br />
∂y x
1.4. FIRST LAW: ENERGY. 15<br />
where in the not<strong>at</strong>ion for the partial deriv<strong>at</strong>ives the variable which is kept constant<br />
is also indic<strong>at</strong>ed. This is always very useful in thermodynamics, because<br />
one often changes variables. There would be no problems if quantities were defined<br />
directly like f(x, y) = x + y. In thermodynamics, however, most quantities<br />
are defined by or measured via small changes in a system. Hence, suppose the<br />
change in a quantity g is rel<strong>at</strong>ed to changes in the st<strong>at</strong>e variables x and y via<br />
¯dg = h(x, y)dx + k(x, y)dy (1.3)<br />
Is the quantity g a st<strong>at</strong>e function, in other words is g uniquely determined<br />
by the st<strong>at</strong>e of a system or does it depend on the history, on how the system<br />
got into th<strong>at</strong> st<strong>at</strong>e? It turns out th<strong>at</strong> a necessary and sufficient condition for g<br />
to be a st<strong>at</strong>e function is th<strong>at</strong><br />
( )<br />
∂h<br />
=<br />
∂y x<br />
( )<br />
∂k<br />
∂x y<br />
(1.4)<br />
everywhere in the x-y space. The necessity follows immedi<strong>at</strong>ely from 1.2, as<br />
long as we assume th<strong>at</strong> the partial deriv<strong>at</strong>ives in 1.4 exist and are continuous.<br />
This is because in second order deriv<strong>at</strong>ives we can interchange the order of the<br />
deriv<strong>at</strong>ives under such conditions. Th<strong>at</strong> it is sufficient can be shown as follows.<br />
Consider a p<strong>at</strong>h (x, y) = (ϕ(t), ψ(t)) from (x1, y1) <strong>at</strong> t1 to (x2, y2) <strong>at</strong> t2 and<br />
dt ,<br />
integr<strong>at</strong>e ¯ dg , using dx = dϕ<br />
dt<br />
Define<br />
∫ t2<br />
t1<br />
dt , dy = dψ<br />
dt<br />
(<br />
h(ϕ(t), ψ(t)) dϕ<br />
+ k(ϕ(t), ψ(t))dψ<br />
dt dt<br />
H(x, y) =<br />
)<br />
dt (1.5)<br />
∫ x<br />
dx<br />
0<br />
′ h(x ′ ∫ y<br />
, y) + dy<br />
0<br />
′ k(0, y ′ ) (1.6)<br />
and H(t) = H(ϕ(t), ψ(t)). It follows th<strong>at</strong><br />
dH<br />
dt =<br />
( )<br />
∂H dϕ<br />
∂x y dt +<br />
( )<br />
∂H dψ<br />
∂y x dt<br />
(1.7)<br />
The partial deriv<strong>at</strong>ives of H are easy to calcul<strong>at</strong>e:<br />
( )<br />
∂H<br />
(x, y) = h(x, y) (1.8)<br />
∂x y<br />
( )<br />
∂H<br />
(x, y) =<br />
∂y x<br />
This implies th<strong>at</strong><br />
∫ x<br />
0<br />
dx ′<br />
dx<br />
0<br />
′<br />
( )<br />
∂h<br />
∂y<br />
( )<br />
∂k<br />
∂x<br />
∫ x<br />
(x<br />
x<br />
′ , y) + k(0, y) =<br />
(x<br />
y<br />
′ , y) + k(0, y) = k(x, y) (1.9)
16 CHAPTER 1. BASIC THERMODYNAMICS.<br />
∫ t2<br />
t1<br />
¯dg =<br />
∫ t2<br />
t1<br />
dH<br />
dt<br />
(1.10)<br />
and hence the integral of ¯ dg is equal to H(t2)−H(t1) which does not depend<br />
on the p<strong>at</strong>h taken between the end-points of the integr<strong>at</strong>ion.<br />
Example.<br />
An example might illustr<strong>at</strong>e this better. Suppose x and y are two st<strong>at</strong>e variables,<br />
and they determine the internal energy completely. If we define changes<br />
in the internal energy via changes in the st<strong>at</strong>e variables x and y via<br />
dU = x 2 ydx + 1<br />
3 x3dy (1.11)<br />
we see immedi<strong>at</strong>ely th<strong>at</strong> this definition is correct, the energy U is a st<strong>at</strong>e function.<br />
The partial deriv<strong>at</strong>ives obey the symmetry rel<strong>at</strong>ion 1.4 and one can simply<br />
integr<strong>at</strong>e dU and check th<strong>at</strong> we get U(x, y) = 1<br />
3x3y + U0.<br />
The changes in he<strong>at</strong> and work are now assumed to be rel<strong>at</strong>ed in the following<br />
way<br />
¯dQ = 1<br />
2 x2 ydx + 1<br />
2 x3 dy (1.12)<br />
¯dW = − 1<br />
2 x2 ydx + 1<br />
6 x3 dy (1.13)<br />
These definitions do indeed obey the first law 1.1. It is also clear using the<br />
symmetry rel<strong>at</strong>ion 1.4 th<strong>at</strong> these two differentials are not exact.<br />
Suppose the system which is described above is originally in the st<strong>at</strong>e (x, y) =<br />
(0, 0). Now we change the st<strong>at</strong>e of the system by a continuous transform<strong>at</strong>ion<br />
from (0, 0) to (1, 1). We do this in two different ways, however. P<strong>at</strong>h one takes<br />
us from (0, 0) to (0, 1) to (1, 1) along two straight line segments, p<strong>at</strong>h two is<br />
similar from (0, 0) to (1, 0) to (1, 1). The integrals for dU, ¯ dQ , and ¯ dW are<br />
easy, since along each part of each p<strong>at</strong>h either dx or dy is zero.<br />
First take p<strong>at</strong>h one.<br />
U(1, 1) − U(0, 0) =<br />
∆Q =<br />
∆W =<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
dy 1<br />
2 (0)3 +<br />
dy 1<br />
6 (0)3 +<br />
dy 1<br />
3 (0)3 +<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
dx 1<br />
2 x21 = 1<br />
6<br />
dxx 2 1 = 1<br />
3<br />
dx(− 1<br />
2 )x21 = − 1<br />
6<br />
(1.14)<br />
(1.15)<br />
(1.16)<br />
First of all, the change in U is consistent with the st<strong>at</strong>e function we found,<br />
U(x, y) = 1<br />
3 x3 y + U0. Second, we have ∆U = ∆Q − ∆W indeed. It is easy to
1.4. FIRST LAW: ENERGY. 17<br />
calcul<strong>at</strong>e th<strong>at</strong> for the second p<strong>at</strong>h we have ∆U = 1<br />
1<br />
1<br />
3 , ∆Q = 2 , and ∆W = 6 .<br />
The change in internal energy is indeed the same, and the first law is again<br />
s<strong>at</strong>isfied.<br />
Importance of Q and W not being st<strong>at</strong>e functions.<br />
Life on earth would have been very different if Q and W would have been<br />
st<strong>at</strong>e variables. Steam engines would not exist, and you can imagine all consequences<br />
of th<strong>at</strong> fact.<br />
Expand on the consequences of Q and W being st<strong>at</strong>e functions<br />
Any engine repe<strong>at</strong>s a certain cycle over and over again. A complete cycle<br />
in our example above might be represented by a series of continuous changes in<br />
the st<strong>at</strong>e variables (x,y) like (0, 0) → (0, 1) → (1, 1) → (1, 0) → (0, 0). After the<br />
completion of one cycle, the energy U is the same as <strong>at</strong> the start of the cycle.<br />
The change in he<strong>at</strong> for this cycle is ∆Q = 1 1 1<br />
6 − 2 = − 3 and the work done on<br />
the environment is ∆W = − 1 1 1<br />
6 − 6 = − 3 . This cycle represents a he<strong>at</strong>er: since<br />
∆Q is neg<strong>at</strong>ive, he<strong>at</strong> is added to the environment and since ∆W is neg<strong>at</strong>ive<br />
the environment does work on the system. Running the cycle in the opposite<br />
direction yields an engine converting he<strong>at</strong> into work. If Q and W would be st<strong>at</strong>e<br />
variables, for each complete cycle we would have ∆Q = ∆W = 0, and no net<br />
change of work into he<strong>at</strong> and vice-versa would be possible!<br />
When was the first steam engine constructed?<br />
Work can be done in many different ways. A change in any of the extensive<br />
st<strong>at</strong>e variables of the system will cause a change in energy, or needs a force<br />
in order th<strong>at</strong> it happens. Consider a system with volume V, surface area A,<br />
polariz<strong>at</strong>ion ⃗ P , magnetic moment ⃗ M, and number of moles of m<strong>at</strong>erial N. The<br />
work done by the system on the environment is<br />
¯dW = pdV − σdA − ⃗ Ed ⃗ P − ⃗ Hd ⃗ M − µdN (1.17)<br />
where the forces are rel<strong>at</strong>ed to the intensive variables pressure p, surface tension<br />
σ, electric field ⃗ E, magnetic field ⃗ H, and chemical potential µ. Note th<strong>at</strong> some<br />
textbooks tre<strong>at</strong> the µdN term in a special way. There is, however, no formal<br />
need to do so. The general form is<br />
¯dW = − ∑<br />
xjdXj<br />
j<br />
(1.18)<br />
where the generalized force xj causes a generalized displacement dXj in the<br />
st<strong>at</strong>e variable Xj.
18 CHAPTER 1. BASIC THERMODYNAMICS.<br />
The signs in work are normally neg<strong>at</strong>ive. If we increase the total magnetic<br />
moment of a sample in an external magnetic field, we have to add energy to<br />
the sample. In other words, an increase in the total magnetic moment increases<br />
the internal energy, and work has to be done on the sample. The work done<br />
by the sample is neg<strong>at</strong>ive. Note th<strong>at</strong> the pdV term has the opposite sign from<br />
all others. If we increase the volume of a sample, we push outside m<strong>at</strong>erial<br />
away, and do work on the outside. A positive pressure decreases the volume,<br />
while a positive magnetic field increases the magnetic magnetiz<strong>at</strong>ion in general.<br />
This difference in sign is on one historical, and is justified by the old, intuitive<br />
definitions of pressure and other quantities. But it also has a deeper meaning.<br />
Volume tells us how much space the sample occupies, while all other extensive<br />
quantities tell us how much of something is in th<strong>at</strong> space. In terms of densities,<br />
volume is in the denomin<strong>at</strong>or, while all other variables are in the numer<strong>at</strong>or.<br />
This gives a change in volume the opposite effect from all other changes.<br />
1.5 Second law: Entropy.<br />
Clausius and Kelvin.<br />
The second law of thermodynamics tells us th<strong>at</strong> life is not free. According<br />
to the first law we can change he<strong>at</strong> into work, apparently without limits. The<br />
second law, however, puts restrictions on this exchange. There are two versions<br />
of the second law, due to Kelvin and Clausius. Clausius st<strong>at</strong>ed th<strong>at</strong> there are<br />
no thermodynamic processes in which the only net change is a transfer of he<strong>at</strong><br />
from one reservoir to a second reservoir which has a higher temper<strong>at</strong>ure. Kelvin<br />
formul<strong>at</strong>ed it in a different way: there are no thermodynamic processes in which<br />
the only effect is to extract a certain amount of he<strong>at</strong> from a reservoir and convert<br />
it completely into work. The opposite is possible, though. These two st<strong>at</strong>ements<br />
are equivalent as we will see.<br />
He<strong>at</strong> is a special form of energy exchange.<br />
The second law singles out he<strong>at</strong> as compared to all other forms of energy.<br />
Since work is defined via a change in the extensive st<strong>at</strong>e variables, we can think<br />
of he<strong>at</strong> as a change of the internal degrees of freedom of the system. Hence<br />
he<strong>at</strong> represents all the degrees of freedom we have swept under the rug when<br />
we limited st<strong>at</strong>e variables to measurable, average macroscopic quantities. The<br />
only reason th<strong>at</strong> we can say anything <strong>at</strong> all about he<strong>at</strong> is th<strong>at</strong> it is connected to<br />
an extremely large number of variables (because of the thermodynamic limit).<br />
In th<strong>at</strong> case the m<strong>at</strong>hem<strong>at</strong>ical laws of large numbers apply, and the st<strong>at</strong>ements<br />
about he<strong>at</strong> become purely st<strong>at</strong>istical. In st<strong>at</strong>istical mechanics we will return<br />
to this point. Note th<strong>at</strong> the second law does not limit the exchange of energy<br />
switching from one form of work to another. In principal we could change
1.5. SECOND LAW: ENTROPY. 19<br />
mechanical work into electrical work without penalty! In practice, he<strong>at</strong> is always<br />
gener<strong>at</strong>ed.<br />
He<strong>at</strong> as a measurable quantity.<br />
One important implicit assumption in these st<strong>at</strong>ements is th<strong>at</strong> a large outside<br />
world does exist. In the first law we define the change of energy via an<br />
exchange of he<strong>at</strong> and work with the outside world. Hence we assume th<strong>at</strong> there<br />
is something outside our system. As a consequence, the second law does not<br />
apply to the universe as a whole. This is actually quite important. In thermodynamics<br />
we discuss samples, we observe sample, and we are on the outside.<br />
We have large reservoirs available to set pressure or temper<strong>at</strong>ure values. Hence<br />
when we take the thermodynamic limit for the sample, we first have to take the<br />
limit for the outside world and make it infinitely large. This point will come<br />
back in st<strong>at</strong>istical mechanics, and the first to draw <strong>at</strong>tention to it was Maxwell<br />
when he deployed his demon.<br />
The definitions of he<strong>at</strong> and entropy in thermodynamics are based on quantities<br />
th<strong>at</strong> we can measure. They are oper<strong>at</strong>ional definitions. The second law is<br />
an experimental observ<strong>at</strong>ion, which has never been falsified in macroscopic experiments.<br />
Maxwell started an important discussion trying to falsify the second<br />
law on a microscopic basis (his famous demon), but th<strong>at</strong> never worked either.<br />
It did lead to important st<strong>at</strong>ements about computing, though!<br />
If the second law is universally valid, it defines a preferred direction of time<br />
(by increasing entropy or energy stored in the unusable internal variables), and<br />
seems to imply th<strong>at</strong> every system will die a he<strong>at</strong> de<strong>at</strong>h. This is not true, however,<br />
because we always invoke an outside world, and <strong>at</strong> some point he<strong>at</strong> will have<br />
to flow from the system to the outside. This is another interesting point of<br />
discussion in the philosophy of science.<br />
In st<strong>at</strong>istical mechanics we can define entropy and energy by considering<br />
the system only, and is seems possible to define the entropy of the universe in<br />
th<strong>at</strong> way. Here one has to keep in mind th<strong>at</strong> the connection between st<strong>at</strong>istical<br />
mechanics and thermodynamics has to be made, and as soon as we make th<strong>at</strong><br />
connection we invoke an outside world. This is an interesting point of deb<strong>at</strong>e,<br />
too, which takes place on the same level as the deb<strong>at</strong>e in quantum mechanics<br />
about the interpret<strong>at</strong>ion of wave functions and changes in the wave functions.<br />
Carnot engine.<br />
We have seen before th<strong>at</strong> machines th<strong>at</strong> run in cycles are useful to do work.<br />
In the following we will consider such a machine. The important aspect of the<br />
machine is th<strong>at</strong> every step is reversible. The second law leads to the important<br />
conclusion th<strong>at</strong> all reversible machines using the same process have the same<br />
efficiency. We want to make a connection with temper<strong>at</strong>ure, and therefore we<br />
define an engine with a cycle in which two parts are <strong>at</strong> constant temper<strong>at</strong>ures, in<br />
order to be able to compare values of these temper<strong>at</strong>ures. The other two parts
20 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Figure 1.1: Carnot cycle in PV diagram .<br />
are simplified by making them adiab<strong>at</strong>ic, so no he<strong>at</strong> is exchanged. Connecting<br />
the workings of such engines with the second law will then allow us to define a<br />
temper<strong>at</strong>ure scale, and also define entropy.<br />
An engine is a system which changes its thermodynamic st<strong>at</strong>e in cycles and<br />
converts he<strong>at</strong> into work by doing so. A Carnot engine is any system repe<strong>at</strong>ing<br />
the following reversible cycle: (1) an isothermal expansion <strong>at</strong> a high temper<strong>at</strong>ure<br />
T1, (2) an adiab<strong>at</strong>ic expansion in which the temper<strong>at</strong>ure is lowered to T2,<br />
(3) an isothermal contraction <strong>at</strong> temper<strong>at</strong>ure T2, and finally (4) an adiab<strong>at</strong>ic<br />
contraction back to the initial st<strong>at</strong>e. In this case work is done using a change<br />
in volume. Similar Carnot engines can be defined for all other types of work. It<br />
is easiest to talk about Carnot engines using the pressure p, the volume V, and<br />
the temper<strong>at</strong>ure T as variables. A diagram of a Carnot engine in the pV plane<br />
is shown in figure 1.1.<br />
The m<strong>at</strong>erial in a Carnot engine can be anything. For practical reasons it<br />
is often a gas. Also, because steps one and three are isothermal, contact with<br />
a he<strong>at</strong> reservoir is required, and the Carnot engine oper<strong>at</strong>es between these two<br />
he<strong>at</strong> reservoirs, by definition. Mechanical work is done in all four parts of the<br />
cycle. We can define Carnot engines for any type of work, but mechanical work is<br />
the easiest to visualize (and construction of Carnot engines based on mechanical<br />
work is also most common). One also recognizes the historical importance of<br />
steam engines; such engines ”drove” the development of thermodynamics!<br />
Carnot engines are the most efficient!<br />
The second law of thermodynamics has a very important consequence for<br />
Carnot engines. One can show th<strong>at</strong> a Carnot engine is the most efficient engine<br />
oper<strong>at</strong>ing between two reservoirs <strong>at</strong> temper<strong>at</strong>ure T1 and T2! This is a very strong<br />
st<strong>at</strong>ement, based on minimal inform<strong>at</strong>ion. The efficiency η is the r<strong>at</strong>io of the<br />
work W performed on the outside world and the he<strong>at</strong> Q1 absorbed by the system
Ì É<br />
1.5. SECOND LAW: ENTROPY. 21<br />
Ï<br />
Ì É<br />
Figure 1.2: Schem<strong>at</strong>ics of a Carnot engine.<br />
in the isothermal step one <strong>at</strong> high temper<strong>at</strong>ure. Remember th<strong>at</strong> in steps two<br />
and four no he<strong>at</strong> is exchanged. The he<strong>at</strong> absorbed from the reservoir <strong>at</strong> low<br />
temper<strong>at</strong>ure in step three is Q2 and the first law tells us th<strong>at</strong> W = Q1 + Q2.<br />
We define the flow of he<strong>at</strong> Qi to be positive when he<strong>at</strong> flows into the system.<br />
In most engines we will, of course, have Q1 > 0 and Q2 < 0. This gives us<br />
η = W<br />
= 1 +<br />
Q1<br />
Q2<br />
Q1<br />
(1.19)<br />
Work is positive when it represents a flow of energy to the outside world. A<br />
Carnot engine in reverse is a he<strong>at</strong>er (or refriger<strong>at</strong>or depending on which reservoir<br />
you look <strong>at</strong>).<br />
Can the efficiency be gre<strong>at</strong>er than one?<br />
A Carnot engine can be represented in as follows, see figure 1.2. In this<br />
figure the arrows point in the direction in which the energy flow is defined to<br />
be positive.<br />
Equivalency of Clausius and Kelvin.<br />
The two formul<strong>at</strong>ions of the second law of Clausius and Kelvin are equivalent.<br />
If a Kelvin engine existed which converts he<strong>at</strong> completely into work, this work<br />
can be transformed into he<strong>at</strong> dumped into a reservoir <strong>at</strong> higher temper<strong>at</strong>ure, in<br />
contradiction with Clausius. If a Clausius process would exist, we can use it to<br />
store energy <strong>at</strong> a higher temper<strong>at</strong>ure. A normal engine would take this amount<br />
of he<strong>at</strong>, dump he<strong>at</strong> <strong>at</strong> the low temper<strong>at</strong>ure again while performing work, and
Ì<br />
22 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Ì É<br />
É ÏÏ É Ï<br />
<br />
É<br />
Figure 1.3: Two engines feeding eachother.<br />
there would be a contradiction with Kelvin’s formul<strong>at</strong>ion of the second law. The<br />
st<strong>at</strong>ement about Carnot engines is next shown to be true in a similar way.<br />
Contradictions if existence of more efficient engine.<br />
Assume th<strong>at</strong> we have an engine X which is more efficient than a Carnot<br />
engine C. We will use this engine X to drive a Carnot engine in reverse, see<br />
figure 1.3. The engine X takes an amount of he<strong>at</strong> QX > 0 from a reservoir <strong>at</strong><br />
high temper<strong>at</strong>ure. It produces an amount of work W = ηXQX > 0 and takes<br />
an amount of he<strong>at</strong> Q2X = (ηX − 1)QX from the reservoir <strong>at</strong> low temper<strong>at</strong>ure.<br />
Notice th<strong>at</strong> we need ηX < 1, (and hence Q2X < 0 ), otherwise we would<br />
viol<strong>at</strong>e Kelvin’s formul<strong>at</strong>ion of the second law. This means th<strong>at</strong> the net flow<br />
of he<strong>at</strong> is towards the reservoir of low temper<strong>at</strong>ure. Now take a Carnot engine<br />
oper<strong>at</strong>ing between the same two reservoirs. This Carnot engine is driven by<br />
the amount of work W, hence the amount of work performed by the Carnot<br />
engine is WC = −W . This Carnot engine takes an amount of he<strong>at</strong> Q1C =<br />
WC ηX = − ηC ηC QX from the reservoir <strong>at</strong> high temper<strong>at</strong>ure and an amount Q2C =<br />
WC − Q1C = ( 1<br />
ηC − 1)ηXQX from the reservoir <strong>at</strong> low temper<strong>at</strong>ure. Now<br />
consider the combin<strong>at</strong>ion of these two engines. This is a machine which takes<br />
an amount of he<strong>at</strong> (1 − ηx<br />
ηc )Qx from the reservoir <strong>at</strong> high temper<strong>at</strong>ure and the<br />
opposite amount from the reservoir <strong>at</strong> low temper<strong>at</strong>ure. Energy is conserved,<br />
but Clausius tells us th<strong>at</strong> the amount of he<strong>at</strong> taken from the high temper<strong>at</strong>ure<br />
reservoir should be positive, or ηx ≤ ηc. Hence a Carnot engine is the most<br />
efficient engine which one can construct!<br />
In a different proof we can combine an engine X and a Carnot engine, but<br />
require Q2X + Q2C = 0. Such an engine produces an amount of work Wnet<br />
which has to be neg<strong>at</strong>ive according to Kelvin.
1.5. SECOND LAW: ENTROPY. 23<br />
Show th<strong>at</strong> this implies ηX ≤ ηC.<br />
All Carnot engines are equally efficient.<br />
One can easily show th<strong>at</strong> all Carnot engines have the same efficiency. Suppose<br />
the efficiencies of Carnot engine one and two are η1 and η2, respectively.<br />
Use one Carnot engine to drive the other in reverse, and it follows th<strong>at</strong> we need<br />
η1 ≤ η2 and also η2 ≤ η1, or η1 = η2. Hence the efficiency of an arbitrary<br />
Carnot engine is ηC. This is independent of the details of the Carnot engine,<br />
except th<strong>at</strong> it should oper<strong>at</strong>e between a reservoir <strong>at</strong> T1 and a reservoir <strong>at</strong> T2.<br />
These are the only two variables which play a role, and the Carnot efficiency<br />
should depend on them only: ηC(T1, T2).<br />
Carnot efficiency can be measured experimentally.<br />
This efficiency function can be determined experimentally by measuring Q<br />
and W flowing in and out of a given Carnot engine. How? Th<strong>at</strong> is a problem.<br />
First, consider the work done. This is the easier part. For example, because of<br />
the work done a weight is lifted a certain distance. This gives us the change<br />
in energy, and hence the work done. In order to use this type of measurement,<br />
however, we need to know details about the type of work. This is essentially<br />
the same as saying th<strong>at</strong> we need to understand the measurements we are doing.<br />
How do we measure he<strong>at</strong>? We need a reference. For example, take a large<br />
closed glass container with w<strong>at</strong>er and ice, initially in a one to one r<strong>at</strong>io. Assume<br />
th<strong>at</strong> the amount of energy to melt a unit mass of ice is our basic energy value.<br />
We can measure the amount of he<strong>at</strong> th<strong>at</strong> went into this reference system by<br />
measuring the change in the volumes of w<strong>at</strong>er and ice. Also, if a sample of<br />
unknown temper<strong>at</strong>ure is brought into contact with the reference system, we can<br />
easily determine whether the temper<strong>at</strong>ure of the sample is higher or lower then<br />
the reference temper<strong>at</strong>ure of the w<strong>at</strong>er and ice system. If it is higher, ice will<br />
melt, if it is lower, w<strong>at</strong>er will freeze. Note th<strong>at</strong> we assume th<strong>at</strong> the temper<strong>at</strong>ure<br />
of the reference system is positive!<br />
Experimental definition of temper<strong>at</strong>ure.<br />
St<strong>at</strong>e variables are average, macroscopic quantities of a system which can be<br />
measured. This is certainly a good definition of variables like volume, pressure,<br />
and number of particles. They are rel<strong>at</strong>ed to basic concepts like length, mass,<br />
charge, and time. Temper<strong>at</strong>ure is a different quantity, however. A practical<br />
definition of the temper<strong>at</strong>ure of an object is via a thermometer. The active<br />
substance in the thermometer could be mercury or some ideal gas. But those<br />
are definitions which already incorpor<strong>at</strong>e some physics, like the linear expansion<br />
of solids for mercury or the ideal gas law for a gas. It is different from the
24 CHAPTER 1. BASIC THERMODYNAMICS.<br />
definitions of length and time in terms of the standard meter and clock. In a<br />
similar vein we would like to define temper<strong>at</strong>ure as the result of a measurement<br />
of a comparison with a standard. Hence we assume th<strong>at</strong> we have a known object<br />
of temper<strong>at</strong>ure T0, similar to the standard meter and clock. An example would<br />
be the container with the w<strong>at</strong>er and ice mixture mentioned above.<br />
Now how do we compare temper<strong>at</strong>ures on a quantit<strong>at</strong>ive level? If we want<br />
to find the temper<strong>at</strong>ure of an object of unknown temper<strong>at</strong>ure T, we take a<br />
Carnot engine and oper<strong>at</strong>e th<strong>at</strong> engine between the object and the standard. We<br />
measure the amount of he<strong>at</strong> Q flowing from the reference system to the Carnot<br />
engine or from the Carnot engine to the reference system. We also measure<br />
the amount of work W done by the Carnot engine. We use the first law to<br />
determine the amount of he<strong>at</strong> flowing out of the high temper<strong>at</strong>ure reservoir, if<br />
needed. The r<strong>at</strong>io of these two quantities is the efficiency of the Carnot engine,<br />
which only depends on the two temper<strong>at</strong>ures.<br />
We first determine if the object has a higher or lower temper<strong>at</strong>ure then<br />
the reference by bringing them in direct contact. If ice melts, the object was<br />
warmer, if ice forms, it was colder.<br />
If the temper<strong>at</strong>ure of the reference system is higher th<strong>at</strong> the temper<strong>at</strong>ure of<br />
the object, we use the reference system as the high temper<strong>at</strong>ure reservoir. We<br />
measure the amount of he<strong>at</strong> Q going out of the reference system and find:<br />
ηC = W<br />
Q<br />
(1.20)<br />
If the temper<strong>at</strong>ure of the reference system is lower th<strong>at</strong> the temper<strong>at</strong>ure of<br />
the object, we use the reference system as the low temper<strong>at</strong>ure reservoir. We<br />
measure the amount of he<strong>at</strong> Q going out of the reference system, which is now<br />
neg<strong>at</strong>ive, since he<strong>at</strong> is actually going in, and find:<br />
ηC = W<br />
W − Q<br />
In the first case we assign a temper<strong>at</strong>ure T to the object according to<br />
T<br />
T0<br />
and in the second case according to<br />
(1.21)<br />
= (1 − ηC) (1.22)<br />
T0<br />
T = (1 − ηC) (1.23)<br />
This is our definition of temper<strong>at</strong>ure on the Carnot scale. It is an important<br />
step forward, based on the unique efficiency of Carnot engines, which in<br />
itself is based on the second law. Theoretically, this is a good definition because<br />
Carnot engines are well-defined. Also, energy is well-defined. The important<br />
question, of course, is how this definition rel<strong>at</strong>es to known temper<strong>at</strong>ure scales.<br />
We will rel<strong>at</strong>e the Carnot temper<strong>at</strong>ure scale to the ideal gas temper<strong>at</strong>ure scale<br />
<strong>at</strong> the end of this chapter.
Ì<br />
1.5. SECOND LAW: ENTROPY.<br />
É<br />
25<br />
Ï<br />
Ì<br />
<br />
Ì É<br />
É<br />
Ï<br />
Figure 1.4: Two Carnot engines in series.<br />
Efficiency for arbitrary temper<strong>at</strong>ures.<br />
We can analyze the general situ<strong>at</strong>ion for a Carnot engine between arbitrary<br />
temper<strong>at</strong>ures as follows. Assume th<strong>at</strong> we have T > T0 > T ′ , all other cases<br />
work similarly. Consider the following couple of Carnot engines (see figure 1.4<br />
) and demand th<strong>at</strong> Q ′ 1 + Q2 = 0 (no he<strong>at</strong> going in or out the reference system).<br />
Argue th<strong>at</strong> this is equivalent to a single Carnot engine working between T and<br />
T ′ .<br />
′<br />
T For this system we have T0 = (1 − η′ T0<br />
C ) and T = (1 − ηC),<br />
′<br />
T or T = (1 −<br />
η ′ C )(1 − ηC). The efficiencies can be expressed in the energy exchanges and we<br />
′<br />
′<br />
T W have T = (1 − Q ′ )(1 −<br />
1<br />
W<br />
Q1 ). But we have Q′ 1 = −Q2 = Q1 − W and hence<br />
T ′<br />
′<br />
′<br />
W W<br />
W W<br />
T = (1 − Q1−W )(1 − ). The right hand side is equal to 1 − − Q1 Q1 Q1−W (1 −<br />
′<br />
W<br />
W W ) = 1 − − . In other words:<br />
Q1 Q1 Q1<br />
T ′<br />
T<br />
′ W + W<br />
= 1 − = 1 − ηC<br />
where the rel<strong>at</strong>ion now holds for arbitrary values of the temper<strong>at</strong>ure.<br />
Carnot cycle again.<br />
Q1<br />
Can we obtain neg<strong>at</strong>ive values of the temper<strong>at</strong>ure?<br />
(1.24)
26 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Using the temper<strong>at</strong>ure scale defined by the Carnot engine, we can reanalyze<br />
the Carnot cycle. The efficiency is rel<strong>at</strong>ed to the he<strong>at</strong> ∆Q1 absorbed in the first<br />
step and ∆Q2 absorbed in the third step (which is neg<strong>at</strong>ive in an engine) by<br />
ηC = 1 + ∆Q2<br />
(1.25)<br />
∆Q1<br />
where we have used a not<strong>at</strong>ion with ∆Q to emphasize the fact th<strong>at</strong> we look<br />
<strong>at</strong> changes. But th<strong>at</strong> is not really essential. From the previous equ<strong>at</strong>ion we find:<br />
∆Q1<br />
+<br />
T1<br />
∆Q2<br />
= 0 (1.26)<br />
T2<br />
Since there is no he<strong>at</strong> exchanged in steps two and four of the Carnot cycle, this<br />
is equivalent to<br />
<br />
C<br />
¯dQ<br />
T<br />
= 0 (1.27)<br />
where the closed contour C specifies the p<strong>at</strong>h of integr<strong>at</strong>ion in the space of st<strong>at</strong>e<br />
variables.<br />
Integral for arbitrary cycles.<br />
Next we consider the combined effect of two Carnot engines, one working<br />
between T1 and T2, the other one between T2 and T3. Now compare this with a<br />
single system which follows the thermodynamic transform<strong>at</strong>ion defined by the<br />
outside of the sum of the two Carnot contours. One can think of the total process<br />
as the sum of the two Carnot steps, introducing an intermedi<strong>at</strong>e reservoir, in<br />
which no net he<strong>at</strong> is deposited. The contour integral of ¯ dQ<br />
T is also zero for the<br />
single process, since the two contributions over the common line are opposite<br />
and cancel. Any general closed p<strong>at</strong>h in the space of st<strong>at</strong>e variables, restricted to<br />
those surfaces which are allowed by the equ<strong>at</strong>ions of st<strong>at</strong>e, can be approxim<strong>at</strong>ed<br />
as the sum of a number of Carnot cycles with temper<strong>at</strong>ure difference ∆T . The<br />
error in this approxim<strong>at</strong>ion approaches zero for ∆T → 0. Hence:<br />
<br />
¯dQ<br />
= 0 (1.28)<br />
T<br />
where R is an arbitrary cyclic, reversible process.<br />
Definition of entropy.<br />
Formula 1.28 has the important consequence th<strong>at</strong><br />
We define a new variable S by<br />
R<br />
∫ 2<br />
S2 = S1 +<br />
1<br />
¯dQ<br />
T<br />
2∫<br />
1<br />
¯dQ<br />
T<br />
is p<strong>at</strong>h independent.<br />
(1.29)
1.5. SECOND LAW: ENTROPY. 27<br />
and because the integral is p<strong>at</strong>h independent S is a st<strong>at</strong>e function. When the<br />
integr<strong>at</strong>ion points are close together we get<br />
T dS = ¯ dQ (1.30)<br />
in which dS is an exact differential. The quantity S is called the entropy. In<br />
thermodynamics we define the entropy from a purely macroscopic point of view.<br />
It is rel<strong>at</strong>ed to infinitesimally small exchanges of thermal energy by requiring<br />
th<strong>at</strong> the differential ¯ dQ can be transformed into an exact differential by multiplying<br />
it with a function of the temper<strong>at</strong>ure alone. One can always transform a<br />
differential into an exact differential by multiplying it with a function of all st<strong>at</strong>e<br />
variables. In fact, there are an infinite number of ways to do this. The restriction<br />
th<strong>at</strong> the multiplying factor only depends on temper<strong>at</strong>ure uniquely defines<br />
this factor, apart from a constant factor. One could also define 5T dS = ¯ dQ ,<br />
which would simply re-scale all temper<strong>at</strong>ure values by a factor five.<br />
First law in exact differentials.<br />
The first law of thermodynamics in terms of changes in the entropy is<br />
dU = T dS − ¯ dW (1.31)<br />
For example, if we consider a system where the only interactions with the outside<br />
world are a possible exchange of he<strong>at</strong> and mechanical work, changes in the<br />
internal energy are rel<strong>at</strong>ed to changes in the entropy and volume through<br />
dU = T dS − pdV (1.32)<br />
An important note <strong>at</strong> this point is th<strong>at</strong> we often use the equ<strong>at</strong>ion 1.32 as a<br />
model. It does indeed exemplify some basic concepts, but for real applic<strong>at</strong>ions<br />
it is too simple. The simplest form of the first law th<strong>at</strong> has physical meaning is<br />
the following:<br />
Entropy is extensive.<br />
dU = T dS − pdV + µdN (1.33)<br />
In the definition of the Carnot temper<strong>at</strong>ure of an object the size of the object<br />
does not play a role, only the fact th<strong>at</strong> the object is in thermal equilibrium. As<br />
a consequence the temper<strong>at</strong>ure is an intensive quantity. On the other hand, if<br />
we compare the he<strong>at</strong> absorbed by a system during a thermodynamic process<br />
with the he<strong>at</strong> absorbed by a similar system which is α times larger, it is not<br />
hard to argue th<strong>at</strong> the amount of he<strong>at</strong> exchanged is α times larger as well. As<br />
a consequence, the entropy S is an extensive st<strong>at</strong>e variable.
28 CHAPTER 1. BASIC THERMODYNAMICS.<br />
N<strong>at</strong>ural variables for the internal energy are all extensive st<strong>at</strong>e<br />
variables.<br />
Changes in the internal energy U are rel<strong>at</strong>ed to changes in the extensive<br />
st<strong>at</strong>e variables only, since the amount of work done is determined by changes<br />
in extensive st<strong>at</strong>e variables only and S is extensive. In this sense, the n<strong>at</strong>ural<br />
set of variables for the st<strong>at</strong>e function U is the set of all extensive variables. By<br />
n<strong>at</strong>ural we mean the set of variables th<strong>at</strong> show as differentials in the first law,<br />
hence small changes are directly rel<strong>at</strong>ed.<br />
Importance of thermodynamic limit.<br />
Equ<strong>at</strong>ion 1.31 has an interesting consequence. Suppose th<strong>at</strong> we decide th<strong>at</strong><br />
another extensive st<strong>at</strong>e variable is needed to describe the st<strong>at</strong>e of a system.<br />
Hence we are adding a term xdX to ¯ dW . This means th<strong>at</strong> the number of internal<br />
degrees of freedom is reduced by one, since we are specifying one additional<br />
combin<strong>at</strong>ion of degrees of freedom via X. This in its turn indic<strong>at</strong>es th<strong>at</strong> the<br />
entropy should change, since it is a represent<strong>at</strong>ion of the internal degrees of<br />
freedom of the system. The definition of the entropy would therefore depend<br />
on the definition of work, which is an unacceptable situ<strong>at</strong>ion. Fortun<strong>at</strong>ely, the<br />
thermodynamic limit comes to rescue here. Only when the number of degrees of<br />
freedom is infinitely large, the change by one will not alter the entropy. Hence<br />
the entropy is only well-defined in the thermodynamic limit.<br />
Independent and dependent variables.<br />
and<br />
From equ<strong>at</strong>ion 1.32 we find immedi<strong>at</strong>ely th<strong>at</strong><br />
( )<br />
∂U<br />
T =<br />
∂S V<br />
p = −<br />
( )<br />
∂U<br />
∂V S<br />
(1.34)<br />
(1.35)<br />
which shows th<strong>at</strong> in the set of variables p,V,T,S only two are independent. If<br />
we know the basic physics of the system, we know the st<strong>at</strong>e function U(S, V ) and<br />
can derive the values for p and T according to the two st<strong>at</strong>e functions defined by<br />
the partial deriv<strong>at</strong>ives. Functions of the form T = f(S, V ) and p = g(S, V ) are<br />
called equ<strong>at</strong>ions of st<strong>at</strong>e. More useful forms elimin<strong>at</strong>e the entropy from these<br />
equ<strong>at</strong>ions and lead to equ<strong>at</strong>ions of st<strong>at</strong>e of the form p = h(T, V ). The rel<strong>at</strong>ion<br />
U = u(S, V ) is called an energy equ<strong>at</strong>ion and is not an equ<strong>at</strong>ion of st<strong>at</strong>e, since<br />
it defines an energy as a function of the independent st<strong>at</strong>e variables. Such<br />
rel<strong>at</strong>ions are the basis for equ<strong>at</strong>ions of st<strong>at</strong>e, but we use equ<strong>at</strong>ion of st<strong>at</strong>e only<br />
when we describe st<strong>at</strong>e variables th<strong>at</strong> occur in pairs, like T and S, or p and V .
1.5. SECOND LAW: ENTROPY. 29<br />
Equ<strong>at</strong>ions of st<strong>at</strong>e give dependent st<strong>at</strong>e variables of this n<strong>at</strong>ure as a function of<br />
independent ones.<br />
In the next chapter we will discuss how to change variables and make combin<strong>at</strong>ions<br />
like T and V independent and the others dependent.<br />
Entropy in terms of work.<br />
We now return to the definition of entropy according to equ<strong>at</strong>ion 1.29 above.<br />
If we apply the first law we get<br />
∫ 2<br />
dU +<br />
S2 = S1 +<br />
1<br />
¯ dW<br />
T<br />
and if we express work in its generalized form according to 1.18 we see<br />
∫ 2<br />
S2 = S1 +<br />
1<br />
dU + ∑<br />
j xjdXj<br />
T<br />
(1.36)<br />
(1.37)<br />
which shows th<strong>at</strong> entropy is defined based on basic properties of the system,<br />
which can be directly measured. There is no device th<strong>at</strong> measures entropy<br />
directly, and in th<strong>at</strong> sense it is different from all other st<strong>at</strong>e variables. But<br />
it is possible to perform processes in which the entropy does not change, and<br />
hence we do have control over changes in the entropy. Since entropy is based<br />
on changes in other st<strong>at</strong>e variables and the internal energy, it is well defined<br />
in reversible processes. The second law singles out he<strong>at</strong> as a more restricted<br />
change of energy, and this has consequences for entropy. We now discuss those<br />
consequences.<br />
Change in entropy in an irreversible process.<br />
Up to this point we only considered the entropy in connection with reversible<br />
processes. Different rules follow for irreversible processes. Consider a general<br />
process in which he<strong>at</strong> is transferred from a reservoir <strong>at</strong> high temper<strong>at</strong>ure to a<br />
reservoir <strong>at</strong> low temper<strong>at</strong>ure (and hence Q1 > 0). The efficiency of this process<br />
is <strong>at</strong> most equal to the Carnot efficiency, and hence<br />
W<br />
Q1<br />
For such a general process we have<br />
= 1 + Q2<br />
≤ ηC = 1 −<br />
Q1<br />
T2<br />
T1<br />
Q1<br />
T1<br />
+ Q2<br />
T2<br />
(1.38)<br />
≤ 0 (1.39)<br />
Suppose th<strong>at</strong> the he<strong>at</strong> is taken from the reservoir <strong>at</strong> high temper<strong>at</strong>ure T1<br />
in a general process, but dumped in the reservoir <strong>at</strong> low temper<strong>at</strong>ure T2 in a<br />
reversible way. Suppose we only consider small changes in energy. For comparison,<br />
consider a completely reversible way of transferring the he<strong>at</strong> from one
30 CHAPTER 1. BASIC THERMODYNAMICS.<br />
reservoir to another. In this process he<strong>at</strong> is transferred reversibly from the high<br />
temper<strong>at</strong>ure reservoir to the engine. This leads to<br />
∆Q g<br />
1<br />
T1<br />
≤ − ∆Qrev 2<br />
T2<br />
= ∆Qrev 1<br />
T1<br />
= ∆S1<br />
(1.40)<br />
For a general process <strong>at</strong> a temper<strong>at</strong>ure T, in which the amount of he<strong>at</strong> added<br />
to the system is ∆Q and the change in entropy of the system is ∆S, the changes<br />
are rel<strong>at</strong>ed by<br />
T ∆S ≥ ∆Q (1.41)<br />
The equal sign holds for all reversible processes only. Hence in an irreversible<br />
process the change in entropy accompanying the absorption of a certain amount<br />
of he<strong>at</strong> is larger than necessary. The change in entropy is minimal for a reversible<br />
process. An engine in which all processes are reversible will return to its initial<br />
st<strong>at</strong>e without a change in entropy. If, however, some processes are not reversible,<br />
the entropy of the engine will increase after a complete cycle. It does not return<br />
to the same initial st<strong>at</strong>e, but to a st<strong>at</strong>e of higher entropy. The internal energy<br />
of the system decreases more than necessary, and we lose energy, which can be<br />
seen from<br />
∆U = ∆Q − ∆W ≤ T ∆S − ∆W (1.42)<br />
The price in energy we pay to transform a system from one st<strong>at</strong>e to another<br />
is higher than needed if the transform<strong>at</strong>ion is irreversible.<br />
Maximum entropy principle.<br />
If we wait long enough, a completely isol<strong>at</strong>ed system will end up in equilibrium.<br />
If this system was not in equilibrium to start with, a spontaneous process<br />
will take it there. Since the system is completely isol<strong>at</strong>ed, there is no exchange<br />
of he<strong>at</strong> with the outside world, and <strong>at</strong> each time T ∆S ≥ 0. In reaching equilibrium<br />
by a spontaneous process, the entropy of the system has increased. This<br />
is true for any non-equilibrium initial st<strong>at</strong>e. Hence we find the following very<br />
important rule:<br />
The equilibrium st<strong>at</strong>e of a completely isol<strong>at</strong>ed system has maximal<br />
entropy.<br />
By completely isol<strong>at</strong>ed system we mean th<strong>at</strong> the internal energy U, the volume<br />
V, the amount of m<strong>at</strong>erial N, and all other extensive st<strong>at</strong>e variables are kept<br />
constant. The way to change the entropy is to change some internal variables,<br />
hence the entropy is a maximum as a function of the internal variables only.<br />
This is quite a strong st<strong>at</strong>ement based on some minimal input (the second law<br />
of thermodynamics).
1.6. THIRD LAW OF THERMODYNAMICS. 31<br />
1.6 Third law of thermodynamics.<br />
Using equ<strong>at</strong>ion 1.29 we can find the entropy difference between two arbitrary<br />
st<strong>at</strong>es. This still leaves us with one constant of integr<strong>at</strong>ion, which could be<br />
system dependent. The third law of thermodynamics sets this constant to zero<br />
<strong>at</strong> zero temper<strong>at</strong>ure for all systems. The short form of the third law is<br />
S(T = 0) = 0 (1.43)<br />
but this is not completely correct. First of all, we cannot reach zero temper<strong>at</strong>ure,<br />
and second we need to take the thermodynamic limit. Therefore, the correct<br />
formul<strong>at</strong>ion of the third law is:<br />
lim<br />
N→∞ lim<br />
S(T )<br />
= 0 (1.44)<br />
T →0 N<br />
Note th<strong>at</strong> we cannot interchange the order of the limits, we have to take the limit<br />
for the temper<strong>at</strong>ure first. This is a general characteristic of the thermodynamic<br />
limit, we have to take this limit last, after we have done all our calcul<strong>at</strong>ions!<br />
This is an important point to remember, and a source of errors if done wrong.<br />
One very important aspect of the third law is th<strong>at</strong> the value of the entropy is<br />
zero (in the sense discussed above) <strong>at</strong> zero temper<strong>at</strong>ure, no m<strong>at</strong>ter which values<br />
the other st<strong>at</strong>e variables have. Th<strong>at</strong> is quite restrictive. Another aspect of the<br />
third law is th<strong>at</strong> it does not tell us how close to zero the temper<strong>at</strong>ure has to<br />
be. There are systems in which the entropy appears to approach a non-zero<br />
value, until we reach a very low temper<strong>at</strong>ure, <strong>at</strong> which point the entropy does<br />
suddenly drop.<br />
Response <strong>at</strong> zero temper<strong>at</strong>ure.<br />
At zero temper<strong>at</strong>ure the entropy becomes infinitesimally small compared to<br />
the number of degrees of freedom in the thermodynamic limit. In the language of<br />
quantum mechanics, the degeneracy of the ground st<strong>at</strong>e is very small compared<br />
to the number of degrees of freedom. The third law seems to hold for all physical<br />
systems. It has an interesting consequence. Suppose we change the st<strong>at</strong>e of a<br />
system by changing the st<strong>at</strong>e variable X, but keep the temper<strong>at</strong>ure <strong>at</strong> zero.<br />
Since S(T = 0, X) = 0, we find th<strong>at</strong> the partial deriv<strong>at</strong>ive of S with respect to<br />
X has to be 0, or<br />
lim<br />
N→∞ lim<br />
T →0<br />
( )<br />
1 ∂S<br />
= 0 (1.45)<br />
N ∂X T<br />
This seems to hold for all experiments, and has important consequences for the<br />
values of response functions <strong>at</strong> low values of the temper<strong>at</strong>ure. It also allows us<br />
to check if we reached low enough temper<strong>at</strong>ures. If the entropy seems constant,<br />
but partial deriv<strong>at</strong>ives like the ones above are not zero, we are not <strong>at</strong> sufficiently<br />
low temper<strong>at</strong>ures yet.
32 CHAPTER 1. BASIC THERMODYNAMICS.<br />
He<strong>at</strong> capacity <strong>at</strong> zero temper<strong>at</strong>ure.<br />
An example of a simple system is a system where the only thermodynamic<br />
variables are S, T, p, and V. The internal energy follows from dU = T dS −pdV .<br />
If we work <strong>at</strong> constant volume, dV = 0, and the first law leads to<br />
S(V, T1) − S(V, T2) =<br />
∫ T1<br />
T2<br />
dU<br />
T =<br />
∫ T1<br />
dT<br />
T2 T<br />
( )<br />
∂U<br />
∂T V<br />
(1.46)<br />
Since according to the third law the left hand side exist for T2 → 0, the integral<br />
on the right hand side must exist, which implies<br />
lim<br />
T →0<br />
This is also observed experimentally.<br />
T=0 unreachable.<br />
( )<br />
∂U<br />
∂T V<br />
= 0 (1.47)<br />
Another consequence of the third law is the impossibility of reaching T = 0K<br />
in a finite number of steps using a reversible process. Problem 7 illustr<strong>at</strong>es this<br />
with an example.<br />
1.7 Ideal gas and temper<strong>at</strong>ure.<br />
There are several temper<strong>at</strong>ure scales used in daily life, like Celcius, Fahrenheit,<br />
Reamur, and Kelvin. In thermodynamics we define the temper<strong>at</strong>ure first, via<br />
an experiment. In st<strong>at</strong>istical mechanics one defines the entropy first (technically<br />
such a definition is called an entropy analogue), and derives the temper<strong>at</strong>ure<br />
from this. Since we cannot measure the entropy directly this is not a good<br />
oper<strong>at</strong>ional definition.<br />
Equivalency of temper<strong>at</strong>ure scales.<br />
There are many ways to measure temper<strong>at</strong>ure, and experimentally they have<br />
all been shown to be equivalent. Therefore, we need only to equ<strong>at</strong>e the formal<br />
definition of temper<strong>at</strong>ure via Carnot engines to one type of measurement in order<br />
to show th<strong>at</strong> the Carnot temper<strong>at</strong>ure is really the temper<strong>at</strong>ure we measure in<br />
real experiments.<br />
Ideal gas temper<strong>at</strong>ure.<br />
One very important standard of temper<strong>at</strong>ure is based on the ideal gas. The<br />
energy equ<strong>at</strong>ion and the equ<strong>at</strong>ion of st<strong>at</strong>e for a mono<strong>at</strong>omic ideal gas are
1.7. IDEAL GAS AND TEMPERATURE. 33<br />
and<br />
U = 3<br />
pV (1.48)<br />
2<br />
pV = NRT id<br />
(1.49)<br />
The second equ<strong>at</strong>ion of st<strong>at</strong>e is used to define the ideal gas temper<strong>at</strong>ure. If<br />
a known amount of an ideal gas is in equilibrium we measure its volume and<br />
pressure, and then evalu<strong>at</strong>e the temper<strong>at</strong>ure.<br />
Ideal gas cannot be <strong>at</strong> zero temper<strong>at</strong>ure.<br />
Note th<strong>at</strong> for the ideal gas we have U = 3<br />
2NRT , and hence ( )<br />
∂U<br />
∂T V<br />
= 3<br />
2 NR.<br />
This does not approach zero in the limit T → 0. The ideal gas law U = 3<br />
2 pV<br />
is therefore in contradiction with the third law. The ideal gas law cannot be<br />
applied <strong>at</strong> low temper<strong>at</strong>ures. This is not surprising, since all m<strong>at</strong>erials transform<br />
to a liquid <strong>at</strong> sufficiently low temper<strong>at</strong>ures.<br />
Carnot cycle for ideal gas.<br />
We now construct a Carnot engine based on an ideal gas. The equ<strong>at</strong>ions<br />
of st<strong>at</strong>e show th<strong>at</strong> the internal energy of the ideal gas is determined by the<br />
ideal gas temper<strong>at</strong>ure only, and hence does not change in an isothermal process.<br />
Therefore in steps one and three of the Carnot cycle all the he<strong>at</strong> absorbed by<br />
the system is changed into work. This is not in contradiction with the second<br />
law, because in each separ<strong>at</strong>e step of a Carnot cycle the system also changes.<br />
Note the word ”only” in the second law! The Carnot cycle is given by<br />
(p1, V1, T id<br />
1 ) → (p2, V2, T<br />
one id<br />
1 ) → (p3, V3, T<br />
two id<br />
2 ) → (p4, V4, T<br />
three id<br />
2 ) → (p1, V1, T<br />
four id<br />
1 )<br />
(1.50)<br />
During steps one and three the volume and pressure are rel<strong>at</strong>ed via the<br />
equ<strong>at</strong>ion of st<strong>at</strong>e, and the first law then st<strong>at</strong>es th<strong>at</strong> the work done is equal to<br />
the he<strong>at</strong> absorbed, since the energy is constant:<br />
∆W12 = ∆Q12 =<br />
∆W34 = ∆Q34 =<br />
∫ V2<br />
V1<br />
∫ V4<br />
V3<br />
pdV = NRT id<br />
∫ V2 dV<br />
1<br />
V1 V<br />
pdV = NRT id<br />
∫ V4<br />
dV<br />
2<br />
V3 V<br />
= NRT id<br />
= NRT id<br />
1 ln( V2<br />
V1<br />
2 ln( V4<br />
V3<br />
) (1.51)<br />
) (1.52)<br />
Steps two and four are adiab<strong>at</strong>ic, and there is no he<strong>at</strong> exchange. The change<br />
in energy is the work done by the system.
34 CHAPTER 1. BASIC THERMODYNAMICS.<br />
∆W23 = 3 id<br />
NR(T1 − T<br />
2 id<br />
2 ) = −∆W41<br />
(1.53)<br />
∆Q23 = ∆Q41 = 0 (1.54)<br />
We are now able to calcul<strong>at</strong>e the efficiency of this engine, which is wh<strong>at</strong> we<br />
need in order to compare the temper<strong>at</strong>ure of the ideal gas with the theoretical<br />
temper<strong>at</strong>ure. The efficiency is given by:<br />
which leads to<br />
Connecting the elements.<br />
η = ∆W12 + ∆W23 + ∆W34 + ∆W41<br />
∆Q12<br />
η = 1 +<br />
id T2 T id<br />
ln(<br />
1<br />
V4<br />
V3 )<br />
ln( V2<br />
V1 )<br />
(1.55)<br />
(1.56)<br />
In order to analyze this further, we need to calcul<strong>at</strong>e the entropy of an ideal<br />
gas, which is constant during the adiab<strong>at</strong>ic steps. This will then allow us to<br />
connect the values of the volume in st<strong>at</strong>es 2 and 3. Using the two equ<strong>at</strong>ions<br />
describing the ideal gas and the first law in terms of the entropy and volume<br />
only, we find th<strong>at</strong> small changes in the entropy are rel<strong>at</strong>ed to changes in the<br />
temper<strong>at</strong>ure and volume by<br />
T id dS = 3<br />
3<br />
d(pV ) + pdV =<br />
2 2 NRdT id +<br />
id NRT<br />
dV (1.57)<br />
V<br />
After dividing both sides by T id , it is not hard to find th<strong>at</strong> one can integr<strong>at</strong>e<br />
dS. The entropy for an ideal gas is<br />
S(T id , V, N) = NRln((T id ) 3<br />
2 V ) + S0(N) (1.58)<br />
Changes in the entropy are rel<strong>at</strong>ed to he<strong>at</strong> exchange. In an adiab<strong>at</strong>ic process<br />
there is no he<strong>at</strong> exchange, and hence an adiab<strong>at</strong>ic process is a process <strong>at</strong> constant<br />
entropy. In the adiab<strong>at</strong>ic steps two and four of the Carnot cycle changes in the<br />
volume are rel<strong>at</strong>ed to changes in the temper<strong>at</strong>ure according to<br />
( id<br />
V3 T1 =<br />
V2 T id<br />
)<br />
2<br />
3<br />
2<br />
= V4<br />
V1<br />
which shows th<strong>at</strong> the efficiency is given by<br />
η = 1 −<br />
T id<br />
2<br />
T id<br />
1<br />
The Carnot temper<strong>at</strong>ures were defined using this efficiency. We have<br />
(1.59)<br />
(1.60)
1.7. IDEAL GAS AND TEMPERATURE. 35<br />
T2<br />
T1<br />
= 1 − η =<br />
T id<br />
2<br />
T id<br />
1<br />
(1.61)<br />
and hence the two temper<strong>at</strong>ure scales are proportional to each other. This<br />
means th<strong>at</strong> they are really the same, since any temper<strong>at</strong>ure scale can always<br />
be modified by a constant scale factor. A similar analysis can be applied to all<br />
different kinds of thermometers. Standard thermometers are equivalent to the<br />
Carnot scale.<br />
Gibbs paradox.<br />
Suppose we have a system with N moles of m<strong>at</strong>erial in a volume V <strong>at</strong> temper<strong>at</strong>ure<br />
T and we add a similar system. If we use 1.58 the total entropy of the<br />
combined system is:<br />
which has to be equal to<br />
and hence we need<br />
which leads to<br />
Stotal(T id , 2V, 2N) = 2NR ln((T id ) 3<br />
2 V ) + 2S0(N) (1.62)<br />
S(T id , 2V, 2N) = 2NR ln((T id ) 3<br />
2 2V ) + S0(2N) (1.63)<br />
2S0(N) = S0(2N) + 2NR ln(2) (1.64)<br />
S0(N) = −NR ln(N) + NRc (1.65)<br />
where c is some constant. Therefore, the entropy of an ideal gas is<br />
S(T id (<br />
, V, N) = NR ln((T id ) 3<br />
)<br />
V<br />
2 ) + c<br />
N<br />
(1.66)<br />
Hence the dependence of the entropy on N is completely fixed by requiring<br />
th<strong>at</strong> the entropy is an extensive quantity. The constant c can be found from<br />
st<strong>at</strong>istical mechanical calcul<strong>at</strong>ions, and the expression for c contains the constant<br />
¯h. This is quite remarkable, since this is a direct manifest<strong>at</strong>ion of a quantum<br />
mechanical quantity in a macroscopic equ<strong>at</strong>ion!<br />
Without the constant of integr<strong>at</strong>ion S0(N) the entropy is not extensive, and<br />
using the first part of 1.58 only gives many problems. For example, consider a<br />
container of volume V with an amount NA of m<strong>at</strong>erial A and a similar container<br />
of volume V with an amount NB of m<strong>at</strong>erial B, all <strong>at</strong> temper<strong>at</strong>ure T. If we add<br />
these containers together, the new entropy is<br />
Stotal = S(T id , 2V, NA) + S(T id , 2V, NB) (1.67)<br />
since each gas now occupies twice the volume. The change in entropy is
36 CHAPTER 1. BASIC THERMODYNAMICS.<br />
∆S = Stotal − S(T id , V, NA) − S(T id , V, NB) =<br />
S(T id , 2V, NA) − S(T id , V, NA) + S(T id , 2V, NB) − S(T id , V, NB) (1.68)<br />
which gives<br />
∆S = (NA + NB)Rln(2) (1.69)<br />
which is called the entropy of mixing. Therefore, if we mix two different gases<br />
the entropy will increase. If we ignore the constant of integr<strong>at</strong>ion in equ<strong>at</strong>ion<br />
1.58 we get exactly the same answer if we make the m<strong>at</strong>erials A and B the same<br />
(Gibbs paradox), which cannot be true, since in this case there should not be a<br />
change in entropy. Including the constant of integr<strong>at</strong>ion resolves this paradox,<br />
as we demonstr<strong>at</strong>ed when we constructed the analytical form of this constant.<br />
The first steam engine<br />
One can argue about the exact definition of a steam engine, but it is fair to<br />
say th<strong>at</strong> Hero already constructed a steam engine more than 2000 years ago.<br />
The Romans did not have the analytical methods (m<strong>at</strong>hem<strong>at</strong>ics) to analyze<br />
this new toy, and never saw any use for it. Can you imagine wh<strong>at</strong> would have<br />
happened if the Romans, who were masters of engineering, would have had good<br />
smooth roads and steam driven cars?<br />
1.8 Extra questions.<br />
1. How do we measure temper<strong>at</strong>ure?<br />
2. How large does a thermometer need to be?<br />
3. How do we measure pressure?<br />
4. Wh<strong>at</strong> is the difference between direct and indirect measurements?<br />
5. Why is characteriz<strong>at</strong>ion in terms of st<strong>at</strong>e variables important?<br />
6. How realistic are reversible processes?<br />
7. Wh<strong>at</strong> happens when a process is not reversible?<br />
8. Comment on time-scales and equilibrium.<br />
9. Wh<strong>at</strong> is the law of large numbers, how does it play a role?<br />
10. Would pV = (NRT ) 2 be a possible equ<strong>at</strong>ion of st<strong>at</strong>e?<br />
11. Is pdV an exact differential?<br />
12. How is the internal energy defined in thermodynamics?
1.8. EXTRA QUESTIONS. 37<br />
13. Why is the internal energy a st<strong>at</strong>e function?<br />
14. Why would it be bad is Q and W were st<strong>at</strong>e functions?<br />
15. How many work terms can there be?<br />
16. Why would one tre<strong>at</strong> the µdN term differently?<br />
17. Why worry about measuring he<strong>at</strong> via Carnot engines?<br />
18. Why can the efficiency not be gre<strong>at</strong>er than one?<br />
19. Why is the maximum entropy principle important?<br />
20. Wh<strong>at</strong> is the thermodynamical basis for the maximum entropy principle?<br />
21. Wh<strong>at</strong> is the microscopic basis for the maximum entropy principle?<br />
22. Give a possible diagram for an irreversible engine.<br />
23. How do we measure entropy?<br />
24. How are the laws of thermodynamics obtained?<br />
25. Can we reach neg<strong>at</strong>ive temper<strong>at</strong>ures?<br />
26. Would a measure of temper<strong>at</strong>ure via T −1 be better?<br />
27. Give an example of a function where limits cannot be interchanged.<br />
28. Explain why we cannot reach T=0 (driving force to zero).<br />
29. Wh<strong>at</strong> is an ideal gas?<br />
30. Why is the calcul<strong>at</strong>ion in sec 1.7 important?<br />
31. Using dU = T dS − pdV find the equ<strong>at</strong>ions of st<strong>at</strong>e.<br />
32. How many of the variables T,S,p,V are independent?<br />
33. Suppose th<strong>at</strong> we have for an ideal gas pV = NRT and U = Nf(T ). Wh<strong>at</strong><br />
are the restrictions on f(T )?<br />
34. In a p-V-T-S system, suppose I know T=F(p,V), U=G(p,V). Can I find<br />
S? Do so.
38 CHAPTER 1. BASIC THERMODYNAMICS.<br />
1.9 Problems for chapter 1<br />
Problem 1.<br />
A system is characterized by the st<strong>at</strong>e variables x and y. The following<br />
functions are defined via differential changes in these st<strong>at</strong>e variables:<br />
(1) ¯ df = x 2 dx + y 2 dy<br />
(2) ¯ dg = y 2 dx + x 2 dy<br />
(3) ¯ dh = 3x 2 y 2 dx + (2x 3 y + y 4 )dy<br />
For each of these functions f,g, and h determine if they correspond to a st<strong>at</strong>e<br />
function. Express each st<strong>at</strong>e function in terms of x and y.<br />
Problem 2.<br />
The quantity q rel<strong>at</strong>ed to the st<strong>at</strong>e variables x and y is defined by ¯ dq =<br />
xydx + 1<br />
xdy. Show th<strong>at</strong> q is not a st<strong>at</strong>e function. The st<strong>at</strong>e function s is defined<br />
by ds = f(x) ¯ dq where f is a function of x only. Find f(x)<br />
Problem 3.<br />
The st<strong>at</strong>e of a system is determined by the st<strong>at</strong>e variables x and y. The<br />
actual st<strong>at</strong>e of the system can be changed experimentally by changing x and<br />
y. The energy U of the system is a function of x and y defined via dU =<br />
sin(x + y)dx + sin(x + y)dy. The work done on the outside world by the system<br />
when x and y are changed is given by ¯ dW = sin(x) cos(y)dx.<br />
(a) Show th<strong>at</strong> U is a st<strong>at</strong>e function and find U(x,y).<br />
(b) Show th<strong>at</strong> W is not a st<strong>at</strong>e function.<br />
(c) Give an expression for the differential change in he<strong>at</strong> ¯ dQ for this system.<br />
The system is cycled through st<strong>at</strong>es (x, y) in a reversible process. The cycle is<br />
given by (0, 0) → (απ, 0) → (απ, απ) → (0, απ) → (0, 0). α is a parameter with<br />
values restricted to [0, 2].<br />
(d) Calcul<strong>at</strong>e the work done on the environment in this cycle and the amount<br />
of he<strong>at</strong> taken up by the system as a function of α.<br />
(e) For which values of α is this process a he<strong>at</strong> engine?
1.9. PROBLEMS FOR CHAPTER 1 39<br />
Problem 4.<br />
An ideal gas is in equilibrium and has volume V0, temper<strong>at</strong>ure T0, and<br />
pressure p0. These are the only three st<strong>at</strong>e variables we can vary. The amount<br />
of m<strong>at</strong>erial is kept constant <strong>at</strong> N moles. This gas is cycled through the following<br />
process. First it expands <strong>at</strong> constant pressure to twice the volume. Next, it is<br />
cooled <strong>at</strong> constant volume to half the temper<strong>at</strong>ure it had <strong>at</strong> the beginning of<br />
this step. Then its volume is reduced to the original value <strong>at</strong> constant pressure.<br />
Finally, it is he<strong>at</strong>ed <strong>at</strong> constant volume to the original st<strong>at</strong>e. During this process<br />
an amount of he<strong>at</strong> Q1 > 0 flows into the system <strong>at</strong> steps one and four combined,<br />
and an amount of he<strong>at</strong> Q2 > 0 flows out of the system during steps two and<br />
three combined. The total amount of work performed on the environment is<br />
W. Calcul<strong>at</strong>e the efficiency for this process, defined by the r<strong>at</strong>io of the work W<br />
done on the environment and the amount of he<strong>at</strong> Q1 flowing into the system<br />
during of the process. Compare this value with the efficiency of a Carnot engine<br />
oper<strong>at</strong>ing between the highest and lowest temper<strong>at</strong>ure th<strong>at</strong> occur in the process<br />
described above.<br />
Problem 5.<br />
The internal energy of a system as a function of the entropy S and the volume<br />
V is given by U(S, V ) = S2<br />
V . The only other st<strong>at</strong>e variables of importance in<br />
this problem are the temper<strong>at</strong>ure T and the pressure p.<br />
(a) Using dU = T dS − pdV , calcul<strong>at</strong>e p(S, V ) and T (S, V ) for equilibrium<br />
st<strong>at</strong>es of the system.<br />
(b) Calcul<strong>at</strong>e the efficiency of a Carnot cycle for this system as a function of<br />
the temper<strong>at</strong>ures of the two reservoirs between which the Carnot engine is<br />
oper<strong>at</strong>ing. Calcul<strong>at</strong>e all contributions to the exchange of he<strong>at</strong> and work,<br />
and show th<strong>at</strong> the temper<strong>at</strong>ure defined in this problem is indeed equal to<br />
the Carnot temper<strong>at</strong>ure.<br />
Problem 6.<br />
A magnetic substance has temper<strong>at</strong>ure T and magnetiz<strong>at</strong>ion M. The first<br />
law in this case is dU = T dS +HdM, where H is the applied magnetic field. The<br />
internal energy U is a function of T only and does not depend on M, U = Nf(T ).<br />
The equ<strong>at</strong>ion of st<strong>at</strong>e is M = CH<br />
T , where C is a constant proportional to the<br />
size of the system. Calcul<strong>at</strong>e the efficiency of a Carnot cycle for an engine<br />
using a magnetic substance which is doing work on the outside world when<br />
reducing the magnetiz<strong>at</strong>ion for this substance, by calcul<strong>at</strong>ing all contributions<br />
to the exchange of he<strong>at</strong> and work, and show th<strong>at</strong> the temper<strong>at</strong>ure defined in<br />
this problem is indeed equal to the Carnot temper<strong>at</strong>ure.
40 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Problem 7.<br />
This problem is an example of a consequence of the third law: it is impossible<br />
to reach absolute zero temper<strong>at</strong>ure in a finite number of steps if a reversible<br />
process is used. The entropy of a system is given by S = αV T , where α is a<br />
positive constant and V is the volume of the system. The system is originally<br />
<strong>at</strong> temper<strong>at</strong>ure T0 and volume V0. A reversible two-step process is used to cool<br />
this system. In the first part of the process the system expands adiab<strong>at</strong>ically<br />
to a volume 2V0. In the second part of the process the system is compressed<br />
isothermally back to V0. Calcul<strong>at</strong>e the temper<strong>at</strong>ure Tn of the system after this<br />
two-step process has been applied n times.<br />
Problem 8.<br />
The internal energy of systems 1 and 2 is given by Ui = CiT , where the<br />
constants C1 and C2 are both positive. Originally system 1 is <strong>at</strong> temper<strong>at</strong>ure<br />
T1 and system 2 <strong>at</strong> temper<strong>at</strong>ure T2. These systems are brought into thermal<br />
contact. This establishes an infinitesimally slow flow of energy between the<br />
systems, but the two systems are completely isol<strong>at</strong>ed from the rest of the world<br />
and energy cannot flow out of the combined system. The individual systems are<br />
always in equilibrium. The total system is not in equilibrium just after contact<br />
is established, but finally will reach equilibrium <strong>at</strong> a common temper<strong>at</strong>ure Tf .<br />
(a) Calcul<strong>at</strong>e Tf .<br />
(b) Calcul<strong>at</strong>e the change in entropy during this spontaneous process.<br />
(c) Show th<strong>at</strong> the entropy did increase.<br />
Problem 9.<br />
A removable partition divides a container in two parts with volumes V1 and<br />
V2. The container is filled with an ideal gas. The temper<strong>at</strong>ure and the pressure<br />
in each part are equal to T and p, respectively.<br />
(a) Calcul<strong>at</strong>e the amount of m<strong>at</strong>erial N1 and N2 in each part of the system.<br />
(b) Calcul<strong>at</strong>e the entropy of the combined system by adding the entropies of<br />
the two parts.<br />
The partition is now removed. Since the pressure and temper<strong>at</strong>ure of both parts<br />
were already equal, the combined system is already in equilibrium.<br />
(c) Calcul<strong>at</strong>e the total entropy of the system directly as a function of N =<br />
N1 + N2 and V = V1 + V2.
1.9. PROBLEMS FOR CHAPTER 1 41<br />
(d) Show th<strong>at</strong> the results of b and c are different.<br />
Since the removal of the partition does not cause any spontaneous process, the<br />
entropy before and after should be the same. The problem is rel<strong>at</strong>ed to our<br />
deriv<strong>at</strong>ion of the entropy for an ideal gas. The constant S0 does not depend on<br />
T or V, but could still depend on N.<br />
(e) Assuming th<strong>at</strong> S0 is a function of N, derive its functional form from the<br />
requirement th<strong>at</strong> the entropy calcul<strong>at</strong>ed in b and c is the same.<br />
Problem 10.<br />
An engine oper<strong>at</strong>es between a high temper<strong>at</strong>ure reservoir and a low temper<strong>at</strong>ure<br />
reservoir. Everything in this engine is ideal, except th<strong>at</strong> from each amount<br />
of he<strong>at</strong> Q taken by the engine from the high temper<strong>at</strong>ure reservoir an amount<br />
αQ with 0 < α < 1 flows <strong>at</strong> the same time directly into the low temper<strong>at</strong>ure<br />
reservoir through a leak.<br />
(a) How much entropy is cre<strong>at</strong>ed in this leak?<br />
(b) Calcul<strong>at</strong>e the efficiency of this engine.<br />
Problem 11.<br />
A Stirling engine uses the following cycle (all steps are reversible):<br />
(1) Isothermal expansion, absorbing an amount of he<strong>at</strong> Qh <strong>at</strong> Th.<br />
(2) Isochoric (constant volume) cooling to Tl, giving up an amount of he<strong>at</strong><br />
Qc.<br />
(3) Isothermal contraction, giving off an amount of he<strong>at</strong> Ql <strong>at</strong> Tl.<br />
(4) Isochoric warming to Th, absorbing an amount of he<strong>at</strong> Qa.<br />
Using these non-standard definitions all values of Qi are positive. Assume th<strong>at</strong><br />
Qa = Qc and th<strong>at</strong> this energy is recycled, th<strong>at</strong> is the he<strong>at</strong> given up in step 2<br />
returns to the system in step 4. Show th<strong>at</strong> for an ideal gas the efficiency of the<br />
Stirling cycle is less than the Carnot efficiency.<br />
Problem 12.<br />
The temper<strong>at</strong>ure is an intensive st<strong>at</strong>e variable since the definition is independent<br />
of the size of the reservoir. Show th<strong>at</strong> the entropy is extensive, in two<br />
different ways, using:
42 CHAPTER 1. BASIC THERMODYNAMICS.<br />
(a) Carnot engines which are working in parallel between the same reservoirs.<br />
(b) the knowledge th<strong>at</strong> the internal energy U is extensive.<br />
Problem 13.<br />
Car<strong>at</strong>heodory’s formul<strong>at</strong>ion of the second law is: There exists arbitrarily<br />
close to any given st<strong>at</strong>e of a system other st<strong>at</strong>es which cannot be reached from it<br />
by adiab<strong>at</strong>ic processes alone. In order to show th<strong>at</strong> this formul<strong>at</strong>ion is equivalent<br />
to Kelvin’s, we use a simple system where only mechanical work can be done.<br />
Hence an adiab<strong>at</strong>ic process is described by dU = −pdV .<br />
(a) Show th<strong>at</strong> in the U-V plane all points th<strong>at</strong> are connected by adiab<strong>at</strong>ic<br />
processes alone define a set of non-intersecting curves.<br />
These curves are labelled by a parameter S, with S(U,V) continuous and differentiable.<br />
(b) Show th<strong>at</strong> Car<strong>at</strong>heodory’s formul<strong>at</strong>ion of the second law implies th<strong>at</strong><br />
this is possible<br />
Define T −1 = ( )<br />
∂S<br />
∂U V .<br />
(c) Use the first law to show th<strong>at</strong> identifying S with the entropy and T with<br />
temper<strong>at</strong>ure is consistent with the standard definitions.<br />
(d) Is the identific<strong>at</strong>ion in part c the only possible one?<br />
Problem 14.<br />
An ideal gas is defined by the equ<strong>at</strong>ions of st<strong>at</strong>e pV = NRT , U = 3<br />
2NRT ,<br />
and S = NR ln(T √ T V<br />
N ). It is used in an engine following the Otto cycle:<br />
(1) Adiab<strong>at</strong>ic compression from volume V1 to a smaller volume V2, with temper<strong>at</strong>ure<br />
changing from Tc to T ′ .<br />
(2) He<strong>at</strong>ing <strong>at</strong> constant volume to a temper<strong>at</strong>ure Th.<br />
(3) Adiab<strong>at</strong>ic expansion V2 → V1 (power stroke). The temper<strong>at</strong>ure changes<br />
to T ′′ .<br />
(4) Cooling <strong>at</strong> constant volume to Tc.<br />
The amount of he<strong>at</strong> absorbed by the gas in step two is Qin, the amount of<br />
he<strong>at</strong> absorbed by the gas in step four is Qout (which is neg<strong>at</strong>ive). Calcul<strong>at</strong>e<br />
the efficiency of the Otto cycle and show th<strong>at</strong> it is less than the efficiency of a<br />
Carnot cycle oper<strong>at</strong>ing between Th and Tc.
1.9. PROBLEMS FOR CHAPTER 1 43<br />
Problem 15.<br />
In this problem we use dS = 1 p<br />
T dU + T dV . The internal energy is given by<br />
U(T, V ) = cV T 4 , with c a constant. Since the entropy is a st<strong>at</strong>e function, dS is<br />
exact. Use this to show th<strong>at</strong> p(T, V ) = 1<br />
3cT 4 + f(V )T , where f(V ) is a function<br />
of volume only. Calcul<strong>at</strong>e the entropy S(T, V ) and show th<strong>at</strong> f(V ) ≡ 0.<br />
Problem 16.<br />
Consider the following differential:<br />
¯df = g(x, y, z)dx + h(x, y, z)dy + k(x, y, z)dz<br />
Wh<strong>at</strong> are the necessary and sufficient conditions for ¯ df to be exact?<br />
Problem 17.<br />
Consider the following examples:<br />
¯df = 3x 2 y 2 zdx + 2x 3 yzdy + 6x 3 y 2 dz<br />
¯dg = 2xy 2 z 2 dx + 2x 2 yz 2 dy + 2x 2 y 2 zdz<br />
Are these differentials exact? If so, find the corresponding st<strong>at</strong>e function.<br />
Problem 18.<br />
A system is completely described by the four st<strong>at</strong>e variables T,S,p, and V.<br />
1. Show th<strong>at</strong> the general form of the energy equ<strong>at</strong>ion is U = V f( S<br />
V ).<br />
2. If U = S α V 1−α find the equ<strong>at</strong>ions of st<strong>at</strong>e p(S,V) and T (S,V).<br />
3. The entropy S is not always easy to deal with, and we often elimin<strong>at</strong>e this<br />
variable. In the previous case, solve for p(T,V). Wh<strong>at</strong> is special in this<br />
equ<strong>at</strong>ion, and why is th<strong>at</strong> the case?<br />
4. Calcul<strong>at</strong>e the efficiency of a Carnot engine using a m<strong>at</strong>erial with the energy<br />
equ<strong>at</strong>ion U = S α V 1−α .<br />
Problem 19.<br />
Consider the following cyclic process (Brayton), with an ideal mono-<strong>at</strong>omic<br />
gas as the working m<strong>at</strong>erial :
44 CHAPTER 1. BASIC THERMODYNAMICS.<br />
1. Adiab<strong>at</strong>ic expansion from pA, V1, T1 to pB, V2, T2.<br />
2. Isobaric expansion from pB, V2, T2 to pB, V3, T3.<br />
3. Adiab<strong>at</strong>ic contraction from pB, V3, T3 to pA, V4, T4.<br />
4. Isobaric contraction from pA, V4, T4 to pA, V1, T1.<br />
Calcul<strong>at</strong>e the efficiency in terms of pA and pB. Compare this with the<br />
Carnot efficiency (indic<strong>at</strong>e the temper<strong>at</strong>ures used, and why you choose these<br />
references).<br />
Problem 20.<br />
In an Ericsson cycle isothermal expansion and contraction is altern<strong>at</strong>ed with<br />
expansion and contraction <strong>at</strong> constant pressure. Use an ideal gas, pV = NRT<br />
and U = 3<br />
2pV , to show th<strong>at</strong> the r<strong>at</strong>io of the total work done to the he<strong>at</strong> in <strong>at</strong><br />
the isothermal expansion <strong>at</strong> high temper<strong>at</strong>ure is the same as th<strong>at</strong> for a Carnot<br />
engine oper<strong>at</strong>ing between the same two reservoirs.<br />
Problem 20.
1.9. PROBLEMS FOR CHAPTER 1 45<br />
The figure above shows the experimental results th<strong>at</strong> led to the most recent<br />
Nobel prize. Left are the d<strong>at</strong>a from Fert on multilayers, right the d<strong>at</strong>a from<br />
Grünberg for a double layer. At zero field the direction of the magnetic moments<br />
of the iron layers altern<strong>at</strong>es, <strong>at</strong> high field they all point in the same direction.<br />
They measured the resistance to currents parallel to the layers, and found th<strong>at</strong> it<br />
changed as a function of field. Based on the reading in the notes, sections 1.1-3,<br />
wh<strong>at</strong> can you say about the difference and similarities between these figures?<br />
Problem 21.<br />
Consider a system where we have<br />
and we are given th<strong>at</strong><br />
dU = pdV − HdM<br />
2 M<br />
p(V, M) = π<br />
V 2<br />
Find H(V, M), if we know th<strong>at</strong> H = 0 ⇒ M = 0. Find M(V, H).<br />
Problem 22.<br />
Derive an expression for ( )<br />
∂U<br />
∂V T,N th<strong>at</strong> contains ( )<br />
∂S<br />
∂V .<br />
T,N<br />
Calcul<strong>at</strong>e the<br />
l<strong>at</strong>ter deriv<strong>at</strong>ive in the case th<strong>at</strong> p = RT g( N<br />
V ), with an unknown function g(x).<br />
Use these results to show th<strong>at</strong> the internal energy has the form U = Nh(T ).<br />
Problem 23.<br />
You have measured a m<strong>at</strong>erials property, call it incognito, measured in<br />
ploughs, as a function of temper<strong>at</strong>ure, in Kelvin, and found the following d<strong>at</strong>a<br />
Temper<strong>at</strong>ure (Kelvin) Incognito (ploughs)<br />
0 0.000<br />
100 0.048<br />
200 0.091<br />
300 0.130<br />
400 0.167<br />
500 0.200<br />
600 0.231<br />
700 0.259<br />
800 0.286<br />
900 0.310<br />
1000 0.333<br />
Find a rel<strong>at</strong>ion th<strong>at</strong> gives incognito as a function of temper<strong>at</strong>ure reasonably<br />
well for this measurement.<br />
Next, you measure incognito for another m<strong>at</strong>erial, and you find
46 CHAPTER 1. BASIC THERMODYNAMICS.<br />
Temper<strong>at</strong>ure (Kelvin) Incognito (ploughs)<br />
0 0.000<br />
100 0.833<br />
200 0.909<br />
300 0.938<br />
400 0.952<br />
500 0.962<br />
600 0.968<br />
700 0.972<br />
800 0.976<br />
900 0.978<br />
1000 0.980<br />
Find a rel<strong>at</strong>ion th<strong>at</strong> gives incognito as a function of temper<strong>at</strong>ure reasonably<br />
well for this measurement.<br />
But now you realize th<strong>at</strong> these two m<strong>at</strong>erials are really similar in the kind<br />
of physics they display. They are different only because the temper<strong>at</strong>ure has<br />
different effects. Assume th<strong>at</strong> there is a typical temper<strong>at</strong>ure T0 associ<strong>at</strong>ed with<br />
each m<strong>at</strong>erial, a m<strong>at</strong>erials parameter. Find a rel<strong>at</strong>ion as a function of temper<strong>at</strong>ure<br />
which includes this typical temper<strong>at</strong>ure T0 as the only unknown parameter,<br />
which can explain both d<strong>at</strong>a sets. Find the value of T0 for each m<strong>at</strong>erial.<br />
Problem 24.<br />
A gas is contained in a cylinder with a piston th<strong>at</strong> can move. We can measure<br />
the pressure on the piston, the volume of the gas, and its temper<strong>at</strong>ure. We find<br />
the following set of d<strong>at</strong>a:<br />
Day Time p (<strong>at</strong>m) T (Cent) V (liters)<br />
Monday, March 16 11am 1.0 22.2 3.1<br />
Tuesday, March 17 11am 1.0 22.2 3.1<br />
Wednesday, March 18 11am 1.0 22.2 3.1<br />
Thursday, March 19 3pm 1.0 22.2 2.7<br />
Friday, March 20 11am 1.0 22.2 2.7<br />
S<strong>at</strong>urday, March 21 11am 1.0 22.2 2.7<br />
Sunday, March 22 11am 1.0 22.2 2.7<br />
Give an explan<strong>at</strong>ion for these observ<strong>at</strong>ions, which is consistent with the<br />
physics we have discussed, and does not assume irr<strong>at</strong>ional behavior by the observer.<br />
Problem 25.<br />
Consider a system which is electrically polarizable where we have<br />
dU = pdV − EdP
1.9. PROBLEMS FOR CHAPTER 1 47<br />
and we are given th<strong>at</strong><br />
p(V, P ) = ϵ0 P<br />
2<br />
2<br />
V 2<br />
Find U(V, P ), if we know th<strong>at</strong> E = 0 ⇒ P = 0.<br />
Problem 26.<br />
We all know th<strong>at</strong> it is more expensive to oper<strong>at</strong>e a refriger<strong>at</strong>or in summer<br />
than in winter. Give three reasons why this might be the case, and rank them<br />
according to how important they are. Give a clear justific<strong>at</strong>ion for your answer.
48 CHAPTER 1. BASIC THERMODYNAMICS.
Chapter 2<br />
Thermodynamic potentials<br />
and response functions.<br />
In the previous chapter we had a lot of discussion about the meaning of various<br />
quantities. Th<strong>at</strong> is important, because it does give us some idea wh<strong>at</strong> th<strong>at</strong><br />
mysterious quantity entropy is. The m<strong>at</strong>hem<strong>at</strong>ical deriv<strong>at</strong>ions were minimal,<br />
which is out of necessity. We first need to have definitions.<br />
One can also take a very practical approach, however, and simply st<strong>at</strong>e th<strong>at</strong><br />
the first law of thermodynamics is given by dU = T dS − pdV + µdN for the<br />
simple systems we considered. It is understood th<strong>at</strong> additional work terms can<br />
be added when needed. The second law transl<strong>at</strong>es into the st<strong>at</strong>ement th<strong>at</strong> for an<br />
isol<strong>at</strong>ed system the equilibrium st<strong>at</strong>e is th<strong>at</strong> st<strong>at</strong>e th<strong>at</strong> maximizes the entropy as<br />
a function of internal variables. These two st<strong>at</strong>ements are sufficient for starting<br />
the current chapter. Of course, there is some additional semantics th<strong>at</strong> needs<br />
to be referred to, and it will be important to go back to the first chapter when<br />
it is not quite clear wh<strong>at</strong> certain wording means.<br />
One other idea from the first chapter should not get lost, however. The<br />
fundamental definition of thermodynamical processes requires th<strong>at</strong> the system<br />
we consider is very large on some scale, but th<strong>at</strong> there always has to be an<br />
even larger outside world th<strong>at</strong> allows us to make contact with the system and<br />
measure properties of the system. <strong>Thermodynamics</strong> applies to systems looked<br />
<strong>at</strong> from the outside. The n<strong>at</strong>ure of the first law demands this. If we try to relax<br />
this idea, and use one part of the system to measure the other and vice-versa,<br />
we will run into some real problems. In such a case we have lost our common<br />
reference. So we are free to add the entropy of two blocks of iron, but we are<br />
not free to add the entropy of all separ<strong>at</strong>e blocks of m<strong>at</strong>erial in the universe<br />
to get the entropy of the universe. The st<strong>at</strong>ement th<strong>at</strong> based on the second<br />
law of thermodynamics the universe will keep getting warmer and warmer is<br />
an incorrect applic<strong>at</strong>ion of thermodynamics. We need to apply fundamental<br />
theories of gravity to answer the question if the universe is warming up or not.<br />
We start this chapter with a discussion of energy. At first, th<strong>at</strong> may seem<br />
49
50 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
redundant, because introductory physics has us well prepared to think about<br />
energy. The problem is th<strong>at</strong> in thermodynamical processes we have many different<br />
variables th<strong>at</strong> we can change of keep constant. So work done depends<br />
on the type of process. The mechanical work done on the outside world in a<br />
process with constant amounts of m<strong>at</strong>erial is not the same as the work done in<br />
a process where the chemical potential is kept constant. Since work is rel<strong>at</strong>ed<br />
to a change in available energy th<strong>at</strong> means th<strong>at</strong> in different processes different<br />
amounts of energy are available, or free, to do work.<br />
The introduction of different free energies allows us to change the maximum<br />
entropy st<strong>at</strong>ement in more useful terms. Maximum entropy turns out to be<br />
equivalent to minimum energy. Since we have several free energies we can now<br />
make st<strong>at</strong>ements about equilibrium for systems th<strong>at</strong> are not isol<strong>at</strong>ed.<br />
In the sections th<strong>at</strong> follow after th<strong>at</strong> we will explore the m<strong>at</strong>hem<strong>at</strong>ical consequences<br />
of the first law as st<strong>at</strong>ed above. The theory of multi-variable calculus<br />
leads to some important restrictions on partial deriv<strong>at</strong>ives, and the distinction<br />
between intensive and extensive variables gives us an Euler rel<strong>at</strong>ion. First order<br />
partial deriv<strong>at</strong>ives are fundamental connection between theory and experiment<br />
in thermodynamics, because they represent the linear response of systems. Experiment<br />
gives us values of partial deriv<strong>at</strong>ives, and theory tells us th<strong>at</strong> they are<br />
somehow rel<strong>at</strong>ed, and hence gives important checks on experimental d<strong>at</strong>a.<br />
The minimum energy principle gives us inform<strong>at</strong>ion about the signs of second<br />
order deriv<strong>at</strong>ives of the free energy. Those represent stability criteria, since a<br />
system th<strong>at</strong> is not in equilibrium will want to go back to equilibrium. As a<br />
consequence, we will find for example th<strong>at</strong> he<strong>at</strong> capacities have to be positive.<br />
It is truly remarkable th<strong>at</strong> the basic formul<strong>at</strong>ion of the first and second law,<br />
combined with pure m<strong>at</strong>hem<strong>at</strong>ics, gives us so many conclusions we can draw<br />
about thermodynamical processes. This chapter is general, and the conclusions<br />
hold for any system. We do not need to know the details of the system. The<br />
conclusions are independent of the equ<strong>at</strong>ions of st<strong>at</strong>e. In fact, they give us<br />
restrictions on wh<strong>at</strong> we can use for equ<strong>at</strong>ions of st<strong>at</strong>e! To me, this is the beauty<br />
of thermodynamics.<br />
In the last two sections we take two small excursions to show the applicability<br />
of wh<strong>at</strong> we found. In magnetic systems response functions th<strong>at</strong> should be<br />
positive can be neg<strong>at</strong>ive, which seems a contradiction. Th<strong>at</strong> contradiction is easily<br />
resolved, however, by correctly including the energy stored in the magnetic<br />
fields. Similar observ<strong>at</strong>ions hold for electric fields. Finally, we analyze in more<br />
detail the n<strong>at</strong>ure of fluctu<strong>at</strong>ions. Here we do introduce an additional element,<br />
based on ideas from st<strong>at</strong>istical mechanics. We propose th<strong>at</strong> the probability of a<br />
given fluctu<strong>at</strong>ion is rel<strong>at</strong>ed to the change in entropy for th<strong>at</strong> fluctu<strong>at</strong>ion, with<br />
the system tre<strong>at</strong>ed as being internally in equilibrium, only not with the outside<br />
world. We will come back to fluctu<strong>at</strong>ions when we discuss transport.
2.1. INTERNAL ENERGY. 51<br />
2.1 Internal energy.<br />
Potential energy.<br />
When a mass is dropped from a certain height, it will gain kinetic energy.<br />
If it hits the floor, this kinetic energy will be transformed into he<strong>at</strong> and/or do<br />
work. The amount of work this mass is able to do is completely determined<br />
by the difference in potential energy between the initial position and the floor.<br />
Hence the gravit<strong>at</strong>ional potential energy is a measure of the amount of energy<br />
stored in this mass initially, and this energy can be released to do work.<br />
Energy stored in a gas.<br />
Energy is also stored in a gas. If we allow a gas to expand, it will do work<br />
on the outside world. The amount of energy which can be released, however,<br />
depends on the n<strong>at</strong>ure of the process by which the gas is doing this work. In<br />
most cases some of the thermodynamic variables are kept constant. A standard<br />
example is the expansion of a gas <strong>at</strong> constant temper<strong>at</strong>ure or pressure. The<br />
energy which can be released depends on the values of these constant variables.<br />
For each situ<strong>at</strong>ion one defines an appropri<strong>at</strong>e thermodynamic potential or free<br />
energy. This makes thermodynamics slightly more complic<strong>at</strong>ed than ordinary<br />
mechanics. One has to remember which thermodynamic potential applies in<br />
which case.<br />
Why does the energy given up by a gas depend on the process?<br />
Internal energy is a free energy.<br />
The easiest thermodynamic potential pertains to a process which is adiab<strong>at</strong>ic.<br />
In an adiab<strong>at</strong>ic process there is no he<strong>at</strong> exchange and the entropy is<br />
constant. The first law is in this case ∆U = −∆W . In other words:<br />
The internal energy measures the amount of work a system is able<br />
to do in an reversible adiab<strong>at</strong>ic process.<br />
This tells us th<strong>at</strong> the internal energy is the appropri<strong>at</strong>e thermodynamic potential<br />
or free energy for an adiab<strong>at</strong>ic process. As usual, in reality the amount of work<br />
done is always less then the change in internal energy due to inefficiencies in the<br />
process. But those details are of a technical n<strong>at</strong>ure, and can be minimized by<br />
improving the design.
52 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Internal energy and extensive st<strong>at</strong>e variables.<br />
Consider a completely isol<strong>at</strong>ed or closed system. All the extensive variables<br />
of the system have either a fixed value or interact with a generalized external<br />
force which is zero. In a closed system the amount of m<strong>at</strong>erial N is constant<br />
and typically the volume V is constant. We might, however, not be able to<br />
set all extensive st<strong>at</strong>e variables. For example, the magnetic moment ⃗ M might<br />
freely find its optimal value. This is only true if the external magnetic fields are<br />
small enough. If th<strong>at</strong> is the case, the magnetic moment is an unimportant st<strong>at</strong>e<br />
variable and we ignore it in the description of the system. One only includes<br />
st<strong>at</strong>e variables which can be measured, and if there are no external magnetic<br />
fields we cannot measure the magnetic moment. One has to include the entropy<br />
though. Since the system cannot exchange he<strong>at</strong> with the outside world, the<br />
entropy is constant, but its actual value will determine the value of the internal<br />
energy. In our closed system <strong>at</strong> constant volume and number of particles the<br />
internal energy is a function of the entropy, the volume, and the amount of<br />
m<strong>at</strong>erial only, when the system is in equilibrium. Hence we have U(S,V,N).<br />
Also, the temper<strong>at</strong>ure T and the pressure p, and the chemical potential µ are<br />
completely determined by the values of S, V, and N. The n<strong>at</strong>ural set of st<strong>at</strong>e<br />
variables to describe a completely isol<strong>at</strong>ed system are all the extensive st<strong>at</strong>e<br />
variables which cannot be changed without doing work on the outside world.<br />
The free energy for such a system is the internal energy.<br />
Minimum energy principle.<br />
The number of parameters to describe a macroscopic system is infinitely<br />
large. We could decide to measure some of those by using an appropri<strong>at</strong>e external<br />
probe. For example, we could divide the total system in two parts, separ<strong>at</strong>ed<br />
by a movable wall (no friction, but the wall has mass), which is an ideal thermal<br />
conductor.<br />
The pressure in the left part is pl and in the right part pr. Assume th<strong>at</strong><br />
the left and right part of the system are both in equilibrium, but not necessarily<br />
with each other. The parameter pl is an additional parameter needed to<br />
describe this system. This is a macroscopic parameter, since it pertains to a<br />
macroscopic part of the system. Once we know pl, we know Vl (equilibrium)<br />
and hence Vr and pr. In equilibrium we obviously need pl = pr (no net force<br />
on the partition), otherwise the wall would move. Suppose the system is not in<br />
equilibrium, however, and th<strong>at</strong> pl < pr. The opposite case can be tre<strong>at</strong>ed in a<br />
similar way.<br />
No friction.
2.1. INTERNAL ENERGY. 53<br />
ÔÐ ÔÖ<br />
Figure 2.1: Container with piston as internal divider.<br />
First assume th<strong>at</strong> this is the complete system, and th<strong>at</strong> there is no friction.<br />
It is easy to see wh<strong>at</strong> will happen. The wall will start to move to the left, and<br />
gain kinetic energy. At the point where pl = pr the wall is moving, and will keep<br />
moving to the left until the pressure on the left is wh<strong>at</strong> the original pressure was<br />
on the right. Then the process will continue in reverse. The wall will oscill<strong>at</strong>e!<br />
Since there is no exchange with the outside world and since there is no friction,<br />
for the complete system we have ∆V = ∆N = ∆S = 0 and indeed the internal<br />
energy does not change. The internal energy has the form:<br />
U = Ul + Ur + 1<br />
2 Mv2<br />
(2.1)<br />
where the first two terms represent the internal energy of the left and right side,<br />
and the last term the kinetic energy of the wall. When the pressure left and<br />
right are the same, the wall has the largest kinetic energy, and in th<strong>at</strong> case the<br />
sum of internal energies is smallest.<br />
The motion of the wall is an organized motion, and this kinetic energy could<br />
be extracted.<br />
With friction.<br />
Can you design an experiment th<strong>at</strong> will do this?<br />
In the second case suppose there is friction and the wall loses energy to the<br />
gas when it is moving. The wall will again oscill<strong>at</strong>e, but now this oscill<strong>at</strong>ion<br />
is damped, and the final st<strong>at</strong>e has the wall <strong>at</strong> a position with pl = pr. The<br />
mechanical energy of the partition has been transformed into thermal energy of
54 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
<br />
Å<br />
ÔÐ ÔÖ<br />
Figure 2.2: Container where the internal degree of freedom becomes external<br />
and hence can do work.<br />
the gases! Since the whole system is isol<strong>at</strong>ed we have ∆Q = 0 and hence for<br />
this process we find from 1.41 ∆S ≥ 0. This agrees with the maximum entropy<br />
principle, the equilibrium st<strong>at</strong>e corresponds to maximum entropy as far as the<br />
internal coordin<strong>at</strong>es are concerned. The maximum entropy principle tells us<br />
th<strong>at</strong> the organized energy of the wall has gone into the internal energy of the<br />
two parts of the system, and th<strong>at</strong> increase in energy is directly rel<strong>at</strong>ed to the<br />
increase in entropy.<br />
Wall corresponds to st<strong>at</strong>e variable.<br />
In the third case we connect the movable wall with a massless cord to some<br />
object with mass M (see figure). The value of this mass M is adjusted in such<br />
a way th<strong>at</strong> Mg = (pr − pl)A(1 − ϵ) is always true, where A is the area of the<br />
movable wall. We also take the limit ϵ → 0, which allows the wall to move in a<br />
quasi-st<strong>at</strong>ic manner. Therefore, in this system there is no net force acting on the<br />
wall, or the generalized force connected with the st<strong>at</strong>e variable corresponding<br />
to the wall position is always zero. We also assume th<strong>at</strong> there is no friction.<br />
At this point the total system is in equilibrium. Now we make the mass M<br />
smaller by very slowly taking very small pieces of th<strong>at</strong> mass away. At every point<br />
during this process the total system is in equilibrium, and the whole process is<br />
reversible. The wall will move slowly to the left, and the mass M will move<br />
upwards. Hence the system is doing work on this mass. At the end, the mass is<br />
zero and the wall does not interact with the outside any more. The total system<br />
is in equilibrium, as is the closed system itself.<br />
In this process there is always equilibrium and because the wall is a thermal<br />
conductor we always have Tl = Tr. Since there is no he<strong>at</strong> exchange with the<br />
outside world we have Tl∆Sl = ∆Ql = −∆Qr = −Tr∆Sr, and hence ∆Sl =<br />
−∆Sr. For the total system we have therefore ∆S = 0, as it should be for a
2.1. INTERNAL ENERGY. 55<br />
reversible process.<br />
Originally, the closed system was not in equilibrium. The internal energy of<br />
the original system was higher than in the final equilibrium situ<strong>at</strong>ion, since during<br />
the process of reaching equilibrium mass was lifted to a higher gravit<strong>at</strong>ional<br />
potential energy. The internal energy of the system is now lower. We arrive <strong>at</strong><br />
the important conclusion:<br />
The internal energy of a closed system is a minimum as a function<br />
of all parameters which are not fixed by external constraints.<br />
Explain carefully the difference between these three scenarios.<br />
Formal deriv<strong>at</strong>ion of minimum energy principle.<br />
This minimum energy st<strong>at</strong>ement also follows from the rel<strong>at</strong>ion between energy,<br />
entropy, and work, T ∆S ≥ ∆U + ∆W . In our example of case three<br />
we have a closed system, and the parameters S, V, and N are fixed, but the<br />
internal coordin<strong>at</strong>e pl is allowed to vary. No work is done on the outside world<br />
by changing the extensive parameters V and N, and neither is he<strong>at</strong> exchanged,<br />
which keeps S constant. Hence for a spontaneous process bringing us from<br />
the initial non-equilibrium situ<strong>at</strong>ion to the final equilibrium situ<strong>at</strong>ion, we have<br />
0 ≥ ∆U + 0. The internal energy decreases and is minimal <strong>at</strong> equilibrium. Note<br />
th<strong>at</strong> this minimum principle does not hold for those variables which are fixed<br />
by non-zero external constraints, since a change in those variables involves work<br />
done on the outside world and the n<strong>at</strong>ure of the outside world has to be taken<br />
into account.<br />
If a system is kept <strong>at</strong> constant entropy, volume, etc, it will still be able<br />
to exchange energy with the outside world. In the earlier thought experiment<br />
we had one of the internal degrees of freedom do work. In general, th<strong>at</strong> is<br />
not possible. But a system <strong>at</strong> constant entropy is still able to exchange he<strong>at</strong><br />
with the outside world! The loss in internal energy is due to he<strong>at</strong> dumped<br />
to the outside world. Remember, th<strong>at</strong> only for reversible processes we have<br />
T ∆S = ∆Q. If a system spontaneously goes to equilibrium, the process is by<br />
definition irreversible.<br />
Difference between minimum energy and maximum entropy.<br />
In the previous chapter we considered a system <strong>at</strong> constant energy, volume,<br />
amount of m<strong>at</strong>erial, etc. All the extensive variables th<strong>at</strong> would be able to do<br />
work are kept constant, and hence no work is done. The internal energy is also<br />
constant, and hence no he<strong>at</strong> is transferred either. We want to find the entropy.
56 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Suppose we have a model th<strong>at</strong> allows us to calcul<strong>at</strong>e S(U, V, N, Xint) where Xint<br />
represents the internal degrees of freedom. The maximum entropy principle tells<br />
us th<strong>at</strong> we can find the values of these internal coordin<strong>at</strong>es (coordin<strong>at</strong>es th<strong>at</strong><br />
are not connected to terms in the work differential) by maximizing the function<br />
S. In the current chapter we consider a system <strong>at</strong> constant entropy, volume,<br />
amount of m<strong>at</strong>erial, etc. In this case there is again no work done, but he<strong>at</strong><br />
can be exchanged. Suppose we now have a model th<strong>at</strong> allows us to calcul<strong>at</strong>e<br />
U(S, V, N, Xint). The minimum energy principle tells us th<strong>at</strong> we can find the<br />
values of these internal coordin<strong>at</strong>es.<br />
Example.<br />
A typical applic<strong>at</strong>ion of this minimum principle would be the determin<strong>at</strong>ion<br />
of the magnetic moment of a gas. Consider the variable ⃗ M as an internal<br />
parameter. A typical form of the internal energy of a gas with magnetic moment<br />
⃗M would be U(S, V, N; ⃗ M) = U(S, V, N;⃗0) + 1<br />
2V M 2 . The last term represents<br />
the self-energy needed to cre<strong>at</strong>e a magnetic moment. The minimum principle<br />
tells us th<strong>at</strong> the magnetic moment has to be zero in equilibrium in this case.<br />
Additional interactions are needed to cre<strong>at</strong>e a magnetic moment, like applying<br />
a non-zero magnetic field.<br />
Internal energy and adiab<strong>at</strong>ic processes.<br />
The internal energy measures the amount of work a system can do in an<br />
adiab<strong>at</strong>ic process. Consider a thermally isol<strong>at</strong>ed system which can change its<br />
volume by expanding. An example is a gas in an completely insul<strong>at</strong>ed cylinder<br />
with a piston. The system starts in equilibrium with the piston fixed by an<br />
external force. In the next step, the piston is allowed to move and work is<br />
performed on the outside world. The piston stops again, and the volume of the<br />
system is increased. Is the amount of work done equal to the change in internal<br />
energy? The answer is no, since the final st<strong>at</strong>e might not be in equilibrium. If<br />
the final st<strong>at</strong>e is not in equilibrium, its value of the internal energy is too high,<br />
and hence the amount of work done is less than expected! The final st<strong>at</strong>e will go<br />
to equilibrium, but since the piston is now fixed, there is no mechanical contact<br />
with the outside world anymore. The only way to get rid of the excess energy is<br />
to dump he<strong>at</strong> into the outside world. The internal energy measures the amount<br />
of work a system can do in a reversible, adiab<strong>at</strong>ic process. The amount of work<br />
done in an irreversible process, starting from an initial st<strong>at</strong>e in equilibrium, is<br />
always less, because he<strong>at</strong> is gener<strong>at</strong>ed. If we assume th<strong>at</strong> the outside world is<br />
always in thermal equilibrium, this indic<strong>at</strong>es th<strong>at</strong> the entropy of the outside<br />
world must increase.<br />
The irreversible expansion is described in two steps. We start <strong>at</strong> volume Vi<br />
and entropy Si, with the system in equilibrium and Ui = U(Si, Vi, N). Next,<br />
the system expands to a volume Vf and entropy Sf . The amount of work done<br />
is ∆W . The system is not in equilibrium, however, and the internal energy
2.2. FREE ENERGIES. 57<br />
is NOT equal to U(Sf , Vf , N), but is equal to Ui − ∆W . The system now<br />
follows a spontaneous process and reaches equilibrium. This is a process <strong>at</strong><br />
constant S and V, and in the end the energy is Uf = U(Sf , Vf , N). The change<br />
in energy is ∆U = Uf − Ui. The value of ∆W depends on the n<strong>at</strong>ure of the<br />
irreversible process, and is in general hard to calcul<strong>at</strong>e. We only know for sure<br />
th<strong>at</strong> Ui −∆W ≥ Uf or ∆W ≤ −∆U. Note th<strong>at</strong> the change in entropy is hard to<br />
predict, since we have in general no formulas to describe non-equilibrium st<strong>at</strong>es.<br />
It will depend on the changes in the temper<strong>at</strong>ure in the second step. Our final<br />
conclusion is th<strong>at</strong>:<br />
The internal energy measures the amount of work a system is able<br />
to do in a reversible adiab<strong>at</strong>ic process. It gives an upper-bound for<br />
the amount of work done in irreversible processes.<br />
2.2 Free energies.<br />
The problem of having too many variables.<br />
The thermodynamic potential governing the adiab<strong>at</strong>ic expansion of a gas is<br />
the internal energy. Other measures of the free energy are needed in different<br />
types of processes. Wh<strong>at</strong> do we need when the system is able to exchange<br />
he<strong>at</strong> or particles with the outside world? In other words, which variables are<br />
controlled? Do we keep the volume or pressure constant?<br />
A construction of the appropri<strong>at</strong>e free energy is straightforward, and will be<br />
illustr<strong>at</strong>ed via an example system. Suppose the only extensive st<strong>at</strong>e variables<br />
which we can measure or change are the volume V and the amount of m<strong>at</strong>erial<br />
N. The work done in such changes depends on the pressure p and the chemical<br />
potential µ. We can write<br />
¯dW = pdV − µdN (2.2)<br />
Note again the difference in sign between the mechanical work term and all others.<br />
In this formula for the work done we always see a combin<strong>at</strong>ion of the differential<br />
of an extensive variable and the corresponding intensive variable. How<br />
can we change th<strong>at</strong> to the product of an extensive variable and the differential<br />
of the corresponding intensive st<strong>at</strong>e variable? The answer is found by considering<br />
the product of corresponding extensive and intensive st<strong>at</strong>e variables. The<br />
differential of such a product is easy to calcul<strong>at</strong>e, since d(xy) = xdy + ydx. This<br />
suggests th<strong>at</strong> we have to add to or subtract from the internal energy terms of the<br />
form pV and µN. Such transform<strong>at</strong>ions are called Legendre transform<strong>at</strong>ions.<br />
Helmholtz free energy.
58 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
As a first example we consider the Helmholtz free energy<br />
F = U − T S (2.3)<br />
By definition, F is a st<strong>at</strong>e function and its differential is exact. Note th<strong>at</strong> some<br />
books use the not<strong>at</strong>ion A for the Helmholtz free energy. The combin<strong>at</strong>ion T S is<br />
a mixture of an intensive variable, T, and the corresponding extensive variable,<br />
S. A small change in the Helmholtz energy is rel<strong>at</strong>ed to changes in he<strong>at</strong> and<br />
work via the first law:<br />
dF = dU − d(T S) = ( ¯ dQ − T dS ) − SdT − ¯ dW (2.4)<br />
In a reversible process ¯ dQ − T dS = 0, and now the interpret<strong>at</strong>ion of the<br />
formula for dF is easy. If we consider a thermodynamic process <strong>at</strong> constant<br />
temper<strong>at</strong>ure, the amount of work done on the outside world is equal to the<br />
decrease in Helmholtz free energy. Hence:<br />
The Helmholtz free energy measures the amount of work a system<br />
can do in an isothermal, reversible process!<br />
If a process is not reversible, the first term in 2.4 does not vanish. For an<br />
irreversible process we have ¯ dQ < T dS and hence for an irreversible isothermal<br />
process dF < − ¯ dW . Hence, like in the case of the internal energy, the<br />
Helmholtz free energy measures the maximal amount of work a system can do<br />
in an isothermal process. In order to do this maximal amount of work, the<br />
process has to be reversible. In addition, for a spontaneous process in which no<br />
work is done, dF < 0. Hence we arrive <strong>at</strong> the following minimum principle:<br />
The Helmholtz free energy of a system <strong>at</strong> constant temper<strong>at</strong>ure is<br />
a minimum as a function of all parameters which are not fixed by<br />
external constraints.<br />
Again, this is very similar to the minimum principle for the internal energy.<br />
The big difference is th<strong>at</strong> we are now considering a process in which we keep the<br />
temper<strong>at</strong>ure constant. This requires in general th<strong>at</strong> the entropy has to change,<br />
which means th<strong>at</strong> he<strong>at</strong> will flow. Therefore, the amount of work th<strong>at</strong> can be<br />
done is different than th<strong>at</strong> can be done <strong>at</strong> constant entropy!<br />
N<strong>at</strong>ural variables.<br />
The change in Helmholtz free energy in a reversible process is given by<br />
dF = −SdT − pdV + µdN (2.5)<br />
for a system which interacts with the outside world through he<strong>at</strong> exchange<br />
(dT), mechanical work (dV), and chemical work (dN). This form suggests th<strong>at</strong>
2.2. FREE ENERGIES. 59<br />
the n<strong>at</strong>ural variables for an expression of the Helmholtz free energy are T, V,<br />
and N. Similarly, for the internal energy U the n<strong>at</strong>ural variables are S, V, and<br />
N.<br />
Why are n<strong>at</strong>ural variables more useful than other combin<strong>at</strong>ions?<br />
Other free energies.<br />
Using the idea of Legendre transform<strong>at</strong>ions many more free energies can be<br />
constructed. Three very useful forms are:<br />
Gibbs potential G = U − T S + pV (2.6)<br />
Enthalpy H = U + pV (2.7)<br />
Grand potential Ω = U − T S − µN (2.8)<br />
All these potentials are st<strong>at</strong>e functions and their differentials are all exact.<br />
The Gibbs potential measures the amount of work a system can do <strong>at</strong> constant<br />
temper<strong>at</strong>ure and pressure. Since we work <strong>at</strong> constant pressure, the volume has<br />
to change and our system will do mechanical work on the outside world. This<br />
amount of energy cannot be used, because it is needed to keep the pressure<br />
constant. Similarly, the system is kept <strong>at</strong> constant temper<strong>at</strong>ure and he<strong>at</strong> will<br />
flow to the outside world. Again, this is energy which cannot be used. Hence<br />
the Gibbs potential measures the amount of non-mechanical (not-pdV ) work a<br />
system can do on the outside world <strong>at</strong> constant pressure and temper<strong>at</strong>ure. If<br />
µdN is the only other term in the expression for ¯ dW , then the Gibbs free energy<br />
is a measure of the amount of chemical work a system is able to do.<br />
The enthalpy measures the amount of work a system can do in an adiab<strong>at</strong>ic,<br />
isobaric process. Again, this is the amount of non-mechanical work, since the<br />
pressure has to be kept constant. The original interpret<strong>at</strong>ion of the enthalpy is<br />
th<strong>at</strong> it measures the amount of he<strong>at</strong> th<strong>at</strong> can be released in a process <strong>at</strong> constant<br />
pressure and constant amount of m<strong>at</strong>erial. The grand potential pertains to processes<br />
<strong>at</strong> constant temper<strong>at</strong>ure and chemical potential. Therefore, it measures<br />
non-chemical work, and in a simple system this is only mechanical, pdV type<br />
of work. The Gibbs potential and the enthalpy are typically used to describe<br />
solids, where one cannot change the volume directly but only through changes<br />
in the pressure. The grand potential is useful in chemistry, where the number of<br />
particles can change due to chemical reactions and where this number can only<br />
be determined by specifying the chemical potential. All these potentials obey a<br />
minimum principle similar to those we saw before.<br />
Summary of five free energies.
60 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
In summary, we introduced five free energies (others can be constructed in<br />
a similar manner):<br />
Symbol Differential (reversible) N<strong>at</strong>ural Name<br />
variables<br />
U dU = T dS − pdV + µdN S,V,N Internal energy<br />
F = U − T S dF = −SdT − pdV + µdN T,V,N Helmholtz free energy<br />
G = F + pV dG = −SdT + V dp + µdN T,p,N Gibbs free energy<br />
H = U + pV dH = T dS + V dp + µdN S,p,N Enthalpy<br />
Ω = F − µN dΩ = −SdT − pdV − Ndµ T,V,µ Grand potential<br />
The internal energy is the only free energy which is expressed in changes<br />
of extensive variables only. All the other energies involve <strong>at</strong> least one intensive<br />
variable. If we prescribe a given set of n<strong>at</strong>ural variables, the equilibrium st<strong>at</strong>e of<br />
a system can be found by minimizing the appropri<strong>at</strong>e free energy as a function<br />
of all internal variables. Which set of n<strong>at</strong>ural variables we use depends on the<br />
system we want to describe.<br />
In a given experiment, independent st<strong>at</strong>e variables are those variables whose<br />
values we can control. This can mean two things. First, it could be th<strong>at</strong> we are<br />
able to set the value of the variable <strong>at</strong> any value we want (often in a certain<br />
range only, though). For example, if we have a container with a piston we<br />
can set the volume of the container <strong>at</strong> arbitrary values by fixing the position<br />
of the piston. A more common way of controlling a st<strong>at</strong>e variable, however,<br />
is to prevent its value from changing! A gas in a closed container has a fixed<br />
volume. Any st<strong>at</strong>e variable which we can control by either setting its value or by<br />
preventing changes in its value is an independent variable. These independent<br />
variables are the variables we select as n<strong>at</strong>ural variables.<br />
General formalism.<br />
Is it possible to write H(T, V, N)?<br />
The general form for the differential of the internal energy is<br />
dU = T dS + ∑<br />
xjdXj<br />
j<br />
(2.9)<br />
which follows from 1.18. A general Legendre transform<strong>at</strong>ion pertaining to an<br />
adiab<strong>at</strong>ic environment would single out the values of j included in a set J and<br />
define<br />
YJ = U − ∑<br />
j∈J<br />
Xjxj<br />
(2.10)
2.3. EULER AND GIBBS-DUHEM RELATIONS. 61<br />
with a corresponding differential<br />
dYJ = T dS − ∑<br />
Xjdxj + ∑<br />
xjdXj<br />
j∈J<br />
j /∈J<br />
(2.11)<br />
In order to change to the temper<strong>at</strong>ure as a n<strong>at</strong>ural variable we also subtract a<br />
term T S. A typical example might be the discussion of magnetism in a solid. We<br />
have no way to fix the total magnetic moment ⃗ M other than using a magnetic<br />
field ⃗ H. Therefore, ⃗ H is the n<strong>at</strong>ural variable for this system, and typically we<br />
use a Gibbs-like energy G, defined by G = U − T S + pV − ⃗ M · ⃗ H, to describe<br />
this system. The differential for G is dG = −SdT + V dp + µdN − ⃗ M · d ⃗ H.<br />
2.3 Euler and Gibbs-Duhem rel<strong>at</strong>ions.<br />
Scaling of the internal energy.<br />
Changes in the internal energy are rel<strong>at</strong>ed to changes in the extensive variables,<br />
and the set of all extensive variables is the n<strong>at</strong>ural set of variables to<br />
describe U. In a general case we have U(S, V, N, X) where X represents all<br />
other extensive variables which can be measured and specified to determine<br />
the n<strong>at</strong>ure of the system. If we consider a homogeneous system, in which all<br />
extensive variables are scaled by a value λ, we can derive the following rel<strong>at</strong>ion:<br />
U(λS, λV, λN, λX) = λU(S, V, N, X) (2.12)<br />
This holds for all values of λ! This rel<strong>at</strong>ion can be differenti<strong>at</strong>ed with respect to<br />
λ. The result involves partial deriv<strong>at</strong>ives of U, and since we have so many choices<br />
for our independent variables, we will always specify the constant variables in<br />
the partial deriv<strong>at</strong>ives. We find<br />
( ) ( ) ( ) ( )<br />
∂U<br />
∂U<br />
∂U<br />
∂U<br />
S +<br />
V +<br />
N +<br />
X = U (2.13)<br />
∂S V,N,X ∂V S,N,X ∂N S,V,X ∂X S,V,N<br />
where we have set λ equal to one.<br />
Since the partial deriv<strong>at</strong>ives involve the n<strong>at</strong>ural st<strong>at</strong>e variables for U, they<br />
can be obtained from the differential for U:<br />
( )<br />
∂U<br />
= T (2.14)<br />
∂S V,N,X<br />
( )<br />
∂U<br />
= −p (2.15)<br />
∂V S,N,X<br />
( )<br />
∂U<br />
= µ (2.16)<br />
∂N<br />
S,V,X
62 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
( )<br />
∂U<br />
= x (2.17)<br />
∂X S,V,N<br />
where x is the generalized force corresponding to X. Therefore we find the important<br />
rel<strong>at</strong>ion (Euler equ<strong>at</strong>ion)<br />
U = T S − pV + µN + xX (2.18)<br />
The other free energies obey similar rel<strong>at</strong>ions, which can be derived immedi<strong>at</strong>ely<br />
from this expression. For example, the Gibbs energy for a system which<br />
can only exchange he<strong>at</strong>, do mechanical work (pdV ) and chemical work (µdN)<br />
is G = µN. This is a very simple expression indeed. In such a case it is often<br />
easier to work with the chemical potential then with the Gibbs energy itself.<br />
Note th<strong>at</strong> expression 2.18 implies th<strong>at</strong> we work in the thermodynamic limit.<br />
We need the freedom to choose x in 2.18 in an appropri<strong>at</strong>e way. This choice<br />
should not change the definition of S, and hence the number of internal degrees<br />
of freedom should be very large. In th<strong>at</strong> case a change in the number of internal<br />
degrees of freedom by a few does not m<strong>at</strong>ter.<br />
Independent variables.<br />
Equ<strong>at</strong>ions 2.14 through 2.17 show th<strong>at</strong> for a system in equilibrium the<br />
values of the temper<strong>at</strong>ure, pressure, etc are completely determined once we<br />
specify all the extensive st<strong>at</strong>e variables. In general, the set of st<strong>at</strong>e variables<br />
{S, T, p, V, µ, N, X, x} contains only four independent variables. The physics of<br />
the system is completely determined by the rel<strong>at</strong>ion U(S,V,N,X) and everything<br />
else follows. Obviously, we can change variables if we want. For example, we<br />
could use {T, V, N, X} as independent variables. The n<strong>at</strong>ural energy to determine<br />
in this case is the Helmholtz free energy F. The partial deriv<strong>at</strong>ives of F<br />
with respect to {T, V, N, X} are straightforward and easy to obtain from the<br />
expression for the differential of F. Since we switch variables so often, we always<br />
want to add the constants to the not<strong>at</strong>ion of the partial deriv<strong>at</strong>ives.<br />
Not all intensive variables are independent.<br />
The differential for dU is rel<strong>at</strong>ed to the changes in all extensive variables. If<br />
we take the differential of 2.18 we also find terms pertaining to changes in the<br />
intensive variables. These additional terms have to sum to zero, and hence<br />
SdT − V dp + Ndµ + Xdx = 0 (2.19)<br />
This is called the Gibbs-Duhem rel<strong>at</strong>ion. In words, not all changes in the intensive<br />
variables are independent. This means th<strong>at</strong> if we change basic variables in<br />
the set {S, T, p, V, µ, N, X, x} it is not possible to choose {T, p, µ, x} as our set of<br />
independent variables. Every other choice is allowed, though. It is not hard to<br />
understand this restriction. The intensive variables are independent of the size
2.3. EULER AND GIBBS-DUHEM RELATIONS. 63<br />
of the system, and cannot specify the size of the system. We need <strong>at</strong> least one<br />
measure of how large the system is. If all intensive variables were independent,<br />
we could, however, choose them as a basic set of st<strong>at</strong>e variables, which is in<br />
contradiction with the previous sentence.<br />
Formul<strong>at</strong>ion in terms of densities.<br />
Another way of looking <strong>at</strong> the dependency of the intensive variables is working<br />
with densities in stead of extensive variables. For example, we could specify<br />
the energy per unit volume or per mole. If we define all extensive variables per<br />
unit volume<br />
we find<br />
s = S<br />
V<br />
, u = U<br />
V<br />
N X<br />
, n = , X =<br />
V V<br />
(2.20)<br />
u = T s − p + µn + xX (2.21)<br />
and using V du + udV = d(uV ) = dU = T d(sV ) − pdV + µd(nV ) + xd(X V ),<br />
which is equal to (T s − p + µn + xX ) dV + (T ds + µdn + xdX ) V , we derive<br />
du = T ds + µdn + xdX (2.22)<br />
Similar formulas hold for densities per particle or other combin<strong>at</strong>ions. Hence<br />
one of the extensive variables can always be scaled out, and does not contribute<br />
to the basic physics of the system. It merely sets the size and is, of course,<br />
needed in th<strong>at</strong> respect.<br />
The obvious extensive variable to use for scale is volume. Volume is somewh<strong>at</strong><br />
special because it is a quantity directly rel<strong>at</strong>ed to the sp<strong>at</strong>ial geometry, and<br />
not to an amount of a physical quantity. The pdV work term has the opposite<br />
sign of all others. Hence there is something fundamentally different in volume<br />
compared to other extensive quantities. But one can also scale out amount of<br />
m<strong>at</strong>erial, and in many applic<strong>at</strong>ions we want to follow a given amount of m<strong>at</strong>erial<br />
and see wh<strong>at</strong> happens. We will return to this in the transport chapter.<br />
Using a Legendre transform<strong>at</strong>ion.<br />
Yet another way of looking <strong>at</strong> this problem is to define the following free<br />
energy via a Legendre transform<strong>at</strong>ion:<br />
Y = U − T S − µN − xX = −pV (2.23)<br />
The free energy Y is a st<strong>at</strong>e function and an extensive quantity. Its n<strong>at</strong>ural<br />
variables are {T, µ, x, V }. Hence we have<br />
Y (T, µ, x, V ) = −V p(T, µ, x, V ) (2.24)
64 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
If we scale the volume to αV and use the extensive n<strong>at</strong>ure of Y, we find<br />
p(T, µ, x, αV ) = p(T, µ, x, V ) (2.25)<br />
for all values of α. If we now take the limit α → 0, we see th<strong>at</strong> the pressure<br />
does not depend on the volume, and th<strong>at</strong> it is a function of the other intensive<br />
variables only.<br />
2.4 Maxwell rel<strong>at</strong>ions.<br />
Basic principle.<br />
One should not memorize Maxwell rel<strong>at</strong>ions, but in stead recognize the basic<br />
principle which determines all these rel<strong>at</strong>ions. If a function of many variables is<br />
twice differentiable, and if the second order deriv<strong>at</strong>ives are continuous, the order<br />
in which we take the second order deriv<strong>at</strong>ives is not important. We apply this<br />
rule to all possible free energies and all possible combin<strong>at</strong>ions of independent<br />
variables. This yields a large number of rel<strong>at</strong>ions, called Maxwell rel<strong>at</strong>ions.<br />
The most important occur when we take each free energy as a function of its<br />
n<strong>at</strong>ural variables, since in th<strong>at</strong> case the first order deriv<strong>at</strong>ives are simple st<strong>at</strong>e<br />
variables. These simple st<strong>at</strong>e variables can be found immedi<strong>at</strong>ely by inspecting<br />
the form of the differential. Since so many combin<strong>at</strong>ions of st<strong>at</strong>e variables are<br />
possible, we always write the constant st<strong>at</strong>e variables in the not<strong>at</strong>ion of the<br />
partial deriv<strong>at</strong>ives.<br />
Simple example.<br />
As an example consider F(T,V,N), with dF = −SdT − pdV + µdN. There<br />
are three choices for the mixed second order deriv<strong>at</strong>ives, leading to the three<br />
Maxwell equ<strong>at</strong>ions:<br />
( ) ( ) ( )<br />
2 ∂ F<br />
∂S<br />
∂p<br />
= −<br />
= −<br />
(2.26)<br />
∂T ∂V N ∂V T,N ∂T V,N<br />
( ) ( ) ( )<br />
2 ∂ F<br />
∂S ∂µ<br />
= −<br />
=<br />
(2.27)<br />
∂T ∂N ∂N ∂T<br />
( ∂ 2 F<br />
∂N∂V<br />
)<br />
V<br />
T<br />
=<br />
( ∂µ<br />
∂V<br />
)<br />
N,T<br />
T,V<br />
= −<br />
( ∂p<br />
∂N<br />
)<br />
V,N<br />
V,T<br />
(2.28)<br />
These rel<strong>at</strong>ions are very important, since they connect changes in one st<strong>at</strong>e<br />
variables to changes in another st<strong>at</strong>e variable in a different process. They limit<br />
the freedom of a system in equilibrium to respond to external causes. Maxwell<br />
rel<strong>at</strong>ions are easy to derive if you know which st<strong>at</strong>e variables are your basic set,<br />
which free energy belongs to th<strong>at</strong> set, and which form the differential has.
2.5. RESPONSE FUNCTIONS. 65<br />
Which rel<strong>at</strong>ions to use?<br />
It is not hard to find out if a Maxwell rel<strong>at</strong>ion can be found, as long as we use<br />
the convention of indic<strong>at</strong>ing) all variables th<strong>at</strong> are kept constant. For example,<br />
consider the deriv<strong>at</strong>ive . The set of variables used is {S, N, V } and this<br />
( ∂p<br />
∂S<br />
N,V<br />
is indeed the set of n<strong>at</strong>ural variables for a free energy, in this case the internal<br />
energy U. In general, we have a good set of variables if from each pair {T, S},<br />
{p, V }, {µ, N}, and {x, X} exactly one variable is used.<br />
The variable which is differenti<strong>at</strong>ed is one of the remaining variables (else<br />
the deriv<strong>at</strong>ive is trivial), and can be found by inspecting the differential of the<br />
free energy for which the set of variables is the set of n<strong>at</strong>ural variables. In our<br />
example we have p = − ( )<br />
∂U<br />
∂V and hence<br />
S,N<br />
( ) ( ) ( )<br />
2 ∂p<br />
∂ U<br />
∂T<br />
= −<br />
= −<br />
(2.29)<br />
∂S N,V ∂V ∂S N ∂V N,S<br />
2.5 Response functions.<br />
Linear response.<br />
How do we investig<strong>at</strong>e the physics of some system? How do we perform<br />
a basic measurement? We apply an external field, and measure the response<br />
of the system. For example, we change the temper<strong>at</strong>ure of a metal rod and<br />
measure the change in length. Life would be simple, and thermodynamics easy,<br />
if all responses were linear. This is not true, unfortun<strong>at</strong>ely. A linear rel<strong>at</strong>ion<br />
between cause and effect is only true in the limit of small changes of the causing<br />
agents, when we can write<br />
( )<br />
∂A<br />
∆A =<br />
∆B (2.30)<br />
∂B C,D,..<br />
for a change in B, keeping C,D,... constant. We do not know how small ∆B<br />
has to be in order for a linear rel<strong>at</strong>ion to be a good approxim<strong>at</strong>ion. Ohm’s law<br />
holds for metals for large changes in the applied voltage, but for semiconductors<br />
the linear range is very small. If we measure the magnetic susceptibility of iron,<br />
a change in temper<strong>at</strong>ure of 1K <strong>at</strong> room temper<strong>at</strong>ure is small, because the Curie<br />
temper<strong>at</strong>ure of iron is very large. Near the Curie point, however, changes are<br />
more rapid since some of the deriv<strong>at</strong>ives start to diverge.<br />
When is a linear approxim<strong>at</strong>ion to the response of a system good enough?<br />
The linear rel<strong>at</strong>ion 2.30 holds for small changes in B, and by measuring A we<br />
can find the partial deriv<strong>at</strong>ive. Since the partial deriv<strong>at</strong>ive measures the response
66 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
in A of the system to a small change in B, it is called a response function. The<br />
most important way we obtain inform<strong>at</strong>ion about thermodynamic systems is by<br />
measuring response functions. Hence measurements give us response functions,<br />
and the analysis of the behavior of response functions then will allow us to draw<br />
conclusions about the physics of the system!<br />
He<strong>at</strong> capacities.<br />
He<strong>at</strong> capacities are one of the most important thermodynamic response functions<br />
to measure. The tell us how much he<strong>at</strong> the system will take up if we increase<br />
its temper<strong>at</strong>ure. These measurements are always performed as a function<br />
of temper<strong>at</strong>ure, but can be done either <strong>at</strong> fixed pressure or volume. The other<br />
extensive thermodynamic variables kept constant are denoted by X. Hence we<br />
have<br />
( ) ( )<br />
∆Q<br />
∂S<br />
CV (T, V, X) =<br />
= T<br />
(2.31)<br />
∆T<br />
∂T<br />
Cp(T, p, X) =<br />
( ∆Q<br />
∆T<br />
)<br />
V,X<br />
p,X<br />
= T<br />
( ∂S<br />
∂T<br />
)<br />
V,X<br />
p,X<br />
(2.32)<br />
The he<strong>at</strong> capacities are rel<strong>at</strong>ed to the partial deriv<strong>at</strong>ives of the entropy, since<br />
the entropy governs he<strong>at</strong> exchange. The he<strong>at</strong> capacity <strong>at</strong> constant volume is<br />
directly rel<strong>at</strong>ed to the internal energy. Since we know th<strong>at</strong> for small changes<br />
∆U = T ∆S −p∆V +x∆X, we can write for a process in which the temper<strong>at</strong>ure<br />
changes by a small amount ∆T , but the volume V and the variables X are kept<br />
constant,<br />
∆U<br />
∆T ⎪V,X = T ∆S<br />
∆T ⎪V,X In the limit ∆T → 0 the r<strong>at</strong>ios become partial deriv<strong>at</strong>ives, and we have<br />
( )<br />
∂U<br />
CV (T, V, X) =<br />
∂T<br />
V,X<br />
(2.33)<br />
(2.34)<br />
Similarly, for a process <strong>at</strong> constant pressure and X, the differential of U<br />
yields<br />
and hence<br />
∆U<br />
∆T ⎪p,X Cp(T, p, X) =<br />
= T ∆S<br />
∆T ⎪<br />
− p<br />
p,X<br />
∆V<br />
∆T ⎪p,X ( ) ( )<br />
∂U<br />
∂V<br />
+ p<br />
∂T p,X ∂T p,X<br />
(2.35)<br />
(2.36)<br />
Since most m<strong>at</strong>erials expand when he<strong>at</strong>ed, the volume change in the last case<br />
will cause the system to do work and we have to supply this energy. We expect
2.5. RESPONSE FUNCTIONS. 67<br />
therefore Cp > CV . Also, in general we need he<strong>at</strong> to increase the temper<strong>at</strong>ure, so<br />
we also expect CV > 0. These st<strong>at</strong>ements will be proven l<strong>at</strong>er in the this chapter,<br />
where we talk about equilibrium conditions. He<strong>at</strong> capacities are extensive st<strong>at</strong>e<br />
functions. One often defines a specific he<strong>at</strong> by using the he<strong>at</strong> capacity per unit<br />
volume or per unit mass. Unfortun<strong>at</strong>ely, the word specific he<strong>at</strong> is used for both<br />
these r<strong>at</strong>ios, so make sure th<strong>at</strong> you understand from the context of the problem<br />
which one applies.<br />
Compressibilities.<br />
Another set of response functions which are often measured are rel<strong>at</strong>ed to<br />
volume changes, and called compressibilities. We apply pressure to a system<br />
and measure the change in volume. A measurement like this is done either<br />
adiab<strong>at</strong>ically or isothermally, and the two relevant compressibilities are<br />
κT (T, p, X) = − 1<br />
V<br />
κS(S, p, X) = − 1<br />
V<br />
( )<br />
∂V<br />
∂p T,X<br />
( )<br />
∂V<br />
∂p<br />
S,X<br />
(2.37)<br />
(2.38)<br />
We expect κT > κS > 0 because <strong>at</strong> constant entropy the internal variables<br />
of the system are essentially only scaled to a smaller volume, but <strong>at</strong> constant<br />
temper<strong>at</strong>ure they can also rearrange to allow for an even smaller volume. Again,<br />
these inequalities follow from the equilibrium conditions l<strong>at</strong>er in the this chapter.<br />
The minus sign is included to indic<strong>at</strong>e th<strong>at</strong> the volume decreases with increasing<br />
pressure. The size of the system is scaled out by dividing by the volume.<br />
Coefficient of thermal expansion.<br />
Finally, we often use the coefficient of thermal expansion<br />
α(T, p, X) = 1<br />
V<br />
( )<br />
∂V<br />
∂T p,X<br />
(2.39)<br />
Again, we divide by the volume to scale out the system size. We also expect<br />
α to be positive. This is, however, not always true. There are m<strong>at</strong>erials with<br />
a neg<strong>at</strong>ive coefficient of thermal expansion. These m<strong>at</strong>erials are exceptions,<br />
however.<br />
Response functions for ideal gas.<br />
It is useful to calcul<strong>at</strong>e the values of the response functions for a mono<strong>at</strong>omic<br />
ideal gas. This gives us a standard, a reference for comparison if we<br />
consider other systems. Using U = 3<br />
2pV and pV = NRT we find
68 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
CV = 3<br />
NR (2.40)<br />
2<br />
Cp = 5<br />
NR (2.41)<br />
2<br />
κT = 1<br />
p<br />
κS = 3<br />
5p<br />
α = 1<br />
T<br />
(2.42)<br />
(2.43)<br />
(2.44)<br />
The formula for κS follows from the formula for the entropy of an ideal gas<br />
3 − 1.58, which implies th<strong>at</strong> <strong>at</strong> constant entropy V ∝ T 2 . Together with the ideal<br />
3 − gas law this gives V ∝ p 5 . The ideal gas values indeed s<strong>at</strong>isfy κT > κS > 0<br />
and Cp > CV > 0.<br />
Compressibilities are often given in a different form, by st<strong>at</strong>ing the rel<strong>at</strong>ion<br />
between pressure and volume in the form pV γ = c, where c is a constant and<br />
the value of γ depends on the process we consider. In an isothermal process <strong>at</strong><br />
constant N we see from pV = NRT th<strong>at</strong> γ = 1, and in an adiab<strong>at</strong>ic process <strong>at</strong><br />
constant N we have γ = 5<br />
3 .<br />
There are some obvious problems with these numbers. We have seen before<br />
th<strong>at</strong> <strong>at</strong> low temper<strong>at</strong>ure the he<strong>at</strong> capacities should approach zero, clearly not<br />
the case for the ideal gas. Also, <strong>at</strong> low temper<strong>at</strong>ure the expansion coefficient<br />
becomes infinite. This is due to the fact th<strong>at</strong> the volume approaches zero.<br />
Also, the compressibilities diverge <strong>at</strong> zero pressure. All these facts point <strong>at</strong> the<br />
limit<strong>at</strong>ions of the model of an ideal gas. At low temper<strong>at</strong>ure this model is not<br />
valid.<br />
Why is the ideal gas not a good model <strong>at</strong> low temper<strong>at</strong>ure?<br />
2.6 Rel<strong>at</strong>ions between partial deriv<strong>at</strong>ives.<br />
Two deriv<strong>at</strong>ives.<br />
<strong>Thermodynamics</strong> makes extensive use of the theory of calculus of many<br />
variables. This is a very powerful m<strong>at</strong>hem<strong>at</strong>ical theory! It limits the changes<br />
in dependent variables. For example, response functions are partial deriv<strong>at</strong>ives,<br />
and they are rel<strong>at</strong>ed via Maxwell rel<strong>at</strong>ions if they are also the second order
2.6. RELATIONS BETWEEN PARTIAL DERIVATIVES. 69<br />
deriv<strong>at</strong>ives of a free energy. M<strong>at</strong>hem<strong>at</strong>ics proves other rel<strong>at</strong>ions between partial<br />
deriv<strong>at</strong>ives too. For example, assume we have a st<strong>at</strong>e function z(x, y). It is<br />
possible to define a function f(x, y, z) in such a way th<strong>at</strong> f(x, y, z) = 0 leads<br />
exactly to this function. Hence f(x, y, z(x, y)) ≡ 0 for all values of x and y. The<br />
partial deriv<strong>at</strong>ive of this identity with respect to x is<br />
0 =<br />
( )<br />
∂f<br />
(x, y, z(x, y)) +<br />
∂x y,z<br />
( )<br />
∂f<br />
(x, y, z(x, y))<br />
∂z x,y<br />
( )<br />
∂z<br />
(x, y) (2.45)<br />
∂x y<br />
In a similar way, we can use f(x, y, z) = 0 to define a function x(y, z), with<br />
f(x(y, z), y, z) ≡ 0. The partial deriv<strong>at</strong>ive of this equ<strong>at</strong>ion with respect to z is:<br />
0 =<br />
( )<br />
∂f<br />
(x(y, z), y, z)<br />
∂x y,z<br />
( )<br />
∂x<br />
(y, z) +<br />
∂z y<br />
( )<br />
∂f<br />
(x(y, z), y, z) (2.46)<br />
∂z x,y<br />
Comparing these two rel<strong>at</strong>ions shows th<strong>at</strong> for any point (x0, y0, z0) obeying<br />
f(x0, y0, z0) = 0 we have<br />
( ) ( )<br />
∂z<br />
∂x<br />
(x0, y0) (y0, z0) = 1 (2.47)<br />
∂x y ∂z y<br />
when the partial deriv<strong>at</strong>ives of f(x, y, z) are zero in isol<strong>at</strong>ed points only, and<br />
the partial deriv<strong>at</strong>ives of x and z are continuous. Hence given a st<strong>at</strong>e function<br />
z(x, y) we can use z as a st<strong>at</strong>e variable and change variables to {y, z} , which<br />
defines a new st<strong>at</strong>e function x(y, z). Partial deriv<strong>at</strong>ives of these two functions<br />
are rel<strong>at</strong>ed.<br />
Three deriv<strong>at</strong>ives.<br />
A more surprising result is obtained when all three combin<strong>at</strong>ions of independent<br />
variables are used. Using {x, y} as independent variables we have as<br />
before<br />
−<br />
( )<br />
∂f<br />
(x, y, z(x, y)) =<br />
∂x y,z<br />
In a similar way we obtain<br />
−<br />
−<br />
( )<br />
∂f<br />
(x(y, z), y, z) =<br />
∂y x,z<br />
( )<br />
∂f<br />
(x, y(x, z), z) =<br />
∂z x,y<br />
( )<br />
∂f<br />
(x, y, z(x, y))<br />
∂z x,y<br />
( )<br />
∂f<br />
(x(y, z), y, z)<br />
∂x y,z<br />
( )<br />
∂f<br />
(x, y(x, z), z)<br />
∂y x,z<br />
( )<br />
∂z<br />
(x, y) (2.48)<br />
∂x y<br />
( )<br />
∂x<br />
(y, z) (2.49)<br />
∂y z<br />
( )<br />
∂y<br />
(x, z) (2.50)<br />
∂z x
70 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Multiplying these equ<strong>at</strong>ions leads to<br />
( ) ( ) ( )<br />
∂x<br />
∂y<br />
∂z<br />
(y0, z0) (x0, z0) (x0, y0) = −1 (2.51)<br />
∂y z ∂z x ∂x y<br />
for all points (x0, y0, z0) obeying f(x0, y0, z0) = 0, with similar assumptions one<br />
the partial deriv<strong>at</strong>ives as before. This rel<strong>at</strong>ion is surprising because of the minus<br />
sign. It is often found in the form<br />
( ∂x<br />
∂y<br />
)<br />
(y0, z0) = −<br />
z<br />
( )<br />
∂x<br />
(y0, z0)<br />
∂z y<br />
( )<br />
∂z<br />
(x0, y0) (2.52)<br />
∂y x<br />
This rule can be generalized to arbitrary numbers of variables. It is not hard<br />
to show th<strong>at</strong> a minus sign always occurs when the total number of variables is<br />
odd.<br />
Difference of he<strong>at</strong> capacities.<br />
There are subtle differences in semantics between questions posed by physicists<br />
and m<strong>at</strong>hem<strong>at</strong>icians, and it is important to realize them. If a physicist<br />
asks the question wh<strong>at</strong> is U(T, p, N) if U(T, V, N) = N( 3 N<br />
2RT + V (T − T0)) it is<br />
understood th<strong>at</strong> we try to find a way th<strong>at</strong> allows us to find first V from T, p, N<br />
and then use the formula above. We are looking for two different mappings, and<br />
need to find (T, p, N) → U if we know (T, V, N) → U. We denote the function<br />
by the outcome, which is the same in both cases. M<strong>at</strong>hem<strong>at</strong>icians would simply<br />
respond U(T, p, N) = N( 3 N<br />
2RT + p (T − T0)), since it is a simple substitution of<br />
variables. Like always, we ask questions but assume a specific context. In the<br />
following we will be clear, though, and use the m<strong>at</strong>hem<strong>at</strong>ical distinction, which<br />
says find U2(T, p, N) if we know U1(T, V, N). It is useful, believe me.<br />
As an example of changing variables, consider the st<strong>at</strong>e variables temper<strong>at</strong>ure<br />
T, pressure p, volume V, and the extensive variable(s) X. Considering<br />
the entropy as a function of {T, V, X} gives S1(T, V, X) and considering the<br />
entropy as a function of {T, p, X} gives a different function S2(T, p, X). These<br />
two functions are clearly rel<strong>at</strong>ed by<br />
S2(T, p, X) = S1(T, V (T, p, X), X) (2.53)<br />
where V(T,p,X) is the function for the equilibrium volume with {T, p, X} as independent<br />
variables. Remember th<strong>at</strong> all these rel<strong>at</strong>ions only hold for equilibrium<br />
systems! Therefore the partial deriv<strong>at</strong>ives are rel<strong>at</strong>ed by<br />
( ∂S2<br />
∂T<br />
)<br />
p,X<br />
=<br />
Using the Maxwell rel<strong>at</strong>ion<br />
( )<br />
∂S<br />
∂V<br />
T,X<br />
( ∂S1<br />
∂T<br />
)<br />
V,X<br />
+<br />
( ∂S1<br />
∂V<br />
)<br />
( ) 2 ∂ F<br />
= −<br />
=<br />
∂V ∂T X<br />
T,X<br />
( ∂V<br />
∂T<br />
( )<br />
∂p<br />
∂T V,X<br />
)<br />
p,X<br />
(2.54)<br />
(2.55)
2.6. RELATIONS BETWEEN PARTIAL DERIVATIVES. 71<br />
and the chain rule<br />
we then derive<br />
( ) ( ) ( )<br />
∂p<br />
∂p ∂V<br />
= −<br />
∂T V,X ∂V T,X ∂T p,X<br />
V T α2<br />
Cp − CV =<br />
which tells us th<strong>at</strong> Cp > CV indeed if κT > 0.<br />
Important points.<br />
κT<br />
(2.56)<br />
(2.57)<br />
In this deriv<strong>at</strong>ion we did two things. First, we write these equ<strong>at</strong>ions in<br />
terms of functions, leaving out the variables, since they are all contained in<br />
the partial deriv<strong>at</strong>ives. For example, ( )<br />
∂S<br />
∂V T,X clearly is rel<strong>at</strong>ed to S1(T, V, X),<br />
)<br />
while would imply S2(T, p, X). Even though in the partial deriv<strong>at</strong>ives<br />
( ∂S<br />
∂p<br />
T,X<br />
we write the same symbol S in the top part, which is correct since in both cases<br />
the result of the function is the entropy, the functional form is understood to be<br />
different since the variables are different. This is probably the most confusing<br />
aspect of the not<strong>at</strong>ion used in calculus of many variables. Second, there are<br />
several ways to get from the beginning to the end. We started by using 2.55<br />
since often it is a good idea to get rid of the entropy. In order to do th<strong>at</strong> we<br />
either use the definition of he<strong>at</strong> capacity, or a Maxwell rel<strong>at</strong>ion.<br />
R<strong>at</strong>ios of response functions.<br />
In a similar way we find<br />
V T α2<br />
κT − κS =<br />
Cp<br />
and from the r<strong>at</strong>io of these expressions we find<br />
Cp<br />
CV<br />
= κT<br />
κS<br />
(2.58)<br />
(2.59)<br />
which is a remarkably simple rel<strong>at</strong>ion between very different response functions.<br />
Note th<strong>at</strong> the last three formulas hold for all systems, temper<strong>at</strong>ures, pressures,<br />
etc. Of course, on has to measure Cp and CV <strong>at</strong> the same temper<strong>at</strong>ure and pressure<br />
in order to be able to compare them. These rel<strong>at</strong>ions are experimentally<br />
confirmed. The most important message here is th<strong>at</strong> many response functions<br />
are dependent and such dependencies give a good consistency check for experimental<br />
results. The ideal gas values do obey these rel<strong>at</strong>ions, as can be easily<br />
checked. The fact th<strong>at</strong> such rel<strong>at</strong>ions exist is based on the fact th<strong>at</strong> the function
72 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
U(S, V, N, X) gives a complete description of the system and th<strong>at</strong> all other st<strong>at</strong>e<br />
variables follow from this function.<br />
Check th<strong>at</strong> 2.59 holds for the ideal gas.<br />
Equ<strong>at</strong>ion 2.59 has an interesting consequence. Suppose we have an ideal gas,<br />
but this gas is not mono-<strong>at</strong>omic. In th<strong>at</strong> case we still have pV = NRT , but the<br />
energy versus temper<strong>at</strong>ure rel<strong>at</strong>ion is more complic<strong>at</strong>ed. The isothermal compressibility<br />
κT ia still equal to 1<br />
p , which means th<strong>at</strong> the adiab<strong>at</strong>ic compressibility<br />
is given by:<br />
or<br />
− 1<br />
V<br />
( )<br />
∂V<br />
∂p S,N<br />
= κS = 1 CV<br />
p Cp<br />
( )<br />
∂V<br />
= −<br />
∂p S,N<br />
V CV<br />
p Cp<br />
(2.60)<br />
(2.61)<br />
Suppose th<strong>at</strong> we have a range of volumes for which the temper<strong>at</strong>ure changes<br />
while compressing are such th<strong>at</strong> the r<strong>at</strong>io CV remains constant. Such a range<br />
Cp<br />
is often found for poly<strong>at</strong>omic gases. For example, <strong>at</strong> some intermedi<strong>at</strong>e temper<strong>at</strong>ures<br />
all rot<strong>at</strong>ional degrees of freedom might be active, but the vibr<strong>at</strong>ional<br />
are not. In th<strong>at</strong> case we find for adiab<strong>at</strong>ic compression th<strong>at</strong> p ∝ V −γ with<br />
γ = Cp<br />
. The coefficient of thermal expansion for this poly-<strong>at</strong>omic gas is still<br />
CV<br />
determined by pV = NRT , and hence α = 1<br />
T . Equ<strong>at</strong>ion 2.57 then tells us th<strong>at</strong><br />
Cp − CV = NR, and hence γ = 1 + NR . Therefore, measuring the compress-<br />
CV<br />
ibility gives us the he<strong>at</strong> capacity, from which we can derive inform<strong>at</strong>ion about<br />
the internal degrees of freedom of the molecules!<br />
2.7 Conditions for equilibrium.<br />
P<strong>at</strong>h to equilibrium.<br />
Wh<strong>at</strong> happens when two equilibrium systems are brought into contact? Obviously,<br />
the combined system will try to reach thermal equilibrium. Wh<strong>at</strong> are<br />
the conditions for this total equilibrium? These conditions are easy to derive<br />
if the combined system is completely isol<strong>at</strong>ed or closed, and cannot exchange<br />
energy in the form of work with the rest of the world. He<strong>at</strong> exchange is possible,<br />
though. This means th<strong>at</strong> the total volume, amount of m<strong>at</strong>erial, etc. is constant.<br />
We also assume th<strong>at</strong> we can keep the entropy of the system constant. This does<br />
not conflict with the st<strong>at</strong>ement th<strong>at</strong> he<strong>at</strong> transfer is possible, since we consider<br />
non-equilibrium systems!
2.7. CONDITIONS FOR EQUILIBRIUM. 73<br />
The two subsystems are able to exchange energy by changing their individual<br />
volumes (⇒ mechanical work), amount of m<strong>at</strong>erial (⇒ chemical work), or<br />
entropies (⇒ he<strong>at</strong> exchange). We will label the two subsystems l for left and r<br />
for right. The total internal energy is<br />
with<br />
Ut = Ur + Ul<br />
(2.62)<br />
Ut = U(S, V, N, X) (2.63)<br />
Ul = U(Sl, Vl, Nl, Xl) (2.64)<br />
Ur = U(S − Sl, V − Vl, N − Nl, X − Xl) (2.65)<br />
where we use the symbol U for the function th<strong>at</strong> gives the values Ul, Ur, Ut for<br />
the (sub)systems, and where X stands for all other extensive variables. Since<br />
the total system is closed, the total internal energy in equilibrium will be a<br />
minimum. Also the total sum of each extensive variable cannot change, and<br />
we have S = Sl + Sr, etc. The equilibrium st<strong>at</strong>e follows by minimizing the<br />
internal energy as a function of the four internal (for the total system) variables<br />
{Sl, Vl, Nl, Xl}. This leads to four equ<strong>at</strong>ions of the form<br />
( )<br />
∂U<br />
(Sl, Vl, Nl, Xl) −<br />
∂S V,N,X<br />
( )<br />
∂U<br />
(S − Sl, V − Vl, N − Nl, X − Xl) = 0<br />
∂S V,N,X<br />
(2.66)<br />
Using the expression for the differential dU we then see immedi<strong>at</strong>ely th<strong>at</strong><br />
the conditions for equilibrium are:<br />
Tl = Tr, pl = pr, µl = µr, xl = xr<br />
(2.67)<br />
All intensive variables have to be the same. This is of course exactly wh<strong>at</strong> we<br />
expected.<br />
Using the minimum principle.<br />
At this point we only used the fact th<strong>at</strong> the internal energy is an extremum<br />
<strong>at</strong> equilibrium. We also know th<strong>at</strong> is should be a minimum, and this gives a<br />
number of very useful inequalities for the second order deriv<strong>at</strong>ives. Assume th<strong>at</strong><br />
both subsystems are identical. Assume th<strong>at</strong> the total entropy is 2S, the total<br />
volume 2V, etc. Using identical systems is just for convenience, but does not<br />
affect the general conclusion. The entropy of the system on the left is S + ∆S<br />
and for the system on the right S − ∆S. The volume and all other extensive<br />
quantities are defined in a similar way. The difference in internal energy of the<br />
total system between equilibrium (∆S = 0, etc) and non-equilibrium is:
74 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
∆U = U(S + ∆S, V + ∆V, N + ∆N, X + ∆X)+<br />
U(S − ∆S, V − ∆V, N − ∆N, X − ∆X) − 2U(S, V, N, X) (2.68)<br />
The changes ∆S, ∆V , ∆N, and ∆X in this formula are independent and<br />
can be chosen arbitrarily. The total energy is a minimum as a function of the<br />
internal variables {∆S, ∆V, ∆N, ∆X}, and hence ∆U ≥ 0. Since the values of<br />
{S, V, N, X} are arbitrary, this means th<strong>at</strong> U(S, V, N, X) is a convex function.<br />
Consequences for second order deriv<strong>at</strong>ives.<br />
The previous discussion was true for any value of the parameters ∆S, etc.<br />
When these changes are small, however, we can use a Taylor series expansion for<br />
the energy of the subsystems. The zeroth order term vanishes by construction.<br />
The first order term gives us the equilibrium conditions for the intensive st<strong>at</strong>e<br />
variables we derived before. Hence we can write T for the temper<strong>at</strong>ure of the<br />
whole system and do not have to specify left or right. The same holds for all<br />
other intensive variables. The second order terms form an expression of the<br />
form<br />
( 2 ∂ U<br />
∆U =<br />
∂S2 )<br />
(∆S)<br />
V,N,X<br />
2 ( ) 2 ∂ U<br />
+<br />
∆S∆V + · · · (2.69)<br />
∂S∂V N,X<br />
This expression has to be convex. By changing only one extensive variable <strong>at</strong> a<br />
time we find the necessary inequalities on the st<strong>at</strong>e function U(S, V, N, X) of a<br />
single system:<br />
( ) 2 ∂ U<br />
≥ 0 (2.70)<br />
∂S 2<br />
( ) 2 ∂ U<br />
∂V 2<br />
( ) 2 ∂ U<br />
∂N 2<br />
( 2 ∂ U<br />
∂X 2<br />
)<br />
V,N,X<br />
S,N,X<br />
S,V,X<br />
S,V,N<br />
≥ 0 (2.71)<br />
≥ 0 (2.72)<br />
≥ 0 (2.73)<br />
This is not all, however. Suppose we only change S and V, in which case we<br />
have<br />
∆U =<br />
( 2 ∂ U<br />
∂S2 )<br />
(∆S)<br />
V,N,X<br />
2 ( ) 2 ∂ U<br />
+2<br />
∆S∆V +<br />
∂S∂V N,X<br />
( 2 ∂ U<br />
∂V 2<br />
)<br />
(∆V )<br />
S,N,X<br />
2 ≥ 0<br />
(2.74)<br />
If we divide by (∆V ) 2 we have an inequality of the form ax 2 + bx + c ≥ 0, which<br />
has to hold for all values of x. We already determined th<strong>at</strong> a ≥ 0 and c ≥ 0.
2.7. CONDITIONS FOR EQUILIBRIUM. 75<br />
The form ax 2 + bx + c has two distinct roots when b 2 > 4ac. In our case we<br />
therefore need b 2 ≤ 4ac, or<br />
( ( ) ) 2<br />
2 ∂ U<br />
≤<br />
∂S∂V N,X<br />
He<strong>at</strong> capacities are positive.<br />
( 2 ∂ U<br />
∂S2 )<br />
V,N,X<br />
( 2 ∂ U<br />
∂V 2<br />
)<br />
S,N,X<br />
(2.75)<br />
These inequalities have ( a number of consequences for the signs of response<br />
2<br />
∂ U<br />
functions. For example, ∂S2 )<br />
=<br />
V,N,X<br />
( )<br />
∂T<br />
T<br />
∂S = . Since the tempera-<br />
V,N,X CV<br />
ture is positive (the existence of neg<strong>at</strong>ive temper<strong>at</strong>ure values viol<strong>at</strong>es the second<br />
law), we find<br />
CV ≥ 0 (2.76)<br />
or the specific he<strong>at</strong> <strong>at</strong> constant volume must be positive. This can be rel<strong>at</strong>ed<br />
to a condition for local stability. Consider a small volume in a large system in<br />
equilibrium. Due to some fluctu<strong>at</strong>ion the energy inside this volume is larger<br />
than average. If we would have CV < 0, the temper<strong>at</strong>ure of this volume would<br />
be less than average! Due to the second law, energy would flow into this volume,<br />
making this fluctu<strong>at</strong>ion in energy even larger. Therefore a system with CV < 0 is<br />
unstable against fluctu<strong>at</strong>ions in the energy. Stability therefore requires positive<br />
he<strong>at</strong> capacities <strong>at</strong> constant volume.<br />
Compressibilities are positive.<br />
Similarly, we find from the condition for<br />
(<br />
2<br />
∂ U<br />
∂V 2<br />
)<br />
S,N,X th<strong>at</strong><br />
κS ≥ 0 (2.77)<br />
Compressibilities also have to be positive because of a local stability argument.<br />
Suppose a small volume element has a pressure less than average. The outside<br />
environment has a larger pressure and hence will the imbalance of forces will<br />
reduce the volume of the small element. If the compressibility would be neg<strong>at</strong>ive,<br />
the inside pressure would reduce even more and the process would run<br />
away. Therefore, a neg<strong>at</strong>ive compressibility implies an instability against small<br />
fluctu<strong>at</strong>ions in the pressure.<br />
The more m<strong>at</strong>erial we have, the harder is it to add some.<br />
The stability requirement for the chemical potential is<br />
( )<br />
∂µ<br />
∂N S,V<br />
≥ 0 (2.78)
76 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
which implies th<strong>at</strong> the energy µ it takes to add one unit mass of m<strong>at</strong>erial to<br />
a system increases when the amount of m<strong>at</strong>erial increases. It is harder to add<br />
m<strong>at</strong>erial to a system which already contains a lot! This is again connected with<br />
a local stability requirement for fluctu<strong>at</strong>ions in the number of particles.<br />
Finally we can use equ<strong>at</strong>ion 2.75 to get<br />
T<br />
V κSCV<br />
≥<br />
( ( ) ) 2 ( ( ) ) 2<br />
2 ∂ U<br />
∂T<br />
=<br />
∂S∂V N,X ∂V S,N,X<br />
The chain rule gives<br />
( ) ( )<br />
∂T<br />
∂T<br />
= −<br />
∂V S,N,X ∂S V,N,X<br />
( )<br />
∂S<br />
∂V T,N,X<br />
(2.79)<br />
(2.80)<br />
The first factor on the right hand side is T , while the second is transformed<br />
CV<br />
according to the Maxwell equ<strong>at</strong>ion<br />
( ∂S<br />
∂V<br />
)<br />
T,N,X<br />
Another applic<strong>at</strong>ion of the chain rule gives<br />
( ) ( )<br />
∂p<br />
∂p<br />
= −<br />
∂T<br />
∂V<br />
or<br />
V,N,X<br />
Combining everything gives<br />
=<br />
( )<br />
∂p<br />
∂T V,N,X<br />
T,N,X<br />
( )<br />
∂V<br />
∂T p,N,X<br />
(2.81)<br />
(2.82)<br />
( )<br />
∂p<br />
= −<br />
∂T V,N,X<br />
−1<br />
αV (2.83)<br />
V κT<br />
T<br />
V κSCV<br />
(<br />
αT<br />
≥<br />
CV κT<br />
) 2<br />
(2.84)<br />
Multiplying these inequality by C2 V V<br />
T , which is positive, and using 2.57 leads to<br />
CV<br />
≥<br />
κS<br />
Cp − CV<br />
κT<br />
With the help of 2.76 and 2.59 we then find<br />
and hence due to 2.57<br />
and using 2.58<br />
(2.85)<br />
κT ≥ 0 (2.86)<br />
Cp ≥ CV ≥ 0 (2.87)<br />
κT ≥ κS ≥ 0 (2.88)
2.7. CONDITIONS FOR EQUILIBRIUM. 77<br />
The inequalities for the he<strong>at</strong> capacities and compressibilities hold for all systems.<br />
They have been derived without any knowledge of the function U(S,V,N,X),<br />
apart from the requirements of continuity of the second order deriv<strong>at</strong>ives and<br />
the restrictions due to the laws of thermodynamics. These are remarkable consequences<br />
indeed, based on minimal input.<br />
Would it have been possible to derive these inequalities using the maximum<br />
entropy principle?<br />
Note th<strong>at</strong> we can write the difference between the he<strong>at</strong> capacities in the<br />
following form:<br />
( ) ( )<br />
∂U ∂U<br />
Cp − CV =<br />
−<br />
+ αpV (2.89)<br />
∂T p,X ∂T V,X<br />
and this expression is always positive, even if the coefficient of thermal expansion<br />
α is neg<strong>at</strong>ive! In such a case the internal energy <strong>at</strong> constant pressure has to<br />
increase rapidly as a function of temper<strong>at</strong>ure. This has consequences for the<br />
microscopic description of neg<strong>at</strong>ive thermal expansion m<strong>at</strong>erials!<br />
On paradoxes.<br />
Consider a system where a thermally insul<strong>at</strong>ed piston can move back and<br />
forth in a closed cylinder. We have identical amounts of m<strong>at</strong>erial in both parts<br />
of the cylinder, but they start out <strong>at</strong> different temper<strong>at</strong>ures. At the start, both<br />
parts of the system are in thermal equilibrium, but they are not in equilibrium<br />
with each other. Wh<strong>at</strong> do you expect th<strong>at</strong> the equilibrium st<strong>at</strong>e is? The common<br />
answer is in the middle, with the temper<strong>at</strong>ures and pressures on both sides equal<br />
to each other. Well, people have even done simul<strong>at</strong>ions showing th<strong>at</strong> this is the<br />
case, so it must be true! But if you read the system description very carefully<br />
you will see th<strong>at</strong> this is not true. Both parts of the system are thermally<br />
isol<strong>at</strong>ed. Assume th<strong>at</strong> we take a p<strong>at</strong>h from the initial st<strong>at</strong>e to the final st<strong>at</strong>e<br />
along which each part of the system is always in equilibrium. The initial values<br />
of the entropy of the left and right sides are Sl and Sr, and these could be<br />
different. Equilibrium is obtained when<br />
pl = −<br />
( )<br />
( )<br />
∂U<br />
∂U<br />
(Vl, Sl, N) = pr = − (Vr, Sr, N) (2.90)<br />
∂V S,N<br />
∂V S,N<br />
with V = Vl + Vr. Only when Sl = Sr does this lead to a stable situ<strong>at</strong>ion with<br />
the piston in the middle.<br />
A paradox is a result which seems to contradict common experience. They<br />
often arise from either not giving all inform<strong>at</strong>ion (like in our example), or by<br />
making a mistake in the analysis. Careful analysis will always show where we
78 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
went wrong. For example, in the Gibbs paradox we obtain wrong results if<br />
we omit the N dependence of the constant of integr<strong>at</strong>ion. Unfortun<strong>at</strong>ely, in<br />
thermodynamics we always deal with many variables, and the opportunity to<br />
cre<strong>at</strong>e paradoxes is large. But as long as we always first determine the set of<br />
independent st<strong>at</strong>e variables for a given problem, we will avoid many of these<br />
paradoxes.<br />
2.8 Stability requirements on other free energies.<br />
The conditions for equilibrium discussed in the previous section can be generalized<br />
to all thermodynamic potentials as a function of extensive variables. It<br />
does not hold for intensive variables, though. As an example, consider the Gibbs<br />
free energy G(T,p,N). It is easy to show th<strong>at</strong><br />
( 2 ∂ G<br />
∂N 2<br />
)<br />
≥ 0 (2.91)<br />
using the same thought experiment as before. But these experiments do not<br />
work for intensive variables. It is not possible, for example, to exchange an<br />
amount of temper<strong>at</strong>ure. On the other hand, one can calcul<strong>at</strong>e the second order<br />
deriv<strong>at</strong>ives in this case directly:<br />
( ) 2 ∂ G<br />
∂T 2<br />
( ∂ 2 G<br />
∂p 2<br />
)<br />
p,N<br />
T,N<br />
p,T<br />
( )<br />
∂S<br />
= −<br />
∂T<br />
( )<br />
∂V<br />
=<br />
∂p<br />
p,N<br />
T,N<br />
= − Cp<br />
T<br />
≤ 0 (2.92)<br />
= −V κT ≤ 0 (2.93)<br />
and the signs are opposite from wh<strong>at</strong> we find for the extensive variables. On<br />
the other hand<br />
( ) ( )<br />
2 ∂ G ∂V<br />
=<br />
= αV (2.94)<br />
∂p∂T N ∂T p,N<br />
and this leads to<br />
( 2 ∂ G<br />
∂T 2<br />
)<br />
p,N<br />
V κT Cp<br />
T<br />
( 2 ∂ G<br />
∂p2 ) (( ) )2<br />
2 ∂ G<br />
−<br />
=<br />
T,N ∂p∂T N<br />
− α 2 V 2 = V κT CV<br />
T<br />
≥ 0 (2.95)<br />
where we used 2.57. As a result we find th<strong>at</strong> G(p,T,N) is concave as a function<br />
of p and T, but convex as a function of N. Note th<strong>at</strong> these conclusions are<br />
independent of the minimum property of the Gibbs free energy. In th<strong>at</strong> case we<br />
have the st<strong>at</strong>ement th<strong>at</strong> <strong>at</strong> given values of p,T, and N the Gibbs free energy in<br />
equilibrium is a minimum as a function of the internal degrees of freedom!
2.8. STABILITY REQUIREMENTS ON OTHER FREE ENERGIES. 79<br />
Entropy rel<strong>at</strong>ions.<br />
We can also perform the thought experiment for a completely closed system,<br />
in which no he<strong>at</strong> is allowed to go to the external world. In th<strong>at</strong> case we specify<br />
U,V,N,X for the total system. The equilibrium st<strong>at</strong>e is found by maximizing the<br />
entropy. Because the n<strong>at</strong>ure of the extremum has now changed to a maximum,<br />
this implies th<strong>at</strong> S(U,V,N,X) is a concave function. As a consequence we have<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
≤ 0 (2.96)<br />
V,N,X<br />
( ∂ 2<br />
∂V 2<br />
)<br />
U,N,X<br />
≤ 0 (2.97)<br />
( 2 ∂ S<br />
∂N 2<br />
)<br />
≤ 0 (2.98)<br />
V,U,X<br />
( 2 ∂ S<br />
∂X 2<br />
)<br />
≤ 0 (2.99)<br />
U,V,N<br />
with opposite signs from wh<strong>at</strong> we found for the second order deriv<strong>at</strong>ives of the<br />
internal energy. But we still have the additional requirements of the form<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
V,N,X<br />
( 2 ∂ S<br />
∂V 2<br />
) ( ( ) ) 2<br />
2 ∂ S<br />
≥<br />
U,N,X ∂U∂V N,X<br />
(2.100)<br />
as before. Here the sign does not change because these requirements are rel<strong>at</strong>ed<br />
to a no-root condition.<br />
No inversion for second order deriv<strong>at</strong>ives.<br />
It is interesting to compare the deriv<strong>at</strong>ives containing S and U. We derived<br />
in general th<strong>at</strong><br />
( )<br />
∂S<br />
∂U V,N,X<br />
( )<br />
∂U<br />
= 1 (2.101)<br />
∂S V,N,X<br />
Such a simple rel<strong>at</strong>ion does not hold for the second order deriv<strong>at</strong>ives. From<br />
the inequalities we find<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
V,N,X<br />
( 2 ∂ U<br />
∂S2 )<br />
≤ 0 (2.102)<br />
V,N,X
80 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
2.9 A magnetic puzzle.<br />
Wh<strong>at</strong> is wrong?<br />
Consider a magnetic system, in which the basic extensive variables are the<br />
entropy S, volume V, amount of m<strong>at</strong>erial N, and total magnetic moment M.<br />
The magnetic work term is given by<br />
¯dWmag = −HdM (2.103)<br />
where we assume th<strong>at</strong> the direction of the applied field and the direction of the<br />
resulting moment are always the same. We define the magnetic susceptibilities<br />
by<br />
( )<br />
∂M<br />
χT =<br />
(2.104)<br />
∂H T<br />
and<br />
χS =<br />
( )<br />
∂M<br />
∂H S<br />
(2.105)<br />
In problem 7 we analyze this, and we find th<strong>at</strong> the normal stability requirements<br />
demand th<strong>at</strong><br />
χT ≥ χS ≥ 0 (2.106)<br />
for all m<strong>at</strong>erials, all values of the temper<strong>at</strong>ure, pressure, etc.<br />
This is very similar to the case of compressibilities, and in experiments compressibilities<br />
are always positive. Wh<strong>at</strong> could go wrong here? Well, there are<br />
many diamagnetic m<strong>at</strong>erials, for which the susceptibilities are neg<strong>at</strong>ive!<br />
Possible solutions.<br />
One possible answer is th<strong>at</strong> thermodynamics is wrong. Before we take this<br />
drastic step, however, it is advisable to consider other possibilities. Could the<br />
experiments be wrong? No, they are not. Is there some physics missing? Is<br />
there anything th<strong>at</strong> takes up energy? Yes! In order to apply a field we cre<strong>at</strong>e a<br />
field in the m<strong>at</strong>erial, and this magnetic field also needs energy! Therefore, the<br />
work term is incomplete. We need to include an extra term th<strong>at</strong> accounts for<br />
the energy stored in the field.<br />
Some basic E&M.<br />
In order to understand the effects of electric and magnetic fields on a sample,<br />
we divide space into two parts, separ<strong>at</strong>ed by a surface A. This surface is just
2.9. A MAGNETIC PUZZLE. 81<br />
outside the sample, so th<strong>at</strong> the enclosed volume is equal to the volume V of the<br />
sample. The only interactions through this surface are via electric and magnetic<br />
fields. Outside the sample we have applied electric and magnetic fields ⃗ E and<br />
⃗H, gener<strong>at</strong>ed by charges and currents. Outside the sample these fields are<br />
equivalent to the ⃗ D and ⃗ B fields, since there is no m<strong>at</strong>ter outside the sample.<br />
In real life one might have to correct for the gas in the sample chamber, but<br />
th<strong>at</strong> is not important here. The energy flow through a surface is given by the<br />
Poynting vector, ⃗ S, and the increase in internal energy of the sample is<br />
<br />
∆U = −∆t ⃗S · d 2 A ⃗ (2.107)<br />
where we have a minus sign because the surface element points away from the<br />
sample. Using the divergence theorem:<br />
∫<br />
∆U = −∆t ⃗∇ · ⃗ Sd 3 V (2.108)<br />
In MKSA units we have<br />
and hence<br />
∫<br />
∆U = −∆t<br />
V<br />
V<br />
A<br />
⃗S = ⃗ E × ⃗ H (2.109)<br />
⃗∇ · ⃗ E × ⃗ Hd 3 V (2.110)<br />
We now apply a product rule, of the form ⃗ ∇ · ⃗ A × ⃗ B = ⃗ B · ⃗ ∇ × ⃗ A − ⃗ A · ⃗ ∇ × ⃗ B<br />
to get<br />
∫ [ ]<br />
∆U = −∆t ⃗H · ∇ ⃗ × E⃗ − E ⃗ · ∇ ⃗ × H⃗<br />
d<br />
V<br />
3 V (2.111)<br />
At this point we can use Maxwell’s equ<strong>at</strong>ions to simplify m<strong>at</strong>ters. We have<br />
and<br />
⃗∇ × ⃗ H = ⃗j + ∂ ⃗ D<br />
∂t<br />
⃗∇ × ⃗ E = − ∂ ⃗ B<br />
∂t<br />
(2.112)<br />
(2.113)<br />
where the current density represents the free currents in the sample. Because<br />
we have magnetic fields th<strong>at</strong> can change, Eddy currents can be induced. We<br />
now have an expression for the energy change in terms of changing fields and<br />
currents:<br />
∫ [<br />
∆U = −∆t ⃗H · (−<br />
V<br />
∂ ⃗ B<br />
∂t ) − ⃗ E · (⃗j + ∂ ⃗ D<br />
∂t )<br />
]<br />
(2.114)
82 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
The middle term we recognize as the Joule he<strong>at</strong>ing taking place in the sample.<br />
This is he<strong>at</strong> th<strong>at</strong> is gener<strong>at</strong>ed inside the system, and hence does indeed<br />
add to the internal energy. We will get back to this term l<strong>at</strong>er in Chapter 6. In<br />
equilibrium thermodynamics we have no net currents, though, and so for now<br />
we ignore this term.<br />
To simplify m<strong>at</strong>ters we assume th<strong>at</strong> the fields are all uniform in the sample,<br />
so the volume integrals can be done. We also integr<strong>at</strong>e over a small time and<br />
find the changes in the fields. Th<strong>at</strong> leads to<br />
∆U = V<br />
[ ]<br />
⃗H · ∆B ⃗ + E ⃗ · ∆D⃗ (2.115)<br />
The B field and the H field are rel<strong>at</strong>ed via the magnetic moment density<br />
according to<br />
and similarly for the electric quantities<br />
⃗B = µ0( ⃗ H + 1<br />
V ⃗ M) (2.116)<br />
⃗D = ϵ0 ⃗ E + 1<br />
V ⃗ P (2.117)<br />
Note th<strong>at</strong> in Electricity and Magnetism the quantities in the previous equ<strong>at</strong>ions<br />
contain magnetiz<strong>at</strong>ion and polariz<strong>at</strong>ion density. In <strong>Thermodynamics</strong>, on<br />
the other hand, we use extensive quantities for these properties, and hence ⃗ M<br />
is the total magnetiz<strong>at</strong>ion and ⃗ P the total polariz<strong>at</strong>ion.<br />
Energy stored in the fields.<br />
This leaves us with four terms in the energy expression. Two depend on the<br />
fields only, they are:<br />
[<br />
∆Uf = V µ0 ⃗ H · ∆ ⃗ H + ϵ0 ⃗ E · ∆ ⃗ ]<br />
E<br />
(2.118)<br />
These two terms we recognize as the energy stored in the fields. If we change<br />
fields inside a sample it changes the m<strong>at</strong>erials properties, but we also need to<br />
cre<strong>at</strong>e the fields. Since the fields are not part of the sample, we could ignore<br />
them, but th<strong>at</strong> is part of the problem.<br />
Energy rel<strong>at</strong>ed to m<strong>at</strong>ter.<br />
The remaining two terms give the energy of the sample in a magnetic field. If<br />
we change the sample properties in a fixed field, we need more energy, according<br />
to<br />
[<br />
∆Um = µ0 ⃗ H · ∆ ⃗ M + ⃗ E · ∆ ⃗ ]<br />
P<br />
(2.119)
2.9. A MAGNETIC PUZZLE. 83<br />
Units are not always obvious in E&M.<br />
In thermodynamics we are used to seeing the formulas for magnetic work<br />
without the constants µ0. This is not hard to accomplish, as long as we remember<br />
to use the correct units. To obtain Gaussian units we make the following<br />
replacements:<br />
H →<br />
1<br />
√ H (2.120)<br />
4πµ0<br />
B → µ0<br />
√4π B (2.121)<br />
√<br />
4π<br />
M → √ M (2.122)<br />
µ0<br />
E →<br />
1<br />
√ E (2.123)<br />
4πϵ0<br />
D → ϵ0<br />
√4π D (2.124)<br />
P → √ 4πϵ0P (2.125)<br />
and in Gaussian units we have, assuming th<strong>at</strong> the polariz<strong>at</strong>ions are parallel to<br />
the fields,<br />
Interpret<strong>at</strong>ion.<br />
dU = V V<br />
HdH + EdE + HdM + EdP (2.126)<br />
4π 4π<br />
In the remainder of this section we focus on the magnetic part of the energy,<br />
and assume th<strong>at</strong> we apply external magnetic fields only. Th<strong>at</strong> is not essential<br />
for the discussion, though. The work term in 2.126 includes a term th<strong>at</strong> does<br />
not depend on the magnetiz<strong>at</strong>ion of the m<strong>at</strong>erial. If we do an experiment to<br />
measure the magnetic susceptibility and leave out the sample, we would still<br />
measure an effect, since we are storing energy in the magnetic field in the region<br />
where the sample would have been. The first term in 2.126 is really not a term<br />
connected with a sample, and should be subtracted. This is, however, never<br />
done in practice, and one always measure the total response of sample and field<br />
together, and defines the susceptibility from th<strong>at</strong> total response. If we would<br />
subtract the energy stored in the field, susceptibilities would be positive indeed.<br />
Experimentally measured susceptibilities.<br />
A simple rel<strong>at</strong>ion between the B and H fields is given by B = µH, which<br />
combines with H = B − 4π<br />
V M (Gaussian units!) gives
84 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
(µ − 1)H = 4π<br />
M (2.127)<br />
V<br />
from which we derive for the work done, which is the opposite from the change<br />
in energy,<br />
¯dW = − µ<br />
HdM (2.128)<br />
µ − 1<br />
or by defining M ′ = µ<br />
µ−1M we have in this new variable<br />
¯dW = −HdM ′<br />
(2.129)<br />
exactly as we had before we introduced the energy in the field. In terms of this<br />
new variable M ′ we have therefore:<br />
or<br />
or<br />
But we also have using 2.127<br />
which leads to<br />
or<br />
χ =<br />
( 2 ∂ U<br />
∂M ′2<br />
)<br />
≥ 0 (2.130)<br />
(<br />
∂H<br />
∂M ′<br />
)<br />
≥ 0 (2.131)<br />
µ − 1<br />
µ<br />
1<br />
≥ 0 (2.132)<br />
χ<br />
( )<br />
∂M µ − 1<br />
= V<br />
∂H 4π<br />
4π<br />
µV<br />
(2.133)<br />
≥ 0 (2.134)<br />
µ ≥ 0 (2.135)<br />
In other words, the permeability is a positive quantity. The consequence for the<br />
susceptibility is then from 2.133 th<strong>at</strong><br />
χ ≥ − V<br />
4π<br />
which is indeed observed experimentally.<br />
(2.136)
2.10. ROLE OF FLUCTUATIONS. 85<br />
2.10 Role of fluctu<strong>at</strong>ions.<br />
In this section we take a first look <strong>at</strong> the role of fluctu<strong>at</strong>ions in m<strong>at</strong>erials. A<br />
n<strong>at</strong>ural explan<strong>at</strong>ion of phenomena rel<strong>at</strong>ed to fluctu<strong>at</strong>ions is given in st<strong>at</strong>istical<br />
mechanics, where we consider the <strong>at</strong>omic n<strong>at</strong>ure of m<strong>at</strong>erials on a microscopic<br />
scale. In this section we look <strong>at</strong> fluctu<strong>at</strong>ions on a larger scale, and derive some<br />
conclusions which should always be valid, independent of the <strong>at</strong>omic n<strong>at</strong>ure of<br />
the m<strong>at</strong>erial. This again gives us some basic rel<strong>at</strong>ions which a true microscopic<br />
theory should obey. This is another example of the limits set by thermodynamics<br />
on microscopic theories.<br />
It is important to realize th<strong>at</strong> the way we will be describing fluctu<strong>at</strong>ions<br />
here is on a macroscopic level. We assume th<strong>at</strong> the laws of thermodynamics<br />
are obeyed, and th<strong>at</strong> st<strong>at</strong>e variables can be defined. In st<strong>at</strong>istical mechanics we<br />
will look <strong>at</strong> microscopic st<strong>at</strong>es, and see how the average of those microscopic<br />
st<strong>at</strong>es lead to the thermodynamic averages. Because of the fact th<strong>at</strong> in this section<br />
we consider thermodynamic fluctu<strong>at</strong>ions, we will always average over many<br />
microscopic st<strong>at</strong>es. In st<strong>at</strong>istical mechanics we will discuss how to perform such<br />
averages, which need to be consistent with the choice of independent variables<br />
for our thermodynamic system.<br />
A small subsystem.<br />
Consider a system with volume VL, amount of m<strong>at</strong>erial NL (in moles!), and<br />
energy UL. These quantities are fixed and they completely define the system.<br />
In other words, the temper<strong>at</strong>ure TL, the pressure pL, and the chemical potential<br />
µL of the system are completely defined in equilibrium. Also, the entropy SL<br />
of the system in equilibrium is determined.<br />
Consider a small part of this system. In order to define a small part, we<br />
need <strong>at</strong> least on extensive variable. The n<strong>at</strong>ural variable is to use the volume,<br />
although one could also use the amount of m<strong>at</strong>erial. Hence we consider a small<br />
volume V inside the system. We know th<strong>at</strong> on average we have for the energy<br />
U and the amount of m<strong>at</strong>erial N in this subsystem:<br />
and<br />
V<br />
U = UL<br />
VL<br />
V<br />
N = NL<br />
VL<br />
(2.137)<br />
(2.138)<br />
The energy and the amount of m<strong>at</strong>erial in the subsystem can fluctu<strong>at</strong>e by<br />
amounts ∆U and ∆N, however, and as derived before we have for the entropy<br />
of the subsystem:<br />
S(U + ∆U, V, N + ∆N) = S(U, V, N) + 1<br />
TL<br />
∆U − µL<br />
∆N+<br />
TL
86 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
1<br />
2<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
(∆U)<br />
V,N<br />
2 ( ) 2 ∂ S<br />
+<br />
∆U∆N +<br />
∂U∂N V<br />
1<br />
( 2 ∂ S<br />
2 ∂N 2<br />
)<br />
(∆N)<br />
U,V<br />
2 + · · ·<br />
(2.139)<br />
which is exactly the rel<strong>at</strong>ion we used to determine the equilibrium conditions<br />
for a stable m<strong>at</strong>erial. Note th<strong>at</strong> we assume th<strong>at</strong> the m<strong>at</strong>erial inside V is always<br />
in thermal equilibrium with itself, but not necessarily with the environment.<br />
Th<strong>at</strong> is why we can use the entropy formula. A discussion of more complic<strong>at</strong>ed<br />
fluctu<strong>at</strong>ions is beyond whet we can do here, because we need a definition of<br />
thermal equilibrium. One can make more progress in st<strong>at</strong>istical mechanics,<br />
since the microscopic details will play a role. But even there one has to be<br />
careful, and not go against basic theoretical assumptions.<br />
Note th<strong>at</strong> in our discussion we start with extensive variables, since these<br />
quantities can be measured directly. This is the logical first step. Quantities like<br />
pressure and temper<strong>at</strong>ure will also fluctu<strong>at</strong>e, but in order to find the fluctu<strong>at</strong>ions<br />
in intensive quantities we will need to invoke thermal equilibrium.<br />
The discussion we had before used fluctu<strong>at</strong>ions to derive criteria for stability,<br />
by requiring th<strong>at</strong> the entropy in equilibrium is maximal. This discussion does<br />
not answer the question how likely these fluctu<strong>at</strong>ions are to happen. In other<br />
words, wh<strong>at</strong> is the probability W (∆U, ∆N) of such a fluctu<strong>at</strong>ion? Which terms<br />
in the entropy should play a role?<br />
Subtracting the reference values.<br />
The first term in 2.139 simply sets the value of the entropy as expected from<br />
equilibrium, when the total system is completely homogeneous. This is just a<br />
reference value. The linear terms determine the conditions for equilibrium, and<br />
show th<strong>at</strong> T = TL and µ = µL are needed. Th<strong>at</strong> is why we use TL and µL in<br />
equ<strong>at</strong>ion 2.139 and not the actual values, since the difference has to show up in<br />
the second order terms. In equilibrium both the temper<strong>at</strong>ure and the chemical<br />
potential are uniform. Therefore we consider the function:<br />
R(U, V, N, ∆U, ∆N) = S(U + ∆U, V, N + ∆N) − S(U, V, N)−<br />
( ) ( )<br />
∂S<br />
∂S<br />
∆U −<br />
∆N (2.140)<br />
∂U V,N ∂N U,V<br />
where by definition we do not subtract terms with fluctu<strong>at</strong>ions in the volume,<br />
since the volume is considered to be a constant. Another motiv<strong>at</strong>ion for subtracting<br />
the linear terms is th<strong>at</strong> these terms are equal to the change in entropy<br />
of the remainder of the system. Since the remainder is extremely large, th<strong>at</strong><br />
change in entropy only contains the first order terms. Hence the function R<br />
really represents the change in entropy for the whole system! We have to add<br />
the changes in entropy of the rest of the system, but since the changes in U and<br />
N are opposite, we have to subtract the two terms in the equ<strong>at</strong>ion above.
2.10. ROLE OF FLUCTUATIONS. 87<br />
This function R contains all higher order terms in the fluctu<strong>at</strong>ions. Also,<br />
this function is neg<strong>at</strong>ive, since in equilibrium the entropy of the complete system<br />
is a maximum. Therefore we have th<strong>at</strong> 0 ≤ e R ≤ 1 and this could perhaps be<br />
used for a probability. Another reason is th<strong>at</strong> the law of large numbers tells<br />
us th<strong>at</strong> the probability as a function of the fluctu<strong>at</strong>ions should be a Gaussian<br />
distribution for small fluctu<strong>at</strong>ions, and e R has th<strong>at</strong> form. It also incorpor<strong>at</strong>es<br />
the fact th<strong>at</strong> large fluctu<strong>at</strong>ions are less likely, in an exponential manner.<br />
We have introduced a new element here. Based on wh<strong>at</strong> we know from<br />
st<strong>at</strong>istical mechanics we can say th<strong>at</strong> probabilities are rel<strong>at</strong>ed to the exponent<br />
of the entropy. We note here th<strong>at</strong> a function of the n<strong>at</strong>ure e R shows the right<br />
behavior, so we will simply postul<strong>at</strong>e th<strong>at</strong> it indeed represents the probabilities<br />
we need. It is also the n<strong>at</strong>ural extension in a st<strong>at</strong>istical mechanical theory.<br />
Finally, the consequences are supported by experimental evidence.<br />
Two new variables.<br />
and<br />
In order to make the m<strong>at</strong>h easier, we introduce two new variables:<br />
XU = 1<br />
TL<br />
XN = − µL<br />
TL<br />
(2.141)<br />
(2.142)<br />
If we assume th<strong>at</strong> the m<strong>at</strong>erial in the volume V is in equilibrium with the total<br />
sample when there are no fluctu<strong>at</strong>ions, we can then rewrite the function R in<br />
the form:<br />
R(U, V, N, ∆U, ∆N) = S(U + ∆U, V, N + ∆N) − S(U, V, N) − XU∆U − XN ∆N<br />
(2.143)<br />
The entropy is the extensive st<strong>at</strong>e variable describing the internal structure<br />
of a m<strong>at</strong>erial. It is therefore reasonable to use the entropy as the st<strong>at</strong>e function<br />
th<strong>at</strong> governs the probability of the fluctu<strong>at</strong>ions. Using the fact th<strong>at</strong> for small<br />
fluctu<strong>at</strong>ions in the thermodynamic limit (V large, but always small compared<br />
to the total volume VL) the probabilities need to be Gaussian we define for the<br />
probability W of a fluctu<strong>at</strong>ion in U and V:<br />
W (∆U, ∆N) = Ae ωR(U,V,N,∆U,∆N)<br />
(2.144)<br />
This definition contains two parameters, but the value of A is easily fixed by<br />
normaliz<strong>at</strong>ion:<br />
∫<br />
Ae ωR(U,V,N,∆U,∆N) d∆Ud∆N = 1 (2.145)<br />
The remaining question is how to find the value of ω. This requires microscopic<br />
theory or experiment. It turns out th<strong>at</strong> ωkB = 1, where kB is the
88 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Boltzmann constant, rel<strong>at</strong>ed to the molar gas constant R by R = NAkB, with<br />
NA Avogadro’s number. This result easily explained in st<strong>at</strong>istical mechanics.<br />
This observ<strong>at</strong>ion also allows us to connect macroscopic experiments with microscopic<br />
theory, since we can measure probability distributions of macroscopic<br />
fluctu<strong>at</strong>ions!<br />
In equ<strong>at</strong>ion 2.144, wh<strong>at</strong> is really new and unexpected? The basic form can be<br />
derived from the law of large numbers, which st<strong>at</strong>es th<strong>at</strong> in the thermodynamic<br />
limit the probability distribution has to be Gaussian. This requires th<strong>at</strong> the<br />
leading term in the exponent should be of the form<br />
−A (∆U) 2 − B∆U∆N − C (∆N) 2<br />
(2.146)<br />
and th<strong>at</strong> A > 0, C > 0 , and B 2 ≤ 4AC. It does not require any further rel<strong>at</strong>ion<br />
between the values of A, B, and C, though. It also does not say anything<br />
about the higher order terms. Equ<strong>at</strong>ion 2.144 limits how these coefficients<br />
are rel<strong>at</strong>ed. This has consequences for experimental observ<strong>at</strong>ions, and rel<strong>at</strong>ion<br />
based on 2.144 have been confirmed both theoretically (st<strong>at</strong>istical mechanics)<br />
and experimentally.<br />
Averages.<br />
The use of equ<strong>at</strong>ion 2.144 follows from the calcul<strong>at</strong>ion of average values.<br />
Suppose we have a st<strong>at</strong>e function f th<strong>at</strong> depends on the amount of m<strong>at</strong>erial and<br />
the energy in the volume V. We assume th<strong>at</strong> the fluctu<strong>at</strong>ions are slow, and th<strong>at</strong><br />
the m<strong>at</strong>erial in V is always in equilibrium internally (but not with the outside<br />
world, of course). We can then define an average of this function by<br />
∫<br />
⟨f(∆U, ∆N)⟩ = Ae ωR(U,V,N,∆U,∆N) f(∆U, ∆N)d∆Ud∆N (2.147)<br />
Obviously we need to have ⟨∆U⟩ = 0 and ⟨∆N⟩ = 0.<br />
Ignore higher order terms.<br />
Why is this obvious?<br />
The definition 2.144 includes third order terms. Because of these terms, the<br />
probability distribution is not symmetric around the maximum value, and as a<br />
result a calcul<strong>at</strong>ion of ⟨∆U⟩ using 2.144 might not produce zero. This seems<br />
to be a problem, until we realize th<strong>at</strong> we have to take the thermodynamic<br />
limit (this thing is hidden everywhere!) and as a consequence the effects of the<br />
higher order terms become much smaller than those of the second order terms.<br />
Therefore we approxim<strong>at</strong>e the probability by
2.10. ROLE OF FLUCTUATIONS. 89<br />
W (∆U, ∆N) = Ae ω<br />
( (<br />
1 ∂<br />
2<br />
S<br />
2 ∂U 2<br />
)<br />
V,N (∆U)2 ( ) (<br />
∂<br />
2<br />
S<br />
1 ∂<br />
2<br />
S<br />
+ ∂U∂N ∆U∆N+<br />
V<br />
2 ∂N 2<br />
)<br />
U,V (∆N)2<br />
)<br />
(2.148)<br />
Using this form in the thermodynamic limit will give the same results as using<br />
the original definition.<br />
Consequences.<br />
As an example we study the case where only energy can fluctu<strong>at</strong>e, and hence<br />
the amount of m<strong>at</strong>erial in V is constant, or ∆N = 0. Because ( )<br />
∂S<br />
1<br />
∂U = V,N T we<br />
have<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
V,N<br />
= − 1<br />
T 2<br />
( )<br />
∂T<br />
∂U V,N<br />
= − 1<br />
T 2CV (2.149)<br />
and the probability for a fluctu<strong>at</strong>ion in the energy is<br />
ω(∆U)2<br />
−<br />
2T W (∆U) = Ae 2CV (2.150)<br />
The calcul<strong>at</strong>ion of the average mean square fluctu<strong>at</strong>ion is easy, and we find<br />
⟨(∆U) 2 2 T<br />
⟩ =<br />
ω CV<br />
(2.151)<br />
This formula has some interesting consequences. Since the right hand side<br />
is extensive, the left hand side is proportional to N, and we have<br />
√<br />
⟨(∆U) 2 ⟩<br />
U<br />
∝ 1<br />
√ N<br />
(2.152)<br />
In other words, in the thermodynamic limit the rel<strong>at</strong>ive magnitude of the fluctu<strong>at</strong>ions<br />
goes to zero!<br />
A second interesting observ<strong>at</strong>ion is th<strong>at</strong> fluctu<strong>at</strong>ions are rel<strong>at</strong>ed to the corresponding<br />
response function. M<strong>at</strong>erials with a large value of the he<strong>at</strong> capacity<br />
show large fluctu<strong>at</strong>ions in the energy. In other words, if we have a m<strong>at</strong>erial with<br />
a large he<strong>at</strong> capacity a small fluctu<strong>at</strong>ion in the temper<strong>at</strong>ure will give a large<br />
fluctu<strong>at</strong>ion in the energy. Such m<strong>at</strong>erials apparently respond more dram<strong>at</strong>ically<br />
to temper<strong>at</strong>ure fluctu<strong>at</strong>ions.<br />
Both previous observ<strong>at</strong>ions hold in general, and are useful to remember.<br />
Fluctu<strong>at</strong>ions are correl<strong>at</strong>ed.<br />
Wh<strong>at</strong> is the value of ⟨∆U∆N⟩? Naively one might say zero, but th<strong>at</strong> is not<br />
true since we assume th<strong>at</strong> the m<strong>at</strong>erial in V is always in equilibrium. A change<br />
in U will change the chemical potential, and the effect of changes in N are now
90 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
dependent on how U is changing. This is rel<strong>at</strong>ed to the presence of the mixed<br />
term in the exponent in the definition of the probability. The form is<br />
( ) 2 ∂ S<br />
= −<br />
∂U∂N<br />
1<br />
T 2<br />
( )<br />
∂T<br />
(2.153)<br />
∂N<br />
V<br />
which indeed rel<strong>at</strong>es changed in N to changes in T and hence changes in U.<br />
A calcul<strong>at</strong>ional trick.<br />
We use 2.144 in connection with 2.143, and consider XU and XN to be<br />
independent variables, which in equilibrium will take the appropri<strong>at</strong>e values<br />
rel<strong>at</strong>ed to the big system. This leads to<br />
( )<br />
∂W<br />
= −ω∆UW (2.154)<br />
∂XU<br />
and<br />
This leads to<br />
U,V<br />
( )<br />
∂W<br />
= −ω∆NW (2.155)<br />
∂XN<br />
⟨∆U∆N⟩ = − 1<br />
ω ⟨∆U<br />
( )<br />
∂W 1<br />
⟩ (2.156)<br />
∂XN W<br />
Using the definition of an average we find<br />
⟨∆U∆N⟩ = − 1<br />
∫ (<br />
∂W<br />
∆U<br />
ω ∂XN<br />
)<br />
d∆Ud∆N (2.157)<br />
where the 1<br />
W cancels the probability distribution W needed in the integral.<br />
Using integr<strong>at</strong>ion by parts<br />
⟨∆U∆N⟩ = − 1<br />
ω<br />
∂<br />
∂XN<br />
∫<br />
∆UW d∆Ud∆N + 1<br />
ω<br />
∫ ( ∂∆U<br />
∂XN<br />
)<br />
W d∆Ud∆N<br />
(2.158)<br />
The first term is zero because ∫ ∆UW d∆Ud∆N = 0. The second term<br />
contains<br />
( ) ( )<br />
∂∆U ∂(Uinstantaneous − U)<br />
=<br />
∂XN<br />
∂XN<br />
( )<br />
∂U<br />
= −<br />
∂XN<br />
(2.159)<br />
since the instantaneous energy is an independent variable. This expression is<br />
also independent of the fluctu<strong>at</strong>ion, and we obtain<br />
⟨∆U∆N⟩ = − 1<br />
( )<br />
∂U<br />
(2.160)<br />
ω ∂XN<br />
V,XU
2.10. ROLE OF FLUCTUATIONS. 91<br />
In a similar manner, elimin<strong>at</strong>ing ∆U, we could have obtained<br />
⟨∆U∆N⟩ = − 1<br />
( )<br />
∂N<br />
ω ∂XU<br />
V,XN<br />
(2.161)<br />
which shows a type of rel<strong>at</strong>ion between partial deriv<strong>at</strong>ives which is similar to<br />
the Maxwell rel<strong>at</strong>ions discussed in this chapter.<br />
We now use the last equ<strong>at</strong>ion and use th<strong>at</strong> XU is in equilibrium determined<br />
by TL via XU = 1 and obtain:<br />
TL<br />
( )<br />
2 T ∂N<br />
⟨∆U∆N⟩ = (2.162)<br />
ω ∂T V,XN<br />
where the partial deriv<strong>at</strong>ive is calcul<strong>at</strong>ed <strong>at</strong> equilibrium values corresponding<br />
to V, T = TL, µ = µL. The only point of concern is th<strong>at</strong> the partial deriv<strong>at</strong>ive<br />
is calcul<strong>at</strong>ed <strong>at</strong> constant XN, or <strong>at</strong> constant µ<br />
T . Using th<strong>at</strong> in this case<br />
N(T, V, µ) = N(T, V, XN T ) we derive<br />
( ) ( ) ( )<br />
∂N<br />
∂N<br />
∂N<br />
= + XN<br />
(2.163)<br />
∂T<br />
∂T V,µ ∂µ T,V<br />
V,XN<br />
and we have for our final result<br />
⟨∆U∆N⟩ = T<br />
[<br />
T<br />
ω<br />
( ) ( ) ]<br />
∂N<br />
∂N<br />
+ µ<br />
∂T V,µ ∂µ T,V<br />
(2.164)<br />
which indic<strong>at</strong>es th<strong>at</strong> the fluctu<strong>at</strong>ions in U and N are correl<strong>at</strong>ed, unless the<br />
partial deriv<strong>at</strong>ives cancel in a special way.<br />
If we would have used the first formula for this correl<strong>at</strong>ion, we would have<br />
found<br />
⟨∆U∆N⟩ = − 1<br />
(<br />
∂U<br />
ω ∂(− µ<br />
T )<br />
)<br />
V, 1<br />
T<br />
= T<br />
ω<br />
( )<br />
∂U<br />
∂µ V,T<br />
(2.165)<br />
and using the energy rel<strong>at</strong>ions dU = T dS + µdN, where there is no dV term<br />
since the volume is constant, we find<br />
⟨∆U∆N⟩ = T<br />
( ( ) ( ) )<br />
∂S<br />
∂N<br />
T + µ<br />
(2.166)<br />
ω ∂µ V,T ∂µ V,T<br />
and Mr. Maxwell tells us th<strong>at</strong> this is indeed the same as our first equ<strong>at</strong>ion for<br />
the correl<strong>at</strong>ion between fluctu<strong>at</strong>ions.<br />
Fluctu<strong>at</strong>ions in U again.<br />
In a similar way we derive th<strong>at</strong><br />
⟨(∆U) 2 ⟩ =<br />
T 2<br />
ω<br />
( )<br />
∂U<br />
∂T V,XN<br />
(2.167)
92 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
and this differs from the form we derived before, because in the partial deriv<strong>at</strong>ive<br />
XN is held constant. This difference can be explained very easily. In the current<br />
calcul<strong>at</strong>ion we allow both U and N to fluctu<strong>at</strong>e, and we know th<strong>at</strong> these are correl<strong>at</strong>ed.<br />
In the case we considered before we demanded th<strong>at</strong> ∆N = 0. The space<br />
over which the average value is calcul<strong>at</strong>ed is therefore different, and the result<br />
will be different! As usual, this shows how important it is in thermodynamic<br />
processes to specify the conditions under which we do an experiment!<br />
Fluctu<strong>at</strong>ions in other variables.<br />
In our approach to fluctu<strong>at</strong>ions we assume th<strong>at</strong> the m<strong>at</strong>erial inside the volume<br />
V is always in thermal equilibrium with itself, and th<strong>at</strong> the fluctu<strong>at</strong>ions are<br />
therefore very slow. In th<strong>at</strong> case it is easy to calcul<strong>at</strong>e the average root mean<br />
square fluctu<strong>at</strong>ions in other variables. For example, for small fluctu<strong>at</strong>ions we<br />
have<br />
( ) ( )<br />
∂T<br />
∂T<br />
∆T =<br />
∆U +<br />
∆N (2.168)<br />
∂U V,N ∂N U,V<br />
and this gives<br />
⟨(∆T ) 2 ⟩ =<br />
( ( ) ) 2<br />
∂T<br />
⟨(∆U)<br />
∂U V,N<br />
2 ⟩+<br />
( ) ( )<br />
( ( ) ) 2<br />
∂T ∂T<br />
∂T<br />
2<br />
⟨∆U∆N⟩ +<br />
⟨(∆N)<br />
∂U V,N ∂N U,N<br />
∂N U,N<br />
2 ⟩ (2.169)<br />
An interesting cross correl<strong>at</strong>ion.<br />
We now evalu<strong>at</strong>e the following correl<strong>at</strong>ion<br />
( )<br />
( )<br />
∂T<br />
∂T<br />
⟨∆T ∆N⟩ =<br />
⟨∆U∆N⟩ +<br />
∂U<br />
∂N<br />
V,N<br />
U,V<br />
⟨(∆N) 2 ⟩ (2.170)<br />
In order to evalu<strong>at</strong>e this expression we need ⟨(∆N) 2 ⟩. Similar to the deriv<strong>at</strong>ion<br />
before, we find<br />
⟨∆N∆N⟩ = − 1<br />
∫ ( )<br />
∂W<br />
∆N d∆Ud∆N (2.171)<br />
ω ∂XN<br />
and<br />
⟨∆N∆N⟩ = − 1<br />
ω<br />
∂<br />
∂XN<br />
∫<br />
∆NW d∆Ud∆N + 1<br />
ω<br />
∫ ( ∂∆N<br />
∂XN<br />
)<br />
W d∆Ud∆N<br />
(2.172)
2.10. ROLE OF FLUCTUATIONS. 93<br />
which leads to<br />
or<br />
Hence we find<br />
⟨∆T ∆N⟩ =<br />
( )<br />
∂T T<br />
∂U V,N ω<br />
⟨∆N∆N⟩ = − 1<br />
( )<br />
∂N<br />
ω ∂XN V,XU<br />
⟨∆N∆N⟩ = + T<br />
ω<br />
( )<br />
∂N<br />
∂µ V,T<br />
( ) ( )<br />
∂U ∂T<br />
+<br />
∂µ V,T ∂N U,V<br />
T<br />
ω<br />
( )<br />
∂N<br />
∂µ V,T<br />
(2.173)<br />
(2.174)<br />
(2.175)<br />
which can be calcul<strong>at</strong>ed using the following observ<strong>at</strong>ion. Suppose we have the<br />
temper<strong>at</strong>ure T as a function of N, V, U, and want to calcul<strong>at</strong>e the change in T<br />
as a function of µ while we keep the variables V and X constant, where X is to<br />
be specified. We have<br />
( ) ( )<br />
∂T ∂T<br />
=<br />
∂µ V,X ∂U V,N<br />
( ) ( )<br />
∂U ∂T<br />
+<br />
∂µ V,X ∂N V,U<br />
( )<br />
∂N<br />
∂µ V,X<br />
(2.176)<br />
which is exactly the form of the right hand side of the previous equ<strong>at</strong>ion, with<br />
X = ) T . But if we keep T constant, it does not change, and the partial deriv<strong>at</strong>ive<br />
is zero! Hence we find an important result<br />
( ∂T<br />
∂µ<br />
V,T<br />
⟨∆T ∆N⟩ = 0 (2.177)<br />
which means th<strong>at</strong> fluctu<strong>at</strong>ions in temper<strong>at</strong>ure and amount of m<strong>at</strong>erial are uncorrel<strong>at</strong>ed!<br />
It can be shown th<strong>at</strong> this is true in general for the correl<strong>at</strong>ion<br />
between fluctu<strong>at</strong>ions in an extensive st<strong>at</strong>e variable and an intensive st<strong>at</strong>e variable,<br />
as long as they are not a conjug<strong>at</strong>e pair. Hence also ⟨∆p∆N⟩ = 0. For<br />
correl<strong>at</strong>ions between the members of a conjug<strong>at</strong>e pair we find the following<br />
⟨∆µ∆N⟩ =<br />
( )<br />
∂µ T<br />
∂U V,N ω<br />
( ) ( )<br />
∂U ∂µ<br />
+<br />
∂µ V,T ∂N U,V<br />
and now the right hand side involves the partial deriv<strong>at</strong>ive<br />
hence<br />
T<br />
ω<br />
( )<br />
∂N<br />
∂µ V,T<br />
( ∂µ<br />
∂µ<br />
)<br />
V,T<br />
(2.178)<br />
= 1, and<br />
⟨∆µ∆N⟩ = T<br />
ω = kBT (2.179)<br />
and again one can show th<strong>at</strong> this is true for all correl<strong>at</strong>ions between members<br />
of conjug<strong>at</strong>e pairs. With the exception of the pressure and volume pair, since
94 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
we have defined our system to be <strong>at</strong> constant volume cells. If we would have<br />
worked with constant amount of m<strong>at</strong>erial cells, then we would have had this<br />
rel<strong>at</strong>ion to be valid for the pressure volume combin<strong>at</strong>ion, too.<br />
Non-equilibrium fluctu<strong>at</strong>ions.<br />
In the previous discussion we have always assumed th<strong>at</strong> the volume we are<br />
considering is in local equilibrium. Wh<strong>at</strong> if th<strong>at</strong> is not the case. We will first<br />
look <strong>at</strong> a simple example. Suppose the volume is divided into two parts, both in<br />
local equilibrium, but not in equilibrium with each other. For example, assume<br />
th<strong>at</strong> the pressure in the two parts is different, but the temper<strong>at</strong>ure is the same.<br />
We can define the entropy of this system, it is the sum of the entropies of the<br />
two parts. The entropy is lower than the entropy of the same system after<br />
it came into equilibrium, and hence the probability of such a non-equilibrium<br />
fluctu<strong>at</strong>ion is smaller (exponentially!) than the local equilibrium fluctu<strong>at</strong>ion. In<br />
this example we could define the temper<strong>at</strong>ure but not the pressure of the nonequilibrium<br />
fluctu<strong>at</strong>ion, but the temper<strong>at</strong>ure is not the same as the temper<strong>at</strong>ure<br />
of the equilibrium fluctu<strong>at</strong>ion!<br />
We can generalize this to the following case. Suppose we have a nonequilibrium<br />
fluctu<strong>at</strong>ion in which we can divide the volume into smaller cells,<br />
all in local equilibrium, but not in equilibrium with each other. The entropy<br />
of such a fluctu<strong>at</strong>ion can be measured, and is lower than the entropy of the<br />
corresponding equilibrium fluctu<strong>at</strong>ion. Hence the probability of such a nonequilibrium<br />
fluctu<strong>at</strong>ion is much smaller. Also, the values of intensive st<strong>at</strong>e<br />
variables are generally not defined for such a fluctu<strong>at</strong>ion.<br />
It is important to realize th<strong>at</strong> in order to measure the entropy we need thermal<br />
equilibrium and the measuring device. So there will be many fluctu<strong>at</strong>ions<br />
for which we cannot measure the entropy, and for which we hence cannot say<br />
anything quantit<strong>at</strong>ively about the probability of it happening. But if we can<br />
describe such a fluctu<strong>at</strong>ion, this means th<strong>at</strong> we can define some set of internal<br />
parameters characterizing this fluctu<strong>at</strong>ion, and we could include these in our<br />
thermodynamic description to some approxim<strong>at</strong>ion. Hence we can approxim<strong>at</strong>e<br />
the entropy and again we know th<strong>at</strong> it will be even lower than before. Hence<br />
the probability of such a non-equilibrium fluctu<strong>at</strong>ion will be even smaller!<br />
2.11 Extra questions.<br />
1. If the only st<strong>at</strong>e variables are T, S, p,and V,wh<strong>at</strong> is the meaning of U −<br />
T S + pV ?<br />
2. Why is there a difference in useful work done between a constant N and a<br />
constant /mu process?<br />
3. Explain the three cylinder scenarios.<br />
4. Give examples for the use of each free energy.
2.11. EXTRA QUESTIONS. 95<br />
5. The n<strong>at</strong>ural variables for U are S, V, and N. Why does th<strong>at</strong> make sense?<br />
6. So we have U = T S − pV + µN and also U = T S − pV + µN + MH. How<br />
can th<strong>at</strong> be?<br />
7. Wh<strong>at</strong> is the importance of Euler and Gibbs-Duhem?<br />
8. How many of the Maxwell rel<strong>at</strong>ions should you memorize?<br />
9. Should α always be positive?<br />
10. For a neg<strong>at</strong>ive thermal expansion m<strong>at</strong>erial, would Cp < CV ?<br />
11. Which additional response functions could one define in a magnetic system?<br />
How many he<strong>at</strong> capacities?<br />
12. Which Maxwell to use?<br />
13. Wh<strong>at</strong> are the signs of an ideal gas model failing <strong>at</strong> low T?<br />
14. Find a Maxwell rel<strong>at</strong>ion to change ( )<br />
∂S<br />
∂T V,N<br />
15. Show th<strong>at</strong> α = 1<br />
( ) ( )<br />
2<br />
∂ G<br />
1 ∂S<br />
V ∂p∂T = − V ∂p<br />
N N<br />
16. Minus sign in three deriv<strong>at</strong>ive rel<strong>at</strong>ion Why also in five, seven?<br />
17. Why is the r<strong>at</strong>io rel<strong>at</strong>ion for he<strong>at</strong> capacities, compressibilities important?<br />
18. Explain why a neg<strong>at</strong>ive compressibility leads to an instability<br />
19. f(x, y, z) = x2 + y2 + z2 − R2 . x = √ R2 − y2 − z2 ( )<br />
∂x y<br />
. ∂y = − x<br />
z<br />
chain rule OK?<br />
20. Why is there a minimum only in the extensive variables?<br />
21. Are there more rel<strong>at</strong>ions to make sure th<strong>at</strong> the entropy is a maximum?<br />
22. Wh<strong>at</strong> is the difference between H and T?<br />
Is the<br />
23. Which m<strong>at</strong>erials have χ = − V<br />
4π ?<br />
)<br />
24. Is always positive, no m<strong>at</strong>er which variables are held constant?<br />
( ∂µ<br />
∂N<br />
??<br />
25. Wh<strong>at</strong> kind of fluctu<strong>at</strong>ions can we describe?<br />
26. Why use the entropy to characterize fluctu<strong>at</strong>ions?<br />
27. Which R is in ωR = 1?<br />
28. Why do higher order terms drop out in the TD limit?<br />
29. Wh<strong>at</strong> are uses of the rel<strong>at</strong>ion between < (∆U) 2 > and CV ?<br />
30. If the correl<strong>at</strong>ion between fluctu<strong>at</strong>ions cancel, wh<strong>at</strong> is the condition? (<br />
µ ∝ T !!)
96 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
2.12 Problems for chapter 2<br />
Problem 1.<br />
A container with volume V contains N moles of an ideal gas. The container is<br />
divided into two parts by a movable partition. This partition can move without<br />
any friction. The volume of the left part is V1. The number of moles in the left<br />
part is N1. The partition always moves infinitesimally slow, and the ideal gas<br />
in each part is always in equilibrium. The temper<strong>at</strong>ure is T everywhere in the<br />
system. T is kept constant. M<strong>at</strong>erial cannot go through the partition, hence<br />
N1 remains constant. V1 can vary, though, and hence we consider V1 to be an<br />
internal parameter.<br />
(a) Which thermodynamic potential does one have to minimize as a function<br />
of V1 in order to find the equilibrium st<strong>at</strong>e of the total system?<br />
(b) Show th<strong>at</strong> the minimum occurs when the pressure left and right is the<br />
same. Note: do not only show th<strong>at</strong> it is an extremum, but show th<strong>at</strong> it is<br />
not a maximum!<br />
Problem 2.<br />
Show th<strong>at</strong> κT − κS =<br />
Problem 3.<br />
V T α2<br />
Cp<br />
(equ<strong>at</strong>ion 2.58 ).<br />
In first approxim<strong>at</strong>ion, the electrons in a conductor can be modelled by a<br />
gas of particles <strong>at</strong> a given temper<strong>at</strong>ure T in a volume V <strong>at</strong> a chemical potential<br />
µ. The chemical potential is set by connecting the conductor to a b<strong>at</strong>tery, the<br />
other terminal of the b<strong>at</strong>tery is grounded.<br />
(a) Which thermodynamic potential is needed to describe this system?<br />
The number of electrons in the conductor is measured as a function of the<br />
temper<strong>at</strong>ure. Experimentally one finds the rel<strong>at</strong>ion N(T, V, µ) = n0V (1 − RT<br />
µ )<br />
for a range of temper<strong>at</strong>ures with RT ≪ µ.<br />
(b) Using a Maxwell rel<strong>at</strong>ion, calcul<strong>at</strong>e how the entropy S depends on µ.<br />
(c) Show th<strong>at</strong> lim<br />
µ→0 S(T, V, µ) = ∞.<br />
The last result seems to indic<strong>at</strong>e th<strong>at</strong> lim<br />
T →0 S(T, V, 0) = ∞, in contradiction<br />
with the third law.
2.12. PROBLEMS FOR CHAPTER 2 97<br />
(d) Argue th<strong>at</strong> this experiment does not invalid<strong>at</strong>e the third law.<br />
Problem 4.<br />
Show th<strong>at</strong><br />
to show th<strong>at</strong><br />
Problem 5.<br />
( )<br />
∂µ<br />
∂N<br />
( ) S,V<br />
∂µ<br />
∂N<br />
T,V<br />
≤ CV<br />
T<br />
≥ 0.<br />
( ( ) ) 2<br />
∂T<br />
∂N . Use this expression for µ(S, V, N)<br />
S,V<br />
Show th<strong>at</strong> the energy function U(S, V ) = S2<br />
V s<strong>at</strong>isfies the criteria for stability.<br />
Calcul<strong>at</strong>e CV , Cp, κS, κT , and α for this system as a function of the temper<strong>at</strong>ure<br />
T.<br />
Problem 6.<br />
Calcul<strong>at</strong>e<br />
(<br />
2<br />
∂ S<br />
∂U 2<br />
)<br />
V,N,X<br />
and<br />
(<br />
2<br />
∂ U<br />
∂S2 )<br />
V,N,X<br />
in terms of response functions dis-<br />
cussed in this chapter. Use these results to show th<strong>at</strong> the product of these two<br />
deriv<strong>at</strong>ives is neg<strong>at</strong>ive.<br />
Problem 7.<br />
A magnetic system is characterized by the st<strong>at</strong>e variables T, S, M, H, µ, and<br />
N. N is always constant in this problem. All partial deriv<strong>at</strong>ives are <strong>at</strong> constant<br />
N. The magnetic susceptibilities are defined by χT = ( )<br />
∂M<br />
∂H T and χS = ( )<br />
∂M<br />
∂H S<br />
.<br />
(a) Show th<strong>at</strong> the he<strong>at</strong> capacities <strong>at</strong> constant field and <strong>at</strong> constant moment<br />
are rel<strong>at</strong>ed to the susceptibility χT by CH − CM = T<br />
(( ) )<br />
∂M 2.<br />
χT ∂T H<br />
(b) Show th<strong>at</strong> the susceptibilities <strong>at</strong> constant temper<strong>at</strong>ure and entropy obey<br />
χT − χS = T<br />
(( ) )<br />
∂M 2.<br />
CH ∂T H<br />
(c) Show th<strong>at</strong> χT ≥ χS ≥ 0 and th<strong>at</strong> CH ≥ CM ≥ 0.<br />
(d) Calcul<strong>at</strong>e CH, CM, χT , and χS for a paramagnetic gas containing N moles<br />
of m<strong>at</strong>erial with U = 3<br />
2<br />
Problem 8.<br />
NRT and M = CNH<br />
T .<br />
Show th<strong>at</strong> ⟨(∆M) 2 ⟩ = T<br />
ω χV when ∆U = ∆N = 0.
98 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Problem 9.<br />
A closed system (no exchange of he<strong>at</strong> and work with the outside world)<br />
contains an ideal gas. The volume V of the total system is constant. An internal<br />
partition divides the system in two parts. There is no flow of m<strong>at</strong>erial through<br />
this partition. The volumes of the two parts are V1 and V − V1. The parameter<br />
V1 is an internal parameter. He<strong>at</strong> can flow back and forth through the partition.<br />
As a consequence, the temper<strong>at</strong>ure T is always the same left and right, and is<br />
equal to T (V1). The partition always moves in a quasi-st<strong>at</strong>ic manner and each<br />
part of the system is always in thermal equilibrium. The internal energy of<br />
the system is the sum of the internal energy of the parts plus the mechanical<br />
potential energy of the partition (the pressures in both parts are not the same!).<br />
The total entropy of the system is constant.<br />
(a) Use the entropy equ<strong>at</strong>ion for an ideal gas to find<br />
V1.<br />
( )<br />
∂T as a function of<br />
∂V1<br />
(b) Use this result to show th<strong>at</strong> in equilibrium the temper<strong>at</strong>ure of a system<br />
with given S, V, and N is a minimum.<br />
(c) The minimum energy principle in connection with U = 3<br />
2NRT seems to<br />
lead to the same conclusion. Wh<strong>at</strong> is wrong in th<strong>at</strong> argument?<br />
Problem 10.<br />
A gas has the equ<strong>at</strong>ion of st<strong>at</strong>e pV = NRT , but does not have to be ideal,<br />
hence U does not have to be pV . The volume expands by a small amount ∆V<br />
<strong>at</strong> constant temper<strong>at</strong>ure and constant number of particles.<br />
(A) Use a Maxwell rel<strong>at</strong>ion to calcul<strong>at</strong>e ( )<br />
∂S<br />
∂V T N .<br />
(B) Calcul<strong>at</strong>e the amount of work ∆W done during this process, in terms of<br />
st<strong>at</strong>e variables and ∆V .<br />
(C) Calcul<strong>at</strong>e the amount of he<strong>at</strong> ∆Q entering the system during this process,<br />
in terms of st<strong>at</strong>e variables and ∆V .<br />
(D) Show th<strong>at</strong> ∆U = 0 during this process.<br />
(E) If (D) is true, show th<strong>at</strong> U is a function of T and N only.<br />
(F) If (E) is true, show th<strong>at</strong> U has the form U = Nf(T ).<br />
Problem 11.
2.12. PROBLEMS FOR CHAPTER 2 99<br />
Two containers are connected by a valve. The first container, volume V,<br />
contains a gas <strong>at</strong> temper<strong>at</strong>ure T, pressure p. The second container is empty<br />
(vacuum). The total volume of both containers is V’. Both containers are completely<br />
isol<strong>at</strong>ed from the environment. The valve is opened, and the gas in the<br />
first container expands freely, but irreversibly, and finally fills both containers.<br />
In the end we have again thermodynamic equilibrium, with st<strong>at</strong>e variables V’,<br />
T’, and p’. Calcul<strong>at</strong>e T’ in the following two cases: (a) the gas is ideal (b) the<br />
2<br />
N gas has equ<strong>at</strong>ions of st<strong>at</strong>e (p + a V 2 )(V − Nb) = NRT , CV = 3<br />
2NR. Problem 12.<br />
Show th<strong>at</strong> the osmotic pressure p exerted by N moles of solute in a very<br />
dilute solution of temper<strong>at</strong>ure T and volume V is given by p = NRT<br />
V . Use<br />
entropy arguments.<br />
Problem 13.<br />
The function U is given by U(S, V, N) = 1<br />
N (S2 +S −V 2 ). Give three reasons<br />
why this is not a correct function describing the internal energy of a m<strong>at</strong>erial.<br />
Problem 14.<br />
The internal energy of system is given by U(S, V, N) = S α V β N γ .<br />
(a) Using the fact th<strong>at</strong> U is extensive, find a rel<strong>at</strong>ion between α, β, and γ.<br />
(b) Calcul<strong>at</strong>e the pressure p, temper<strong>at</strong>ure T, and chemical potential µ as a<br />
function of S, V, and N.<br />
(c) Calcul<strong>at</strong>e the he<strong>at</strong> capacity <strong>at</strong> constant volume V (and constant N).<br />
(d) Calcul<strong>at</strong>e the adiab<strong>at</strong>ic compressibility (<strong>at</strong> constant N).<br />
(e) Based on the sign of these response functions, find inequalities for α and<br />
β.<br />
Problem 15.<br />
Use a Maxwell rel<strong>at</strong>ion to rel<strong>at</strong>e the coefficient of thermal expansion to a<br />
partial deriv<strong>at</strong>ive of the entropy and use th<strong>at</strong> to show th<strong>at</strong> the coefficient of<br />
thermal expansion approaches zero when the temper<strong>at</strong>ure goes to zero.<br />
Problem 16.
100 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
A system is characterized by four st<strong>at</strong>e variables, S, T, p, and V. The internal<br />
energy is given by<br />
with c being a constant.<br />
(A) Calcul<strong>at</strong>e ( )<br />
∂S<br />
∂T V .<br />
(B) Calcul<strong>at</strong>e S(T, V ).<br />
(C) Calcul<strong>at</strong>e p(T, V ).<br />
Problem 17.<br />
U(T, V ) = cV T 2<br />
The equ<strong>at</strong>ions of st<strong>at</strong>e for an ideal gas are pV = NRT and U = 3<br />
2NRT .<br />
(A) Show th<strong>at</strong> CV = 3<br />
2 NR.<br />
(B) Calcul<strong>at</strong>e the entropy difference S(T, V, N) − S(T0, V, N).<br />
(C) Wh<strong>at</strong> happens in the limit T → 0? Which conclusion can you draw from<br />
this?<br />
(D) Calcul<strong>at</strong>e the enthalpy H.<br />
(E) Use the enthalpy to show th<strong>at</strong> Cp = 5<br />
2 NR.<br />
Problem 18.<br />
The chemical potential of an ideal gas with density n = N<br />
V<br />
µ = RT ln( n<br />
nG<br />
)<br />
is given by<br />
with nG = cT 3<br />
2 , where c is a constant with the appropri<strong>at</strong>e dimensions. This<br />
gas is put in a uniform gravit<strong>at</strong>ional filed, acceler<strong>at</strong>ion g. Calcul<strong>at</strong>e the pressure<br />
in this isothermal gas as a function of height.<br />
Problem 19.<br />
A steam engine works changing w<strong>at</strong>er to steam, and using the steam’s pressure<br />
to do mechanical work. Consider a steam engine pulling a train of cars (no<br />
friction), holding w<strong>at</strong>er and fuel. At the beginning the mass of the empty train<br />
is M, the mass of the w<strong>at</strong>er is Mw<strong>at</strong>er, and the mass of the fuel is Mfuel. A<br />
unit mass of fuel releases J Joules upon burning. This energy evapor<strong>at</strong>es the<br />
w<strong>at</strong>er and brings the steam to 200C. The he<strong>at</strong> per mole needed to do this is H.<br />
Mechanical energy is gained from the steam by adiab<strong>at</strong>ically cooling it to 100C.<br />
The adiab<strong>at</strong>ic index γ of w<strong>at</strong>er is 4<br />
3 .
2.12. PROBLEMS FOR CHAPTER 2 101<br />
(a) Calcul<strong>at</strong>e the fraction f of the fuel’s energy th<strong>at</strong> is available to do mechanical<br />
work.<br />
(b) The train moves up a hill on a 10km stretch with a 1% slope. Calcul<strong>at</strong>e<br />
the amount of fuel used in this trip, assuming it is much less than the<br />
total mass of the train.<br />
(c) Calcul<strong>at</strong>e the amount of w<strong>at</strong>er used during this trip, assuming th<strong>at</strong> this,<br />
too, is much less than the mass of the train.<br />
Problem 20.<br />
Each member of a family of 4 opens the refriger<strong>at</strong>or 8 times a day. Every<br />
time it is opened, 75% of the 1 m 3 of air in the refriger<strong>at</strong>or is replaced by air<br />
from the room. In addition to the air, the refriger<strong>at</strong>or contains different things,<br />
equivalent to 200 liters of w<strong>at</strong>er. The temper<strong>at</strong>ure inside the refriger<strong>at</strong>or is 2C,<br />
the room is 20C. No he<strong>at</strong> leaks out when the door is closed. The refriger<strong>at</strong>or is<br />
cooled by an ideal Carnot machine. The vapor pressure of w<strong>at</strong>er <strong>at</strong> 20C is 0.1<br />
<strong>at</strong>m, <strong>at</strong> 2C it can be ignored. The he<strong>at</strong> of vaporiz<strong>at</strong>ion of w<strong>at</strong>er is 2.26 MJ/kg.<br />
R = 8.31 J/Kmol. A mole of gas occupies 22.4 liters <strong>at</strong> 0C.<br />
(a) Wh<strong>at</strong> are the molar he<strong>at</strong> capacities cV and cp for air?<br />
(b) Wh<strong>at</strong> is the average power needed to oper<strong>at</strong>e the refriger<strong>at</strong>or when the<br />
air in the house is dry?<br />
(c) Wh<strong>at</strong> is the average power needed to oper<strong>at</strong>e the refriger<strong>at</strong>or when the<br />
air in the house is s<strong>at</strong>ur<strong>at</strong>ed with w<strong>at</strong>er vapor?<br />
Problem 21.<br />
The he<strong>at</strong> capacity of a solid is C = AT 3 , where A is a constant. This solid is<br />
the low temper<strong>at</strong>ure reservoir of a reversible refriger<strong>at</strong>or. The high temper<strong>at</strong>ure<br />
reservoir is <strong>at</strong> room temper<strong>at</strong>ure. The solid is cooled from room temper<strong>at</strong>ure<br />
to almost absolute zero.<br />
(a) Find an expression for the amount of work required to cool this solid.<br />
(b) Wh<strong>at</strong> is the decrease in entropy of the solid?<br />
(c) Wh<strong>at</strong> is the decrease in internal energy of the solid?<br />
(d) Wh<strong>at</strong> are the increases in entropy and internal energy of the high temper<strong>at</strong>ure<br />
reservoir?
102 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
Problem 22.<br />
The equ<strong>at</strong>ion of st<strong>at</strong>e of an imperfect gas is given by p = N<br />
V RT ( 1 + B(T ) N<br />
)<br />
V<br />
where B(T ) is some function of the temper<strong>at</strong>ure. This gas undergoes a small<br />
isothermal expansion ∆V <strong>at</strong> constant N. The amount of he<strong>at</strong> going into the<br />
system is ∆Q.<br />
(a) Rel<strong>at</strong>e ∆Q to ∆V , keeping only linear terms.<br />
(b) Evalu<strong>at</strong>e the partial deriv<strong>at</strong>ives in (a).<br />
(c) Interpret the result.<br />
Problem 23.<br />
A well insul<strong>at</strong>ed steel tank has volume V . The tank contains air <strong>at</strong> pressure<br />
p1 and temper<strong>at</strong>ure T1. It is <strong>at</strong>tached to a compressed air line which supplies<br />
air <strong>at</strong> steady pressure p ′ and temper<strong>at</strong>ure T ′ . Initially the valve between the<br />
tank and the line is closed. The valve is then opened long enough to change the<br />
pressure in the tank to p2, and it is then closed again.<br />
Show th<strong>at</strong> in transport of this kind the energy balance is given by N2(u2 −<br />
h ′ ) − N1(u1 − h ′ ) = Q, where u is the molar internal energy, h’ is the molar<br />
enthalpy h’=u’+p’v’ where v’ is the molar volume occupied by air in the line,<br />
and Q is the he<strong>at</strong> added to the air.<br />
How many moles N2 are in the tank <strong>at</strong> th<strong>at</strong> moment, and wh<strong>at</strong> is the temper<strong>at</strong>ure<br />
T2?<br />
Assume th<strong>at</strong> (1) the tank and the air are always <strong>at</strong> the same temper<strong>at</strong>ure,<br />
(2) Air is a perfect gas with molar specific he<strong>at</strong>s cp = 7<br />
2 R and cV = 5<br />
For numbers use V1 = 20m 3 , p1 = 1bar , T1 = 22C, p ′ = 5bar, T ′ = 30C,<br />
p2 = 3bar. The mass of the steel tank is mt 1000 kg, the he<strong>at</strong> capacity of steel<br />
cs is 100 cal/kg/K and the gas constant is R=8.3 x 10 −5 m 3 bar/mol/K.<br />
Problem 24.<br />
A thermodynamic model for gravit<strong>at</strong>ional collapse can be constructed by<br />
adding the gravit<strong>at</strong>ional self energy to the ideal gas. A spherical mass M of a<br />
cloud of dust with radius R has gravit<strong>at</strong>ional energy Ug = − 3M 2 G<br />
5R .<br />
(a) Find the equ<strong>at</strong>ion of st<strong>at</strong>e of the cloud.<br />
(b) Find an expression for the isothermal compressibility.<br />
(c) Discuss the conditions under which the cloud is no longer thermally stable.<br />
Problem 25.<br />
2 R.
2.12. PROBLEMS FOR CHAPTER 2 103<br />
The he<strong>at</strong> capacity <strong>at</strong> constant pressure for a certain solid is γT with γ =<br />
5mcal/degree 2 /mol. At T=300K the solid changes into a classical ideal gas<br />
which is kept <strong>at</strong> constant pressure. The he<strong>at</strong> of this transition is 150 cal/mol.<br />
The entropy of this gas <strong>at</strong> T=600K is 5.5 cal/degree/mol. Is this gas mono<strong>at</strong>omic<br />
or di<strong>at</strong>omic? Use R=2cal/degree/mol.<br />
Problem 26.<br />
A substance has the following properties:<br />
(I) The amount of m<strong>at</strong>erial is fixed <strong>at</strong> N moles.<br />
(II) At constant temper<strong>at</strong>ure T0 the work W done by it on expansion from a<br />
volume V0 to a volume V is<br />
(III) The entropy is given by<br />
W = NRT0 ln( V<br />
S = NR V0<br />
V<br />
( T<br />
T0<br />
V0<br />
)<br />
) x<br />
In these expressions T0, V0, and x are fixed constants.<br />
(A) Calcul<strong>at</strong>e the Helmholtz free energy of the system.<br />
(B) Find the equ<strong>at</strong>ion of st<strong>at</strong>e for the pressure p in terms of V , T , and N.<br />
(C) Find the work done on expansion <strong>at</strong> constant T, where T is arbitrary.<br />
Problem 27.<br />
The he<strong>at</strong> capacity <strong>at</strong> constant pressure is given by<br />
( )<br />
∂S<br />
Cp = T<br />
∂T<br />
and the coefficient of thermal expansion is given by<br />
α = 1<br />
( )<br />
∂V<br />
V ∂T<br />
The amount of m<strong>at</strong>erial N is always kept constant.<br />
If we consider S(T, p) we have<br />
( ) ( )<br />
∂S ∂S<br />
dS = dT + dp<br />
∂T p ∂p T<br />
p<br />
p
104 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
(A) Show th<strong>at</strong> this leads to<br />
T dS = CpdT − αT V dp<br />
One can show th<strong>at</strong> (but you do NOT have to!)<br />
where X is a finite constant.<br />
αV<br />
lim<br />
T →0 Cp<br />
= X<br />
(B) Show th<strong>at</strong> this has the consequence th<strong>at</strong> in an adiab<strong>at</strong>ic process one can<br />
never reach T = 0 by a finite change in pressure.<br />
Problem 28.<br />
A container of volume V0 is divided into two equal parts by a removable<br />
wall. There are N0 moles of gas in equilibrium <strong>at</strong> temper<strong>at</strong>ure Ti on the left,<br />
the right side is empty. The container is isol<strong>at</strong>ed, no he<strong>at</strong> goes through. The<br />
internal wall is suddenly removed and the gas can expand freely until it finally<br />
fills up the whole container.<br />
(A) Is the gas in equilibrium just after the wall has been removed?<br />
(B) Is the free expansion a reversible process?<br />
(C) Does the entropy of the gas increase, decrease, or stay the same?<br />
(D) Does the energy of the gas increase, decrease, or stay the same?<br />
Assume th<strong>at</strong> we have a different, reversible process th<strong>at</strong> brings a gas <strong>at</strong><br />
constant internal energy from a temper<strong>at</strong>ure Ti and volume 1<br />
2V0 to a volume V0<br />
and temper<strong>at</strong>ure Tf . At each stage of this reversible process the gas is described<br />
by an internal energy function of the form U = u(T, V, N).<br />
(E) Is this reversible process adiab<strong>at</strong>ic?<br />
(F) Express the response function ( )<br />
∂T<br />
∂V<br />
the energy function u(T, V, N).<br />
U,N<br />
in terms of partial deriv<strong>at</strong>ives of<br />
(G) A mono-<strong>at</strong>omic ideal gas has u(T, V, N) = 3<br />
2NRT . How much does the<br />
temper<strong>at</strong>ure change in a free expansion from volume 1<br />
2V0 to V0?<br />
Problem 29.<br />
Under adiab<strong>at</strong>ic compression a gas changes its temper<strong>at</strong>ure when the pressure<br />
is changed, in a reversible manner. The response function describing this<br />
situ<strong>at</strong>ion is Ω = ( )<br />
∂T<br />
∂P S,N .
2.12. PROBLEMS FOR CHAPTER 2 105<br />
T V α<br />
Show th<strong>at</strong> Ω = Cp<br />
( )<br />
1 ∂V<br />
with α = − V ∂T p,N and Cp<br />
constant pressure.<br />
the he<strong>at</strong> capacity <strong>at</strong><br />
For an ideal mono-<strong>at</strong>omic gas αT = 1 and Cp = 5<br />
2NR. Find the function f<br />
defined by T = f(p) in this adiab<strong>at</strong>ic process.<br />
Problem 30.<br />
The Jacobian J is defined by<br />
J(x, y; u, v) =<br />
∂(x, y)<br />
∂(u, v) =<br />
Show th<strong>at</strong> J(T, S; p, V ) = 1.<br />
Problem 31.<br />
( )<br />
∂x<br />
∂u v<br />
( )<br />
∂y<br />
−<br />
∂v u<br />
( )<br />
∂x<br />
∂v u<br />
( )<br />
∂y<br />
∂u v<br />
A fluid undergoes an adiab<strong>at</strong>ic throttling process. This means th<strong>at</strong> the fluid<br />
is pressed through a porous plug, with initial pressure p1 and final pressure p2.<br />
(A) Show th<strong>at</strong> the enthalpy of the fluid is constant in this process.<br />
As a result of this process, the temper<strong>at</strong>ure changes (remember when you<br />
pump air in your bicycle tire, or when you let the air escape?). The amount of<br />
change is measured by the Joule-Thomson coefficient ηJT , defined by<br />
( )<br />
∂T<br />
ηJT =<br />
∂p<br />
(B) Show th<strong>at</strong> ηJT = V (T α − 1)<br />
Cp<br />
(C) Wh<strong>at</strong> is the value for an ideal gas?<br />
Problem 32.<br />
2NRTf<br />
5V<br />
The pressure of a gas of fermions <strong>at</strong> low temper<strong>at</strong>ure is given by p =<br />
[<br />
1 + 5π2<br />
12<br />
( T<br />
Tf<br />
H<br />
) 2 ]<br />
, where Tf is a m<strong>at</strong>erial dependent parameter called the<br />
Fermi temper<strong>at</strong>ure. One finds experimentally th<strong>at</strong> Tf = c ( N<br />
V<br />
constant. One also finds th<strong>at</strong> CV = 1<br />
2 π2 NR T<br />
Tf<br />
Helmholtz free energy of this gas.<br />
Problem 33.<br />
Assume the he<strong>at</strong> capacity <strong>at</strong> constant volume is given by<br />
) 2<br />
3 where c is a<br />
. Calcul<strong>at</strong>e the entropy and the
106 CHAPTER 2. THERMODYNAMIC POTENTIALS AND<br />
T 2<br />
CV = NR<br />
T 2 + T 2 0<br />
Show th<strong>at</strong> this implies th<strong>at</strong> the coefficient of thermal expansion is equal to<br />
zero.<br />
Problem 34.<br />
The internal energy of a system is given by U = cV T 4 , where c is a constant.<br />
Find the entropy, pressure, and chemical potential for this system.<br />
Problem 35.<br />
Suppose we know th<strong>at</strong><br />
F (T, V, N) = N<br />
∞∑<br />
( ) n<br />
N<br />
fn(T )<br />
V<br />
n=0<br />
Find the corresponding expansions for S(T, V, N), p(T, V, N), and U(T, V, N)<br />
Problem 36.<br />
Define a st<strong>at</strong>e function A by A = U − cT N, where c is a constant with the<br />
appropri<strong>at</strong>e dimensions to make the equ<strong>at</strong>ion dimensionally correct. Assume<br />
th<strong>at</strong> we have A(T, p, N), hence A as a function of the variables T , p , and N.<br />
Show th<strong>at</strong><br />
( )<br />
∂A<br />
= Cp − pV α − cN<br />
∂T p,N<br />
Problem 37.<br />
In cases where chemical reactions occur quickly, we can discuss the change<br />
in volume as a function of pressure in an adiab<strong>at</strong>ic process where we change the<br />
amount of m<strong>at</strong>erial or the chemical potential. The ) chemical volume expansion<br />
. The adiab<strong>at</strong>ic r<strong>at</strong>e of<br />
coefficient αchem is defined by αchem = 1<br />
V<br />
( ∂V<br />
∂µ<br />
change of amount ) of m<strong>at</strong>erial as a function of chemical potential is defined by<br />
. Show th<strong>at</strong><br />
RS = 1<br />
N<br />
( ∂N<br />
∂µ<br />
S,p<br />
κS,µ − κS,N = V<br />
N<br />
α 2 chem<br />
RS<br />
S,p
Chapter 3<br />
Phase transitions.<br />
In the first two chapters the philosophical and m<strong>at</strong>hem<strong>at</strong>ical basis for thermodynamics<br />
was developed. The notion of thermal equilibrium plays a very<br />
important role, and helps us to define the stable st<strong>at</strong>e of the system. Thermal<br />
equilibrium is defined between the system and the outside world, in cases where<br />
work can be done or he<strong>at</strong> can be exchanged. For example, if we say th<strong>at</strong> a<br />
system is <strong>at</strong> a given temper<strong>at</strong>ure we mean th<strong>at</strong> the average temper<strong>at</strong>ure of the<br />
system is equal to the temper<strong>at</strong>ure of some outside reservoir. In order for this<br />
st<strong>at</strong>ement to be meaningful, the system has to be very large, so rel<strong>at</strong>ive fluctu<strong>at</strong>ions<br />
are small. In addition, the fluctu<strong>at</strong>ions in temper<strong>at</strong>ure of the reservoir<br />
have to be much smaller, and hence the reservoir has to be much larger than<br />
the system.<br />
Thermodynamical equilibrium also implies th<strong>at</strong> different parts of the system<br />
are in equilibrium with each other. On average, different parts have the same<br />
density, and even though m<strong>at</strong>erial can flow back and forth, there are no net<br />
macroscopic currents.<br />
N<strong>at</strong>ure does cre<strong>at</strong>e extra complic<strong>at</strong>ions, though, and they are by far the most<br />
interesting. Wh<strong>at</strong> happens if a system can change its st<strong>at</strong>e abruptly? This is<br />
a combin<strong>at</strong>ion of the two ideas described above. In phase equilibrium we have<br />
two parts of the same system th<strong>at</strong> look quite different. M<strong>at</strong>erial can go from<br />
one phase to another. So on one hand we have two separ<strong>at</strong>e systems, but on<br />
the other hand they are just two parts of the same system.<br />
We can apply the ideas developed in the first two chapters to describe phase<br />
transitions. Th<strong>at</strong> is the topic of this chapter. In most cases we can change the<br />
phase of a system by applying a temper<strong>at</strong>ure or pressure change. Therefore,<br />
the free energy th<strong>at</strong> we need is the Gibbs free energy. The minimum energy<br />
principle for the Gibbs free energy then allows us to predict which phase is<br />
stable. We start this chapter by describing the prototype phase diagram for a<br />
system with variables p, V, T, S, N, µ. As a function of intensive st<strong>at</strong>e variables<br />
we can move continuously, but as a function of extensive st<strong>at</strong>e variables we have<br />
gaps. All these properties can be derived from the minimal energy principle.<br />
We can also follow phase equilibrium along phase boundaries, and this leads to<br />
107
108 CHAPTER 3. PHASE TRANSITIONS.<br />
some restrictions as well. Next, we discuss briefly the n<strong>at</strong>ure of binary phase<br />
diagrams, and how they can be discussed in the same manner. This is a very<br />
important and practical applic<strong>at</strong>ion of thermodynamics.<br />
So far we have assumed th<strong>at</strong> we have two different forms of the free energy,<br />
and we compare them to get the stable st<strong>at</strong>e. One way to connect the free<br />
energies of the two different phases is via the use of equ<strong>at</strong>ions of st<strong>at</strong>e. The<br />
most important one is the van der Waals equ<strong>at</strong>ion. We will use this equ<strong>at</strong>ion to<br />
show how a liquid to gas phase transition can be modelled. Integr<strong>at</strong>ions along<br />
isotherms allow us to connect the free energy of the two phases, and analyze<br />
phase equilibrium.<br />
The n<strong>at</strong>ure of this chapter is quite basic. We focus on a description of phase<br />
equilibrium and some simple consequences based on the assumption th<strong>at</strong> one<br />
can analyze the system by dividing it into two parts, each of which is internally<br />
in thermal equilibrium. We do allow net macroscopic flows of m<strong>at</strong>ter from one<br />
phase to another, though, during the transition. But <strong>at</strong> a given set of external<br />
conditions the thermodynamics will describe exactly how much m<strong>at</strong>erial is in<br />
phase one as compared to phase two. The r<strong>at</strong>io of these two amounts can be<br />
changed by changing temper<strong>at</strong>ure or pressure. We will analyze these r<strong>at</strong>ios <strong>at</strong><br />
the end of the chapter.<br />
3.1 Phase diagrams.<br />
Wh<strong>at</strong> happens when I suddenly change my st<strong>at</strong>e dram<strong>at</strong>ically?<br />
Study of rel<strong>at</strong>ions.<br />
In the previous two chapters we have studied the m<strong>at</strong>hem<strong>at</strong>ical description<br />
of thermal systems, i.e. systems which have a large number of internal degrees of<br />
freedom. We found the existence of many rel<strong>at</strong>ions between reversible changes<br />
in in st<strong>at</strong>e variables. Note th<strong>at</strong> the study of irreversible changes is much harder,<br />
but certainly not less important. Th<strong>at</strong> is for a different course, however.<br />
Entropy.<br />
The fundamental question in thermodynamics is: Wh<strong>at</strong> is entropy? Is it<br />
simply a quantity th<strong>at</strong> represents the energy stored in the internal degrees of<br />
freedom, or is it more? This question can not be answered in thermodynamics,<br />
and we have to apply st<strong>at</strong>istical mechanics to obtain one possible answer. In<br />
thermodynamics we simply use entropy as a st<strong>at</strong>e variable, and find ways to<br />
measure it.
3.1. PHASE DIAGRAMS. 109<br />
Consequences of rel<strong>at</strong>ions.<br />
Based on the laws of thermodynamics and the m<strong>at</strong>hem<strong>at</strong>ics of functions<br />
of many variables we were able to draw some very strong conclusions about<br />
response functions and stability. The rel<strong>at</strong>ions between response functions have<br />
important technical consequences. For example, if we try to design a m<strong>at</strong>erial<br />
with a large coefficient of thermal expansion α we know th<strong>at</strong> we always have<br />
α 2 T V = κT (Cp − CV ) and hence the right hand side should be large. But in<br />
a solid Cp − CV is always small and hence we need a m<strong>at</strong>erial with a large<br />
isothermal compressibility! And then the next question is: does th<strong>at</strong> have<br />
neg<strong>at</strong>ive consequences for the applic<strong>at</strong>ions of our new m<strong>at</strong>erial?<br />
Discontinuities.<br />
In this chapter we will focus on another important general aspect of the<br />
equ<strong>at</strong>ions of thermodynamics. We will investig<strong>at</strong>e the effects of discontinuities in<br />
the equ<strong>at</strong>ions of st<strong>at</strong>e. These discontinuities manifest themselves in the physical<br />
world as phase transitions. Wh<strong>at</strong> happens when a m<strong>at</strong>erial abruptly changes<br />
its properties?<br />
Wh<strong>at</strong> is the most important phase transition in the world?<br />
For example, zirconia is a very important ceramic m<strong>at</strong>erial. It has a high<br />
melting point, is very hard, and is inert. In short, it is very useful in many<br />
applic<strong>at</strong>ions. The problem, however, is th<strong>at</strong> <strong>at</strong> around 1100 centigrade zirconia<br />
transforms from a tetragonal high temper<strong>at</strong>ure phase into a monoclinic low temper<strong>at</strong>ure<br />
phase. This transition is very disruptive, and fractures the m<strong>at</strong>erial.<br />
The more we know about wh<strong>at</strong> happens <strong>at</strong> this phase transition, the more we<br />
can control it and make the consequences more benign.<br />
Practical importance.<br />
The study of phase transitions is of enormous practical importance. When<br />
does a m<strong>at</strong>erial become superconductive? How does magnetism change? In this<br />
and the following chapters we will study phase transitions from a macroscopic<br />
point of view. This is important, because this rel<strong>at</strong>es directly to the way phase<br />
transitions are observed. In order to explain why phase transitions happen, however,<br />
we need to look <strong>at</strong> the microscopic n<strong>at</strong>ure of the m<strong>at</strong>erial, and hence those<br />
answers are only found in st<strong>at</strong>istical mechanics. Nevertheless, thermodynamics<br />
will provide us again with important boundary conditions th<strong>at</strong> all m<strong>at</strong>erials will<br />
have to obey.<br />
Model system.
110 CHAPTER 3. PHASE TRANSITIONS.<br />
In order to make the discussion less cumbersome, we will consider a model<br />
system in which the only relevant st<strong>at</strong>e variables are {p, V, µ, N, S, T } . There<br />
is only one m<strong>at</strong>erial present. These restrictions are, however, not essential, and<br />
other systems can be discussed in a similar way. If we ignore chemical reactions<br />
and their influence on phase transitions, we can assume th<strong>at</strong> the amount of<br />
m<strong>at</strong>erial, N, is fixed. The three st<strong>at</strong>e variables which are easiest to change<br />
and control are the pressure p, the volume V, and the temper<strong>at</strong>ure T. Since we<br />
already prescribed the value of N, only two of the variables in the set {p, V, T } are<br />
independent. Since N is an extensive variable, we can choose any combin<strong>at</strong>ion<br />
of two variables from {p, V, T } . Systems like this are called pVT systems. Note<br />
th<strong>at</strong> we assume th<strong>at</strong> the system is in equilibrium, else equ<strong>at</strong>ions of st<strong>at</strong>e like<br />
p = f(V, T ) would not exist.<br />
Phase diagram.<br />
The variables p and T are intensive variables, and their values are chosen<br />
by bringing the system in thermal and mechanical contact with appropri<strong>at</strong>e<br />
reservoirs of pressure p and temper<strong>at</strong>ure T. From an experimental point of view,<br />
these are good handles on the system, and often easy to use. These reservoirs<br />
are infinitely large systems in thermodynamic equilibrium. Any choice for the<br />
values of p and T is possible, and hence we can give our model system arbitrary<br />
values of the pressure and temper<strong>at</strong>ure. Of course, we assume positive values<br />
for both p and T. Hence for any combin<strong>at</strong>ion of p and T we are able to specify<br />
the volume and all other aspects of the system. The system is represented by a<br />
plot in the p-T diagram, called a phase diagram. A very general phase diagram<br />
looks like figure 3.1. The solid lines indic<strong>at</strong>e phase boundaries. Crossing such<br />
a line in the p-T diagram will cause a jump in the value of the volume, and<br />
in the values of other extensive st<strong>at</strong>e variables. Sometimes the jump is only in<br />
the value of the deriv<strong>at</strong>ives, though. Most simple pVT systems have a phase<br />
diagram of the form shown in figure 3.1.<br />
Give an example of a solid to gas transition.<br />
An<strong>at</strong>omy of a phase diagram.<br />
The solid lines separ<strong>at</strong>e the well-known phases of most m<strong>at</strong>erials. At low<br />
temper<strong>at</strong>ure and high pressure the system is in a solid st<strong>at</strong>e. At higher temper<strong>at</strong>ure<br />
it either becomes a gas or a liquid. At the phase boundary the two phases<br />
can coexist. The points t and c are special. At the triple point t all three phases<br />
can coexist. At the critical point c the phase boundary between liquid and gas<br />
vanishes. The phase boundary between solid and liquid never vanishes, <strong>at</strong> least<br />
as far as we know. This implies th<strong>at</strong> gas and liquid are more alike than gas and<br />
solid, which makes sense intuitively.
3.1. PHASE DIAGRAMS. 111<br />
Figure 3.1: Model phase diagram for a simple model system.<br />
Critical point.<br />
The interpret<strong>at</strong>ion of a critical point is as follows. If the temper<strong>at</strong>ure is in<br />
the range (Tt, Tc), an increase in pressure will cause a phase transition between<br />
two st<strong>at</strong>es. At low pressure the st<strong>at</strong>e of the system resembles a gas, <strong>at</strong> high<br />
pressure a liquid. Since these two st<strong>at</strong>es are separ<strong>at</strong>ed by a phase transition,<br />
we can clearly identify them. If the temper<strong>at</strong>ure is above Tc, however, this<br />
clear separ<strong>at</strong>ion vanishes. At low pressure the system still behaves like a gas,<br />
and <strong>at</strong> high temper<strong>at</strong>ure like a liquid. The change from one to the other is<br />
continuous and it is not possible to uniquely identify where we change from one<br />
type of behavior to another. This continuous change is not possible from liquid<br />
to solid, as far as we know from experiments.<br />
There is some semantics which is useful to follow. Technically we make a<br />
distinction between liquid, gas, and vapor. All three phases are rel<strong>at</strong>ed, but can<br />
be distinguished experimentally by the following. If we are working <strong>at</strong> a fixed<br />
temper<strong>at</strong>ure and increase the pressure, we either see a transition or we do not.<br />
If we see a transition, we call those phases liquid (<strong>at</strong> high pressure) and vapor<br />
(<strong>at</strong> low pressure). If we do not see a transition, we call th<strong>at</strong> phase the gas phase.<br />
Hence we have a liquid if T < Tc and p > pt, a vapor if T < Tc and p < pt, and<br />
a gas if T > Tc.<br />
critical points in a solid.<br />
A famous example of a critical point in a solid is the α to γ phase transition in<br />
solid Ce. The phase diagram looks like figure 3.2. If we cool Ce <strong>at</strong> <strong>at</strong>mospheric<br />
pressure we have to go through the phase boundary, and the volume changes in<br />
a discontinuous manner. If the whole m<strong>at</strong>erial were uniform, this would not give<br />
any problems. But any imperfections in the m<strong>at</strong>erial cause some parts of the
112 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.2: Phase diagram for solid Ce.<br />
m<strong>at</strong>erial to transform before other parts, causing large strains in the m<strong>at</strong>erial.<br />
As a result, large single crystals of α-Ce are hard to make. The solution to this<br />
problem is to go around the critical point. First increase the pressure, than<br />
decrease the temper<strong>at</strong>ure, and finally decrease the pressure again.<br />
Energy consider<strong>at</strong>ions.<br />
At a phase boundary two distinct phases of the m<strong>at</strong>erial can coexist. This is<br />
only possible if these two phases have the same energy <strong>at</strong> the phase boundary.<br />
If not, one phase would be preferred. The question is, which energy plays a<br />
role. Since the phase boundary is determined by specific values of p and T, and<br />
since N is kept constant, the energy of interest is the Gibbs energy G(T, p, N).<br />
Therefore, for a p<strong>at</strong>h in p-T space crossing a phase boundary, the Gibbs free<br />
energy is continuous, but its deriv<strong>at</strong>ives are not. The reason for this is th<strong>at</strong> the<br />
Gibbs free energy is minimal as a function of all internal parameters <strong>at</strong> a given<br />
value of p and T. The internal parameters define the phase we are in. Therefore<br />
phase transitions are found by studying singular behavior in the deriv<strong>at</strong>ives of<br />
the Gibbs free energy.<br />
Discontinuity in the volume.<br />
The Gibbs energy is continuous <strong>at</strong> a phase boundary. If the Gibbs free<br />
energy were not continuous <strong>at</strong> a phase boundary, a picture of the Gibbs en-<br />
ergy as ) a function of pressure would look like figure 3.3 for example (note th<strong>at</strong><br />
≤ 0 because of stability). One could then follow the lower branch<br />
( ∂ 2 G<br />
∂p 2<br />
T,N<br />
across the phase boundary, and have a st<strong>at</strong>e of lower energy, which by definition<br />
would be the stable st<strong>at</strong>e. This would simply indic<strong>at</strong>e th<strong>at</strong> we marked the<br />
phase boundary <strong>at</strong> the wrong place. Hence the Gibbs free energy as a function
3.1. PHASE DIAGRAMS. 113<br />
Figure 3.3: Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure,<br />
wrong picture.<br />
of pressure <strong>at</strong> constant temper<strong>at</strong>ure and N looks like 3.4.<br />
Once we have the Gibbs energy) as a function of pressure, the volume follows<br />
autom<strong>at</strong>ically through V =<br />
and this is shown in figure 3.5. First<br />
( ∂G<br />
∂p<br />
T,N<br />
of all, because of the curv<strong>at</strong>ure of the Gibbs free energy the volume decreases<br />
with pressure. Second, <strong>at</strong> the phase transition the volume jumps down in a<br />
discontinuous manner, it never jumps up.<br />
Is it possible for the volume to be continuous as a function of pressure <strong>at</strong> the<br />
phase transition?<br />
Is it possible for the volume to decrease as a function of temper<strong>at</strong>ure <strong>at</strong> a phase<br />
transition?<br />
Other pictures of the phase transition, VT diagram.<br />
In the V-T plane this phase diagram looks like figure 3.6. In this diagram<br />
we see regions which are not accessible, because the volume of the system jumps<br />
<strong>at</strong> the phase boundary. There is no continuous p<strong>at</strong>h from the solid phase to<br />
the others, but there is a continuous p<strong>at</strong>h from liquid to gas, around the critical<br />
point. Inaccessible regions only show up in diagrams when extensive st<strong>at</strong>e<br />
variables are involved. Wh<strong>at</strong> happens when the system is prepared <strong>at</strong> a given<br />
volume V0 <strong>at</strong> a high temper<strong>at</strong>ure and then cooled down to a temper<strong>at</strong>ure which
114 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.4: Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure,<br />
correct picture.<br />
Figure 3.5: Volume versus pressure <strong>at</strong> the phase transition.
3.1. PHASE DIAGRAMS. 115<br />
would force the st<strong>at</strong>e of the system to be in an inaccessible region? The answer<br />
is simple, the system will phase separ<strong>at</strong>e. Part of it will have a small volume,<br />
part of it a large volume, and both phases are in equilibrium with each other.<br />
The amount of each phase is determined by where in the inaccessible region the<br />
point (T, V0) is situ<strong>at</strong>ed.<br />
Figure 3.6: Model phase diagram for a simple model system in V-T space.<br />
Phase separ<strong>at</strong>ion analyzed.<br />
As an example, consider th<strong>at</strong> the system has volume V0 and is cooled down<br />
<strong>at</strong> constant volume to a temper<strong>at</strong>ure T below Tc, but above Tt. The volume<br />
per mole (specific volume) of the gas and the liquid <strong>at</strong> this temper<strong>at</strong>ure and<br />
<strong>at</strong> a pressure p are vg(p, T ) and vl(p, T ). Because gas and liquid will be in<br />
equilibrium, they will be <strong>at</strong> the same pressure. We need to determine the<br />
amounts of m<strong>at</strong>erial Ng and Nl in the gas and liquid phase. We obviously have:<br />
and<br />
N = Ng + Nl<br />
(3.1)<br />
V0 = Ngvg(p, T ) + Nlvl(p, T ) (3.2)<br />
This seems to be incomplete, but we have to keep in mind th<strong>at</strong> the pressure is<br />
determined from the phase diagram 3.1. Hence we know exactly how the system<br />
phase separ<strong>at</strong>es.<br />
We can combine these equ<strong>at</strong>ions by using the volume per mole v0 and write<br />
which is equivalent to<br />
(Ng + Nl)v0 = Ngvg(p, T ) + Nlvl(p, T ) (3.3)
116 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.7: Model phase diagram for a simple model system in p-V space.<br />
or<br />
Ng(vg(p, T ) − v0) = Nl(v0 − vl(p, T )) (3.4)<br />
Ng<br />
Nl<br />
= v0 − vl(p, T )<br />
vg(p, T ) − v0<br />
(3.5)<br />
which is called the lever rule. The phase with a molar volume most different<br />
from the actual volume has the smallest amount of m<strong>at</strong>erial in it.<br />
If we have our initial volume larger than the critical volume, V0 > Vc and<br />
we cool the system, we see th<strong>at</strong> <strong>at</strong> the temper<strong>at</strong>ure when we hit the forbidden<br />
region droplets of m<strong>at</strong>erial with a small volume (liquid) will form. The more<br />
we lower the temper<strong>at</strong>ure, the more liquid we get. On the other hand, if the<br />
original volume is less than the critical volume, V0 < Vc, and we cool down, we<br />
will see droplets of the high volume phase, gas, form in the liquid. The more<br />
we cool, the more gas we get! Th<strong>at</strong> seems counterintuitive, but we have to<br />
remember th<strong>at</strong> when we cool the mixture, the pressure decreases, and <strong>at</strong> lower<br />
pressure we will see more gas phase!<br />
The same system in a p-V diagram looks like figure 3.7. Notice again the<br />
forbidden regions. It is also possible to draw the three dimensional surface<br />
V (p, T, N) for fixed N, but th<strong>at</strong> is beyond my capabilities and artistic talents.<br />
3.2 Clausius-Clapeyron rel<strong>at</strong>ion.<br />
So far we made some general st<strong>at</strong>ements about wh<strong>at</strong> is happening <strong>at</strong> a phase<br />
boundary. Our next task is to give a more quantit<strong>at</strong>ive description of the phase<br />
boundaries. <strong>Thermodynamics</strong> limits the freedom we have in drawing phase<br />
diagrams!
3.2. CLAUSIUS-CLAPEYRON RELATION. 117<br />
basic energy.<br />
Wh<strong>at</strong> is the best way to describe m<strong>at</strong>hem<strong>at</strong>ically wh<strong>at</strong> is happening? Since<br />
in terms of p and T all values of these st<strong>at</strong>e variables are possible, we will use<br />
the pressure and temper<strong>at</strong>ure as independent variables. The free energy in this<br />
case is the Gibbs potential G = U − T S + pV . The Gibbs free energy is a<br />
minimum as a function of all internal parameters when p, T, and N are fixed.<br />
Hence the st<strong>at</strong>e with the lowest Gibbs free energy is stable in a p-T diagram.<br />
In our simple system the Euler rel<strong>at</strong>ion tells us th<strong>at</strong> G = µN. Since N is fixed,<br />
the chemical potential in this case is equal to the Gibbs free energy per mole,<br />
and is often used in stead of the Gibbs free energy in this case. To be precise,<br />
however, we will use the Gibbs energy as our basic energy.<br />
Phase boundary.<br />
In order to determine the phase boundary between to phases we need an<br />
analytical form of the Gibbs free energy for both phases. Assume th<strong>at</strong> we have<br />
expressions for this energy in each phase, which we can extend to all values in<br />
p-T space by analytic continu<strong>at</strong>ion. The phase boundary then follows from<br />
G1(p, T, N) = G2(p, T, N) (3.6)<br />
and since N is fixed, this yields an equ<strong>at</strong>ion p = pco(T ) for the pressure as a<br />
function of temper<strong>at</strong>ure. This rel<strong>at</strong>ion describes the coexistence curve of the<br />
two phases.<br />
The slope) of the Gibbs free energy as a function op pressure is equal to the<br />
volume, = V , and hence as a function of the pressure the Gibbs free<br />
( ∂G<br />
∂p<br />
T,N<br />
energy is always increasing (volumes are positive by definition). The second )<br />
order deriv<strong>at</strong>ive of the Gibbs free energy with pressure is equal to ,<br />
( ∂V<br />
∂p<br />
and is always neg<strong>at</strong>ive because of stability requirements. Therefore the slope of<br />
the Gibbs free energy as a function of pressure is decreasing. A picture of the<br />
Gibbs free energy of the two phases which we study might look like the figure<br />
3.8. The lowest curve gives the stable st<strong>at</strong>e for each value of the pressure. At<br />
the phase transition the slope of the Gibbs free energy jumps to a smaller value,<br />
and hence the volume jumps to a smaller value.<br />
In a picture of volume versus pressure, why can the volume not have the same<br />
value for two different values of the pressure.<br />
Following the coexistence curve <strong>at</strong> both sides.<br />
The differential for the Gibbs free energy is given by<br />
T,N
118 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.8: Gibbs energy across the phase boundary <strong>at</strong> constant temper<strong>at</strong>ure<br />
for both phases.<br />
dG = V dp − SdT + µdN (3.7)<br />
but since N is constant, the last term is zero. The partial deriv<strong>at</strong>ives of the<br />
Gibbs free energy give us V(p,T,N) and S(p,T,N) for both phases. If we are <strong>at</strong><br />
a point on the coexistence curve, and make small changes in the temper<strong>at</strong>ure<br />
and pressure in such a manner th<strong>at</strong> we remain on the coexistence curve, the<br />
Gibbs free energy of the two phases remains the same (by definition), and hence<br />
along the coexistence curve we find:<br />
V1dp − S1dT = V2dp − S2dT (3.8)<br />
or<br />
( )<br />
dp<br />
=<br />
dT co<br />
S1(p, T ) − S2(p, T )<br />
(3.9)<br />
V1(p, T ) − V2(p, T )<br />
which is the famous Clausius-Clapeyron rel<strong>at</strong>ion. The pressure and temper<strong>at</strong>ure<br />
on the right hand side are rel<strong>at</strong>ed by p = pco(T ). Both the numer<strong>at</strong>or and the<br />
denomin<strong>at</strong>or on the right hand side are often divided by N, yielding a r<strong>at</strong>io of<br />
differences in molar volume and molar entropy.<br />
L<strong>at</strong>ent he<strong>at</strong>.<br />
The l<strong>at</strong>ent he<strong>at</strong> of a phase transform<strong>at</strong>ion is the amount of he<strong>at</strong> needed to<br />
transform the system <strong>at</strong> a given pressure and temper<strong>at</strong>ure along the coexistence<br />
curve completely from one phase to the other. The l<strong>at</strong>ent he<strong>at</strong> is positive if we<br />
transform m<strong>at</strong>erial from a low temper<strong>at</strong>ure phase to a high temper<strong>at</strong>ure phase.<br />
We indeed need energy to melt ice and boil w<strong>at</strong>er! The fact th<strong>at</strong> the l<strong>at</strong>ent<br />
he<strong>at</strong> is positive is easy to prove, similar to how we showed th<strong>at</strong> the volume
3.2. CLAUSIUS-CLAPEYRON RELATION. 119<br />
decreases ( when we go to higher pressure. In the previous chapter we saw th<strong>at</strong><br />
2<br />
∂ G<br />
∂T 2<br />
)<br />
≤ 0 and hence the picture of G versus T looks qualit<strong>at</strong>ively like figure<br />
p,N<br />
3.8. Since S = − ( )<br />
∂G<br />
∂T a decrease in the slope of G versus T is a jump down<br />
p,N<br />
in the value of -S and hence an increase in S.<br />
The l<strong>at</strong>ent he<strong>at</strong> is denoted by L1→2(T ) for the process going from phase 1<br />
to phase 2 <strong>at</strong> a temper<strong>at</strong>ure T and a corresponding value of the pressure along<br />
the coexistence curve. Since he<strong>at</strong> is rel<strong>at</strong>ed to changes in the entropy, we have<br />
L1→2(T ) = T (S2(pco(T ), T, N) − S1(pco(T ), T, N)) (3.10)<br />
and the Clausius-Clapeyron rel<strong>at</strong>ion takes the form<br />
( )<br />
dp<br />
l1→2(T )<br />
=<br />
dT co T (v2(p, T ) − v1(p, T ))<br />
(3.11)<br />
where we use the lower case characters to indic<strong>at</strong>e the molar volume and molar<br />
l<strong>at</strong>ent he<strong>at</strong> (i.e. the quantities are per mole).<br />
Consequence for phase boundary.<br />
The Clausius-Clapeyron rel<strong>at</strong>ion 3.9 is still in its general form. Note th<strong>at</strong> if 2<br />
denotes the high temper<strong>at</strong>ure phase, and if the volume of the high temper<strong>at</strong>ure<br />
phase is largest, both the numer<strong>at</strong>or and the denomin<strong>at</strong>or are positive. Hence<br />
the coexistence curve pco(T ) always has a positive slope in such a case. On the<br />
other hand, if we consider the phase boundary between w<strong>at</strong>er and ice, we know<br />
th<strong>at</strong> ice has a larger volume per mole, and hence the denomin<strong>at</strong>or is neg<strong>at</strong>ive.<br />
In this case the slope of the coexistence curve is neg<strong>at</strong>ive!<br />
Example.<br />
As a specific example take the transition from liquid to gas. In this case<br />
the volume of the liquid is much smaller than the volume of the gas and can be<br />
ignored. Using the ideal gas law gives:<br />
( )<br />
dp<br />
≈<br />
dT co<br />
pco(T )l1→2(T )<br />
RT 2<br />
(3.12)<br />
In many cases the l<strong>at</strong>ent he<strong>at</strong> is only very weakly temper<strong>at</strong>ure dependent. In<br />
th<strong>at</strong> case we can do the integr<strong>at</strong>ion and obtain<br />
pco(T ) ≈ p0e l (<br />
1→2 1<br />
R T −<br />
0 1<br />
)<br />
T<br />
(3.13)<br />
which indeed has a positive slope. This is a useful formula for a number of<br />
applic<strong>at</strong>ions. The limit T→ ∞ of pco(T ) is finite. Consequently, this rel<strong>at</strong>ion<br />
does not hold for the complete solid-liquid coexistence curve (but might be close<br />
for part of it). For such a phase boundary we need to take the values of the
120 CHAPTER 3. PHASE TRANSITIONS.<br />
volumes of both phases into account. Also, this rel<strong>at</strong>ion does not hold near a<br />
critical point, where the l<strong>at</strong>ent he<strong>at</strong> goes to zero, and the molar volumes become<br />
equal.<br />
To continue the study of this example, consider a closed volume V containing<br />
N moles of w<strong>at</strong>er. The temper<strong>at</strong>ure T is the control variable. If we are <strong>at</strong> room<br />
temper<strong>at</strong>ure, and the volume is chosen in such a manner th<strong>at</strong> the pressure is<br />
around <strong>at</strong>mospheric pressure, we will see w<strong>at</strong>er <strong>at</strong> the bottom, an interface, and<br />
vapor <strong>at</strong> the top.<br />
Which implicit assumption did we make?<br />
This system lends itself to an easy experiment. By measuring the level of the<br />
interface, we can determine the volume of the liquid and gas, Vl and Vg = V −Vl.<br />
By doing some optical absorption measurement we can determine the density<br />
of the vapor and hence we find the amount of vapor Ng and hence we know the<br />
amount of w<strong>at</strong>er Nl = N − Ng. If the gas can be described by the ideal gas law,<br />
we find pco(T ) = NgRT<br />
. Vg<br />
Up or down or wh<strong>at</strong>??<br />
Wh<strong>at</strong> happens if we increase the temper<strong>at</strong>ure? Will the interface go up or<br />
down? The answer is, th<strong>at</strong> depends on the volume V. The rel<strong>at</strong>ive volumes of<br />
the liquid and gas phase follow from<br />
V = Nlvl(T, pco(T )) + Ngvg(T, pco(T )) (3.14)<br />
for all temper<strong>at</strong>ures as long as we are below the critical temper<strong>at</strong>ure Tc. If<br />
the volume is larger than the critical volume Vc the phase diagram 3.6 shows<br />
th<strong>at</strong> <strong>at</strong> some temper<strong>at</strong>ure below Tc all m<strong>at</strong>erial will be vapor, and hence we see<br />
the interface drop to the bottom of the container. If the volume V is less than<br />
the critical volume Vc, on the other hand, all m<strong>at</strong>erial will transform to liquid,<br />
and the interface will actually rise! This seems counter-intuitive. How can we<br />
increase the temper<strong>at</strong>ure and see more liquid form? The answer is found in<br />
the fact th<strong>at</strong> the total volume is constant, and th<strong>at</strong> the increase in temper<strong>at</strong>ure<br />
comes with an increase in pressure which is sufficient to condens<strong>at</strong>e the vapor<br />
into liquid if the volume is small enough (i.e. smaller than Vc).<br />
Wh<strong>at</strong> happens if V = Vc? In this case the interface does not go to the top<br />
or the bottom of the container, but ends up somewhere in the middle. So wh<strong>at</strong><br />
happens when T reaches Tc. The interface is supposed to disappear, and it does<br />
indeed. If we reach Tc in this experiment, the properties of the liquid phase and<br />
the vapor phase become more and more similar, and <strong>at</strong> the critical interface<br />
they are identical. We only see the interface between w<strong>at</strong>er and vapor because<br />
the index of refraction is different in the two phases. At the critical point the
3.3. MULTI-PHASE BOUNDARIES. 121<br />
indices of refraction become identical, and the interface disappears from our<br />
view. The same holds for any type of measurement.<br />
3.3 Multi-phase boundaries.<br />
At the triple point in the phase diagram three phases coexist, and hence the<br />
equ<strong>at</strong>ions determining the triple point are<br />
G1(p, T, N) = G2(p, T, N) = G3(p, T, N) (3.15)<br />
Since the total amount of m<strong>at</strong>erial is fixed, there are two equ<strong>at</strong>ions for two<br />
unknowns. In general, this set of equ<strong>at</strong>ions has one unique solution. Therefore,<br />
we only find one triple point.<br />
Critical point.<br />
It is also easy to see th<strong>at</strong> in general we have only one critical point. The<br />
coexistence line is determined by one equ<strong>at</strong>ion for the free energy, and the critical<br />
point also st<strong>at</strong>es th<strong>at</strong> a parameter like the volume has to be the same.<br />
In this case the simultaneous equ<strong>at</strong>ions are G1(p, T, N) = G2(p, T, N) and<br />
V1(p, T, N) = V2(p, T, N).<br />
Quadruple point.<br />
Suppose in our model system we would have four different phases. The solid<br />
could, for example, change its structure <strong>at</strong> a certain temper<strong>at</strong>ure, or become<br />
magnetic. Would it be possible to have all four phases coexist? The answer is<br />
clearly no for our simple case. Since the only two independent variables are p and<br />
T, the set of three equ<strong>at</strong>ions determining this special point is over-complete and<br />
in general no solution exists. If we want to solve three simultaneous equ<strong>at</strong>ions,<br />
we need three independent variables. One could allow N to change by bringing<br />
the system into contact with a reservoir of constant chemical potential. This is<br />
not possible, however, since in th<strong>at</strong> case the system would be described by the<br />
set of st<strong>at</strong>e variables {p, T, µ} and these are all intensive. Therefore, we have<br />
to introduce another st<strong>at</strong>e variable. For example, take a (possibly) magnetic<br />
system with total magnetic moment M. For simplicity, we assume th<strong>at</strong> the<br />
direction of this moment is not important. The corresponding intensive st<strong>at</strong>e<br />
variable is the magnetic field H. In this case our system is described by the four<br />
variables {p, T, H, N} and we now use the modified, magnetic Gibbs free energy<br />
G(p, T, H, N) = U − T S + pV − HM (3.16)<br />
which is still equal to µN, since we now have to add one term to the Euler<br />
rel<strong>at</strong>ion. A point where all four phases coexist is now described by certain<br />
values {pq, Tq, Hq} and if such a quadruple point exists, it needs in general a
122 CHAPTER 3. PHASE TRANSITIONS.<br />
non-zero applied magnetic field. Note th<strong>at</strong> we ignored the energy stored in the<br />
field (see previous chapter). This energy is independent of the particular phase,<br />
however, and cancels in the equ<strong>at</strong>ions.<br />
Multi-component systems.<br />
If M+1 phases coexist for a certain system, we need to solve M equ<strong>at</strong>ions.<br />
The number of unknowns cannot be smaller than M. Hence we arrive <strong>at</strong> the<br />
general conclusion th<strong>at</strong><br />
The number of phases which can coexist for a system described by<br />
M independent intensive st<strong>at</strong>e variables is <strong>at</strong> most equal to M+1.<br />
For example, we have a system with r different types of m<strong>at</strong>erial, with number<br />
of moles Ni, i = 1 . . . r, for each type. Assume th<strong>at</strong> the total amount of<br />
m<strong>at</strong>erial is fixed, hence we have something like ∑ wiNi = N is constant, where<br />
the weights wi depend on the chemical reactions th<strong>at</strong> are described. In th<strong>at</strong><br />
case, using also p and T as variables, we have r+1 independent variables th<strong>at</strong><br />
can change, and hence we have <strong>at</strong> most r+2 phases th<strong>at</strong> can coexist <strong>at</strong> one time.<br />
Applic<strong>at</strong>ion to chemical reactions.<br />
In a chemical experiment one might look <strong>at</strong> reactions between two species<br />
A and B, according to A + B ↔ AB. The total amount of moles of A and B,<br />
either free or combined, are fixed. The rel<strong>at</strong>ive number can change. Therefore,<br />
this only leaves one st<strong>at</strong>e variable to describe the st<strong>at</strong>e of the reaction, and<br />
we could take this variable NAB, the number of molecules AB. It is better to<br />
use the corresponding chemical potential µAB, however, since th<strong>at</strong> is in general<br />
the st<strong>at</strong>e variable which is controlled in chemical experiments. The other st<strong>at</strong>e<br />
variable we can easily vary is the temper<strong>at</strong>ure T. Since there are two independent<br />
st<strong>at</strong>e variables, <strong>at</strong> most three phases can coexist. One phase is easily identified<br />
as A and B free, the other as A and B bound in AB. Requiring these two<br />
phases to coexist elimin<strong>at</strong>es one independent st<strong>at</strong>e variable, and leads to an<br />
equ<strong>at</strong>ion for the coexistence curve of the form NAB(T ). Hence the st<strong>at</strong>e of the<br />
reaction is completely determined by the temper<strong>at</strong>ure. If we also allow the AB<br />
m<strong>at</strong>erial to solidify, we introduce a third phase. The AB m<strong>at</strong>erial is either in<br />
solution or precipit<strong>at</strong>ed out. There is only one temper<strong>at</strong>ure Tt <strong>at</strong> which all three<br />
phases coexist. Interesting problems arise when one assumes th<strong>at</strong> only the AB<br />
molecules in solution are able to dissoci<strong>at</strong>e within the time of the experiment.<br />
An efficient reaction p<strong>at</strong>h then often requires manipul<strong>at</strong>ing both µAB and T to<br />
obtain a complete reaction.
3.4. BINARY PHASE DIAGRAM. 123<br />
3.4 Binary phase diagram.<br />
Basic variables.<br />
A very important applic<strong>at</strong>ion is found in binary phase diagrams. Suppose<br />
we have two m<strong>at</strong>erials, A and B, and each can dissolve in the other. Suppose we<br />
have NA moles of m<strong>at</strong>erial A with B dissolved in it, and NB moles of m<strong>at</strong>erial B<br />
with A dissolved in it. Since m<strong>at</strong>erial cannot disappear, we have NA + NB = N<br />
is constant. Therefore, <strong>at</strong> most four phases can coexist <strong>at</strong> a given time. In these<br />
solid solutions we typically work <strong>at</strong> <strong>at</strong>mospheric pressure, however, which for<br />
practical purposes is almost equal to zero (for solids!!) and which therefore can<br />
be ignored. The variables th<strong>at</strong> can be changed easily are the temper<strong>at</strong>ure T<br />
and the fraction of m<strong>at</strong>erial B in the total m<strong>at</strong>erial, cB, which is the r<strong>at</strong>io of<br />
the amount of pure m<strong>at</strong>erial B (without any A dissolved in it) to N. In such a<br />
phase diagram <strong>at</strong> most three phases can coexist.<br />
N<strong>at</strong>ure of cB.<br />
Although the variable cB is intensive, it is only so because we divided by<br />
N to get a density. The variable cB really measures an amount of m<strong>at</strong>erial,<br />
and really plays the role of an extensive st<strong>at</strong>e variable. The true intensive st<strong>at</strong>e<br />
variable would be the corresponding chemical potential, but th<strong>at</strong> one is hard<br />
to control directly in alloys. Just as we saw in the VT diagram, we will have<br />
regions in the cBT diagram which are not allowed, which do not correspond to<br />
st<strong>at</strong>es of the system.<br />
Typical phase diagram.<br />
A typical phase diagram for a binary alloy looks like figure 3.9. The curv<strong>at</strong>ure<br />
of the phase boundaries is always the same as in this figure, and can be<br />
derived from stability criteria. At low temper<strong>at</strong>ures, for very low or very high<br />
concentr<strong>at</strong>ions cB of B we have one type of m<strong>at</strong>erial. If the concentr<strong>at</strong>ion of B<br />
is in between, the m<strong>at</strong>erial will phase separ<strong>at</strong>e into regions rich in B, consisting<br />
of B(A) m<strong>at</strong>erial where A is dissolved in B, and regions poor in B, consisting<br />
of A(B) m<strong>at</strong>erial. The division between these can easily be calcul<strong>at</strong>ed if we<br />
know where the phase boundaries are. We use a similar lever rule as we used<br />
for the volumes earlier in this chapter. At higher temper<strong>at</strong>ures the system is<br />
either liquid, pure m<strong>at</strong>erial, or a mixture of liquid and pure m<strong>at</strong>erial. There is<br />
exactly one temper<strong>at</strong>ure and concentr<strong>at</strong>ion where all three phases can coexist.<br />
This point is called the eutectic point. Of course, when we have more phases or<br />
ordered structures, we can have more eutectic points.<br />
Crossing the line.
124 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.9: Typical binary phase diagram with regions L=liquid, A(B)=B<br />
dissolved in A, and B(A)=A dissolved in B.<br />
Suppose we prepare a liquid <strong>at</strong> high temper<strong>at</strong>ure containing both elements<br />
A and B and the concentr<strong>at</strong>ion of B is C0 B is low. We now cool the mixture and<br />
end in a stable region of m<strong>at</strong>erial A(B). Wh<strong>at</strong> happens in between and wh<strong>at</strong> is<br />
the result of our experiment? In order to describe this, see figure 3.10.<br />
When we reach the upper boundary of the forbidden region, we have an<br />
equilibrium of solid m<strong>at</strong>erial with a concentr<strong>at</strong>ion of B of CL and liquid with<br />
the original concentr<strong>at</strong>ion. When we lower the temper<strong>at</strong>ure, more solid will<br />
form, and the concentr<strong>at</strong>ion of the solid phase will increase. But since we form<br />
more solid with too small an amount of B, the concentr<strong>at</strong>ion of B in the liquid<br />
will also increase. Near the lower boundary of the forbidden region we have<br />
mainly solid m<strong>at</strong>erial with almost the original concentr<strong>at</strong>ion of B and some<br />
liquid with a high concentr<strong>at</strong>ion of B m<strong>at</strong>erial. Since no m<strong>at</strong>erial is lost, we can<br />
easily find the amounts of m<strong>at</strong>erial Nl in the liquid phase and Ns in the solid<br />
phase:<br />
Ns(c 0 B − c1) = Nl(c2 − c 0 B) (3.17)<br />
This rule is called the lever rule.<br />
In the scenario described above we have made one basic assumption, th<strong>at</strong><br />
the system is always in thermal equilibrium! In other words, we assume th<strong>at</strong><br />
the process is infinitely slow. Wh<strong>at</strong> are the time scales involved? First of all,<br />
the cooling time. The second time scale is rel<strong>at</strong>ed to the diffusion of m<strong>at</strong>erial B<br />
in the liquid and in the solid. The last one is the limiting factor, and this time<br />
can be long. In addition, in order for the solid to start forming we need initial<br />
seeds (nucle<strong>at</strong>ion of m<strong>at</strong>erial). In many cases, this is not a limiting factor.<br />
Suppose we cool the m<strong>at</strong>erial and grains of solid form. Suppose th<strong>at</strong> the<br />
diffusion of B m<strong>at</strong>erial is fast, and we end up with only grains of solid with<br />
the correct composition. Since we are now dealing with a solid, most likely we<br />
will have empty space in between the grains! If the original melt had some
3.4. BINARY PHASE DIAGRAM. 125<br />
Figure 3.10: Solidific<strong>at</strong>ion in a reversible process.<br />
impurities, it is possible for these impurities to collect in these empty spaces.<br />
In general, the m<strong>at</strong>erial <strong>at</strong> the grain boundaries will have a lower density and<br />
the volume of the resulting m<strong>at</strong>erial is too large. Wh<strong>at</strong> can we do about th<strong>at</strong>?<br />
If the m<strong>at</strong>erial is a metal, we can roll it and press it to get rid of these empty<br />
spaces. For ceramics, th<strong>at</strong> is much harder.<br />
But th<strong>at</strong> is not all folks. Typically, the diffusion of B m<strong>at</strong>erial in the solid<br />
is incomplete. Therefore, the grains will have a concentr<strong>at</strong>ion gradient. The<br />
centers are poor in B m<strong>at</strong>erial. Near the grain boundaries we have more or<br />
less the correct concentr<strong>at</strong>ion. The excess of B m<strong>at</strong>erial will be in between the<br />
grains! Wh<strong>at</strong> to do now? Since there is a concentr<strong>at</strong>ion gradient, there is a<br />
difference in chemical potential, and there is a driving force which wants to<br />
make the concentr<strong>at</strong>ion uniform. Since the diffusion is always much better <strong>at</strong><br />
higher temper<strong>at</strong>ure, we want to keep the solid <strong>at</strong> a temper<strong>at</strong>ure just below the<br />
lower bound for a while. This annealing process is very important when making<br />
m<strong>at</strong>erials. Sometimes the times involved are quite long. Sometimes other tricks<br />
have to be applied during the anneal.<br />
The previous discussion shows th<strong>at</strong> phase transitions in solids are quite difficult<br />
to deal with. Th<strong>at</strong> is why we deal with gas to liquid transitions in most<br />
of this chapter. In the next chapter we will discuss magnetic phase transitions,<br />
which also are much easier to deal with. Nevertheless, a study of phase transitions<br />
in solids is of gre<strong>at</strong> practical importance! Just talk with your mechanical<br />
engineering friends.<br />
Eutectic point.<br />
The eutectic point is of gre<strong>at</strong> practical importance. Opposed to the gas-liquid<br />
coexistence, phase separ<strong>at</strong>ion <strong>at</strong> low temper<strong>at</strong>ures is a very slow process, since<br />
it involves <strong>at</strong>omic motion in a solid. It is therefore possible to cre<strong>at</strong>e meta-stable<br />
m<strong>at</strong>erials in the forbidden regions by cooling a sample very rapidly. Typically
126 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.11: Typical binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, with regions L=liquid, A(B)=B dissolved in A, B(A)=A dissolved in<br />
B, and AB= AB with either A or B dissolved.<br />
this is done by dropping a drop of liquid into cold w<strong>at</strong>er. If we are not <strong>at</strong> the<br />
eutectic point, it is almost certain th<strong>at</strong> small amounts of m<strong>at</strong>erial A(B) or B(A)<br />
will form during cooling (when we go through the top forbidden regions) and<br />
the final m<strong>at</strong>erial will not be homogeneous. If we are <strong>at</strong> the eutectic point, this<br />
region where m<strong>at</strong>erial separ<strong>at</strong>es is reached <strong>at</strong> the lowest possible temper<strong>at</strong>ure<br />
(making the process the slowest), furthest from the stable phases (increasing<br />
the energy needed for the process). Meta-stable m<strong>at</strong>erials <strong>at</strong> a concentr<strong>at</strong>ion<br />
corresponding to the eutectic point are therefore easiest to prepare.<br />
Intermedi<strong>at</strong>e compounds.<br />
In many cases there are stable compounds of the m<strong>at</strong>erials A and B present,<br />
and the phase diagram then looks different. A typical example is figure 3.11<br />
where we included a stable m<strong>at</strong>erial AB . The overall structure and interpret<strong>at</strong>ion<br />
is the same, except th<strong>at</strong> we now have a region of AB m<strong>at</strong>erial in which<br />
either A or B is dissolved. Quite often the range of stability of the intermedi<strong>at</strong>e<br />
compound is very small, so small th<strong>at</strong> it collapses to a single line in the diagram,<br />
see figure 3.12. An example of an impossible phase diagram is figure 3.13, where<br />
we have four phases coexisting <strong>at</strong> one point.<br />
3.5 Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
More quantit<strong>at</strong>ive approaches based on energy.<br />
The previous discussion of phase transitions was very general. In order to
3.5. VAN DER WAALS EQUATION OF STATE. 127<br />
Figure 3.12: Typical binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, where the intermedi<strong>at</strong>e region is too small to discern from a line.<br />
Figure 3.13: Impossible binary phase diagram with intermedi<strong>at</strong>e compound<br />
AB, where the intermedi<strong>at</strong>e region is too small to discern from a line.
128 CHAPTER 3. PHASE TRANSITIONS.<br />
give details we need to assume a certain form for the energy as a function of<br />
the st<strong>at</strong>e variables. Energy is the most basic thermodynamic st<strong>at</strong>e variable for<br />
which we can develop models th<strong>at</strong> can easily be interpreted. This will be done<br />
in the next chapter.<br />
More quantit<strong>at</strong>ive approaches based on equ<strong>at</strong>ions of st<strong>at</strong>e.<br />
In a p-V-T system the three st<strong>at</strong>e variables which are easily controlled or<br />
measured are the pressure p, the volume V, and the temper<strong>at</strong>ure T. We assume<br />
th<strong>at</strong> the number of particles N is again fixed, and hence of the three variables<br />
{p, V, T } only two are independent. A m<strong>at</strong>hem<strong>at</strong>ical rel<strong>at</strong>ion describing this<br />
dependency, an equ<strong>at</strong>ion of st<strong>at</strong>e, is needed to show how they are rel<strong>at</strong>ed. The<br />
ideal gas law, pV = NRT , is the easiest example of such an equ<strong>at</strong>ion of st<strong>at</strong>e,<br />
but unfortun<strong>at</strong>ely a system obeying the ideal gas law does not show any phase<br />
transitions. An ideal gas is an approxim<strong>at</strong>ion of a real gas, and describes a real<br />
gas in the limit of very low densities. Therefore, an <strong>at</strong>tempt to improve upon<br />
the ideal gas law would be to assume an equ<strong>at</strong>ion of st<strong>at</strong>e of the form<br />
p<br />
RT =<br />
∞∑<br />
( ) j<br />
N<br />
Bj(T )<br />
V<br />
j=1<br />
(3.18)<br />
with B1(T ) = 1. The value of this first term is determined by the requirement<br />
th<strong>at</strong> in the low density limit the ideal gas law should be obtained. This expansion<br />
is called a virial expansion and is the basis for many <strong>at</strong>tempts to describe a real<br />
gas by some kind of perturb<strong>at</strong>ion theory.<br />
Validity of equ<strong>at</strong>ions of st<strong>at</strong>e.<br />
In our discussion of phase transitions we mentioned th<strong>at</strong> every point in a p-T<br />
diagram corresponds to a st<strong>at</strong>e of the system, but th<strong>at</strong> this is not necessarily<br />
the case for every point in a p-V or T-V diagram. Equ<strong>at</strong>ion 3.18 yields a value<br />
of the pressure for all values of V and T. Hence equ<strong>at</strong>ion 3.18 in a technical<br />
sense only represents an equ<strong>at</strong>ion of st<strong>at</strong>e for those points in a T-V diagram<br />
which are equilibrium or stable st<strong>at</strong>es! The advantage of using an equ<strong>at</strong>ion like<br />
3.18 for all points is th<strong>at</strong> it gives an analytic function connecting all possible<br />
regions in phase space. It has to be supplemented by equilibrium conditions.<br />
The most obvious requirements ) are those rel<strong>at</strong>ed to local stability. If we have<br />
p(T,V) we need ≤ 0. This inequality holds for the ideal gas law for all<br />
( ∂p<br />
∂V<br />
T<br />
temper<strong>at</strong>ures and volumes. All points in a T-V diagram are stable st<strong>at</strong>es for an<br />
ideal gas, and there are no phase transitions. This is not the case, however, for<br />
more general equ<strong>at</strong>ions of st<strong>at</strong>e.<br />
Models of the equ<strong>at</strong>ion of st<strong>at</strong>e.
3.5. VAN DER WAALS EQUATION OF STATE. 129<br />
It is also possible to assume the existence of an equ<strong>at</strong>ion of st<strong>at</strong>e, rel<strong>at</strong>ing the<br />
st<strong>at</strong>e variables describing a system. Van der Waals analyzed such a situ<strong>at</strong>ion.<br />
He was interested in the phase transition from a gas to a liquid and derived a<br />
modified form of the ideal gas law which is valid for both the gas and the liquid<br />
phase. The fe<strong>at</strong>ures of his model equ<strong>at</strong>ion of st<strong>at</strong>e are remarkable general. The<br />
explan<strong>at</strong>ion for this fact is found in a microscopic deriv<strong>at</strong>ion of his equ<strong>at</strong>ion of<br />
st<strong>at</strong>e, which we will ignore <strong>at</strong> this point.<br />
Van der Waals followed a route different from the virial approach when he<br />
investig<strong>at</strong>ed liquid to gas phase transitions. He postul<strong>at</strong>ed th<strong>at</strong> for a not-so-ideal<br />
gas the equ<strong>at</strong>ion of st<strong>at</strong>e should have the form<br />
( ) 2 N<br />
p + a (V − Nb) = NRT (3.19)<br />
V 2<br />
where a and b are two parameters. It is possible to give a physical interpret<strong>at</strong>ion<br />
to these parameters a and b if the Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is derived from<br />
a microscopic model. The parameter b represents the volume of a molecule. The<br />
total volume V has to be larger than the volume of all molecules tightly packed<br />
together. The internal energy of an ideal gas only depends on the temper<strong>at</strong>ure,<br />
and a correction term is needed for higher densities. The parameter a represents<br />
the average interaction energy between two molecules and is positive for an<br />
<strong>at</strong>tractive interaction.<br />
Unique solutions or not.<br />
is<br />
The pressure as a function of volume for the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e<br />
p = NRT<br />
− aN2<br />
V − Nb V 2<br />
(3.20)<br />
Therefore, the pressure is a unique function of the volume and temper<strong>at</strong>ure.<br />
The volume, on the other hand, is not a unique function of the pressure and<br />
temper<strong>at</strong>ure, and the van der Waals equ<strong>at</strong>ion allows for multiple solutions in<br />
certain ranges of pressure and temper<strong>at</strong>ure. Switching from one solution to<br />
another causes a jump in the value of the volume, exactly wh<strong>at</strong> we expect for a<br />
phase transition.<br />
It is also useful to discuss the difference between the pressure in a van der<br />
Waals gas and in an ideal gas. For this difference we find:<br />
2 N<br />
p − pideal =<br />
V 2<br />
(<br />
)<br />
V<br />
bRT − a<br />
(3.21)<br />
V − Nb<br />
which shows th<strong>at</strong> the difference is of second order in the density. This difference<br />
is useful if we want to calcul<strong>at</strong>e the energy difference between a van der Waals<br />
gas and an ideal gas, for example.<br />
Unstable regions.
130 CHAPTER 3. PHASE TRANSITIONS.<br />
( ∂p<br />
∂V<br />
Mechanical ) stability requires th<strong>at</strong> compressibilities are positive, and hence<br />
< 0. This deriv<strong>at</strong>ive for the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is<br />
T<br />
( )<br />
∂p<br />
= −<br />
∂V T,N<br />
NRT<br />
V 2<br />
(<br />
)<br />
1 n<br />
− 2a<br />
(1 − nb) 2 RT<br />
(3.22)<br />
where n = N<br />
V is the molar density of the m<strong>at</strong>erial in the gas or liquid. Instabilities<br />
occur when this deriv<strong>at</strong>ive is positive. This can only happen when a<br />
is positive. In a microscopic interpret<strong>at</strong>ion this means th<strong>at</strong> the interaction between<br />
the molecules has to be <strong>at</strong>tractive. This makes sense, for if the interaction<br />
between the molecules were repulsive, they would not want to condense into a<br />
liquid. The stability condition can be rewritten in the form:<br />
Phase space.<br />
RT<br />
2a ≥ (1 − nb)2 n (3.23)<br />
Phase space for a p-V-T system is given by the octant of three dimensional<br />
{p, V, T } space in which all three variables are positive. The van der Waals<br />
equ<strong>at</strong>ion of st<strong>at</strong>e limits the possible st<strong>at</strong>es of the system to a two-dimensional<br />
surface ) in this space. The lines on this surface where the partial deriv<strong>at</strong>ive<br />
is zero separ<strong>at</strong>es stable and unstable regions. Therefore it is of interest<br />
( ∂p<br />
∂V<br />
T<br />
to find these lines by solving<br />
n(1 − nb) 2 = RT<br />
2a<br />
(3.24)<br />
In order th<strong>at</strong> V > Nb we need n < 1<br />
b , hence we do not include the high density<br />
region in which the <strong>at</strong>oms would have to overlap in order to fit into the volume<br />
V.<br />
Case a=0.<br />
There are no solutions to this equ<strong>at</strong>ion when a = 0. In th<strong>at</strong> case the equ<strong>at</strong>ion<br />
of st<strong>at</strong>e is simply<br />
p = NRT<br />
V − Nb<br />
(3.25)<br />
which is well defined for all volumes V > Nb and has a neg<strong>at</strong>ive deriv<strong>at</strong>ive.<br />
Hence the effect of a finite volume of the molecules in itself is not sufficient to<br />
give rise to phase transitions.<br />
case b=0.
3.5. VAN DER WAALS EQUATION OF STATE. 131<br />
is<br />
Figure 3.14: Graphical solution of equ<strong>at</strong>ion 3.24.<br />
There is only one solution to 3.24 if b = 0. In th<strong>at</strong> case the equ<strong>at</strong>ion of st<strong>at</strong>e<br />
p = n(RT − an) (3.26)<br />
and stability requires n < RT<br />
2a . This system shows a sign of a phase transition.<br />
If we start with a gas <strong>at</strong> a high temper<strong>at</strong>ure, and cool it down, nothing special<br />
happens as long as T > 2an<br />
2an<br />
R . As soon as we reach the temper<strong>at</strong>ure R ,<br />
however, the system becomes mechanically unstable, and the system collapses<br />
to a singular point with V = 0. The <strong>at</strong>tractive interaction <strong>at</strong> low temper<strong>at</strong>ure<br />
is strong enough to force condens<strong>at</strong>ion into a liquid, but the liquid is unstable.<br />
We need the finite volume of the molecules to stabilize the liquid phase.<br />
General case.<br />
The general equ<strong>at</strong>ion 3.24 with both a and b non-zero has either one or three<br />
solutions. In the region where the molecules do not overlap we have n < 1<br />
b . For<br />
very large values of the temper<strong>at</strong>ure there are no instabilities for n < 1<br />
b and<br />
there are no phase transitions as a function of volume or density or pressure.<br />
If the temper<strong>at</strong>ure is very small, the solutions are approxim<strong>at</strong>ely n = RT<br />
2a and<br />
√<br />
RT<br />
nb = 1 ± 2a . Two solutions are in the allowed range nb < 1. If we increase<br />
the temper<strong>at</strong>ure, these two solutions will approach each other and <strong>at</strong> a certain<br />
temper<strong>at</strong>ure Tc they will merge together and disappear. The third solution will<br />
always be in the un-physical region n > 1<br />
b . This is represented in figure 3.14.<br />
If the temper<strong>at</strong>ure T is larger than Tc, all values of n in the range [0, 1<br />
b ]<br />
correspond to mechanically stable st<strong>at</strong>es. On the other hand, if T is below<br />
Tc the values of N in between the two crossings correspond to mechanically<br />
unstable st<strong>at</strong>es, and only low and high density phases are possible.
132 CHAPTER 3. PHASE TRANSITIONS.<br />
critical temper<strong>at</strong>ure.<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e shows no phase transitions for T > Tc<br />
and a region of instability for T < Tc. Therefore Tc is identified as the critical<br />
temper<strong>at</strong>ure, belonging to the critical point for this liquid-gas phase transition.<br />
The deriv<strong>at</strong>ive of the function<br />
is easy to find:<br />
f(n) = n(1 − nb) 2<br />
(3.27)<br />
f ′ (n) = (1 − nb) 2 − 2nb(1 − nb) = (1 − 3nb)(1 − nb) (3.28)<br />
and is zero for nb = 1 (as expected) and 3nb = 1. Therefore 3ncb = 1. This<br />
requires b > 0 as expected. The value of Tc follows from RTc<br />
2a = f(nc) = 4<br />
27b .<br />
The pressure follows directly from the van der Waals equ<strong>at</strong>ion.<br />
Loc<strong>at</strong>ion of critical point.<br />
The critical point is therefore specified by<br />
RTc = 8a<br />
27b<br />
(3.29)<br />
Vc = 3Nb (3.30)<br />
pc = a<br />
27b2 (3.31)<br />
The critical volume is indeed larger than Nb, as required. Its value, however,<br />
is close enough to the high density limit and a strict quantit<strong>at</strong>ive interpret<strong>at</strong>ion<br />
of the parameters a and b in terms of microscopic quantities in invalid. In such<br />
an interpret<strong>at</strong>ion higher order effects will certainly play a role. Therefore, it is<br />
better to consider a and b as adjustable parameters rel<strong>at</strong>ed to the interaction<br />
energy and molecular volume, and find their values from experiment. It is also<br />
clear th<strong>at</strong> the equ<strong>at</strong>ions for the critical point only make sense when a and b are<br />
positive.<br />
pV-curves.<br />
The general shape of the p-V curves is shown in figure 3.15. Note th<strong>at</strong> <strong>at</strong> the<br />
critical point not only the slope ) of the p(V) curve is zero, but also the curv<strong>at</strong>ure!<br />
In addition, we have<br />
> 0 and with increasing temper<strong>at</strong>ure<br />
( ∂p<br />
∂T<br />
V,N<br />
= NR<br />
V −Nb<br />
the curves move always up. This means th<strong>at</strong> the p-V-curves will never cross!<br />
Another interesting question pertains to the situ<strong>at</strong>ion depicted in figure 3.16.<br />
Is it possible to have neg<strong>at</strong>ive values of the pressure? Mechanical instability
3.5. VAN DER WAALS EQUATION OF STATE. 133<br />
Figure 3.15: p-V curves in the van der Waals model.<br />
Figure 3.16: p-V curves in the van der Waals model with neg<strong>at</strong>ive values of<br />
the pressure.
134 CHAPTER 3. PHASE TRANSITIONS.<br />
only requires th<strong>at</strong> the slope of the p(V) curve is neg<strong>at</strong>ive. It does not preclude<br />
a neg<strong>at</strong>ive value of the pressure. Neg<strong>at</strong>ive pressure means th<strong>at</strong> the system gains<br />
energy by expanding. One could simul<strong>at</strong>e this by connecting springs to the<br />
outside of a m<strong>at</strong>erial, which try to stretch the m<strong>at</strong>erial. Also, if we replace p-V<br />
by H-M there is no problem <strong>at</strong> all, neg<strong>at</strong>ive magnetic fields are possible!<br />
Do you expect neg<strong>at</strong>ive pressures in real liquids?<br />
Neg<strong>at</strong>ive pressures occur in the van der Waals model as soon as T < 27<br />
32 Tc,<br />
which means for a large range of values of the temper<strong>at</strong>ure. The equ<strong>at</strong>ion p = 0<br />
leads to<br />
an(1 − nb) = RT (3.32)<br />
with solutions<br />
√<br />
1 ± 1 − 4RT<br />
n = b<br />
a<br />
(3.33)<br />
2b<br />
which are always less than 1<br />
b and hence always occur in the physically allowed<br />
region. The square root has to be real, though, which leads to 4RT b < a, which<br />
gives the inequality for the temper<strong>at</strong>ure cited above.<br />
Finding the minimal energy st<strong>at</strong>e.<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e for a temper<strong>at</strong>ure below the critical<br />
value leads to a p(V) curve of the form shown in figure 3.17. If the value of<br />
the pressure is between pmin and pmax, there are three values of the volume<br />
possible. Since the volume is a st<strong>at</strong>e variable, only one of these values can be<br />
correct. Obviously we have to discard the middle value of the volume, since the<br />
deriv<strong>at</strong>ive of the p(V) curve has the wrong sign in this case. The Gibbs free<br />
energy is needed in order to decide which of the other two solutions is stable.<br />
How to calcul<strong>at</strong>e the energy.<br />
The difference in Gibbs free energy between the point 1 and 2 is found the<br />
change in energy along the p<strong>at</strong>h following the equ<strong>at</strong>ion of st<strong>at</strong>e p(V) <strong>at</strong> this<br />
temper<strong>at</strong>ure. This is not a reversible process, since some of the intermedi<strong>at</strong>e<br />
st<strong>at</strong>es are unstable. These intermedi<strong>at</strong>e st<strong>at</strong>es are special, however, since they<br />
are m<strong>at</strong>hem<strong>at</strong>ically well described. The equ<strong>at</strong>ion of st<strong>at</strong>e is still valid for these<br />
points. At constant temper<strong>at</strong>ure and particle number the differential of the<br />
Gibbs free energy is dG = V dp and therefore<br />
G2 − G1 =<br />
∫ 2<br />
1<br />
V dp = A1 − A2<br />
(3.34)
3.5. VAN DER WAALS EQUATION OF STATE. 135<br />
Figure 3.17: p-V curve in the van der Waals model with areas corresponding<br />
to energies.<br />
where A1 and A2 are the indic<strong>at</strong>ed areas on the left and right. If A1 > A2 the<br />
Gibbs free energy of the system in st<strong>at</strong>e 2 is the highest and the system will be<br />
in st<strong>at</strong>e 1, with a low volume. The opposite is true for A2 > A1. If A1 = A2<br />
the Gibbs free energies of both phases are equal and the liquid and gas coexist.<br />
This condition therefore determines pco(T ).<br />
Is this procedure allowed?<br />
One question th<strong>at</strong> might be asked is if the previous procedure is allowed.<br />
We have followed the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e through unstable regions,<br />
which does not seem to be correct. Fortun<strong>at</strong>ely, as expected, the answer is<br />
reassuring, we can do this. At each point along the van der Waals equ<strong>at</strong>ion<br />
of st<strong>at</strong>e the st<strong>at</strong>e of the system is an extremal st<strong>at</strong>e for the energy, either a<br />
maximum or a local or global minimum. If we follow the van der Waals equ<strong>at</strong>ion<br />
of st<strong>at</strong>e in a reversible process, in theory we can always go back and forth and<br />
the energies calcul<strong>at</strong>ed before correspond to the actual energy differences. Of<br />
course, in practice there would always be small fluctu<strong>at</strong>ions th<strong>at</strong> throw us out<br />
of the unstable st<strong>at</strong>e when the energy is a maximum. Hence the procedure we<br />
followed is not a practical experimental procedure to measure energy differences.<br />
Neg<strong>at</strong>ive pressure never correspond to physical st<strong>at</strong>es.<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e allows for neg<strong>at</strong>ive pressures. The equ<strong>at</strong>ion<br />
p = 0 has two solutions when T < 27<br />
32Tc. The pressure <strong>at</strong> very large volumes<br />
always approaches zero inversely proportional to V. The stable st<strong>at</strong>e for a small,<br />
positive pressure is determined by the equality of the areas A1 and A2. In the<br />
limit for the pressure approaching zero, the area on the right A2 becomes in-
136 CHAPTER 3. PHASE TRANSITIONS.<br />
finitely large. Therefore, we always have A2 > A1 in this limit, and the system<br />
will be found <strong>at</strong> the high volume st<strong>at</strong>e. Hence, the st<strong>at</strong>es with neg<strong>at</strong>ive pressure<br />
are never stable, even though one of them is mechanically stable. Mechanical<br />
stability only involves the deriv<strong>at</strong>ive of pressure versus volume. If a system is<br />
stable with a neg<strong>at</strong>ive pressure, it behaves like an expanded rubber band. We<br />
need to stretch it (neg<strong>at</strong>ive pressure) in order to keep it in th<strong>at</strong> st<strong>at</strong>e. Stable<br />
systems with a neg<strong>at</strong>ive pressure are possible on general thermodynamical<br />
grounds. The specific physics of the system under consider<strong>at</strong>ion will decide if<br />
th<strong>at</strong> system can be stable under neg<strong>at</strong>ive pressures. A liquid-gas system should<br />
not be stable under neg<strong>at</strong>ive pressure, since cooling through forced evapor<strong>at</strong>ion<br />
is possible. The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is consistent with the experimental<br />
facts for a liquid-gas system and does not allow for stable st<strong>at</strong>es with<br />
a neg<strong>at</strong>ive pressure. All in all we conclude th<strong>at</strong> the van der Waals equ<strong>at</strong>ion of<br />
st<strong>at</strong>e is a good model for a liquid to gas phase transition, ending in a critical<br />
point.<br />
Joule-Thomson effect.<br />
If a fluid is pushed through a porous plug, this throttling process gives a<br />
change in temper<strong>at</strong>ure due to the change in pressure <strong>at</strong> constant enthalpy. This<br />
was a problem assigned in the previous chapter. The Joule-Thomson coefficient<br />
describing this process is therefore<br />
( )<br />
∂T<br />
ηJT =<br />
(3.35)<br />
∂p H,N<br />
and can be shown to be equal to<br />
ηJT = V<br />
Cp<br />
(αT − 1) (3.36)<br />
For m<strong>at</strong>erials with αT > 1 the fluid will cool down on the expansion through<br />
the plug, and for m<strong>at</strong>erials with αT < 1 the fluid will he<strong>at</strong> up on this expansion.<br />
For an ideal gas we have αT = 1, so no change in temper<strong>at</strong>ure occurs. But wh<strong>at</strong><br />
is happening in a real gas?<br />
The van der Waals equ<strong>at</strong>ion is a good approxim<strong>at</strong>ion for an ideal gas, so<br />
we will calcul<strong>at</strong>e the coefficient of thermal expansion for the van der Waals gas.<br />
Th<strong>at</strong> is a bit tedious, so we will look <strong>at</strong> two extremes. First of all, wh<strong>at</strong> happens<br />
<strong>at</strong> very high temper<strong>at</strong>ures? In th<strong>at</strong> case the pressure will be very high for a<br />
given volume, and hence we can ignore the a ( )<br />
N 2<br />
V term and use:<br />
which gives<br />
p(V − Nb) ≈ NRT (3.37)<br />
( )<br />
∂V<br />
p ≈ NR (3.38)<br />
∂T p,N
3.5. VAN DER WAALS EQUATION OF STATE. 137<br />
and hence<br />
αT − 1 ≈ NRT<br />
pV<br />
pα ≈ NR<br />
V<br />
− 1 ≈ p(V − Nb)<br />
pV<br />
− 1 = −b N<br />
V<br />
(3.39)<br />
(3.40)<br />
So <strong>at</strong> high temper<strong>at</strong>ures the Joule-Thomson coefficient for a real gas is neg<strong>at</strong>ive,<br />
which means th<strong>at</strong> the gas cools down in a throttling process. But wh<strong>at</strong><br />
happens a a lower temper<strong>at</strong>ure? We will study this in the low density limit,<br />
where we have<br />
p = NRT<br />
V − Nb −a<br />
This gives<br />
0 = NR<br />
V<br />
− NRT<br />
V 2<br />
( ) 2<br />
N<br />
V<br />
0 = NR<br />
V<br />
≈ N<br />
RT (1+bN<br />
V V )−a<br />
( )<br />
∂V<br />
+ bR<br />
∂T p,N<br />
− NRT<br />
V<br />
α + bR<br />
( ) 2<br />
N<br />
V<br />
( ) 2<br />
N<br />
V<br />
( ) 2<br />
N<br />
=<br />
V<br />
NRT<br />
+(bRT −a)<br />
V<br />
2 N<br />
− 2(bRT − a)<br />
V 3<br />
( )<br />
∂V<br />
∂T p,N<br />
− 2(bRT − a)<br />
RT α + 2(bRT − a) N<br />
N<br />
α = R + bR<br />
V V<br />
α =<br />
αT − 1 =<br />
αT − 1 = N<br />
V<br />
R + bR N<br />
V<br />
RT + 2(bRT − a) N<br />
V<br />
bRT N<br />
V<br />
− 2(bRT − a) N<br />
V<br />
RT + 2(bRT − a) N<br />
V<br />
2a − bRT<br />
RT + 2(bRT − a) N<br />
V<br />
N 2<br />
( ) 2<br />
N<br />
V<br />
(3.41)<br />
(3.42)<br />
α (3.43)<br />
V 2<br />
(3.44)<br />
(3.45)<br />
(3.46)<br />
(3.47)<br />
This shows th<strong>at</strong> the Joule-Thomson effect changes sign when we lower the<br />
temper<strong>at</strong>ure. The effect is zero <strong>at</strong> the so-called inversion temper<strong>at</strong>ure Ti and<br />
for temper<strong>at</strong>ures below Ti a gas he<strong>at</strong>s up in a throttling process. We see th<strong>at</strong><br />
the inversion temper<strong>at</strong>ure is given by<br />
Ti = 2a 27<br />
=<br />
bR 4 Tc<br />
(3.48)<br />
which is a temper<strong>at</strong>ure well above the critical temper<strong>at</strong>ure! Hence for almost<br />
all gases <strong>at</strong> room temper<strong>at</strong>ure the Joule-Thomson effect is positive.
138 CHAPTER 3. PHASE TRANSITIONS.<br />
Figure 3.18: Unstable and meta-stable regions in the van der Waals p-V<br />
diagram.<br />
3.6 Spinodal decomposition.<br />
Names of separ<strong>at</strong>ion lines.<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e determines the temper<strong>at</strong>ure as a function<br />
of pressure and volume for every point (st<strong>at</strong>e) in the p-V plane. As we<br />
have seen, ) however, not all st<strong>at</strong>es are stable. The region in the p-V plane<br />
where<br />
> 0 corresponds to mechanically unstable st<strong>at</strong>es. The line<br />
( ∂p<br />
∂V<br />
T,N<br />
which separ<strong>at</strong>es this region from the region of mechanically stable (but thermodynamically<br />
meta-stable) st<strong>at</strong>es is called the spinodal and is determined by<br />
)<br />
(p, V, N) = 0. One has to keep in mind th<strong>at</strong> N is fixed. The line sepa-<br />
( ∂p<br />
∂V<br />
T,N<br />
r<strong>at</strong>ing the thermodynamically stable and meta-stable st<strong>at</strong>es is called a binodal<br />
and is determined by the condition th<strong>at</strong> the Gibbs free energy be the same in<br />
both phases.<br />
Meta-stable regions.<br />
Wh<strong>at</strong> happens when we prepare a system in the mechanically unstable region?<br />
This could be done, for example, by cooling a system with volume Vc<br />
rapidly to a temper<strong>at</strong>ure below Tc. Assume th<strong>at</strong> the pressure for this un-physical<br />
system is p0, and th<strong>at</strong> the system is kept <strong>at</strong> a constant temper<strong>at</strong>ure T0. At first,<br />
the system will be homogeneous. Since it is mechanically unstable, any small<br />
fluctu<strong>at</strong>ion in the density will be amplified. A region which momentarily has a<br />
density which is too high, will increase its density following the van der Waals<br />
curve until it reaches the spinodal, where it becomes mechanically stable. A
3.6. SPINODAL DECOMPOSITION. 139<br />
low density region will follow the van der Waals curve in the opposite direction.<br />
The system will separ<strong>at</strong>e in small regions of high density, N , and small regions<br />
Va<br />
with low density N<br />
Vb . The values of the volumes Va and Vb are found from the<br />
curve in figure 3.18. These are the values of the volume for the whole system, so<br />
they have to be transformed to densities. This regions are all mixed together.<br />
The total volume has to be constant, and hence<br />
Vtotal = Na<br />
N Va + Nb<br />
N Vb<br />
(3.49)<br />
where Vtotal is the initial volume. Also we have N = Na + Nb. These two<br />
rel<strong>at</strong>ions together determine how much m<strong>at</strong>erial will go into each phase.<br />
Spinodal decomposition.<br />
The process described in the previous section is an important one and is<br />
called spinodal decomposition. The end st<strong>at</strong>e is mechanically stable. It is thermodynamically<br />
unstable, though, since pa < pb. Hence the process will continue<br />
by a further expansion of the low density regions and a further contraction of<br />
the high density regions until the pressure in both regions is equal again (and<br />
equal to p0). During this process we again follow a van der Waals curve. But<br />
now time scales become important.<br />
Spinodal decomposition in a liquid-gas system.<br />
If the spinodal decomposition occurred in a gas-liquid mixture, the road to<br />
thermal equilibrium is without obstacles and the final system will be a mixture<br />
of gas and liquid <strong>at</strong> the same pressure p0 and temper<strong>at</strong>ure T0 (corresponding to<br />
points 1 and 2 in figure 3.18). The amount of each m<strong>at</strong>erial is determined by the<br />
original volume of the gas. In free space, this is exactly wh<strong>at</strong> happens. Clouds<br />
and fog are such mixtures of gas and liquid. One way to do an experiment along<br />
these lines is to take a liquid in a cylinder with a piston and suddenly pull the<br />
piston out.<br />
Complic<strong>at</strong>ing factors.<br />
In many experiments gravity plays a role and will try to collect all the liquid<br />
in the bottom of the container. Surface tension is another factor, and has to be<br />
taken into account to find the distribution of the sizes of the liquid droplets in a<br />
fog. Note th<strong>at</strong> it is possible for a system to be in the meta-stable region. W<strong>at</strong>er<br />
can be under-cooled well below 0 C. Any small perturb<strong>at</strong>ion (like an abrupt<br />
movement of the container) will cause it to freeze, however, demonstr<strong>at</strong>ing the<br />
meta-stability.<br />
Example in binary phase diagram.
140 CHAPTER 3. PHASE TRANSITIONS.<br />
In a solid compound AxB1−x the composition x is a st<strong>at</strong>e variable and a<br />
spinodal curve might show up in the T-x plane. Certain compositions are mechanically<br />
unstable in th<strong>at</strong> case, and if we quench a liquid of th<strong>at</strong> combin<strong>at</strong>ion<br />
to a low temper<strong>at</strong>ure, the quenched solid is mechanically unstable. It will phase<br />
separ<strong>at</strong>e following a spinodal decomposition and become a mixture of grains of<br />
the compound with low x and high x. The intensive st<strong>at</strong>e variable corresponding<br />
to the concentr<strong>at</strong>ion x is the chemical potential. Therefore, we have cre<strong>at</strong>ed differences<br />
in chemical potential between the grains, and this will force m<strong>at</strong>erial to<br />
move, so we can follow the equivalent of a p-V curve to arrive <strong>at</strong> the stable st<strong>at</strong>es<br />
for this temper<strong>at</strong>ure. In order to do so, however, m<strong>at</strong>erial needs to diffuse in a<br />
solid, which is a very slow process. When the system has become mechanically<br />
stable, it might try to find other ways to rearrange and become thermodynamically<br />
stable. In a solid, however, the grains of each m<strong>at</strong>erial cannot be deformed<br />
randomly and the elastic properties of the m<strong>at</strong>erial will determine how close the<br />
system is able to approach thermodynamical equilibrium. In most cases, large<br />
internal stresses will persist, which has lots of implic<strong>at</strong>ions for the strength of<br />
the m<strong>at</strong>erial.<br />
Wh<strong>at</strong> are the consequences for the measurement of phase diagrams?<br />
If we cool a binary alloy, with a concentr<strong>at</strong>ion corresponding to a stable st<strong>at</strong>e,<br />
from the liquid, do we get a homogeneous solid?<br />
An energy view of the decomposition<br />
If we follow the Helmholtz free energy as a function of volume <strong>at</strong> a given<br />
temper<strong>at</strong>ure below the critical temper<strong>at</strong>ure, we find a behavior as sketched in<br />
figure 3.19. The slope of the energy curve <strong>at</strong> each point gives us the pressure<br />
<strong>at</strong> th<strong>at</strong> st<strong>at</strong>e (with a minus sign). The Helmholtz energy is calcul<strong>at</strong>ed by the<br />
following integr<strong>at</strong>ion along the equ<strong>at</strong>ion of st<strong>at</strong>e:<br />
∫ V<br />
F (V, T, N) = F0(T, N) −<br />
V0<br />
pdV (3.50)<br />
where the integr<strong>at</strong>ion variable is now V (because we switched to the Helmholtz<br />
free energy).<br />
Suppose we are <strong>at</strong> a volume Vi in between points 1 and 2. In th<strong>at</strong> case the<br />
Helmholtz free energy obtained by following the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e<br />
is larger th<strong>at</strong> the value obtained by drawing a straight line between 1 and 2.<br />
Wh<strong>at</strong> does the l<strong>at</strong>ter mean? It corresponds to cre<strong>at</strong>ing a mixture of m<strong>at</strong>erials
3.6. SPINODAL DECOMPOSITION. 141<br />
F<br />
1<br />
V i<br />
Figure 3.19: Energy versus volume showing th<strong>at</strong> decomposition lowers the<br />
energy.<br />
corresponding to the st<strong>at</strong>es <strong>at</strong> points 1 and 2. The equ<strong>at</strong>ion of the straight line<br />
connecting 1 and 2 is<br />
Fs(V ) =<br />
1<br />
V2 − V1<br />
2<br />
V<br />
((V2 − V )F1 + (V − V1)F2) (3.51)<br />
which corresponds to a fraction x1 = V2−V<br />
V2−V1 of m<strong>at</strong>erial in st<strong>at</strong>e 1, and x2 =<br />
V −V1<br />
V2−V1<br />
in st<strong>at</strong>e 2. It is easy to check th<strong>at</strong> this indeed adds up to the correct<br />
volume, x1V1 + x2V2 = V . Therefore, in any point between 1 and 2 we can<br />
lower the energy of the m<strong>at</strong>erial by phase separ<strong>at</strong>ion.<br />
How do we find the loc<strong>at</strong>ion of points 1 and 2? They need to have the same<br />
slope, hence<br />
( ) ( )<br />
∂F<br />
∂F<br />
(V1) =<br />
(V2) = −pcoex<br />
∂V T,N ∂V T,N<br />
and this slope should be the slope of the connecting line:<br />
which leads to<br />
( )<br />
∂F<br />
(V1) =<br />
∂V T,N<br />
F2 − F1<br />
V2 − V1<br />
pcoex(V2 − V1) = −(F2 − F1) =<br />
∫ 2<br />
1<br />
(3.52)<br />
(3.53)<br />
pdV (3.54)<br />
which is an altern<strong>at</strong>ive formul<strong>at</strong>ion of the st<strong>at</strong>ement th<strong>at</strong> the areas are equal.<br />
This approach is due to Maxwell, and the graphical approach in figure 3.19 is<br />
called the Maxwell construction.
142 CHAPTER 3. PHASE TRANSITIONS.<br />
3.7 Generaliz<strong>at</strong>ions of the van der Waals equ<strong>at</strong>ion.<br />
The van der Waals equ<strong>at</strong>ion takes a very simple form when expressed in values<br />
of the st<strong>at</strong>e variables rel<strong>at</strong>ive to the critical values. Define<br />
P = p<br />
, V = V<br />
, T = T<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e in these variables is<br />
(<br />
P + 3<br />
V2 ) (<br />
V − 1<br />
)<br />
=<br />
3<br />
8T<br />
3<br />
pc<br />
Vc<br />
Tc<br />
(3.55)<br />
(3.56)<br />
which is independent of pc, Vc, and Tc! In rel<strong>at</strong>ive st<strong>at</strong>e variables the van der<br />
Waals equ<strong>at</strong>ion does not contain any inform<strong>at</strong>ion about the specific system and<br />
in these units the binodal coexistence curve T (V) should be the same for all<br />
m<strong>at</strong>erials. This has indeed been observed for a number of different substances<br />
like inert gases. The binodal curves for these substances in rel<strong>at</strong>ive units are<br />
almost identical. This common binodal curve differs somewh<strong>at</strong> from th<strong>at</strong> predicted<br />
by the van der Waals equ<strong>at</strong>ion, however, and corrections to the higher<br />
order virial coefficients are needed. Nevertheless, ignoring these rel<strong>at</strong>ively small<br />
differences, it is amazing how well the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e describes<br />
a liquid to gas phase transition, and predicts the universality of the spinodal<br />
curve. In order to understand the reasons for this success, one has to look <strong>at</strong><br />
the microscopic found<strong>at</strong>ion of the van der Waals equ<strong>at</strong>ion. Also, the notion<br />
th<strong>at</strong> the equ<strong>at</strong>ions of st<strong>at</strong>e in scaled units are the same for similar m<strong>at</strong>erials<br />
introduces for the first time the idea of universality. If the same physics is <strong>at</strong><br />
work in different m<strong>at</strong>erials, it is possible to scale experimental results from one<br />
to the other.<br />
Mixtures.<br />
The van der Waals equ<strong>at</strong>ion can also be extended to mixtures. In this case<br />
it loses many of its simple characteristics. For a gas mixture of M species, the<br />
excluded volumes are bi and the average interactions are aij. The number of<br />
<strong>at</strong>oms of type i is Ni. The simplest generaliz<strong>at</strong>ion of the van der Waals equ<strong>at</strong>ion<br />
for this case is<br />
⎛<br />
⎝p + 1<br />
V 2<br />
⎞ (<br />
∑<br />
M∑<br />
)<br />
aijNiNj ⎠ V − Nibi = NRT (3.57)<br />
i
3.8. EXTRA QUESTIONS. 143<br />
where the functions a and b only depend on the composition. As is turns out,<br />
this generaliz<strong>at</strong>ion is not correct from a microscopic point of view. The change<br />
in the first factor can be explained in terms of particle-particle interactions.<br />
Although the modific<strong>at</strong>ion of the second factor intuitively seems correct, a st<strong>at</strong>istical<br />
mechanical approach yields a more complic<strong>at</strong>ed expression.<br />
3.8 Extra questions.<br />
1. Wh<strong>at</strong> is the most important phase transition?<br />
2. Can we have a solid to gas transition?<br />
3. Why is G continuous?<br />
4. Why does V drop with p?<br />
5. Can V be continuous?<br />
6. Should V drop with T?<br />
7. Consider a container with w<strong>at</strong>er and vapor. Increase T. Wh<strong>at</strong> is formed?<br />
8. Wh<strong>at</strong> happens <strong>at</strong> the critical point in an experiment?<br />
9. Suppose we have two phases coexisting with p1 = p2, T1 = T2, but µ1 ̸= µ2.<br />
Wh<strong>at</strong> happens?<br />
10. Why is it easiest to discuss phase boundaries using the Gibbs free energy?<br />
11. Wh<strong>at</strong> can you tell about the slope of the coexistence curve?<br />
12. Could ice 1 and ice 2 meet by accident <strong>at</strong> the triple point?<br />
13. Apply Clausius-Clapeyron to a superconducting phase transition in B field.<br />
14. Wh<strong>at</strong> are the consequences for the slope of pco <strong>at</strong> the w<strong>at</strong>er-ice boundary?<br />
15. How do you measure phase diagrams?<br />
16. Construct phase diagram with a compound forming <strong>at</strong> low T.<br />
17. Wh<strong>at</strong> determines the curv<strong>at</strong>ure of the lines in a binary phase diagram?<br />
18. Do perfect compounds exist <strong>at</strong> low temper<strong>at</strong>ure?<br />
19. Wh<strong>at</strong> does universality of the van der Waals equ<strong>at</strong>ion mean?<br />
20. Discuss the meaning of the sign of the second virial coefficient.<br />
21. Show th<strong>at</strong> ηJT = − N<br />
Cp B2 for constant B2<br />
22. Give an example of a system which has a neg<strong>at</strong>ive pressure.
144 CHAPTER 3. PHASE TRANSITIONS.<br />
23. How did we get the unstable region in the forbidden region?<br />
24. Why can we measure the spinodal curve?<br />
25. Can the curves p(V, T1) and p(V, T2) cross?<br />
26. Explain equal areas in terms of work done.<br />
27. Why do we need the Gibbs free energy to decide in the vdW equ<strong>at</strong>ion, but<br />
use the Helmholz free energy in the Maxwell construction?<br />
3.9 Problems for chapter 3<br />
Problem 1.<br />
The equ<strong>at</strong>ion of st<strong>at</strong>e for some m<strong>at</strong>erial is given by the first two terms of a<br />
virial expansion,<br />
p<br />
RT<br />
= N<br />
V<br />
(<br />
1 + B2(T ) N<br />
)<br />
V<br />
(3.59)<br />
The second virial coefficient B2(T ) has a very simple form, B2(T ) = R (T −T0).<br />
p0<br />
(a) Make a picture of the isotherms in a p-V diagram.<br />
(b) Sketch the region of points (p,V) which correspond to unstable st<strong>at</strong>es.<br />
(c) Sketch the region of unstable st<strong>at</strong>es in the V-T diagram.<br />
(d) Are there any meta-stable st<strong>at</strong>es?<br />
Problem 2.<br />
The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is not the only form proposed to study<br />
phase transitions. Dietrici investig<strong>at</strong>ed the following expression:<br />
aN −<br />
p(V − Nb) = NRT e RT V (3.60)<br />
(a) Show th<strong>at</strong> in the limit V ≫ N this equ<strong>at</strong>ion of st<strong>at</strong>e becomes equivalent<br />
to the ideal gas law.<br />
(b) Calcul<strong>at</strong>e Vc, Tc, and pc for this equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
(c) Sketch the isotherms in a p-V plane for T > Tc, T = Tc, and T < Tc.
3.9. PROBLEMS FOR CHAPTER 3 145<br />
Problem 3.<br />
Drops of a liquid are in equilibrium with a gas of the same m<strong>at</strong>erial. This<br />
experiment is done in an environment where there is no gravity, so the drops<br />
flo<strong>at</strong> freely in the vapor. The drops are spherical because of surface tension.<br />
For a given drop, the change in energy dU is given by<br />
where A is the surface area of the drop.<br />
dU = T dS − pdV + µdN + σdA (3.61)<br />
(a) Show th<strong>at</strong> for spherical drops this rel<strong>at</strong>ion can be written in the form<br />
dU = T dS − peff dV + µdN (3.62)<br />
(b) How is the pressure inside this spherical drop rel<strong>at</strong>ed to the pressure in<br />
the vapor?<br />
The chemical potential of the liquid is µl(p, T ) and for the gas µg(p, T ).<br />
Assume th<strong>at</strong> both gas and liquid have the same temper<strong>at</strong>ure T.<br />
(c) Wh<strong>at</strong> is the equilibrium condition which determines th<strong>at</strong> in equilibrium<br />
there is no net flow of particles between the liquid and the vapor?<br />
(d) Generalize the Clausius-Clapeyron rel<strong>at</strong>ion for a system of vapor and<br />
spherical droplets, including surface tension.<br />
(e) Show th<strong>at</strong> the coexistence pressure of the vapor <strong>at</strong> a given temper<strong>at</strong>ure<br />
increases when the average radius of the droplets decreases.<br />
Problem 4.<br />
A closed container contains w<strong>at</strong>er and helium gas. Assume th<strong>at</strong> helium does<br />
not dissolve in w<strong>at</strong>er.<br />
(a) Show th<strong>at</strong> <strong>at</strong> a given pressure there is <strong>at</strong> most one temper<strong>at</strong>ure <strong>at</strong> which<br />
w<strong>at</strong>er, ice, and w<strong>at</strong>er vapor coexist.<br />
(b) The triple point for w<strong>at</strong>er is pt, Tt. Calcul<strong>at</strong>e the change in these values<br />
as a function of the Helium pressure. Assume th<strong>at</strong> the chemical potential<br />
of w<strong>at</strong>er is independent of pressure<br />
Problem 5.<br />
Show th<strong>at</strong> the l<strong>at</strong>ent he<strong>at</strong> per mole of the liquid to vapor phase transition<br />
using the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is given by
146 CHAPTER 3. PHASE TRANSITIONS.<br />
Problem 6.<br />
l1→2(T ) = RT ln<br />
( )<br />
V2 − Nb<br />
V1 − Nb<br />
(3.63)<br />
(A) Calcul<strong>at</strong>e the difference in Helmholtz free energy between a van der Waals<br />
gas and an ideal gas using F = −pdV <strong>at</strong> constant temper<strong>at</strong>ure and number<br />
of moles of m<strong>at</strong>erial. Assume th<strong>at</strong> this difference is zero in the limit<br />
V → ∞.<br />
(B) Calcul<strong>at</strong>e the difference in entropy between a van der Waals gas and an<br />
ideal gas <strong>at</strong> constant temper<strong>at</strong>ure and number of moles of m<strong>at</strong>erial. Assume<br />
th<strong>at</strong> this difference is zero in the limit V → ∞.<br />
(C) Show th<strong>at</strong> the constant ”a” in the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e leads<br />
2<br />
N to a term −a V 2 in the internal energy of a van der Waals gas. Interpret<br />
this term in a molecular picture.<br />
Problem 7.<br />
(A) For values of the temper<strong>at</strong>ure below the critical temper<strong>at</strong>ure find an upper<br />
and a lower limit for the transition pressure in a van der Waals gas.<br />
(B) For values of the temper<strong>at</strong>ure below the critical temper<strong>at</strong>ure and values<br />
of the pressure in between these two limits, the van der Waals equ<strong>at</strong>ion<br />
of st<strong>at</strong>e has two mechanically stable solutions with volumes Vl and Vh.<br />
Calcul<strong>at</strong>e the difference in Gibbs free energy between these two st<strong>at</strong>es.<br />
(C) Outline a procedure for calcul<strong>at</strong>ing the coexistence curve pco(T ) numerically.<br />
Problem 8.<br />
The boiling point of Hg is 357C <strong>at</strong> 1 <strong>at</strong>m. and 399C <strong>at</strong> 2 <strong>at</strong>m. The density<br />
of liquid mercury is 13.6 g/cm 3 . The he<strong>at</strong> of vaporiz<strong>at</strong>ion is 270 J/g. Calcul<strong>at</strong>e<br />
the order of magnitude of the volume of 200g Hg vapor <strong>at</strong> 357C and 1 <strong>at</strong>m.<br />
Note, 1 <strong>at</strong>m. is about 10 5 N/m 2 , 0C=273K, and G=U-TS+pV.<br />
Problem 9.<br />
For a certain m<strong>at</strong>erial the gas and liquid phase coexist even <strong>at</strong> 0K. Show<br />
th<strong>at</strong> the third law implies th<strong>at</strong> dpco<br />
dT = 0 <strong>at</strong> T=0.
3.9. PROBLEMS FOR CHAPTER 3 147<br />
Problem 10.<br />
A closed container of volume V0 contains N0 moles of m<strong>at</strong>erial, partly liquid,<br />
partly gas, in equilibrium. The energy is constant, the container is thermally<br />
isol<strong>at</strong>ed. The gas is ideal, with pV = NRT . The Gibbs free energy of the liquid<br />
is Gl(p, T, N) = Nµl(p, T ). The unknown variables are (1) the number of moles<br />
Ng in the gas phase, (2) the number of moles Nl in the liquid phase, (3) the<br />
volume Vg of the gas phase, and (4) the volume Vl of the liquid phase.<br />
(A) Why are the pressure and temper<strong>at</strong>ure in the gas phase the same as in<br />
the liquid phase?<br />
(B) Specify four equ<strong>at</strong>ions which we need to solve to find the four unknown<br />
quantities, given th<strong>at</strong> the pressure in the container is p and the temper<strong>at</strong>ure<br />
is T.<br />
(C) Solve this system of equ<strong>at</strong>ions for Nl<br />
(D) Is there a solution for all values of p and T?<br />
Problem 11.<br />
In the vicinity of the triple point of a m<strong>at</strong>erial the vapor pressure of the<br />
liquid is given by log( p<br />
20T0<br />
) = 10 − p0 T . The vapor pressure of the solid is given<br />
by log( p<br />
25T0<br />
) = 15 − p0 T .<br />
(A) Wh<strong>at</strong> are the temper<strong>at</strong>ure and the pressure of the triple point in terms of<br />
T0 and p0?<br />
(B) Using the Clausius-Clapeyron equ<strong>at</strong>ion, show th<strong>at</strong><br />
d log(pco)<br />
dT = l<br />
pcoT ∆v .<br />
(C) Is the volume per mole of the gas larger or smaller than th<strong>at</strong> of the liquid,<br />
assuming th<strong>at</strong> T0 and p0 are both positive?<br />
(D) Assume th<strong>at</strong> the values of the l<strong>at</strong>ent he<strong>at</strong> are independent of temper<strong>at</strong>ure<br />
near the triple point, and th<strong>at</strong> the molar volume of the gas is much larger<br />
than the molar volume of the liquid and of the solid. Calcul<strong>at</strong>e the r<strong>at</strong>ios<br />
lSL : lLG : lSG, where S,L, and G stand for solid, liquid, and gas.<br />
Problem 12.<br />
Grüneisen parameter.
148 CHAPTER 3. PHASE TRANSITIONS.<br />
In many solids we can approxim<strong>at</strong>e the entropy by S = Nf(T/TD) where<br />
the Debye temper<strong>at</strong>ure TD is a function of the molar volume only, TD = g( V<br />
N ).<br />
The Grüneisen parameter γg is defined by<br />
Show th<strong>at</strong><br />
γg = − V<br />
NTD<br />
γg =<br />
V α<br />
CV κT<br />
dTD<br />
d V<br />
N<br />
For many solids the Grüneisen parameter is almost constant. If we assume<br />
th<strong>at</strong> it is a constant, show th<strong>at</strong> we have the equ<strong>at</strong>ion of st<strong>at</strong>e<br />
P V = γgU + Nh( V<br />
N )<br />
This is the Mie-Grüneisen equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
Problem 13.<br />
Murnaghan equ<strong>at</strong>ion.<br />
The isothermal bulk modulus BT is given by BT = −V<br />
( )<br />
∂p<br />
∂V<br />
T<br />
= 1 . The<br />
κT<br />
bulk modulus can often be approxim<strong>at</strong>ed by BT (p, T ) = BT (0, T ) + βT p where<br />
βT is independent of pressure.<br />
Show th<strong>at</strong> this leads to the equ<strong>at</strong>ion of st<strong>at</strong>e<br />
p(V, T ) = BT<br />
( [V0<br />
(0, T )<br />
V<br />
βT<br />
] β<br />
T<br />
− 1<br />
where V0 is the volume <strong>at</strong> p = 0 and temper<strong>at</strong>ure T.<br />
Problem 14.<br />
Derive the virial coefficients Bj(T ) for the Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
Comment on the sign of the second virial coefficient.<br />
Problem 15.<br />
)
3.9. PROBLEMS FOR CHAPTER 3 149<br />
The Gibbs free energy of pure m<strong>at</strong>erials A and B is given by NAµA(p, T ) and<br />
NBµB(p, T ). The entropy of mixing for m<strong>at</strong>erials A and B is −NAR log( NA<br />
NA+NB )−<br />
NBR log( NB<br />
NA+NB<br />
), while the interaction energy between A and B is given by<br />
A(T, p) NANB<br />
NA+NB . Show th<strong>at</strong> a system where we mix NA moles of A and NB<br />
moles of B is stable for 2RT > A and th<strong>at</strong> there are forbidden regions when<br />
2RT < A.<br />
Problem 16.<br />
The Clausius-Clapeyron rel<strong>at</strong>ion reads dpco<br />
dT = l<br />
T (v2−v1) . The l<strong>at</strong>ent he<strong>at</strong> of<br />
fusion for ice is l = 333 kJ kg−1 , the density of w<strong>at</strong>er is 1000 kg m−3 and the<br />
density of ice is 900 kg m−3 . At normal pressure, 105 N m−2 , the melting point<br />
of ice is 273 K. A sk<strong>at</strong>er of weight 1000 N using sk<strong>at</strong>es with a contact area<br />
of 10−5 m2 sk<strong>at</strong>es well if the ice under the sk<strong>at</strong>es melts. Wh<strong>at</strong> is the lowest<br />
temper<strong>at</strong>ure in degrees centigrade <strong>at</strong> which this sk<strong>at</strong>er will be able to sk<strong>at</strong>e? Do<br />
not use a calcul<strong>at</strong>or, I have changed the numbers slightly so you can do without.<br />
Problem 17.<br />
The boiling point of Hg is 357C <strong>at</strong> 1 <strong>at</strong>m. and 399C <strong>at</strong> 2 <strong>at</strong>m. The density<br />
of liquid mercury is 13.6 g/cm 3 . The he<strong>at</strong> of vaporiz<strong>at</strong>ion is 270 J/g. Calcul<strong>at</strong>e<br />
the order of magnitude of the volume of 200g Hg vapor <strong>at</strong> 357C and 1 <strong>at</strong>m.<br />
Note, 1 <strong>at</strong>m. is about 10 5 N/m 2 , 0C=273K, and G=U-TS+pV.<br />
Problem 18.<br />
The pressure of a gas is measured and it is found th<strong>at</strong> the following equ<strong>at</strong>ion<br />
is a good model for the pressure <strong>at</strong> low densities:<br />
p = NRT<br />
V<br />
3 N<br />
− aN2 + bRT<br />
V 2 V 3<br />
with a and b positive numbers. The model is now used <strong>at</strong> all densities. This<br />
gas shows a phase transition to a liquid st<strong>at</strong>e. Find the critical temper<strong>at</strong>ure of<br />
the transition, in terms of the constants a and b and the gas constant R.<br />
Problem 19.<br />
Near a critical point in the liquid to vapor phase transition we have the following<br />
rel<strong>at</strong>ion between volume and temper<strong>at</strong>ure on the border of the forbidden<br />
region:<br />
Tc − T = A(V − Vc) 2
150 CHAPTER 3. PHASE TRANSITIONS.<br />
A gas is <strong>at</strong> a temper<strong>at</strong>ure T above Tc and a volume V close to but not<br />
equal to V0. It is cooled down. Find the temper<strong>at</strong>ure <strong>at</strong> which we see phase<br />
separ<strong>at</strong>ion, and find the fraction of liquid and vapor below this temper<strong>at</strong>ure as<br />
a function of temper<strong>at</strong>ure.<br />
Problem 20.<br />
According to the Goff-Gr<strong>at</strong>ch equ<strong>at</strong>ion (Smithsonian Tables, 1984) we have<br />
for the vapor pressure pi of w<strong>at</strong>er vapor over ice:<br />
10 log( pi<br />
p0<br />
) = −9.09718 ( Tt<br />
T − 1) − 3.56654 10 log( Tt<br />
T<br />
) + 0.876793 (1 − )<br />
T<br />
with T in [K] and pi in [hPa]. p0 is the pressure <strong>at</strong> the triple point, p0 =<br />
6.1173 hP a. Tt is the temper<strong>at</strong>ure <strong>at</strong> the triple point, Tt = 273.16 K.<br />
Calcul<strong>at</strong>e the he<strong>at</strong> of sublim<strong>at</strong>ion of ice as a function of temper<strong>at</strong>ure. Wh<strong>at</strong><br />
is the r<strong>at</strong>io of the he<strong>at</strong> of sublim<strong>at</strong>ion of ice <strong>at</strong> 0 ◦ C and −10 ◦ C?<br />
PS: 1 <strong>at</strong>m = 1013.25 hPa.<br />
Problem 21.<br />
Consider the phase change of a m<strong>at</strong>erial AcB1−c from liquid to a solid solution,<br />
th<strong>at</strong> is in the solid the m<strong>at</strong>erials can always be mixed <strong>at</strong> any concentr<strong>at</strong>ion.<br />
When <strong>at</strong> a given temper<strong>at</strong>ure and pressure there is equilibrium between liquid<br />
and solid, we need to have th<strong>at</strong> the chemical potential for each <strong>at</strong>om type is the<br />
same in the liquid as in the solid, µl = µs. The chemical potential of <strong>at</strong>om type<br />
A and <strong>at</strong>om type B do not have to be the same. The total number of A <strong>at</strong>oms,<br />
in solid and liquid combined, is given by cN, where N is the total number of<br />
<strong>at</strong>oms. Similar for B <strong>at</strong>oms. The chemical potential for <strong>at</strong>oms A in the liquid<br />
is given by µAl = gAl(p, T ) + RT log(cAl), where cAl is the concentr<strong>at</strong>ion of A<br />
<strong>at</strong>oms in the liquid, the r<strong>at</strong>io of the number of A <strong>at</strong>oms in the liquid divided<br />
by the total number of <strong>at</strong>oms in the liquid. We have a similar expression for all<br />
other three cases.<br />
The solid and the liquid can have different concentr<strong>at</strong>ions of A <strong>at</strong>oms, call<br />
these cs and cl. Find formulas for these concentr<strong>at</strong>ions, rel<strong>at</strong>ing them directly<br />
to the quantities e gAs (p,T )−gAl (p,T )<br />
RT and e gBs (p,T )−gBl (p,T )<br />
RT .<br />
Find the r<strong>at</strong>io of the number of <strong>at</strong>oms in the liquid, Nl, and the solid, Ns.<br />
Problem 22.<br />
We can model a peak in the specific he<strong>at</strong> by the following formula:<br />
CV (T, V, N) = C0(T, V, N) + NRf( n ∆<br />
)T<br />
∆2 + (T − T0) 2<br />
n0<br />
Tt
3.9. PROBLEMS FOR CHAPTER 3 151<br />
where n is the density n = N<br />
V , f(n) is a smooth function of n, n0 is a<br />
reference density, and ∆ is a small parameter, measured in Kelvin. R is the gas<br />
constant.<br />
Is there a peak in the coefficient of thermal expansion? In the compressibility?<br />
Evalu<strong>at</strong>e these quantities!
152<br />
PART II<br />
<strong>Thermodynamics</strong> Advanced<br />
Topics
Chapter 4<br />
Landau-Ginzburg theory.<br />
The discussion of phase transitions in the previous chapter was motiv<strong>at</strong>ed by experimental<br />
observ<strong>at</strong>ions. If we have an equ<strong>at</strong>ion of st<strong>at</strong>e, wh<strong>at</strong> can we conclude<br />
from its behavior when it is singular? How does it lead to phase transitions?<br />
All those observ<strong>at</strong>ions are very important, they set the stage for a discussion of<br />
phase transitions from a more fundamental perspective.<br />
The ultim<strong>at</strong>e basis for a description of phase transitions is found in st<strong>at</strong>istical<br />
mechanics, and we will need a microscopic description for th<strong>at</strong>. From such a<br />
microscopic discussion we obtain free energies. But free energies are the domain<br />
of thermodynamics, and hence we can discuss the consequences of different forms<br />
of the free energy without knowing the microscopic found<strong>at</strong>ion for them. Th<strong>at</strong><br />
is exactly wh<strong>at</strong> we will do in this chapter.<br />
The most important control variable we have for phase transitions is the<br />
temper<strong>at</strong>ure, and therefore we need the free energy as a function of temper<strong>at</strong>ure.<br />
In other words, we need the Helmholtz free energy. Landau gave a very<br />
ingenious discussion of phase transitions in terms of expansions of the Helmholtz<br />
free energy in terms of some important extensive parameter. This analysis forms<br />
the basis for a classific<strong>at</strong>ion of all possible phase transitions, and any form of<br />
the Helmholtz free energy obtained from microscopic theory can be analyzed in<br />
the terms explained in this chapter. Again, this is an example of how thermodynamics<br />
dict<strong>at</strong>es the limit<strong>at</strong>ions of wh<strong>at</strong> can happen in a general theory.<br />
In this chapter we first discuss the very important concept of an order parameter,<br />
the way we can measure the occurrence of a phase transition. Next<br />
we set up Landau’s approach to the analysis of phase transitions. We apply the<br />
theory to the two most common phase transitions, first and second order.<br />
In the final section we step away from uniform behavior and ask the question<br />
wh<strong>at</strong> happens when we make a local change in the system. How far does it<br />
extends? This leads to the introduction of a new concept, the correl<strong>at</strong>ion length.<br />
This turns out to be an extremely important parameter for a description of phase<br />
transitions. It is the macroscopic description of the effectiveness of microscopic<br />
interactions between particles and the range over which they extend. It is the<br />
parameter th<strong>at</strong> describes the effects of the microscopic many body interactions.<br />
153
154 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
It constitutes the key connection between microscopic and macroscopic theory.<br />
4.1 Introduction.<br />
Phase transition in equilibrium thermodynamics.<br />
Equilibrium thermodynamics as presented in the previous chapters gives us<br />
st<strong>at</strong>e functions and free energies. In particular, we should be able to find the<br />
Gibbs free energy G(p,T,N). In Chapter 3 we used these ideas to describe phase<br />
transitions. Suppose a system can exist in two different phases, 1 and 2, and<br />
we have a model for the Gibbs free energy in each phase. For each value of<br />
the temper<strong>at</strong>ure and pressure we can therefore determine the Gibbs free energy<br />
of each phase, and by comparing the values we can determine which phase is<br />
stable and which is meta-stable. Sometimes we do not have the explicit forms<br />
of the free energies, but have a model equ<strong>at</strong>ion of st<strong>at</strong>e (like the van der Waals<br />
equ<strong>at</strong>ion of st<strong>at</strong>e) which allows us to rel<strong>at</strong>e the free energies of the two phases to<br />
each other. Since only the difference in free energy is important, this approach<br />
is equivalent to the one sketched first.<br />
Wh<strong>at</strong> happens <strong>at</strong> the transition?<br />
The approach outlined above is very useful, since it gives us equ<strong>at</strong>ions like the<br />
Clausius-Clapeyron equ<strong>at</strong>ion which tell us how phase boundaries change. This<br />
approach is an either/or approach, however. It allows us to determine which<br />
phase is stable, but it does not say anything about the transition. For example,<br />
the he<strong>at</strong> capacity as a function of temper<strong>at</strong>ure in the van der Waals model looks<br />
like figure 4.1, while experimentally it looks much more like figure 4.2. There<br />
is a critical region in which the he<strong>at</strong> capacity is very large (and perhaps even<br />
divergent). The width of this region could be 10K or 1mK, depending on the<br />
m<strong>at</strong>erial.<br />
Fluctu<strong>at</strong>ions.<br />
The next question is wh<strong>at</strong> is missing from our simple equilibrium thermodynamics<br />
model? Anyone who has seen w<strong>at</strong>er boil knows the answer. There are<br />
fluctu<strong>at</strong>ions in the system. Even below the boiling point vapor bubbles appear<br />
in w<strong>at</strong>er. These instabilities live only a short time, but when we get closer to<br />
the boiling point, the vapor bubbles become larger and live longer. To describe<br />
wh<strong>at</strong> is happening is very difficult, and we need a non-equilibrium theory. It<br />
is very important in practical applic<strong>at</strong>ions, though. Any prepar<strong>at</strong>ion technique<br />
of a m<strong>at</strong>erial th<strong>at</strong> involves a phase transition has to deal with this question,<br />
and understanding and being able to change the mechanism could be worth<br />
millions of dollars for industry. In this chapter we describe a theory developed
4.1. INTRODUCTION. 155<br />
Figure 4.1: He<strong>at</strong> capacity across the phase transition in the van der Waals<br />
model.<br />
Figure 4.2: He<strong>at</strong> capacity across the phase transition in an experiment.
156 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
by Landau and Ginzburg, which is a first approach <strong>at</strong> incorpor<strong>at</strong>ing the effects<br />
of fluctu<strong>at</strong>ions.<br />
4.2 Order parameters.<br />
How do we bring order or organiz<strong>at</strong>ion into chaos?<br />
How do we describe a phase transition?<br />
Phase transitions are observed by doing measurements. The question is,<br />
wh<strong>at</strong> do we measure? The phase transition from w<strong>at</strong>er to ice is easy to observe<br />
experimentally. The properties of w<strong>at</strong>er and ice are very different and we rely<br />
on the fact th<strong>at</strong> these properties have a range of values to distinguish between<br />
ice and w<strong>at</strong>er. Since snow is also frozen w<strong>at</strong>er, but different from ice, we also<br />
know th<strong>at</strong> not all properties of ice are characteristic for frozen w<strong>at</strong>er. We need<br />
to single out one or more parameters which will really tell us which phase we are<br />
dealing with. In a solid the rel<strong>at</strong>ive positions of the molecules are fixed (within<br />
small vibr<strong>at</strong>ions) but in a liquid the molecules are able to move freely. In a liquid<br />
the motion of the molecules is locally correl<strong>at</strong>ed, and the local neighborhood of<br />
a molecule in a liquid is similar to th<strong>at</strong> of a molecule in a solid. There is no<br />
long range order, though. In a gas, even the local correl<strong>at</strong>ion has disappeared,<br />
and the molecules are almost independent.<br />
Requirements for characterizing parameter.<br />
If we want to describe a phase transition in m<strong>at</strong>hem<strong>at</strong>ical terms, we need to<br />
find a parameter which uniquely defines which phase we are in. Therefore, this<br />
parameter has to be a st<strong>at</strong>e variable. In a gas to liquid transition we used the<br />
volume to distinguish between the low and high density phases, but this choice<br />
has some problems. If we use a st<strong>at</strong>e variable to distinguish between different<br />
phases of a system, it has to be measurable. Hence, it must be possible to use<br />
this st<strong>at</strong>e variable to do work. But we also require th<strong>at</strong> a measurement of this<br />
st<strong>at</strong>e variable does not change the phase of the system, hence in principle is<br />
should be measured with a corresponding field which approaches zero. In th<strong>at</strong><br />
case the st<strong>at</strong>e variable we use to describe the phase of the system is considered<br />
an internal parameter. It also has to be an extensive parameter. The volume in<br />
a gas to liquid transform<strong>at</strong>ion could play this role, but the ideal gas law in the<br />
limit p → 0 becomes singular. To describe a gas one needs a non-zero pressure,<br />
and the work done by changing the volume becomes an essential part of the<br />
physics of a gas to liquid phase transition.
4.2. ORDER PARAMETERS. 157<br />
Order parameter.<br />
A st<strong>at</strong>e variable which is used to describe the phase of a system is called<br />
an order parameter. The difference in molar volume between the phase under<br />
consider<strong>at</strong>ion and the gas phase is an example of an order parameter. This order<br />
parameter is zero in the gas phase, but non-zero in the liquid phase. Another<br />
standard example is the magnetic moment ⃗ M of a solid. We often use the molar<br />
density ⃗ M<br />
N in stead. This order parameter is zero in the non-magnetic phase,<br />
but non-zero in the ferro-magnetic phase. There is always a generalized force<br />
associ<strong>at</strong>ed with the order parameter, the corresponding intensive st<strong>at</strong>e variable.<br />
For the liquid to gas transition this is the pressure p and for the magnetic system<br />
it is the magnetic field ⃗ H. This generalized force is wh<strong>at</strong> we need to apply in<br />
order to be able to measure the value of the order parameter.<br />
Conserved or non-conserved order parameters.<br />
Some people make a distinction between two types of order parameters. If<br />
we study a ferromagnetic m<strong>at</strong>erial as a function of temper<strong>at</strong>ure <strong>at</strong> zero applied<br />
magnetic field, we find th<strong>at</strong> below the critical temper<strong>at</strong>ure the magnetiz<strong>at</strong>ion<br />
has a non-zero value. Hence we have an extensive st<strong>at</strong>e variable changing its<br />
value, its ”amount” is not conserved. On the other hand, if we study a fixed<br />
volume of gas and cool it down, we see liquid appear. We have two phases with<br />
different densities, but the total density remains constant, by construction. In<br />
this case the order parameter is called conserved. But it is clear th<strong>at</strong> this is only<br />
a m<strong>at</strong>ter of choosing st<strong>at</strong>e variables. If we cool a gas <strong>at</strong> constant pressure we<br />
are in the same situ<strong>at</strong>ion as the ferromagnet, and the volume is not conserved.<br />
Hence the distinction between two types of order parameters is not fundamental,<br />
but has, of course, very important practical consequences!<br />
Measuring the order parameter.<br />
The external parameters used to determine the st<strong>at</strong>e a solid are often the<br />
temper<strong>at</strong>ure T and the pressure p. For the following discussion, let us assume<br />
th<strong>at</strong> we work <strong>at</strong> <strong>at</strong>mospheric pressure, which is essentially zero for a solid. The<br />
appropri<strong>at</strong>e free energy for a system <strong>at</strong> constant temper<strong>at</strong>ure is the Helmholtz<br />
energy F, which in our case only depends on the st<strong>at</strong>e variables T (determined<br />
by the outside world) and ⃗ M (used to determine the st<strong>at</strong>e of the system). The<br />
Helmholtz free energy of the thermodynamic equilibrium st<strong>at</strong>e is found by minimizing<br />
F as a function of ⃗ M. This is consistent with the following observ<strong>at</strong>ion.<br />
In the process of measuring the order parameter we need to consider it as an<br />
external st<strong>at</strong>e variable. In our case the measurement consists of applying a field<br />
⃗H and the appropri<strong>at</strong>e free energy in th<strong>at</strong> case is the Gibbs like free energy<br />
G(T, ⃗ H) = U − T S − ⃗ M · ⃗ H (4.1)
158 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.3: Continuity of phase transition around critical point in p-T plane.<br />
Measuring the work done on the sample <strong>at</strong> constant temper<strong>at</strong>ure as a function<br />
of field gives this energy, and the partial deriv<strong>at</strong>ive with respect to ⃗ H in the<br />
limit ⃗ H → ⃗0 gives us ⃗ M as required. When we consider the magnetic moment<br />
as an external parameter, it is rel<strong>at</strong>ed to the external field via<br />
( )<br />
∂F<br />
∂ ⃗ M<br />
T<br />
= ⃗ H (4.2)<br />
and this partial deriv<strong>at</strong>ive is indeed zero in the limit ⃗ H → ⃗0. In conclusion, the<br />
free energy used to describe the system and the free energy used to describe the<br />
measurement of the order parameter differ by a simple Legendre transform<strong>at</strong>ion.<br />
Phase transitions with critical point.<br />
In the phase diagram for a PVT-system we find two different types of behavior.<br />
The transition from solid to gas or liquid is described by a coexistence curve<br />
which probably extends to infinity. There is no critical point. The liquid to gas<br />
coexistence curve, on the other hand, ends in a critical point. If we start <strong>at</strong> a<br />
st<strong>at</strong>e of the system <strong>at</strong> a high temper<strong>at</strong>ure, and cool the system <strong>at</strong> low pressure,<br />
we will end up in the gas phase. The same procedure <strong>at</strong> high pressure gives a<br />
liquid. Hence from any given initial st<strong>at</strong>e one can reach a st<strong>at</strong>e just above or<br />
below the coexistence curve in a continuous manner. This is shown in figure 4.3.<br />
One way of defining the order of a phase transition is using this continuity. A<br />
phase transition is called second order if it is possible to find a continuous p<strong>at</strong>h<br />
around a critical point from one phase to the other, and it is called first order<br />
if this is not the case. Note, however, th<strong>at</strong> other definitions of the order of a<br />
phase transition are in use, too.<br />
Branch cuts.
4.2. ORDER PARAMETERS. 159<br />
Figure 4.4: Continuity of phase around singular point.<br />
The coexistence curve plays a role which is very similar to th<strong>at</strong> of a branch<br />
cut in a complex function. The function √ z, for example, shows a similar<br />
picture which a line of discontinuities, see figure 4.4. From any initial value<br />
of z it is possible to go to a point just below or just above the branch cut<br />
in a continuous manner. The branch cut represents a line of discontinuities.<br />
The critical point in the p-T diagram is comparable to the essential singularity<br />
which ends a branch cut. Just like √ z cannot be described by a power series <strong>at</strong><br />
z = 0, the function V(p,T) has an essential singularity <strong>at</strong> pc, Tc. This approach<br />
has been generalized. All we need to know is the exact behavior of V(p,T) <strong>at</strong><br />
the singularity in order to predict wh<strong>at</strong> is happening near this point. This is<br />
very much like the complex analysis of a singular complex function. A magnetic<br />
phase transition is characterized in the same way by a real singular point Tc, Hc.<br />
If the system is symmetric for the change M ⇔ −M, it follows th<strong>at</strong> Hc = 0 and<br />
the singular behavior in the H-T plane is like in figure 4.5.<br />
Wh<strong>at</strong> conclusions can you draw from the similarity between the m<strong>at</strong>hem<strong>at</strong>ical<br />
forms of the free energy and a branch cut?<br />
Symmetry in system.<br />
When the external field coupling to the order parameter is zero, one often<br />
expects the free energy to be symmetric as a function of the order parameter.<br />
This is not true in general, since a vanishing external field only requires th<strong>at</strong> the<br />
first order deriv<strong>at</strong>ive is zero. But if we consider this simple symmetric case the<br />
free energy of the system without field is the same for a given value of the order<br />
parameter and for the opposite value. The most symmetric situ<strong>at</strong>ion therefore
160 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.5: Continuity of phase transition around critical point in H-T<br />
plane.<br />
will have a value of zero for the order parameter. This does often, but not<br />
necessarily, correspond to the st<strong>at</strong>e of the system <strong>at</strong> high temper<strong>at</strong>ures. For<br />
example, ⃗ M = ⃗0 <strong>at</strong> high temper<strong>at</strong>ures. There is no preferred direction of the<br />
magnetiz<strong>at</strong>ion. When we apply a magnetic field we will have in general th<strong>at</strong><br />
⃗M∥ ⃗ H.<br />
No singularities in finite system.<br />
If the equilibrium st<strong>at</strong>e of the system has a non-zero value of the order<br />
parameter, there are <strong>at</strong> least two st<strong>at</strong>es for the system with the same energy,<br />
corresponding to a certain value of the order parameter and to the opposite<br />
value. Therefore in th<strong>at</strong> case the ground st<strong>at</strong>e is degener<strong>at</strong>e. Although in<br />
principle the st<strong>at</strong>e of the system can fluctu<strong>at</strong>e between these degener<strong>at</strong>e ground<br />
st<strong>at</strong>es, the time involved is in practice infinitely long since macroscopic variables<br />
are involved. Quantum mechanical tunnelling is highly improbable in th<strong>at</strong> case.<br />
But we also need to remember th<strong>at</strong> we have to take the thermodynamic limit<br />
N → ∞. In th<strong>at</strong> case the tunnelling time becomes infinitely long, and the<br />
system is found in one st<strong>at</strong>e only. There is a m<strong>at</strong>hem<strong>at</strong>ical st<strong>at</strong>ement th<strong>at</strong> says<br />
th<strong>at</strong> for a finite system there are no singularities in st<strong>at</strong>e functions! Hence sharp<br />
transitions and critical points only occur in infinite systems!<br />
Wh<strong>at</strong> do phase transitions look like in finite systems?<br />
When is a finite system infinite for all practical purposes?
4.2. ORDER PARAMETERS. 161<br />
Broken symmetry.<br />
At the coexistence curve the system chooses one of the possible ground st<strong>at</strong>es,<br />
and breaks the symmetry between the possible values of the order parameter by<br />
choosing a particular one. In a scenario like we sketched here a phase transition<br />
is rel<strong>at</strong>ed to spontaneously broken symmetry in a degener<strong>at</strong>e ground st<strong>at</strong>e. Of<br />
course, the external field used to measure the order parameter, will always<br />
break the symmetry. In th<strong>at</strong> case the broken symmetry is forced by external<br />
constraints.<br />
Supercooling and superhe<strong>at</strong>ing.<br />
If we follow a complex function like √ z in a continuous manner through a<br />
branch cut, we end up on the wrong Riemann sheet. In the case of √ z this<br />
simply means th<strong>at</strong> we end up with the wrong sign, since there are only two<br />
Riemann sheets. If we go continuously through a coexistence curve, we end<br />
up in a meta-stable st<strong>at</strong>e. W<strong>at</strong>er vapor, for example, can be super-s<strong>at</strong>ur<strong>at</strong>ed.<br />
Introducing so-called seeds will then allow droplets of liquid to form around<br />
these seeds, and will force the transition to the stable liquid st<strong>at</strong>e.<br />
Could a free energy have more than two Riemann sheets, and if so, can they be<br />
observed?<br />
Essential singularity.<br />
A branch cut of a complex function end in an essential singularity, like the<br />
point z=0 for the function √ z. Second order phase transitions end in an essential<br />
singularity <strong>at</strong> the critical point. M<strong>at</strong>hem<strong>at</strong>ically, the idea of a critical point and<br />
an essential singularity are closely rel<strong>at</strong>ed.<br />
Model systems.<br />
The gas to liquid phase transition is not easy to describe. The problem is<br />
th<strong>at</strong> a gas has a non-zero volume, and th<strong>at</strong> a non-zero pressure exists, needed<br />
for thermal equilibrium. The driving force is non-zero! Therefore we need to<br />
consider quantities like the difference in volume and the difference in pressure.<br />
A symmetric magnetic phase transition is easier to describe. In one phase we<br />
have M=0, which occurs <strong>at</strong> H=0, and in the other phase we have a finite value<br />
of the magnitude M. Choosing M as the order parameter is therefore equivalent<br />
to choosing half the difference between the volumes in the liquid to gas system<br />
as the order parameter. Since magnetic systems are easier to model, we will use<br />
these systems as examples in the rest of this chapter. A model of a liquid to gas
162 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
transition, however, can be done in a similar manner, and only the m<strong>at</strong>hem<strong>at</strong>ical<br />
description becomes more elabor<strong>at</strong>e.<br />
4.3 Landau theory of phase transitions.<br />
Energy is the basic quantity.<br />
Landau devised a very simple method to describe phase transitions. It is<br />
similar in spirit to the approach of van der Waals, but on a more fundamental<br />
level. Landau recognized th<strong>at</strong> the free energy is the important st<strong>at</strong>e function to<br />
parameterize, since all other rel<strong>at</strong>ions follow from a knowledge of the free energy.<br />
The route van der Waals followed is more empirical. The van der Waals equ<strong>at</strong>ion<br />
of st<strong>at</strong>e works very well for a liquid-gas transition, but is hard to generalize to<br />
other phase transitions. Landau suggested to parameterize the free energy as a<br />
function of the order parameter.<br />
Magnetic systems are good working example.<br />
We will use a magnetic system as an example. We will assume th<strong>at</strong> the<br />
system has rot<strong>at</strong>ional symmetry, and th<strong>at</strong> the free energy does not depend on<br />
the direction of the magnetiz<strong>at</strong>ion ⃗ M, but only on its value M, apart from the<br />
obvious minus sign th<strong>at</strong> reflects up or down. In this discussion we therefore<br />
allow M to be neg<strong>at</strong>ive, which is necessary in order to show the degeneracy of<br />
the ground st<strong>at</strong>e. We assume th<strong>at</strong> we use the temper<strong>at</strong>ure T as an independent<br />
variable, and th<strong>at</strong> the volume V and the amount of m<strong>at</strong>erial N are constant.<br />
Therefore, we need to work with the Helmholtz free energy.<br />
The applied magnetic field follows from ( )<br />
∂F<br />
∂M = H and since we plan<br />
T,V,N<br />
to discuss the stable st<strong>at</strong>e <strong>at</strong> zero applied field, this partial deriv<strong>at</strong>ive is zero. We<br />
now switch our point of view, and consider M to be an internal parameter. The<br />
condition mentioned before is now consistent with the fact th<strong>at</strong> the Helmholtz<br />
free is a minimum as a function of the internal parameters. Since V and N<br />
are constant, we will drop these variables from the discussion, unless explicitly<br />
needed.<br />
Model Helmholtz energy.<br />
Because of the symmetry we imposed we have F (T, M) = F (T, −M) and an<br />
expansion of the free energy F only contains even powers of M. Because of this<br />
symmetry the st<strong>at</strong>e with M=0 always corresponds to zero applied external field.<br />
This st<strong>at</strong>e is therefore an extremum, but could be a maximum. If the minimum<br />
of the free energy occurs <strong>at</strong> a non-zero value of M, there will always be two<br />
minima, +M and -M. These are st<strong>at</strong>es with a spontaneous magnetiz<strong>at</strong>ion and<br />
give rise to the broken symmetry discussed before.
4.3. LANDAU THEORY OF PHASE TRANSITIONS. 163<br />
Landau studied the following form for the free energy:<br />
f(T, m) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
4 c(T )m4 + 1<br />
6 d(T )m6 + · · · (4.3)<br />
, as a function<br />
of the average magnetiz<strong>at</strong>ion density, m = M<br />
V . The advantage of a formul<strong>at</strong>ion<br />
in terms of energy and magnetiz<strong>at</strong>ion densities is th<strong>at</strong> it does not depend on<br />
the size of the system. The coefficients a(T), etc. are intensive quantities. They<br />
take the form as expected in the thermodynamic limit, and are not dependent<br />
on N and V (but could depend on the density n = N<br />
V or equivalently on the<br />
pressure p, which dependence we will ignore here for simplicity).<br />
This specifies the Helmholtz free energy per unit volume, f = F<br />
V<br />
Validity of the expansion.<br />
One problem comes to mind <strong>at</strong> once. Landau assumed th<strong>at</strong> the order parameter<br />
is the same for the whole system. When the system changes phase,<br />
however, there is a coexistence of both phases and the order parameter depends<br />
strongly on position and time. This short-coming of the theory is not essential,<br />
and the improved theory of Ginzburg and Landau takes fluctu<strong>at</strong>ions into<br />
account, as we will discuss l<strong>at</strong>er. But, as we discussed before, <strong>at</strong> Tc there is an<br />
essential singularity, and a power series is not appropri<strong>at</strong>e, because it fails to<br />
converge! Hence Landau’s theory in not able to describe exactly wh<strong>at</strong> is<br />
happening for the phase transition <strong>at</strong> a critical point! In m<strong>at</strong>hem<strong>at</strong>ical<br />
terms, <strong>at</strong> the critical point the power series has a radius of convergence equal<br />
to zero.<br />
Value of the theory.<br />
Landau’s theory is very vers<strong>at</strong>ile for classifying all kinds of phase transitions<br />
by comparing the st<strong>at</strong>es <strong>at</strong> both sides of the transition. Keep in mind<br />
th<strong>at</strong> Landau’s theory was developed in 1936, before many models were solved<br />
analytically. By assuming very general functional forms for the coefficients<br />
a(T),b(T),· · ·, only a few basic types of phase transitions appear. Experimentalists<br />
use Landau theory very often to interpret their d<strong>at</strong>a, because it is a<br />
good macroscopic, first approach. It is therefore a good meeting ground for theory<br />
and experiment. Complex theoretical calcul<strong>at</strong>ions should try to derive the<br />
functional forms of the coefficients in Landau’s expansion, while experimentalists<br />
can measure them. This gives an excellent framework to discuss agreements<br />
and conflicts.<br />
The use of this classific<strong>at</strong>ion of phase transitions has been very helpful in<br />
understanding phase transitions, and is done l<strong>at</strong>er more rigorously in renormaliz<strong>at</strong>ion<br />
group theory. It also is another hint <strong>at</strong> universality, phase transitions<br />
with a similar analytical form of the coefficients behave essentially in the same<br />
way. This forms a basis for scaling theory, which is discussed in the next chapter.
164 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
4.4 Case one: a second order phase transition.<br />
One of the basic phase transitions is found when we assume th<strong>at</strong> b(T ) = b0(T −<br />
Tc) with b0 > 0, and c(T ) = c > 0, d(T ) = d > 0, with all higher order<br />
coefficients vanishing. Hence we study wh<strong>at</strong> happens when the first coefficient<br />
changes sign. The last coefficient, d, has to be positive or else the minimum free<br />
energy would occur <strong>at</strong> infinite magnetiz<strong>at</strong>ion. The partial deriv<strong>at</strong>ives are:<br />
( )<br />
∂f<br />
∂m T<br />
= b0(T − Tc)m + cm 3 + dm 5<br />
(4.4)<br />
( 2 ∂ f<br />
∂m2 )<br />
= b0(T − Tc) + 3cm 2 + 5dm 4<br />
(4.5)<br />
The slope is zero if either<br />
T<br />
m = 0 (4.6)<br />
b0(T − Tc) + cm 2 + dm 4 = 0 (4.7)<br />
In the first case this only corresponds to a minimum in the free energy if T > Tc,<br />
otherwise the curv<strong>at</strong>ure of the free energy 4.5 would be neg<strong>at</strong>ive. If a solution<br />
for m exists in the second case, it always corresponds to a local minimum since<br />
the curv<strong>at</strong>ure 4.5 in th<strong>at</strong> case is 2cm 2 +4dm 4 , which is always positive. In order<br />
to see if this st<strong>at</strong>e is the equilibrium st<strong>at</strong>e we have to compare the free energies.<br />
We have<br />
f(T, m) − f(T, 0) = 1<br />
2 b0(T − Tc)m 2 + 1<br />
4 cm4 + 1<br />
6 dm6 =<br />
1<br />
2 ⟨b0(T − Tc) + cm 2 + dm 4 ⟩m 2 − 1<br />
4 cm4 − 1<br />
3 dm6<br />
(4.8)<br />
and this is always neg<strong>at</strong>ive in the second case. Hence if 4.7 has a solution, this<br />
solution corresponds to the equilibrium st<strong>at</strong>e.<br />
General non-zero solution.<br />
The general solution of 4.7 is<br />
m 2 = 1<br />
(<br />
−c ±<br />
2d<br />
√ c2 )<br />
− 4db0(T − Tc)<br />
(4.9)<br />
If T > Tc both roots are neg<strong>at</strong>ive (remember c > 0) and m is not real. This is<br />
not a physical solution and hence for T > Tc the equilibrium st<strong>at</strong>e of the system<br />
has m = 0. For T < Tc, only one of the roots is positive and we find<br />
m 2 = 1<br />
(√ )<br />
c2 + 4db0(Tc − T ) − c<br />
2d<br />
(4.10)
4.4. CASE ONE: A SECOND ORDER PHASE TRANSITION. 165<br />
or<br />
Figure 4.6: Magnetiz<strong>at</strong>ion versus temper<strong>at</strong>ure.<br />
m 2 = c<br />
(√<br />
2d<br />
1 + 4db0<br />
c2 (Tc<br />
)<br />
− T ) − 1<br />
(4.11)<br />
In the limit T ↑ Tc, we can use √ 1 + x−1 ≈ 1<br />
2x, and the values for the magnetic<br />
moment are<br />
m = ±<br />
√ b0<br />
c<br />
√ Tc − T (4.12)<br />
which is shown in figure 4.6. Experiments show a very similar behavior, except<br />
th<strong>at</strong> the general form is<br />
m ∝ (Tc − T ) β<br />
(4.13)<br />
where typical values of β are around 0.33 in stead of 0.5 found in our Landau<br />
model. Therefore the magnetiz<strong>at</strong>ion is qualit<strong>at</strong>ively correct in the Landau model<br />
(for example, it goes to zero without a jump), but the exponent is wrong. For<br />
such quantit<strong>at</strong>ive details it is clearly important to know wh<strong>at</strong> happens <strong>at</strong> the<br />
phase transition in terms of fluctu<strong>at</strong>ions, and Landau’s theory is too simplistic,<br />
as mentioned before, because it ignores fluctu<strong>at</strong>ions .<br />
Free energy pictures.<br />
A picture of the Helmholtz free energy as a function of m looks like figure<br />
4.7. The analysis of these curves is easy. In the limit of infinite magnetiz<strong>at</strong>ion<br />
the curves go to positive infinity. At m=0 the curves have either a maximum<br />
or a minimum. There are two cases. For T > Tc there is only one extremum,<br />
and this has to be the point m=0, which therefore has to be a minimum. For
166 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.7: Forms of the Helmholtz free energy.<br />
T < Tc there are three extrema, one <strong>at</strong> zero, one positive, and one neg<strong>at</strong>ive.<br />
Therefore, the point m=0 has to be a maximum, and the minima are <strong>at</strong> the<br />
symmetric non-zero points.<br />
Since we are dealing with a fifth order equ<strong>at</strong>ion, could we have the sequence<br />
minimum, maximum, minimum, maximum, minimum for the extrema?<br />
Response functions.<br />
One of the important response function for our simple T system is the he<strong>at</strong><br />
capacity. The he<strong>at</strong> capacity is rel<strong>at</strong>ed to the partial deriv<strong>at</strong>ive of the entropy.<br />
Since volume in our model is not a st<strong>at</strong>e variable which can be varied, there is<br />
only one he<strong>at</strong> capacity and it is rel<strong>at</strong>ed to the free energy via:<br />
C = −V T d2 f<br />
dT 2<br />
(4.14)<br />
Since temper<strong>at</strong>ure is the only external st<strong>at</strong>e variable, there are no partial deriv<strong>at</strong>ives.<br />
The free energy is<br />
T > Tc : f(T ) = a(T ) (4.15)<br />
T < Tc : f(T ) = a(T ) + 1<br />
2 b0(T − Tc)m 2 + 1<br />
4 cm4 + 1<br />
6 dm6<br />
(4.16)<br />
Near Tc both the second and the third term in 4.16 are proportional to (T −Tc) 2 .<br />
Near Tc the higher order terms do not play a role, since they are proportional<br />
to larger powers of T − Tc. Hence f is approxim<strong>at</strong>ed by
4.4. CASE ONE: A SECOND ORDER PHASE TRANSITION. 167<br />
Figure 4.8: Entropy near the critical temper<strong>at</strong>ure.<br />
T ↑ Tc : f(T ) ≈ a(T ) − 1<br />
4<br />
The entropy per unit volume, s = S<br />
V<br />
= − df<br />
dT<br />
b2 0<br />
2<br />
(T − Tc)<br />
c<br />
and follows from<br />
(4.17)<br />
T > Tc : s(T ) = − da<br />
(T ) (4.18)<br />
dT<br />
T < Tc : s(T ) = − da<br />
dT (T ) + b2 0<br />
2c (T − Tc) (4.19)<br />
which has the behavior as shown in figure 4.8. The specific he<strong>at</strong> per unit volume<br />
and is given by<br />
follows from c = C<br />
V<br />
= T ds<br />
dT<br />
T > Tc : c(T ) = −T d2a (T ) (4.20)<br />
dT 2<br />
T < Tc : c(T ) = −T d2a dT 2 (T ) + T b20 (4.21)<br />
2c<br />
which is shown in figure 4.9. This is the same type of behavior as found in the<br />
van der Waals model, where the he<strong>at</strong> capacity also jumps. There is no divergent<br />
behavior, however, as seen in many experiments. This is again rel<strong>at</strong>ed to the<br />
fact th<strong>at</strong> we did not include fluctu<strong>at</strong>ions. We have a homogeneous system with<br />
m independent of time and position.<br />
Rel<strong>at</strong>ions across Tc.<br />
The Landau model rel<strong>at</strong>es the free energy below and above the transition<br />
temper<strong>at</strong>ure, and leads to two distinct analytical forms in these regions, with<br />
discontinuities. This is similar to the van der Waals model of a phase transition.
168 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.9: Specific he<strong>at</strong> near the critical temper<strong>at</strong>ure.<br />
The difference, however, is in the origin of the energy rel<strong>at</strong>ion. In stead of<br />
modelling an equ<strong>at</strong>ion of st<strong>at</strong>e, we have a model for the Helmholtz free energy<br />
th<strong>at</strong> is valid everywhere, but selects a different minimum in the different regions.<br />
This approach can be connected to more detailed models in an easier way, since<br />
it is more fundamental.<br />
The term rel<strong>at</strong>ed to a(T) is always present and gives a contribution to s<br />
and C which is continuous <strong>at</strong> the phase transition. The second term in the free<br />
energy for the ordered phase leads to a discontinuity in the slope of s, and the<br />
he<strong>at</strong> capacity jumps by an amount V Tc b2<br />
0<br />
2c . The entropy has an additional term<br />
b 2<br />
0<br />
2c (T − Tc) for temper<strong>at</strong>ures below Tc. Hence the entropy in the ordered st<strong>at</strong>e<br />
is less than the entropy in the st<strong>at</strong>e with m = 0. This is in agreement with our<br />
intuitive notion of entropy as a measure of the internal randomness of a system.<br />
The ordered st<strong>at</strong>e has lower entropy.<br />
Since the entropy is lower in the ordered st<strong>at</strong>e, why does the maximum entropy<br />
principle not imply th<strong>at</strong> the disordered st<strong>at</strong>e should be the stable st<strong>at</strong>e?<br />
Susceptibility.<br />
Very often it is easier to measure the magnetic susceptibility in stead of the<br />
magnetiz<strong>at</strong>ion density itself. This is due to the fact th<strong>at</strong> it is often easier to do a<br />
differential measurement. By definition χT (T, M) = ( )<br />
∂M<br />
∂H and χ−1<br />
T T (T, M) =<br />
)<br />
. Hence we find th<strong>at</strong><br />
( ∂ 2 F<br />
∂M 2<br />
T<br />
χ −1<br />
T (T, M) = V −1 ( b0(T − Tc) + 3cm 2 + 5dm 4)<br />
(4.22)
4.5. ORDER OF A PHASE TRANSITION. 169<br />
Figure 4.10: Magnetic susceptibility near the critical temper<strong>at</strong>ure.<br />
If we now use the solutions for m(T) we find<br />
T > Tc : χT (T ) =<br />
V<br />
b0(T − Tc)<br />
V<br />
T < Tc : χT (T ) =<br />
2b0(Tc − T )<br />
(4.23)<br />
(4.24)<br />
This response function is sketched in the figure 4.10. Note th<strong>at</strong> the susceptibility<br />
diverges <strong>at</strong> the critical point. The n<strong>at</strong>ure of the divergence is the same <strong>at</strong> each<br />
side of the transition, here a power of minus one in both cases. The exponent<br />
is the same. This is a general characteristic which we will discuss more in the<br />
next chapter. The pre-factors are different, though, and in this case the low<br />
temper<strong>at</strong>ure form has an extra factor of two.<br />
4.5 Order of a phase transition.<br />
One of the basic ways to distinguish phase transitions is by looking <strong>at</strong> the<br />
change in order parameter <strong>at</strong> the critical temper<strong>at</strong>ure. If the value of the order<br />
parameter jumps from one value to another, and hence is discontinuous, the<br />
phase transition is called first order. If this is not the case, the phase transition<br />
is called of second order. It is actually possible to even further distinguish the<br />
order of phase transitions, although perhaps not very useful. One way is to<br />
consider the functions ( ∂ n F<br />
∂T n<br />
)<br />
as a function of temper<strong>at</strong>ure near Tc. The order<br />
of the phase transition is then given by the lowest value of n for which this<br />
partial deriv<strong>at</strong>ive is discontinuous. Our previous example would be of second<br />
order in th<strong>at</strong> case.
170 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
4.6 Second case: first order transition.<br />
General observ<strong>at</strong>ions.<br />
In the general form of the Landau expansion, equ<strong>at</strong>ion 4.3, for a symmetric<br />
model, we can typically assume th<strong>at</strong> the lower order terms are the most important<br />
and have the most important temper<strong>at</strong>ure dependence. The term a(T) is<br />
always present and gives a continuous background contribution. If all the other<br />
terms are positive, then the st<strong>at</strong>e with m=0 is the only st<strong>at</strong>e corresponding to<br />
a minimum of the free energy. Therefore, in order to see some interesting behavior,<br />
<strong>at</strong> least on term should change sign as a function of temper<strong>at</strong>ure, or be<br />
neg<strong>at</strong>ive <strong>at</strong> all temper<strong>at</strong>ures. The easiest case is when the term b(T) changes<br />
sign. Here we study the next important case. Other fun cases are tre<strong>at</strong>ed in the<br />
problems.<br />
Neg<strong>at</strong>ive sign in fourth order term.<br />
This very interesting case is obtained when we assume th<strong>at</strong> b(T ) > 0, c(T ) =<br />
c < 0, d(T ) = d > 0, and all other coefficients are zero. We also assume th<strong>at</strong><br />
the dominant temper<strong>at</strong>ure dependence is in the b(T) term, and have indeed set<br />
the higher order terms to be constant. In order to see interesting behavior like<br />
normally observed, we assume th<strong>at</strong> db<br />
dT<br />
> 0. Since b(T ) > 0 the point m = 0 will<br />
always correspond to a local minimum of the Helmholtz free energy. It might,<br />
however, not be the global minimum. Because c < 0 it is possible th<strong>at</strong> the free<br />
energy curve bends down and goes through a maximum as a function of m. In<br />
th<strong>at</strong> case there has to be another minimum in the free energy as a function of<br />
m, since d > 0 implies th<strong>at</strong> for very large values of m the free energy approaches<br />
plus infinity. Three possible forms of the Helmholtz free energy as a function of<br />
m are shown in figure 4.11.<br />
Why don’t we have the sequence minimum, maximum, minimum in the free<br />
energy as a function of the order parameter?<br />
We again construct the partial deriv<strong>at</strong>ives of the Helmholtz free energy:<br />
( )<br />
∂f<br />
= b(T )m + cm<br />
∂m T<br />
3 + dm 5<br />
(4.25)<br />
( 2 ∂ f<br />
∂m2 )<br />
= b(T ) + 3cm 2 + 5dm 4<br />
(4.26)<br />
The solutions for m are<br />
T<br />
m = 0 (4.27)
4.6. SECOND CASE: FIRST ORDER TRANSITION. 171<br />
Figure 4.11: Three possible forms of the Helmholtz free energy in case 2.<br />
m 2 = −c ± √ c 2 − 4b(T )d<br />
2d<br />
(4.28)<br />
Hence if 4b(T )d > c2 only m = 0 is a solution. In th<strong>at</strong> case the fourth order<br />
term in the expansion of the Helmholtz free energy is not strong enough to pull<br />
the function down to a second minimum. In the other case we find two extrema.<br />
Note th<strong>at</strong> in this case the two possible values for m2 are always real, and hence<br />
the values of m are always real. The argument in the previous paragraph shows<br />
th<strong>at</strong> the extremum closest to m2 = 0 corresponds to a maximum, and the other<br />
one to a minimum. In order to obtain a value for m pertaining to th<strong>at</strong> second<br />
minimum we need to take the plus sign in 4.28. A picture of the solutions for m2 is given in figure 4.12, where the temper<strong>at</strong>ure T0 follows from 4b(T0)d = c2 . The<br />
values for m2 <strong>at</strong> this temper<strong>at</strong>ure are m2 (T0) = −c<br />
2d . Since b(T) is increasing<br />
with temper<strong>at</strong>ure, the non-zero solution only plays a role for T < T0.<br />
Energy difference.<br />
The important question, when there are more minima, is to determine which<br />
st<strong>at</strong>e corresponds to a global minimum. The difference in free energy is again<br />
f(T, m) − f(T, 0) = 1<br />
2 b0(T − Tc)m 2 + 1<br />
4 cm4 + 1<br />
6 dm6 =<br />
1<br />
2 ⟨b0(T − Tc) + cm 2 + dm 4 ⟩m 2 − 1<br />
4 cm4 − 1<br />
3 dm6 (4.29)<br />
and the term in brackets is zero <strong>at</strong> a minimum. Since c < 0 the difference in<br />
free energy can either be positive or neg<strong>at</strong>ive and the sign of the difference is<br />
determined by the sign of c + 4<br />
3dm2 . If the value of m2 for the minimum is<br />
small, the term with c will domin<strong>at</strong>e and the free energy for the m = 0 solution
172 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.12: Values for m corresponding to a minimum in the free energy.<br />
is the lowest. Substituting the solution for m in the difference of the free energy<br />
values we find<br />
∆F = − 1<br />
(<br />
m4 c + 2<br />
12 √ c2 )<br />
− 4b(T )d<br />
(4.30)<br />
The function b(T) is an increasing function of temper<strong>at</strong>ure, and for large values<br />
of the temper<strong>at</strong>ure (but always keeping 4b(T )d < c 2 ) the term in brackets is<br />
determined mainly by c and the difference in the free energy is positive (remember<br />
th<strong>at</strong> c is neg<strong>at</strong>ive). In th<strong>at</strong> case m = 0 is the stable st<strong>at</strong>e. For small<br />
values of the temper<strong>at</strong>ure, if b(T) becomes small enough, the term in brackets<br />
will change sign and the st<strong>at</strong>e with non-zero magnetic moment is stable.<br />
Critical temper<strong>at</strong>ure.<br />
Since we assume th<strong>at</strong> b(T) is monotonous, there will only be one value of<br />
the temper<strong>at</strong>ure for which the difference in free energy is zero. This critical<br />
temper<strong>at</strong>ure is determined by<br />
or<br />
c + 2 √ c 2 − 4b(T )d = 0 (4.31)<br />
b(Tc) = 3c2<br />
16d<br />
(4.32)<br />
At the phase transition the value of the order parameter jumps from zero to mc<br />
which is given by<br />
mc =<br />
√<br />
− 3c<br />
4d<br />
(4.33)
4.6. SECOND CASE: FIRST ORDER TRANSITION. 173<br />
Figure 4.13: Magnetiz<strong>at</strong>ion as a function of temper<strong>at</strong>ure.<br />
This transition is therefore of first order.<br />
Analysis of the transition.<br />
The complete plot of the value of the magnetiz<strong>at</strong>ion density as a function of<br />
temper<strong>at</strong>ure is given in figure 4.13. There are some important conclusions to<br />
be drawn. If we start <strong>at</strong> low temper<strong>at</strong>ure in a magnetized st<strong>at</strong>e and he<strong>at</strong> the<br />
system slowly, we will follow the meta-stable magnetic st<strong>at</strong>e even above Tc if<br />
we do things very carefully. This is because in order to go to the stable nonmagnetic<br />
st<strong>at</strong>e the system first has to increase its energy. There is an energy<br />
barrier. This is very typical for a first order phase transition. A system in this<br />
case can be superhe<strong>at</strong>ed or under-cooled into a meta-stable st<strong>at</strong>e. The further<br />
we are in the meta-stable region, the easier it is to perturb the system and force<br />
it to go to the stable st<strong>at</strong>e. The detailed mechanism of such phase transitions is<br />
often described in the framework of nucle<strong>at</strong>ion and growth. One needs to know<br />
how small seeds of the stable phase can form through fluctu<strong>at</strong>ions and how they<br />
can expand by growing. Clearly, in a solid this is a much slower process than<br />
in a liquid, since the motion of <strong>at</strong>oms in a solid is very slow.<br />
Hysteresis.<br />
When we perform a real experiment in which we bring a system through a<br />
first order phase transition we always do the experiment <strong>at</strong> a measurable speed,<br />
never reversibly slow. Therefore, we will always bring the system into the metastable<br />
st<strong>at</strong>e before it transforms to the stable st<strong>at</strong>e. This is called hysteresis,<br />
see figure 4.14. The amount by which we can bring the system into the metastable<br />
st<strong>at</strong>e depends on the m<strong>at</strong>erial we study. If nucle<strong>at</strong>ion and growth is<br />
slow, this can be quite far. Typically, nucle<strong>at</strong>ion is a thermal process with
174 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.14: Hysteresis loop.<br />
probability inversely proportional to a Boltzmann factor, or exponential in the<br />
energy difference between the st<strong>at</strong>es. Therefore, the probability of transforming<br />
becomes exponentially small near the critical temper<strong>at</strong>ure in such a model. As<br />
a consequence, reducing the speed of the process by a factor of two does not<br />
reduce the width of the hysteresis loop by a factor of two, but only by a small<br />
amount. This is of gre<strong>at</strong> importance in many applic<strong>at</strong>ions rel<strong>at</strong>ed to magnetism<br />
or phase transitions in crystal structures.<br />
Wh<strong>at</strong> do you expect to happen for hysteresis loops for very small samples?<br />
Discontinuities.<br />
In the first order phase transition discussed here the order parameter m<br />
jumps in value <strong>at</strong> the critical point. Since we switch from one minimum to another<br />
in the free energy curve, the change in free energy with temper<strong>at</strong>ure will<br />
in general be different in the two st<strong>at</strong>es. Hence df<br />
dT<br />
will also be discontinuous<br />
<strong>at</strong> the critical points, and hence the entropy jumps in value. This is completely<br />
different in a second order phase transition, where both m and S are continuous.<br />
Since m is continuous, there is also no hysteresis in a second order phase<br />
transition.<br />
Susceptibility.<br />
The susceptibility is obtained from χ −1<br />
(<br />
2<br />
∂ F<br />
T (T, M) = ∂M 2<br />
)<br />
T<br />
and is:
4.6. SECOND CASE: FIRST ORDER TRANSITION. 175<br />
T < Tc : χT (T ) =<br />
T > Tc : χT (T ) = V<br />
b(T )<br />
V<br />
b(T ) + 3cm2 V<br />
=<br />
+ 5dm4 2m2 (T ) (c + 2dm2 (T ))<br />
(4.34)<br />
(4.35)<br />
Since b(T) is positive and increasing, the susceptibility above Tc is decreasing.<br />
Also 4.28 shows th<strong>at</strong> m(T) is a decreasing function of T, and hence the susceptibility<br />
is an increasing function of temper<strong>at</strong>ure below Tc. Note th<strong>at</strong> c + 2dm 2<br />
is always positive.<br />
Entropy.<br />
( )<br />
∂f<br />
∂m<br />
( ) T<br />
∂f<br />
∂m<br />
T<br />
(<br />
∂f<br />
The entropy as a function of temper<strong>at</strong>ure follows from s = −<br />
dm<br />
dT<br />
∂T<br />
)<br />
m −<br />
. The deriv<strong>at</strong>ives of m versus temper<strong>at</strong>ure do not play a role, since<br />
= 0. Hence we find<br />
T > Tc : S(T ) = − da<br />
dT<br />
T < Tc : S(T ) = − da<br />
dT<br />
(4.36)<br />
1 db<br />
−<br />
2 dT m2 (T ) (4.37)<br />
Since b(T) is an increasing function of T, the entropy in the ordered st<strong>at</strong>e is<br />
again lower than the entropy in the disordered st<strong>at</strong>e, as expected. This is not<br />
in disagreement with the st<strong>at</strong>ement in chapter one, where we concluded th<strong>at</strong><br />
the entropy has to be a maximum, since th<strong>at</strong> st<strong>at</strong>ement pertains to a situ<strong>at</strong>ion<br />
of constant internal energy, and here the internal energy is different in the two<br />
st<strong>at</strong>es. If we assume th<strong>at</strong> b is linear in T, the he<strong>at</strong> capacity in the ordered<br />
phase differs from the he<strong>at</strong> capacity in the m = 0 phase by a term − db dm<br />
dT m(T ) dT<br />
and this is a positive term which does not diverge <strong>at</strong> Tc. The behavior of these<br />
quantities near Tc is sketched in figure 4.15.<br />
Basic difference in behavior.<br />
We have studied the description of a first order phase transition and of a<br />
second order phase transition. The stable st<strong>at</strong>e of the system is associ<strong>at</strong>ed with<br />
the global minimum in the free energy. The position of such a minimum depends<br />
on an external control variable, like the temper<strong>at</strong>ure. If as a function of<br />
the temper<strong>at</strong>ure the global minimum changes position by an interchange in a<br />
global-local minimum pair, the transition is first order. On the other hand, in<br />
a second order phase transition the global minimum becomes a maximum and<br />
in this process two minima are cre<strong>at</strong>ed on either side. This gives an important<br />
difference, which we have to keep in mind when we construct models for
176 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Figure 4.15: Critical behavior in first order phase transition.<br />
phase transitions. In a first order transition both st<strong>at</strong>es are available <strong>at</strong> either<br />
side of the transition, but in a second order transition new st<strong>at</strong>es are cre<strong>at</strong>ed.<br />
Therefore, hysteresis never occurs in a second order phase transition!<br />
Combined behavior.<br />
In real m<strong>at</strong>erials we see often combined types of behavior. For example,<br />
a transition could be second order in temper<strong>at</strong>ure like in our magnetic case.<br />
If we now work <strong>at</strong> a temper<strong>at</strong>ure below the critical temper<strong>at</strong>ure and apply a<br />
magnetic field, we see a first order transition in which the magnetiz<strong>at</strong>ion flips<br />
from one direction to the opposite direction. But this first order transition is<br />
clearly rel<strong>at</strong>ed to wh<strong>at</strong> happens <strong>at</strong> the critical point. In fact, the Landau model<br />
for a range around the critical point is given by the model in our first case. It<br />
makes therefore more sense to distinguish continuous phase transitions (anding<br />
in a critical point) from discontinuous transitions, and model the behavior<br />
accordingly.<br />
4.7 Fluctu<strong>at</strong>ions and Ginzburg-Landau theory.<br />
Fluctu<strong>at</strong>ions near a phase transition.<br />
The phenomenological theory of Landau has been very useful in classifying<br />
the order and n<strong>at</strong>ure of phase transitions. But the form 4.3 of the free energy<br />
is incomplete near a phase transition. We know th<strong>at</strong> near a phase transition<br />
fluctu<strong>at</strong>ions are important. For example, if we w<strong>at</strong>ch w<strong>at</strong>er boil, we see th<strong>at</strong><br />
vapor bubbles form near the bottom of a pan. Gravity will send these bubbles<br />
up. Since inside these bubbles the pressure is too low, they will collapse. Closer
4.7. FLUCTUATIONS AND GINZBURG-LANDAU THEORY. 177<br />
to the boiling point we will see both more bubbles forming, and larger bubbles<br />
forming. Especially near the critical point, one finds very large bubbles!<br />
How to deal with these?<br />
In order to study fluctu<strong>at</strong>ions we need non-equilibrium thermodynamics.<br />
We need to study a system th<strong>at</strong> is not uniform in space, and also changes over<br />
time. This is a difficult task. Ginzburg, however, found a very good first step<br />
to include sp<strong>at</strong>ial non-uniformity. He started from Landau’s model. He still<br />
assumed th<strong>at</strong> the problem was st<strong>at</strong>ic, but allowed for sp<strong>at</strong>ial vari<strong>at</strong>ions in the<br />
order parameter. In the approach studied here we do not discuss the origin of<br />
the driving forces towards non-uniformity, we simply assume th<strong>at</strong> they exist. We<br />
also ignore surface tension, which certainly plays a role, and simply assume th<strong>at</strong><br />
the effects of surface tension are incorpor<strong>at</strong>ed in the values of the parameters in<br />
our model.<br />
Divide and conquer.<br />
To understand the origin of our problem, we divide the volume of the total<br />
system in a large number of much smaller systems with a much smaller volume.<br />
On one hand, the number of small sub-systems is very large, but these subsystems<br />
themselves are still large enough to be considered to be macroscopic,<br />
e.g. they contain many <strong>at</strong>oms. All these small sub-systems are in thermal<br />
contact. All sub-systems are in thermal equilibrium. The total system is in<br />
thermal equilibrium and the average of the internal energy of each subsystem<br />
is a fixed value and the same everywhere. But the internal energy for a small<br />
sub-system may vary. The larger the devi<strong>at</strong>ion from the average value, the<br />
more unlikely the fluctu<strong>at</strong>ion. In this sense we consider the total system to be<br />
an ensemble of small systems in thermal contact.<br />
Wh<strong>at</strong> is a vapor bubble containing one <strong>at</strong>om?<br />
Position dependent order parameter.<br />
If the system is far away from a phase transition, all sub-systems will be in<br />
the same phase. If the total system is very close to a phase transition, about<br />
half of the sub-systems will have st<strong>at</strong>e variables which correspond to a metastable<br />
thermodynamical st<strong>at</strong>e. These sub-systems try to change phase until<br />
they have exchanged enough energy with the neighboring sub-systems, in which<br />
case they correspond to a stable st<strong>at</strong>e again. In terms of the order parameter<br />
the following description arises. A subsystem <strong>at</strong> position ⃗r has its own order<br />
parameter m(⃗r). The average value of the order parameter, ∫ m(⃗r)d 3 r, is equal
178 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
to the order parameter M for the total system and can be calcul<strong>at</strong>ed from the<br />
values of the st<strong>at</strong>e variables for the total system. Near a phase transition, there<br />
are large fluctu<strong>at</strong>ions in m(⃗r), however. Because our sub-systems contain many<br />
<strong>at</strong>oms, this implies th<strong>at</strong> vari<strong>at</strong>ions in m(⃗r) are on a length scale much larger<br />
than the inter-<strong>at</strong>omic distance, but possibly small compared to the sample size.<br />
Note th<strong>at</strong> for a homogeneous system we have m(⃗r) = m and M=mV indeed.<br />
How does this description of the order parameter compare with the definition<br />
of fields like ⃗ E(⃗r) in electromagnetism?<br />
Energy consider<strong>at</strong>ions.<br />
Ginzburg and Landau improved the phenomenological theory discussed before<br />
by including these fluctu<strong>at</strong>ions. They assumed th<strong>at</strong> the st<strong>at</strong>es of neighboring<br />
sub-systems are close together and th<strong>at</strong> m(⃗r) varies slowly. In th<strong>at</strong> case the<br />
fluctu<strong>at</strong>ions are taken into account by adding terms in the free energy which<br />
contain deriv<strong>at</strong>ives of m(⃗r). Precisely <strong>at</strong> the phase transition this will still be incorrect,<br />
since the order parameter or its slope varies in a discontinuous manner,<br />
and m(⃗r) is not analytical. An expansion of the free energy in terms of gradients<br />
of the order parameter is not valid near Tc and the improved Ginzburg-Landau<br />
theory is still not strictly valid near Tc. Nevertheless, it is a very useful theory,<br />
because it will be a good description of the st<strong>at</strong>e of a system much closer to the<br />
critical temper<strong>at</strong>ure than the previous Landau theory.<br />
Formal theory.<br />
The Helmholtz free energy is in Ginzburg-Landau theory given by an integr<strong>at</strong>ion<br />
over all space. As an example, we will extend our model of a second<br />
order phase transition to include fluctu<strong>at</strong>ions. In th<strong>at</strong> case we write<br />
∫<br />
F (T, m(⃗r)) =<br />
d 3 (<br />
r a(T ) + 1<br />
2 b0(T − Tc)m 2 (⃗r) + 1<br />
4 cm4 (⃗r) + 1<br />
6 dm6 )<br />
(⃗r)<br />
+ 1<br />
2 f<br />
∫<br />
d 3 <br />
<br />
r ⃗ <br />
<br />
∇m(⃗r)<br />
2<br />
(4.38)<br />
with b0, c, d, and f all positive. When the magnetic moment is homogeneous,<br />
m(⃗r) = M<br />
V , this expression is the same as we studied before, except for a redefinition<br />
of the coefficients a, b, c, and d by factors of powers of the volume. The<br />
last term in 4.38 represents fluctu<strong>at</strong>ions in the order parameter. The coefficient<br />
f has to be positive, else the system would be unstable against fluctu<strong>at</strong>ions.<br />
With a positive value of f we see th<strong>at</strong> it costs energy to cre<strong>at</strong>e a fluctu<strong>at</strong>ion, as
4.7. FLUCTUATIONS AND GINZBURG-LANDAU THEORY. 179<br />
expected. Assuming th<strong>at</strong> we have an isotropic system, the free energy cannot<br />
depend on the direction of ⃗ ∇m, and the form 4.38 is the simplest choice we can<br />
make.<br />
Functional deriv<strong>at</strong>ives derived.<br />
How does the Helmholtz free energy 4.38 change when the order parameter<br />
changes by a small amount δm(⃗r)? We need to calcul<strong>at</strong>e δF (T, m(⃗r)) =<br />
F (T, m(⃗r) + δm(⃗r)) − F (T, m(⃗r)). If the change in the order parameter is small,<br />
we only need to keep first order terms, and obtain<br />
∫<br />
δF (T, m(⃗r)) = d 3 rδm(⃗r) ( b0(T − Tc)m(⃗r) + cm 3 (⃗r) + dm 5 (⃗r) )<br />
∫<br />
+f d 3 r⃗ ∇m(⃗r) · ⃗ ∇δm(⃗r) (4.39)<br />
The last term is equal to<br />
∫<br />
−f d 3 rδm(⃗r) ⃗ ∇ 2 <br />
m(⃗r) + f d 2 σδm(⃗r)ˆn(⃗r) · ⃗ ∇m(⃗r) (4.40)<br />
The surface integral vanishes for a finite sample, since in th<strong>at</strong> case m(⃗r) = 0<br />
outside the sample. This integral also vanishes if we employ periodic boundary<br />
conditions. The fluctu<strong>at</strong>ions in the free energy are therefore:<br />
∫<br />
δF (T, m(⃗r)) =<br />
d 3 (<br />
rδm(⃗r) b0(T − Tc)m(⃗r) + cm 3 (⃗r) + dm 5 (⃗r) − f ⃗ ∇ 2 )<br />
m(⃗r)<br />
Position dependent magnetic field.<br />
(4.41)<br />
If we consider the magnetiz<strong>at</strong>ion as a external st<strong>at</strong>e variable, the magnetic<br />
field needed to cause a magnetiz<strong>at</strong>ion m(⃗r) is found by generalizing H = ( )<br />
∂F<br />
∂M T<br />
to a functional deriv<strong>at</strong>ive<br />
( )<br />
δF<br />
h(⃗r) =<br />
(4.42)<br />
δm T<br />
which is equivalent to<br />
and which leads to<br />
∫<br />
δF = d 3 rδm(⃗r)h(⃗r) (4.43)<br />
h(⃗r) = b0(T − Tc)m(⃗r) + cm 3 (⃗r) + dm 5 (⃗r) − f ⃗ ∇ 2 m(⃗r) (4.44)
180 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
by construction from equ<strong>at</strong>ion 4.41. This equ<strong>at</strong>ion is a typical differential equ<strong>at</strong>ion.<br />
The term with the Laplace oper<strong>at</strong>or occurs in many different places. The<br />
difference with other often studied second order differential equ<strong>at</strong>ions is the fact<br />
th<strong>at</strong> equ<strong>at</strong>ion 4.41 is non-linear.<br />
Spontaneous solutions.<br />
A spontaneous magnetiz<strong>at</strong>ion is possible when this equ<strong>at</strong>ion has solutions<br />
for h(⃗r) = 0, and this gives us exactly the situ<strong>at</strong>ion we discussed before. When<br />
h(⃗r) = 0 fluctu<strong>at</strong>ions will increase the free energy (this follows directly from<br />
equ<strong>at</strong>ion 4.38) and the ground st<strong>at</strong>e corresponds to a homogeneous magnetiz<strong>at</strong>ion<br />
m(⃗r) = M<br />
V . Equ<strong>at</strong>ion 4.44 then reduces to the form we solved before,<br />
leading to the same second order phase transition.<br />
Response to a driving force.<br />
The easiest way to study fluctu<strong>at</strong>ions is to consider m(⃗r) as an external<br />
st<strong>at</strong>e variable and ask the question how the system responds to certain kinds<br />
of external fields h(⃗r). This is like in resonance situ<strong>at</strong>ions, where we ask which<br />
driving force gives the largest response. For example, an interesting situ<strong>at</strong>ion<br />
arises when we use h(⃗r) = h0δ(⃗r). The system is locally perturbed and one<br />
would like to know how far away the effects of this perturb<strong>at</strong>ion can be seen. In<br />
terms of the sub-systems we introduced before, this means th<strong>at</strong> we change the<br />
st<strong>at</strong>e of one sub-system away from the average by applying a external magnetic<br />
field, and we want to know how the other sub-systems respond. The use of a<br />
Dirac delta-function is m<strong>at</strong>hem<strong>at</strong>ically justified if we assume th<strong>at</strong> it is possible<br />
to divide the large total system in a very large number of sub-systems. One<br />
can think of this by considering a delta function in the limit of a very narrow<br />
Gaussian, for example.<br />
Wh<strong>at</strong> could be the cause of such a small external force?<br />
Small external force.<br />
We will assume th<strong>at</strong> h0 is small, and th<strong>at</strong> the resulting magnetiz<strong>at</strong>ion m(⃗r)<br />
is close to the value without an external field, m(⃗r, T ) = m0(T ) + ϕ(⃗r, T ). The<br />
function ϕ will also be small. We are interested in changes in the magnetiz<strong>at</strong>ion<br />
near the critical point, when the magnetiz<strong>at</strong>ion itself is small. To simplify the<br />
m<strong>at</strong>hem<strong>at</strong>ics we will therefore ignore the sixth order term, which is small in this<br />
region. In other words, we set d = 0. We then know th<strong>at</strong><br />
0 = b0(T − Tc)m0(T ) + cm 3 0(T ) (4.45)
4.7. FLUCTUATIONS AND GINZBURG-LANDAU THEORY. 181<br />
√ b0(Tc−T )<br />
which has the solutions m0(T ) = 0 for T > Tc and m0(T ) = c for<br />
T < Tc. This equ<strong>at</strong>ion is subtracted from the equ<strong>at</strong>ion 4.44 determining m(⃗r, T )<br />
and only terms linear in ϕ are retained. Note th<strong>at</strong> m<strong>at</strong>hem<strong>at</strong>ically it is possible<br />
to have solutions with large values of ϕ, essentially given by cϕ3 − f ⃗ ∇2ϕ = 0,<br />
but such solutions also imply fast vari<strong>at</strong>ions of ϕ which is physically not possible<br />
because of the <strong>at</strong>omic structure of m<strong>at</strong>ter. As a result we get<br />
h0δ(⃗r) = b0(T − Tc)ϕ(⃗r, T ) + c3m 2 0(T )ϕ(⃗r, T ) − f ⃗ ∇ 2 ϕ(⃗r, T ) (4.46)<br />
The second term on the right is different above and below Tc. Using the solution<br />
for m0(T ) gives us<br />
T > Tc : ⃗ ∇ 2 ϕ(⃗r, T ) − b0(T − Tc)<br />
f<br />
T < Tc : ⃗ ∇ 2 ϕ(⃗r, T ) + 2 b0(T − Tc)<br />
f<br />
Solution away from origin.<br />
ϕ(⃗r, T ) = − h0<br />
δ(⃗r) (4.47)<br />
f<br />
ϕ(⃗r, T ) = − h0<br />
δ(⃗r) (4.48)<br />
f<br />
First we consider the differential equ<strong>at</strong>ion <strong>at</strong> all points away from the origin.<br />
In th<strong>at</strong> case the equ<strong>at</strong>ion has the form:<br />
⃗∇ 2 ϕ + αϕ = − h0<br />
δ(⃗r) = 0 (4.49)<br />
f<br />
with α < 0, which can be solved easily. The solutions are of the form<br />
1<br />
r ul(r)Ylm(Ω) (4.50)<br />
because the system has rot<strong>at</strong>ional symmetry. Since fluctu<strong>at</strong>ions increase the<br />
energy, the energy will be the lowest when ϕ has spherical symmetry, and hence<br />
we need to exclude all terms but the l=0 term. Therefore, ϕ = u(r)<br />
r (the factor<br />
√<br />
4π from the spherical harmonic is now included in the radial function u) and<br />
we need to solve<br />
u ′′ + αu = 0 (4.51)<br />
which has the solution u = e ±ı√αr . Since the total response has to be calcul<strong>at</strong>ed<br />
from ∫ ϕd3r we discard the increasing solution and we get ϕ = A<br />
r<br />
r<br />
e− ξ , where<br />
ξ √ α = ı.<br />
Effects of the origin.<br />
The effects of the delta function singularity <strong>at</strong> the origin are taken into<br />
account by integr<strong>at</strong>ing the equ<strong>at</strong>ion
182 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
⃗∇ 2 ϕ + αϕ = − h0<br />
δ(⃗r) (4.52)<br />
f<br />
over a sphere of radius R around the origin, and then take the limit R → 0.<br />
Th<strong>at</strong> leaves us with<br />
The first integral is<br />
∫<br />
r
4.8. EXTRA QUESTIONS. 183<br />
The perturb<strong>at</strong>ion is also proportional to the local magnetic field, as expected<br />
in our simple calcul<strong>at</strong>ion where we only included linear terms. For small fluctu<strong>at</strong>ions<br />
this is certainly justified. If the external field is zero, the response is<br />
zero, as expected.<br />
The perturb<strong>at</strong>ion is inversely proportional to f, the factor scaling the energy<br />
rel<strong>at</strong>ed to fluctu<strong>at</strong>ions. When f is large, fluctu<strong>at</strong>ions in the magnetiz<strong>at</strong>ion cost a<br />
lot of energy and are more unlikely. ϕ is indeed small in th<strong>at</strong> case. Amazingly,<br />
the correl<strong>at</strong>ion length is proportional to √ f. If in a system f is large, fluctu<strong>at</strong>ions<br />
are small, but they have a very large range! On the other hand, if f approaches<br />
zero, all fluctu<strong>at</strong>ions are essentially localized since ξ approaches zero, but the<br />
fluctu<strong>at</strong>ions themselves can be very large. In other words, when it is easy to<br />
cause a perturb<strong>at</strong>ion, this perturb<strong>at</strong>ion is also screened rapidly because the<br />
system can also gener<strong>at</strong>e a response easily.<br />
Note th<strong>at</strong> in the Ginzburg-Landau theory we merely explain wh<strong>at</strong> happens to<br />
a system when a fluctu<strong>at</strong>ion occurs. We do not explain why fluctu<strong>at</strong>ions occur,<br />
and how instabilities might arise. We only discovered a theory th<strong>at</strong> shows why<br />
fluctu<strong>at</strong>ions are screened and how we can measure th<strong>at</strong> screening.<br />
Divergence <strong>at</strong> Tc.<br />
The most important fe<strong>at</strong>ure of the equ<strong>at</strong>ions for the correl<strong>at</strong>ion length is<br />
the fact th<strong>at</strong> in the limit T → Tc the correl<strong>at</strong>ion length diverges. Near a phase<br />
transition any small change in the magnetiz<strong>at</strong>ion <strong>at</strong> one point will be noticeable<br />
all over the sample. This is an example of the inherent instability of a system<br />
<strong>at</strong> a phase transition. Note th<strong>at</strong> the n<strong>at</strong>ure of the divergence is the same when<br />
we approach the critical temper<strong>at</strong>ure from below or from above. Both go like<br />
the square root of the temper<strong>at</strong>ure difference. The pre-factors are not the same,<br />
however, and in our case they differ by a factor of √ 2.<br />
4.8 Extra questions.<br />
1. Why do we work with the Helmholtz free energy and not Gibbs?<br />
2. Why is there a peak in the he<strong>at</strong> capacity?<br />
3. Give examples of order parameters.<br />
4. How do you measure the value of the order parameter?<br />
5. Why are there no phase transitions in a finite system?<br />
6. Discuss the rel<strong>at</strong>ion between correl<strong>at</strong>ion length versus system size.<br />
7. Wh<strong>at</strong> is broken symmetry?<br />
8. To get f(t,m) from F(T,M,N), wh<strong>at</strong> did we use?<br />
9. When do power series converge?
184 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
10. Why does Landau’s theory imply some kind of universality?<br />
11. Why is Landau more fundamental than using an EOS?<br />
12. Work out the results for a second order transition with d = 0<br />
13. Would a theory with b(T ) > 0 and c(T ) changing sign give a phase transition?<br />
14. Work out results for b = 0, c changing sign, d positive.<br />
15. Why do we go from something of the form (T,m) to (T), and wh<strong>at</strong> happened<br />
with the additional degree of freedom?<br />
16. Using the free energy pictures, analyze the n<strong>at</strong>ure of phase transitions.<br />
17. Is hysteresis equilibrium?<br />
18. Can we under-cool a gas?<br />
19. If magnetic phase transitions are second order, why do we see hysteresis<br />
loops?<br />
20. Why is the gradient term the lowest order and the only second order term?<br />
21. Wh<strong>at</strong> are functional deriv<strong>at</strong>ives?<br />
22. Wh<strong>at</strong> does it mean th<strong>at</strong> the correl<strong>at</strong>ion length diverges <strong>at</strong> the critical temper<strong>at</strong>ure?<br />
23. Interpret the meaning of formula 4.56<br />
24. Do correl<strong>at</strong>ions decay <strong>at</strong> the critical temper<strong>at</strong>ure?<br />
25. Is it significant th<strong>at</strong> the scaling is the same <strong>at</strong> both sides of the critical<br />
temper<strong>at</strong>ure.<br />
4.9 Problems for chapter 4<br />
Problem 1.<br />
Suppose th<strong>at</strong> in the expansion for the Helmholtz free energy per unit volume<br />
f = F<br />
M<br />
V as a function of the magnetiz<strong>at</strong>ion density m = V a third order term<br />
is present. This is possible in systems consisting of large anisotropic molecules.<br />
The free energy in this case is<br />
f(m, T ) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
3 c(T )m3 + 1<br />
d(T )m4<br />
4<br />
Assume th<strong>at</strong> b(T ) = b0 ∗ (T − T0) and b0 > 0, c(T ) = c < 0, and d(T ) = d > 0.
4.9. PROBLEMS FOR CHAPTER 4 185<br />
(a) Show th<strong>at</strong> this system exhibits a first-order phase transition <strong>at</strong> a temper<strong>at</strong>ure<br />
Tc.<br />
(b) Calcul<strong>at</strong>e Tc in terms of b0, T0, c, and d.<br />
(c) Calcul<strong>at</strong>e lim m(T ) .<br />
T ↑Tc<br />
Problem 2.<br />
In this chapter we have only considered the temper<strong>at</strong>ure as a st<strong>at</strong>e variable<br />
driving a phase transition. Suppose the pressure is also important. The<br />
Helmholtz free energy per unit volume is then given by<br />
f(m, p, T ) = a(p, T ) + 1<br />
2 b(p, T )m2 + 1<br />
4 c(p, T )m4 + 1<br />
d(P, T )m6<br />
6<br />
where we ignored all higher order terms. Assume th<strong>at</strong> d(p, T ) = d > 0.<br />
(a) If c(p, T ) = c > 0, give an expression which determines Tc as a function<br />
of pressure for a second order phase transition.<br />
(b) If c(p, T ) = c < 0, b(p, T ) > 0, and b(p,T) increases as a function of temper<strong>at</strong>ure,<br />
give an expression which determines Tc as a function of pressure<br />
for a first order phase transition.<br />
(c) Suppose b(p, T ) = b0p(T − T0) and c(p, T ) = c0T (p − p0), with b0 > 0 and<br />
c0 > 0. Sketch in a p-T diagram the function Tc(p) for both the first order<br />
and the second order phase transition.<br />
(d) Show th<strong>at</strong> the two curves in part (c) connect <strong>at</strong> one point (called a<br />
tri-critical point) and th<strong>at</strong> the deriv<strong>at</strong>ives of the curves are the same in<br />
this tri-critical point.<br />
Problem 3.<br />
The Helmholtz free energy per unit volume for a two dimensional system is<br />
given by<br />
f(mx, my, T ) = a(T ) + 1<br />
2 b(T )(m2 x + m 2 y) + 1<br />
4 c(T )m2 xm 2 y + 1<br />
4 d(T )(m4 x + m 4 y)<br />
Assume th<strong>at</strong> b(T ) = b0(T − Tc) with b0 > 0, c(T ) = c > 0 and d(T ) = d > 0.<br />
(a) Show th<strong>at</strong> this system has a second order phase transition <strong>at</strong> Tc.
186 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
(b) If c = 2d, show th<strong>at</strong> m(T) is like we derived in the text for a second order<br />
phase transition, but th<strong>at</strong> mx and my cannot be determined apart from<br />
the value for m =<br />
√<br />
m 2 x + m 2 y.<br />
(c) If c ̸= 2d, show th<strong>at</strong> mx = my.<br />
Problem 4.<br />
Discuss wh<strong>at</strong> happens in a Landau theory with f(T ; m) = a(T ) + 1<br />
2b0(T −<br />
T0)m2 + 1<br />
4cm4 + 1<br />
6m6 and b0 > 0 , c < 0, and d > 0.<br />
Problem 5.<br />
Discuss wh<strong>at</strong> happens in a Landau theory with f(T ; m) = a(T ) + 1<br />
4 c0(T −<br />
T0)m 4 + 1<br />
6 m6 and c0 > 0 and d > 0. Calcul<strong>at</strong>e the response functions.<br />
Problem 6.<br />
Consider Ginzburg-Landau in d dimensions. Assume again a small perturb<strong>at</strong>ion<br />
of the form h(⃗r) = h0δ(⃗r) and look for a spherical solution. Show th<strong>at</strong><br />
the solutions for ϕ decay exponentially and th<strong>at</strong> for small r ϕ diverges like r 2−d .<br />
Problem 7.<br />
Calcul<strong>at</strong>e the l<strong>at</strong>ent he<strong>at</strong> for the first order phase transition discussed in the<br />
text.<br />
Problem 8.<br />
A system is characterized by a three dimensional order parameter ⃗ S. The<br />
Landau expansion is<br />
f(T ; ⃗ S) = a(T ) + 1<br />
2 b0(T − T0)S 2 + 1<br />
2 c0(T − 1<br />
2 T0)(ˆn · ⃗ S) 2 + 1<br />
4 dS4<br />
with b0 > 0, c0 > 0, d > 0 and ˆn a constant unit vector. Discuss the phase<br />
transitions in this model.<br />
Problem 9.<br />
The Helmholtz free energy per unit volume is given by
4.9. PROBLEMS FOR CHAPTER 4 187<br />
f(T ; m) = R(T + 1<br />
8 T0) cosh( 2m<br />
) − RT0 cosh( m<br />
)<br />
This models a phase transition. Calcul<strong>at</strong>e Tc. Is the transition first or second<br />
order?<br />
Problem 10.<br />
m0<br />
The Helmholtz free energy per unit volume has the form<br />
f(T ; m) = a(T ) + b0(T − T0)m √ m + 1<br />
4 cm4<br />
with b0 > 0 and c > 0. We have introduced an unusual square root term.<br />
Show th<strong>at</strong> this gives a second order phase transition and calcul<strong>at</strong>e the temper<strong>at</strong>ure<br />
dependence of the order parameter near the critical temper<strong>at</strong>ure.<br />
Problem 11.<br />
Consider a model for a second order phase transition in a magnetic field.<br />
The Helmholtz free energy per unit volume is given by<br />
f(T ; m) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
c(T )m4<br />
4<br />
Assume c(T ) = c > 0, a positive constant, and b(T ) = b0(T −Tc) with b0 > 0.<br />
The magnetiz<strong>at</strong>ion density is still considered to be an internal parameter, which<br />
should be chosen to minimize the Helmholtz free energy. The condition is now<br />
( )<br />
∂f<br />
= H<br />
∂m<br />
Calcul<strong>at</strong>e m(T,H) (hint: use Maple or equivalent) and plot m(H) for various<br />
temper<strong>at</strong>ures and m(T) for various fields.<br />
Problem 12.<br />
Consider the following model of a tri-critical point:<br />
f(m, T, p) = 1<br />
2 a(T, p)m2 + 1<br />
4 b(T, p)m4 + 1<br />
6 cm6<br />
where c is a positive constant. The functions a and b are given by<br />
T − Tc p − pc<br />
a(T, p) = A + B<br />
Tc pc<br />
T − Tc p − pc<br />
b(T, p) = C + D<br />
Tc pc<br />
T<br />
m0
188 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Depending on the p<strong>at</strong>h in p-T space, along p = f(T ) with pc = f(Tc), the<br />
behavior of the response functions and the order parameter takes on a different<br />
form. For example, the behavior of the magnetiz<strong>at</strong>ion m(p, T ) always varies<br />
like (Tc − T ) β , but the value of β depends on the p<strong>at</strong>h taken. The two distinct<br />
regions with different values of β are called critical and tri-critical.<br />
(A) Calcul<strong>at</strong>e the β this system.<br />
(B) Draw the phase diagram in p-T space.<br />
(C) Draw the critical and tri-critical regions in this phase diagram.<br />
(D) Give an explan<strong>at</strong>ion of these results, write it in such a way th<strong>at</strong> a new,<br />
incoming gradu<strong>at</strong>e student could understand it.<br />
Problem 13.<br />
Consider the following model Helmholtz free energy per unit volume:<br />
f(T ; m) = a(T ) + e T 0 −T<br />
T 0 e −m2<br />
+ ce m2<br />
where a(T ) is an arbitrary function of temper<strong>at</strong>ure, and T0 and c are positive<br />
constants. Show th<strong>at</strong> this system undergoes a second order phase transition,<br />
calcul<strong>at</strong>e the critical temper<strong>at</strong>ure Tc, and find the critical exponent β.<br />
Problem 14.<br />
A magnetic system has a spontaneous magnetiz<strong>at</strong>ion below a critical temper<strong>at</strong>ure<br />
Tc, given by m2 = m2 0 Tc−T<br />
. The system is in a st<strong>at</strong>e <strong>at</strong> temper<strong>at</strong>ure<br />
Tc<br />
Ti = 2Tc and magnetiz<strong>at</strong>ion mi = 1<br />
2m0. This system is cooled down to very low<br />
temper<strong>at</strong>ure, but <strong>at</strong> constant magnetiz<strong>at</strong>ion. Describe the st<strong>at</strong>e of the system<br />
<strong>at</strong> this very low temper<strong>at</strong>ure. Calcul<strong>at</strong>e relevant quantities as a function of<br />
temper<strong>at</strong>ure.<br />
Problem 15.<br />
Consider the following Landau model, f(T, m) = a(T ) + b0<br />
2 (T − T0)m2 +<br />
1<br />
2N m2N , with b0 > 0. Find the magnetiz<strong>at</strong>ion as a function of temper<strong>at</strong>ure.<br />
Wh<strong>at</strong> happens to the critical exponent β when N → ∞? Wh<strong>at</strong> is the critical<br />
exponent if we change the model to f(T, m) = a(T ) + b0<br />
2 (T − T0)m2 + 1<br />
2em2 ?<br />
Problem 16.<br />
Consider the equ<strong>at</strong>ion for the perturb<strong>at</strong>ion of the magnetiz<strong>at</strong>ion density<br />
when there is a localized field, but write down the equ<strong>at</strong>ion <strong>at</strong> T = Tc. This
4.9. PROBLEMS FOR CHAPTER 4 189<br />
equ<strong>at</strong>ion is simpler than the one discussed in the notes. Find the behavior<br />
for large values of r. Compare this with the behavior away from the critical<br />
temper<strong>at</strong>ure.<br />
Problem 17.<br />
Consider the Landau model for magnetism in an AB alloy,<br />
f(T, mA, mB) = a(T )+ 1<br />
2 b0(T −Tc)(m 2 A+m 2 B)+bm(T −Tm)mAmB+ 1<br />
4 c0(m 4 A+m 4 B)<br />
where mA is the magnetiz<strong>at</strong>ion density for <strong>at</strong>oms A and mB for <strong>at</strong>oms B.<br />
The constants b0, c0, and bm are positive. Find mA(T ) and mB(T ).<br />
Problem 18.<br />
Consider a Landau model, for a second order phase transition,<br />
f(T, m) = a(T ) + b(T )m 2 + c0m 4<br />
Design coefficient a(T ), b(T ), and c0 th<strong>at</strong> give a second order phase transition<br />
where the magnetiz<strong>at</strong>ion is m(T ) = m0(Tc − T ) β for T < Tc and zero above.<br />
Discuss why you think such a model would be realistic or not.<br />
Problem 19.<br />
Using the general Landau formalism, equ<strong>at</strong>ion (4.3), show th<strong>at</strong> there is never<br />
a peak in the he<strong>at</strong> capacity <strong>at</strong> a phase transition.<br />
Problem 20.<br />
Use a Landau model of the form:<br />
and assume th<strong>at</strong><br />
f(T, m) =<br />
∞∑<br />
cn(T )m 2n<br />
n=0<br />
cn(T ) ≡ 0 , n = 0, · · · , N − 1<br />
cN (T ) = CN (T − Tc)<br />
cn(T ) ≡ Cn , n = N + 1, · · ·<br />
where CN , CN+1, etc are constants independent of the temper<strong>at</strong>ure.
190 CHAPTER 4. LANDAU-GINZBURG THEORY.<br />
Find the critical exponent β for this system<br />
Problem 21.<br />
Consider the following Landau model, f(T, m) = a(T ) + b0<br />
2 (T − T1)(T −<br />
T2)m2 + d<br />
4 m4 , with b0 > 0, T1 > 0, T2 > 0, and d > 0. Find the magnetiz<strong>at</strong>ion<br />
as a function of temper<strong>at</strong>ure. Find the entropy as a function of temper<strong>at</strong>ure.<br />
Problem 22.<br />
δ.<br />
Suppose we have in a Landau model<br />
T − Tc<br />
f(T, m) = a(T ) + b0m<br />
4<br />
4 + 1<br />
6 cm6<br />
with b0 > 0 and c > 0. Find the values of the critical exponents α, β, γ, and<br />
Problem 23.<br />
Suppose we have in a Landau model<br />
f[T, m(⃗r)] = a(T )+ 1<br />
∫ ∫<br />
b(T )<br />
2<br />
m(⃗r)J(⃗r−⃗r ′ )m(⃗r ′ )d 3 rd 3 r ′ + 1<br />
[∫<br />
c(T )<br />
4<br />
m 2 (⃗r)d 3 ] 2<br />
r<br />
Find the differential equ<strong>at</strong>ion for m(⃗r) th<strong>at</strong> yields the minimum for the<br />
functional.<br />
Problem 24.<br />
Suppose we have in a Landau model<br />
f[T, m(⃗r)] =<br />
∫ (<br />
a(T ) + 1<br />
2 b(T )m2 (⃗r) + 1<br />
4 c(T )m4 )<br />
(⃗r) d 3 ∫<br />
r+g(T )<br />
|⃗r× ⃗ ∇m(⃗r)| 2 d 3 r<br />
Find the differential equ<strong>at</strong>ion for m(⃗r) th<strong>at</strong> yields the minimum for the<br />
functional.<br />
Problem 25.<br />
Consider Landau’s model for a first order phase transition, but now including<br />
pressure:<br />
f(T, p; m) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
4 c(p)m4 + 1<br />
6 dm6
4.9. PROBLEMS FOR CHAPTER 4 191<br />
> 0, c(p) < 0, d > 0.<br />
Find an expression for dTc in terms of the coefficients b, c, and d. If the<br />
with b(T ) > 0, db<br />
dT<br />
dp<br />
critical temper<strong>at</strong>ure decreases as a function of pressure, wh<strong>at</strong> can you say about<br />
dc<br />
dp ?<br />
Problem 26.<br />
Consider the Landau model for a second order phase transition. We set<br />
the temper<strong>at</strong>ure equal to the critical temper<strong>at</strong>ure, T = Tc. Of course, we then<br />
find m = 0. Now we apply a small magnetic field H and measure the induced<br />
magnetic moment. We find th<strong>at</strong> m ∝ H 1<br />
δ . Find the value of δ.<br />
Problem 27.<br />
Give a Landau type of expansion for the liquid to vapor phase transition.<br />
Note: you need to define an order parameter first!
192 CHAPTER 4. LANDAU-GINZBURG THEORY.
Chapter 5<br />
Critical exponents.<br />
At first glance the analysis of a phase transition is easy. We have two possible<br />
st<strong>at</strong>es for the system, and the st<strong>at</strong>e with lowest free energy wins. But the observ<strong>at</strong>ions<br />
in the last part of the previous chapter open a new way of discussing<br />
wh<strong>at</strong> happens <strong>at</strong> a phase transition. Suppose we are close to the critical temper<strong>at</strong>ure<br />
of a phase transition. Locally, it is possible to have a thermal fluctu<strong>at</strong>ion<br />
which transforms a small part of the m<strong>at</strong>erial to the unstable phase. This region<br />
will grow until it reaches a size dict<strong>at</strong>ed by the correl<strong>at</strong>ion length, but then will<br />
disappear again because its internal energy is too high. We observe this directly<br />
when we boil w<strong>at</strong>er. Also, <strong>at</strong> very low temper<strong>at</strong>ures thermal fluctu<strong>at</strong>ions are<br />
possibly too small, and quantum fluctu<strong>at</strong>ions near a critical point become dominant.<br />
Transitions where the instabilities are driven by quantum fluctu<strong>at</strong>ions<br />
are called quantum phase transitions.<br />
The observ<strong>at</strong>ions made above lead to another description of the st<strong>at</strong>e of the<br />
system. If we are close in temper<strong>at</strong>ure to the phase transition, it is possible to<br />
consider the st<strong>at</strong>e of the system as a dynamic mixture of the stable phase embedded<br />
in which we have regions of the unstable phase. Such a st<strong>at</strong>e clearly has<br />
a higher internal energy, but since there are many possibilities for the unstable<br />
regions, the entropy in this st<strong>at</strong>e is also higher, and the Helmholtz free energy<br />
will be lower! Therefore, the correct description of the st<strong>at</strong>e of the system close<br />
to the critical point is such a mixed st<strong>at</strong>e!<br />
A microscopic description of such a dynamic mixed st<strong>at</strong>e is extremely difficult,<br />
but in thermodynamics we can derive some general st<strong>at</strong>ements th<strong>at</strong> the<br />
free energy of this st<strong>at</strong>e has to obey. On an absolute scale the free energy of the<br />
mixed st<strong>at</strong>e is always close to the free energy of the stable st<strong>at</strong>e, and the difference<br />
between these two free energies is small. Close to the transition, however,<br />
this difference in free energy is much larger than the difference between the free<br />
energies of the two homogeneous st<strong>at</strong>es, and for such temper<strong>at</strong>ure values this is<br />
the dominant free energy. Since observable variables are determined by slopes of<br />
the free energy, they can be quite different from the values for the homogeneous<br />
st<strong>at</strong>es.<br />
In this chapter we start with a discussion of the phase transition in terms<br />
193
194 CHAPTER 5. CRITICAL EXPONENTS.<br />
of critical exponents, a measurement how response functions diverge near the<br />
critical point. We then calcul<strong>at</strong>e these exponents in a mean field description.<br />
The values we obtain in this way are close to the observed values, but differences<br />
between mean field theory and experiment are significant. Next, we discuss<br />
the free energy of the mixed st<strong>at</strong>e, and describe how we can derive some of<br />
its properties based on the idea of scaling. Here we use the physics th<strong>at</strong> is<br />
important near a phase transition to predict how the general form of the free<br />
energy should be. This gives us several rel<strong>at</strong>ions between critical exponents,<br />
which are all observed experimentally. The scaling idea is then extended to<br />
another form called hyper-scaling. We combine the ideas of scaling with the<br />
theory of Landau and Ginzburg and show the validity of the l<strong>at</strong>ter when we<br />
change the number of dimensions in the system. Finally, we apply the scaling<br />
idea to time as well, and are then able to discuss the critical exponents for<br />
transport parameters.<br />
5.1 Introduction.<br />
Critical point is essential singularity.<br />
In this chapter we will discuss the n<strong>at</strong>ure of second order phase transitions<br />
in more detail. Our language is a bit sloppy here, and by second order we<br />
really mean all phase transitions th<strong>at</strong> end in a critical point. We have argued<br />
before th<strong>at</strong> the behavior <strong>at</strong> the critical point determines wh<strong>at</strong> happens away<br />
from the critical point. Therefore, in this chapter we focus on properties <strong>at</strong><br />
the critical point. Response functions will diverge, and near the critical point<br />
this divergent behavior overshadows any background behavior. It is not clear a<br />
priori, however, wh<strong>at</strong> near is. In terms on temper<strong>at</strong>ure, this could be 100K or<br />
1µK.<br />
Wh<strong>at</strong> causes this divergent behavior?<br />
Wh<strong>at</strong> is meant by background?<br />
Descriptions of behavior <strong>at</strong> the critical point.<br />
In chapter 3 we saw th<strong>at</strong> we can model a phase transition if we know the<br />
Gibbs free energy for the two distinct phases. The coexistence curve was found<br />
by comparing these energies. In this scenario the Gibbs free energies of the two
5.1. INTRODUCTION. 195<br />
phases were quite uncorrel<strong>at</strong>ed, there was no hint th<strong>at</strong> they actually came from<br />
the same system. Near a phase transition, in many cases nothing diverged. We<br />
either were in one type of analytical behavior or in another. Landau theory<br />
improved on this by giving these two free energies a common origin. We defined<br />
fequilibrium(T ) = min<br />
m f(T ; m) (5.1)<br />
which rel<strong>at</strong>ed the free energy in each phase to a common model, which includes<br />
an internal parameter. This procedure again gave two distinct analytical forms<br />
of the free energy, one above and one below the transition. The n<strong>at</strong>ure of the<br />
construction of these model energies, however, now often introduces divergent<br />
behavior <strong>at</strong> the critical point.<br />
Background.<br />
We are now able to define wh<strong>at</strong> we mean by a background. Expressions<br />
like Landau theory give us a uniform value of the order parameter, constant in<br />
time. This would be the correct average behavior of the system far away from<br />
a phase transition. We can use a theoretical expression like Landau theory to<br />
extrapol<strong>at</strong>e this average behavior to the point of the phase transition. This<br />
extrapol<strong>at</strong>ion uses the continuous behavior of thermodynamic functions and<br />
incorpor<strong>at</strong>es all uniform behavior of the system.<br />
At non zero temper<strong>at</strong>ures we will have thermal fluctu<strong>at</strong>ions. Far away from<br />
a phase transition these thermal fluctu<strong>at</strong>ions are small. They average out to<br />
zero, both in space and time. They are, however, time dependent phenomena.<br />
There is energy stored in these fluctu<strong>at</strong>ions, but far from a phase transition this<br />
energy is small. Near the transition, however, this energy becomes large and<br />
eventually domin<strong>at</strong>es the system.<br />
Characteriz<strong>at</strong>ion of the singularities.<br />
Any realistic theory of a second order phase transition has to include the<br />
effects of fluctu<strong>at</strong>ions near the critical point. These fluctu<strong>at</strong>ions give rise to divergent<br />
behavior in response functions, different from mean field theory, and we<br />
therefore need to develop a m<strong>at</strong>hem<strong>at</strong>ical description how to quantify differences<br />
in divergent behavior. The main characteristic of a divergence is the r<strong>at</strong>e with<br />
which certain quantities go to infinity. This idea leads us then autom<strong>at</strong>ically to<br />
the idea of critical exponents, which are the parameters th<strong>at</strong> show how fast the<br />
divergence is.<br />
The four common exponents.<br />
We will define the four common critical exponents in terms of the model of<br />
a magnetic phase transition used before. This is just an easy way to describe<br />
wh<strong>at</strong> is happening, and similar procedures are valid for any second order phase
196 CHAPTER 5. CRITICAL EXPONENTS.<br />
transition. The order parameter for a second order phase transition approaches<br />
zero when the temper<strong>at</strong>ure approaches the critical temper<strong>at</strong>ure. Near the phase<br />
transition we have<br />
T < Tc : m(T, H = 0) ∝ (Tc − T ) β<br />
(5.2)<br />
The power β in this rel<strong>at</strong>ion is called a critical exponent. Note th<strong>at</strong> the rel<strong>at</strong>ion<br />
above only holds when there is no applied external magnetic field.<br />
If we apply a small magnetic field for a temper<strong>at</strong>ure equal to the critical<br />
temper<strong>at</strong>ure, we find another critical exponent δ:<br />
H → 0 : m(T = Tc, H) ∝ H 1<br />
δ (5.3)<br />
The magnetic susceptibility χ = ( )<br />
∂M<br />
∂H T also often diverges <strong>at</strong> Tc, due to the<br />
instability of a system near a phase transition.<br />
Critical exponents γ and γ ′ are defined by:<br />
T > Tc : χ(T, H = 0) ∝ (T − Tc) −γ<br />
T < Tc : χ(T, H = 0) ∝ (Tc − T ) −γ′<br />
(5.4)<br />
(5.5)<br />
Experiments suggest th<strong>at</strong> in general γ = γ ′ . Our example is a magnetic system,<br />
but can be generalized to an arbitrary phase transition by replacing M with the<br />
order parameter for th<strong>at</strong> transition and H by the corresponding force.<br />
The specific he<strong>at</strong> ( the he<strong>at</strong> capacity per unit volume) behaves in a special<br />
way:<br />
T > Tc : C(T, H = 0) ∝ (T − Tc) −α<br />
T < Tc : C(T, H = 0) ∝ (Tc − T ) −α′<br />
Experiment shows again th<strong>at</strong> in general α = α ′ .<br />
Correl<strong>at</strong>ion length.<br />
Are these critical exponents independent variables?<br />
(5.6)<br />
(5.7)<br />
In the previous chapter we have seen th<strong>at</strong> including fluctu<strong>at</strong>ions in a description<br />
of a m<strong>at</strong>erial leads to the idea of a correl<strong>at</strong>ion length ξ. This correl<strong>at</strong>ion<br />
length ξ is associ<strong>at</strong>ed with a parameter ν:<br />
T > Tc : ξ(T ) ∝ (T − Tc) −ν<br />
T < Tc : ξ(T ) ∝ (Tc − T ) −ν′<br />
(5.8)<br />
(5.9)
5.1. INTRODUCTION. 197<br />
Again, we will usually find ν = ν ′ .<br />
Universality.<br />
The powers in all these rel<strong>at</strong>ions are called critical exponents. It turns out<br />
th<strong>at</strong> critical exponents are a unique characteriz<strong>at</strong>ion of phase transitions. If two<br />
seemingly different phase transitions have exactly the same critical exponents,<br />
the underlying physics must be similar. This is an example of universality. We<br />
use the values of the critical exponents to classify second order phase transitions,<br />
and we find th<strong>at</strong> there are only a few groups. For example, the critical exponents<br />
in the magnetic phase transitions for iron and nickel are the same, which tells us<br />
th<strong>at</strong> the same physics is <strong>at</strong> work. Of course, the critical temper<strong>at</strong>ure has a very<br />
different value, which simply reflects the fact th<strong>at</strong> we are dealing with different<br />
m<strong>at</strong>erials. But if we find an other magnetic phase transition with different<br />
values of the critical exponents, we know th<strong>at</strong> something different is going on,<br />
and th<strong>at</strong> we will need different models to describe this new m<strong>at</strong>erial.<br />
Critical exponents for the liquid to gas transition.<br />
Critical exponents can also be defined for a liquid to gas phase transition.<br />
The intensive parameters controlling the phase transition are pressure and temper<strong>at</strong>ure<br />
and Tc in the formulas above must be set equal to the temper<strong>at</strong>ure <strong>at</strong><br />
the critical point. The coexistence curve pco(T ) has the same role as the line<br />
H(T ) = 0 in the magnetic phase transition, it represents a discontinuity. If we<br />
expose a ferromagnetic m<strong>at</strong>erial to a magnetic field, cool it below the critical<br />
temper<strong>at</strong>ure, and then reduce the field to zero, the final result will de different<br />
for positive or neg<strong>at</strong>ive fields. The line (T, H = 0) from T = 0 to T = Tc represents<br />
all points in phase space where this discontinuity occurs. In the same way,<br />
if we apply pressure to a gas/liquid, cool it below the critical temper<strong>at</strong>ure, and<br />
then have the pressure approach pco(T ), we observe a discontinuity depending<br />
on the p<strong>at</strong>h in (p,T) space.<br />
Superconductivity.<br />
A more complex example of a second order phase transition is found in<br />
superconductivity. In this case the order parameter is the pseudo wave function<br />
ψ(⃗r) representing the super-conducting electrons. The density |ψ(⃗r)| 2 represents<br />
the density of super-conducting electrons <strong>at</strong> point ⃗r. There is a bit of a problem,<br />
since we need to know wh<strong>at</strong> the force corresponding to the order parameter is.<br />
If we use the square of the pseudo wave function, the generalized force obviously<br />
is some sort of chemical potential. It is not easy to find out which generalized<br />
force corresponds to the pseudo wave functions itself.<br />
Landau model for superconductivity.
198 CHAPTER 5. CRITICAL EXPONENTS.<br />
Deep inside a superconductor we assume th<strong>at</strong> we can describe the free energy<br />
by<br />
f(T ; ψ) = fnormal(T ) + 1<br />
2 α(T )|ψ(⃗r)|2 + 1<br />
β(T )|ψ(⃗r)|4<br />
4<br />
(5.10)<br />
and in order to model a second order phase transition we assume th<strong>at</strong> α(T )<br />
changes sign <strong>at</strong> Tc and th<strong>at</strong> β(T ) > 0. Therefore we find below Tc th<strong>at</strong><br />
|ψ(⃗r)| 2 α(T )<br />
= −<br />
β(T )<br />
(5.11)<br />
The difference in free energy between the two st<strong>at</strong>es, the normal st<strong>at</strong>e and<br />
the super-conducting st<strong>at</strong>e, is given by<br />
fsc − fnormal = − 1 α<br />
4<br />
2<br />
β<br />
(5.12)<br />
and this energy can be rel<strong>at</strong>ed to the critical field Hc(T ) for a type I superconductor<br />
by setting this equal to the energy density, because a type I superconductor<br />
expels all magnetic fields. This gives<br />
− H2 c (T )<br />
8π<br />
α<br />
= −1<br />
4<br />
2<br />
β<br />
Why is this a good assumption?<br />
(5.13)<br />
Next we assume th<strong>at</strong> in the description of α near Tc we only need to include<br />
the linear terms, and hence th<strong>at</strong> α(T ) ∝ (T − Tc). This gives<br />
which is indeed observed, and<br />
which agrees with the older theory due to London.<br />
Fluctu<strong>at</strong>ions.<br />
Hc(T ) ∝ Tc − T (5.14)<br />
|ψ(⃗r)| 2 ∝ Tc − T (5.15)<br />
Our next task is to include fluctu<strong>at</strong>ions in this model. The pseudo wave<br />
function is described by a Schrödinger type of equ<strong>at</strong>ion, and the corresponding<br />
energy, including a magnetic field (using a vector potential ⃗ A) is<br />
1<br />
2m∗ (<br />
<br />
<br />
¯h<br />
ı ⃗ ∇ − e∗<br />
c ⃗ ) <br />
<br />
A ψ<br />
<br />
2<br />
(5.16)
5.1. INTRODUCTION. 199<br />
where m ∗ is the mass of the super-conducting species (i.e. a Cooper pair) and<br />
e ∗ its charge. Even though deep inside the superconductor there is no magnetic<br />
field, we need to include the vector potential in order to describe the interface<br />
between the super-conducting and normal regions. This form of the extra free<br />
energy term is very similar to the Ginzburg term discussed before.<br />
Next, we separ<strong>at</strong>e the pseudo wave function into a magnitude and phase by<br />
ψ = |ψ|e ıϕ which leads to the following form of the energy term:<br />
1<br />
2m∗ (<br />
¯h 2 ( ) (<br />
2<br />
∇|ψ| ⃗ + ¯h ⃗ ∇ϕ − e∗<br />
c ⃗ ) 2<br />
A |ψ| 2<br />
)<br />
(5.17)<br />
where all cross terms disappear since they are purely imaginary. The first term<br />
depends only on the magnitude of the order parameter and is the same as the<br />
term introduced in the last chapter. The second term reflects a gauge invariance,<br />
and is a pure quantum mechanical effect.<br />
Phase of the pseudo wave function.<br />
If there are no external fields and ⃗ A = 0 the minimum in the free energy<br />
occurs when ⃗ ∇ϕ = 0 or ϕ = constant, which means th<strong>at</strong> ϕ is not important in<br />
this case. In general we have<br />
rel<strong>at</strong>ing the phase and the vector potential.<br />
Correl<strong>at</strong>ion length.<br />
¯h ⃗ ∇ϕ = e∗<br />
c ⃗ A (5.18)<br />
The results from the previous chapter allow us to say th<strong>at</strong> the correl<strong>at</strong>ion<br />
length in this case is given by<br />
√<br />
¯h<br />
ξ(T ) =<br />
2<br />
2m∗α(T )<br />
(5.19)<br />
which diverges in the usual way near the critical temper<strong>at</strong>ure. The interpret<strong>at</strong>ion<br />
of this correl<strong>at</strong>ion length is simple. If we have an interface between a<br />
normal st<strong>at</strong>e and a super-conducting st<strong>at</strong>e, the domain wall separ<strong>at</strong>ing these<br />
two st<strong>at</strong>es is of thickness ξ.<br />
Another important length scale is the length λL(T ) over which a magnetic<br />
field dies out inside the superconductor due to the flux expulsion. A comparison<br />
of the values of λL(T ) and ξ allows us to decide if a superconductor is of type<br />
I or type II.
200 CHAPTER 5. CRITICAL EXPONENTS.<br />
5.2 Mean field theory.<br />
Meaning of different parameters.<br />
In the study of phase transitions we distinguish two types of parameters<br />
th<strong>at</strong> characterize a transition. When we consider the terms in the work done<br />
on a sample, we have products like HM and pV. We also have he<strong>at</strong> exchange,<br />
corresponding to a term TS. A phase transition is first characterized by a critical<br />
value in the temper<strong>at</strong>ure, Tc, below which temper<strong>at</strong>ure the sample is ordered<br />
when other fields are not present, i.e. H=0, p=0, etc. But we need also consider<br />
the roles of the other external controls we have on the sample. For example,<br />
in a magnetic phase transition we also have a critical value of the field, Hc,<br />
which is often equal to zero. The critical point in the (T, H) plane is (Tc, Hc).<br />
If we reduce the temper<strong>at</strong>ure through the critical value with a field slightly<br />
higher than Hc we end up with one type of magnetiz<strong>at</strong>ion, and if we lower the<br />
temper<strong>at</strong>ure with a field slightly below Hc we obtain the opposite magnetiz<strong>at</strong>ion.<br />
If we also allow pressure to play a role, we find th<strong>at</strong> the critical point is given<br />
by three parameters, Tc, Hc, and pc.<br />
The position of the critical point depends on the strength of the interactions,<br />
and on the type of interactions. Even though iron and nickel are both<br />
ferromagnetic m<strong>at</strong>erials, the critical temper<strong>at</strong>ures are very different. In terms<br />
of models, they have different exchange interactions.<br />
Phase transitions are also characterized by critical exponents, as defined<br />
earlier. As it turns out, these critical exponents do not depend on the strength<br />
of the interactions, they only depend on the form of the Hamiltonian, or<br />
on the n<strong>at</strong>ure of the interactions. Therefore, systems with similar values of<br />
the critical exponents are similar in n<strong>at</strong>ure, or have similar physics. Critical<br />
exponents are a tool to classify different types of phase transitions, or to sort<br />
them into different universality classes. In the following we discuss an important<br />
example.<br />
Landau and mean field.<br />
Wh<strong>at</strong> does ”form of the Hamiltonian” mean?<br />
The second order phase transition we have studied as an example in Landau<br />
theory has a special meaning. If we ignore fluctu<strong>at</strong>ions, and only include the<br />
lowest order terms, the external magnetic field is given by (with T < Tc)<br />
H =<br />
( )<br />
∂f<br />
= b(T )m + cm<br />
∂m T<br />
3<br />
(5.20)
5.2. MEAN FIELD THEORY. 201<br />
which can be written as<br />
b0T m = H + m(b0Tc − cm 2 ) (5.21)<br />
and near the critical temper<strong>at</strong>ure this is approxim<strong>at</strong>ely b0T m = H + mb0Tc.<br />
In free space the magnetiz<strong>at</strong>ion and field are proportional. In a paramagnetic<br />
substance the rel<strong>at</strong>ion between the magnetiz<strong>at</strong>ion and the magnetic field is given<br />
by Curie’s law m = α H<br />
T . The equ<strong>at</strong>ion 5.21 looks like Curie’s law, except th<strong>at</strong><br />
the magnetic field is modified. Because a magnetiz<strong>at</strong>ion is present there is an<br />
additional effective magnetic field H ′ = m(b0Tc −cm2 ). This can be interpreted<br />
as follows. Because a magnetiz<strong>at</strong>ion is already present, each sub-system in our<br />
large system experiences an additional magnetic field which is simply rel<strong>at</strong>ed<br />
to the average magnetiz<strong>at</strong>ion in the system as a whole. One can think of the<br />
surroundings of a sub-system providing such a magnetic field, based on the<br />
average value of the magnetiz<strong>at</strong>ion in the system. This is called a mean field<br />
approach. It ignores the effect th<strong>at</strong> there are fluctu<strong>at</strong>ions in the magnetiz<strong>at</strong>ion.<br />
Critical exponents in mean field.<br />
The critical exponents for Landau’s theory in this mean field approach follow<br />
immedi<strong>at</strong>ely from the equ<strong>at</strong>ions we derived before:<br />
χ −1 (T, H) =<br />
m ∝ (Tc − T ) 1<br />
2 ⇒ β = 1<br />
2<br />
(5.22)<br />
( 2 ∂ f<br />
∂m2 )<br />
∝ |T − Tc| ⇒ γ = γ<br />
T<br />
′ = 1 (5.23)<br />
T = Tc, f ∝ m 4 ⇒ δ = 3 (5.24)<br />
The value of α cannot be determined, since we did not specify a(T). Since<br />
a(T) is actually independent of the phase transition, it is reasonable to assume<br />
th<strong>at</strong> it does not diverge <strong>at</strong> Tc, and in th<strong>at</strong> case we find<br />
α = α ′ = 0 (5.25)<br />
Any phase transition with the same critical exponents as we listed here has the<br />
same physics incorpor<strong>at</strong>ed. Any model with these critical exponents is therefore<br />
essentially a mean field model!<br />
Example values.<br />
The following table gives some examples of measured values of critical exponents<br />
and theoretical results:
202 CHAPTER 5. CRITICAL EXPONENTS.<br />
type system Tc α β γ δ<br />
ferromagnetic Ni 632 -0.10 0.33 1.32 4.2<br />
F e 1044 -0.12 0.34 1.33<br />
anti-ferromagnetic RbMnF3 83 -0.14 0.32 1.40<br />
liquid-gas Co2 304 0.12 0.34 1.20 4.2<br />
Xe 290 0.08 0.34 1.20 4.4<br />
ordered alloy CoZn 739 0.31 1.25<br />
mean field 0 0.5 1.0 3.0<br />
Ising, d=2 0 0.125 1.75 15<br />
Ising, d=3 0.12 0.31 1.25 5.0<br />
Heisenberg -0.14 0.38 1.38<br />
If mean field theory would be valid for all systems, all systems would have<br />
the same values for the critical exponents. In some sense this is obeyed. They<br />
are indeed quite close, but distinct from mean field. The fact th<strong>at</strong> the critical<br />
exponents do not vary wildly is due to the fact th<strong>at</strong> in many systems mean field<br />
theory is a good first approxim<strong>at</strong>ion.<br />
A very different set of values is obtained in the 2 dimensional Ising model.<br />
This shows the importance of the dimensionality of the system, and hence we<br />
expect big differences between the properties of thin films and bulk solids, which<br />
is indeed observed. The difference between the 3 dimensional Ising model and<br />
the Heisenberg model is not very large, but large enough th<strong>at</strong> experiments can<br />
distinguish between the validity of the two.<br />
In experimental situ<strong>at</strong>ions we often study thin films, and the question always<br />
arises if these films are two or three dimensional. The answer to this question<br />
is easy. One length scale of importance is the thickness of the film, Tfilm. The<br />
other quantity of interest is the correl<strong>at</strong>ion length ξ(T ). If we have ξ(T ) < Tfilm,<br />
even in the direction perpendicular to the film we have many domains, or regions<br />
of different organized behavior, and the film is three dimensional. On the other<br />
hand, if ξ(T ) > Tfilm the film is always coherent over the whole thickness, and<br />
we have a two dimensional system. In the thermodynamic limit we assume th<strong>at</strong><br />
the extend of the film in the plane is infinite, and in th<strong>at</strong> direction we always have<br />
a correl<strong>at</strong>ion length smaller than the dimensions of the film. This argument<br />
therefore shows th<strong>at</strong> a thin film near the critical temper<strong>at</strong>ure always<br />
behaves as a two dimensional system. This situ<strong>at</strong>ion is more complic<strong>at</strong>ed<br />
when we have two different types of correl<strong>at</strong>ion length to consider, in th<strong>at</strong> case<br />
we have to ask the question large or small for each of those lengths.<br />
Example van der Waals.<br />
Wh<strong>at</strong> is near?<br />
In Chapter 3 we studied a model of the liquid to gas phase transition devel-
5.2. MEAN FIELD THEORY. 203<br />
oped by van der Waals. This model was very instructive, and explained many<br />
characteristics of the phase transition. It essentially captured the behavior using<br />
two parameters, a and b. Based on the values for a and b we found the critical<br />
temper<strong>at</strong>ure and pressure. Hence, it is also possible to use the critical temper<strong>at</strong>ure<br />
and pressure as parameters describing the phase transition. There are no<br />
parameters left, and therefore the critical exponents should all be determined<br />
by the form of the equ<strong>at</strong>ion!<br />
Parameter free van der Waals equ<strong>at</strong>ion.<br />
Once the critical temper<strong>at</strong>ure and pressure are known, the van der Waals<br />
equ<strong>at</strong>ion should be unique. This leads us to the introduction of the following<br />
rel<strong>at</strong>ive variables:<br />
T − Tc<br />
θ =<br />
Tc<br />
p − pc<br />
π =<br />
pc<br />
V − Vc<br />
υ =<br />
and in terms of these variables the van der Waals equ<strong>at</strong>ion is<br />
[<br />
(1 + π) +<br />
3<br />
(1 + υ) 2<br />
Vc<br />
(5.26)<br />
(5.27)<br />
(5.28)<br />
]<br />
[3(1 + υ) − 1] = 8(1 + θ) (5.29)<br />
which is indeed parameter-free. Since we are interested in the behavior near<br />
the critical point, where all scaled parameter values are small, we rewrite this<br />
equ<strong>at</strong>ion as follows by multiplying by (1 + υ) 2 :<br />
or<br />
or<br />
[ (1 + π)(1 + υ) 2 + 3 ] [2 + 3υ] = 8(1 + θ)(1 + υ) 2<br />
(5.30)<br />
[ 4 + 2υ + υ 2 + π + 2πυ + πυ 2 ] [2 + 3υ] = 8(1 + θ)(1 + 2υ + υ 2 ) (5.31)<br />
8+16υ +8υ 2 +2π +7πυ +8πυ 2 +3υ 3 +3πυ 3 = 8+16υ +8υ 2 +8θ +16θυ +8θυ 2<br />
or<br />
π ( 2 + 7υ + 8υ 2 + 3υ 3) + 3υ 3 = θ ( 8 + 16υ + 8υ 2)<br />
(5.32)<br />
(5.33)<br />
which is a good start for the deriv<strong>at</strong>ion of the values of the critical exponents.
204 CHAPTER 5. CRITICAL EXPONENTS.<br />
Reduced pressure expansion.<br />
In order to analyze the behavior near the critical point, we expand the<br />
reduced pressure as a function of reduced temper<strong>at</strong>ure and reduced volume<br />
near the critical point and find th<strong>at</strong>:<br />
π ≈ 4θ − 6θυ + 9θυ 2 − 3<br />
2 υ3<br />
(5.34)<br />
where in the third order volume term the small θ term is ignored, and the<br />
higher order volume terms are ignored. This equ<strong>at</strong>ion gives a good and valid<br />
description of the behavior near the critical point.<br />
The partial deriv<strong>at</strong>ive of the pressure versus volume is:<br />
( )<br />
∂π<br />
≈ −6θ + 18θυ −<br />
∂υ θ<br />
9<br />
2 υ2<br />
(5.35)<br />
and equ<strong>at</strong>ing this to zero has solutions if 18 2 θ 2 ≥ 108θ or θ ≤ 0 for the solution<br />
close to the critical point, as expected.<br />
Order parameter.<br />
For θ ≤ 0 and close to zero we can find the volume of the liquid and the gas<br />
phase from the equ<strong>at</strong>ions:<br />
and<br />
The first equ<strong>at</strong>ion is equivalent to:<br />
π(υl) = π(υg) (5.36)<br />
0 =<br />
∫ υg<br />
υl<br />
υdπ (5.37)<br />
−6θυl + 9θυ 2 l − 3<br />
2 υ3 l = −6θυg + 9θυ 2 g − 3<br />
2 υ3 g<br />
while the last equ<strong>at</strong>ion is equivalent to:<br />
or<br />
or<br />
0 =<br />
0 =<br />
∫ υg<br />
υl<br />
∫ υg<br />
υl<br />
υ<br />
(5.38)<br />
( )<br />
∂π<br />
dυ (5.39)<br />
∂υ θ<br />
υ(−6θ + 18θυ − 9<br />
2 υ2 )dυ (5.40)<br />
−3θυ 2 l + 6θυ 3 l − 9<br />
8 υ4 l = −3θυ 2 g + 6θυ 3 g − 9<br />
8 υ4 g<br />
(5.41)
5.2. MEAN FIELD THEORY. 205<br />
Close to the critical point the middle term in each equ<strong>at</strong>ion can be ignored,<br />
because it is small compared to the first term. We cannot ignore the third term,<br />
since it does not contain the temper<strong>at</strong>ure, and we do not yet know the rel<strong>at</strong>ive<br />
scaling of all quantities. As a consequence, we find th<strong>at</strong> we need to solve:<br />
and<br />
4θυl + υ 3 l = 4θυg + υ 3 g<br />
8θυ 2 l + 3υ 4 l = 8θυ 2 g + 3υ 4 g<br />
(5.42)<br />
(5.43)<br />
How to solve these equ<strong>at</strong>ions? First of all, note th<strong>at</strong> we expect υl < 0 < υg.<br />
The two previous equ<strong>at</strong>ions can be rewritten in the form:<br />
and<br />
4θ(υl − υg) = υ 3 g − υ 3 l<br />
(5.44)<br />
8θ(υ 2 l − υ 2 g) = 3(υ 4 g − υ 4 l ) (5.45)<br />
The first has solutions υl = υg, which is not physical because υl < 0 < υg.<br />
Hence we are left with<br />
The second equ<strong>at</strong>ion has solutions<br />
−4θ = υ 2 g + υlυgυ 2 l<br />
υ 2 l = υ 2 g<br />
(5.46)<br />
(5.47)<br />
or υl = −υg. Together with the first equ<strong>at</strong>ion this gives the solutions according<br />
to<br />
υg = −υl = 2 √ |θ| (5.48)<br />
which are only valid for θ < 0, as expected. The other solution of the second<br />
equ<strong>at</strong>ion is<br />
8θ = 3(υ 2 g + υ 2 l ) (5.49)<br />
which cannot be right since it would imply θ > 0.<br />
The order parameter ∆υ is defined as the difference between the volumes,<br />
and we have<br />
along the coexistence curve given by<br />
He<strong>at</strong> capacity.<br />
∆υ = 4 √ |θ| (5.50)<br />
πcoex = 4θ (5.51)
206 CHAPTER 5. CRITICAL EXPONENTS.<br />
We have seen in chapter 3 th<strong>at</strong> the he<strong>at</strong> capacity as a function of temper<strong>at</strong>ure<br />
always shows a finite jump. There is no divergence, but the values of the he<strong>at</strong><br />
capacity are not the same coming from below or above the critical temper<strong>at</strong>ure.<br />
This is simply rel<strong>at</strong>ed to the fact th<strong>at</strong> we have a continuous p<strong>at</strong>h from the gas<br />
to the liquid st<strong>at</strong>e following a van der Waals curve. The Gibbs free energy of<br />
the gas and liquid are the same <strong>at</strong> the coexistence curve, but the slope as a<br />
function of temper<strong>at</strong>ure changes.<br />
The Gibbs free energy follows from<br />
g = g0(θ) +<br />
which leads to the entropy<br />
−s =<br />
∫ υ<br />
0<br />
υdπ = g0(θ) − 3θυ 2 + 6θυ 3 − 9<br />
8 υ4<br />
( )<br />
∂g<br />
= g<br />
∂θ π<br />
′ 0(θ) − 3υ 2 + 6υ 3 (<br />
+ −6θυ + 18θυ 2 − 9<br />
2 υ3<br />
) ( )<br />
∂υ<br />
∂θ π<br />
and a specific he<strong>at</strong><br />
−c = θ<br />
(<br />
θ −6θ + 36θυ − 27<br />
2 υ2<br />
( )<br />
∂s<br />
= θg<br />
∂θ π<br />
′′<br />
0 (θ) + 2θ ( −6υ + 18υ 2) ( )<br />
∂υ<br />
+<br />
∂θ π<br />
) ( ) 2 (<br />
∂υ<br />
+θ −6θυ + 18θυ<br />
∂θ π<br />
2 − 9<br />
2 υ3<br />
) ( 2 ∂ υ<br />
∂θ2 )<br />
π<br />
(5.52)<br />
(5.53)<br />
(5.54)<br />
which does not diverge, but is discontinuous (i.e. does not have the same value<br />
for υl and υg. Near the transition we can therefore model the he<strong>at</strong> capacity by<br />
θ > 0 : C(θ, π = πcoex) ∝ θ −α<br />
θ < 0 : C(θ, π = πcoex) ∝ |θ| −α′<br />
and the only way to get a discontinuity is to have<br />
Order parameter.<br />
From equ<strong>at</strong>ion 5.50 gives immedi<strong>at</strong>ely th<strong>at</strong><br />
Order parameter versus field.<br />
(5.55)<br />
(5.56)<br />
α = α ′ = 0 (5.57)<br />
β = 1<br />
2<br />
(5.58)
5.2. MEAN FIELD THEORY. 207<br />
This response function follows from (leaving out a factor<br />
not change the value of the critical exponent):<br />
−κ −1 =<br />
which we need to follow along πcoex = 4θ or<br />
( )<br />
∂π<br />
= −6θ + 18θυ −<br />
∂υ θ<br />
9<br />
2 υ2<br />
0 = −6θυ + 9θυ 2 − 3<br />
2 υ3<br />
V pc , which does<br />
Vc<br />
(5.59)<br />
(5.60)<br />
which can be used (after dividing by υ ) to elimin<strong>at</strong>e the last term, and we get<br />
θ < 0 : κ −1 = −12θ + 9θυ (5.61)<br />
in which only the first term is important near the critical point. Similarly, above<br />
the critical temper<strong>at</strong>ure we follow the constant volume curve and find:<br />
which leads immedi<strong>at</strong>ely to<br />
θ > 0 : κ −1 = +6θ (5.62)<br />
γ = γ ′ = 1 (5.63)<br />
Volume versus pressure <strong>at</strong> the critical temper<strong>at</strong>ure.<br />
The equ<strong>at</strong>ion here is simple, we have<br />
which shows immedi<strong>at</strong>ely th<strong>at</strong><br />
Van der Waals is mean field.<br />
π = − 3<br />
2 υ3<br />
(5.64)<br />
δ = 3 (5.65)<br />
We have found th<strong>at</strong> the critical exponents in a van der Waals model of<br />
a phase transition are the same as the mean field exponents. Therefore, the<br />
van der Waals model is a mean field model. This can be explained easily in<br />
a microscopic model, where the parameters a and b only describe the average<br />
interaction between molecules, or the interaction between one molecule and<br />
its average (mean) environment. The van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is no<br />
exception. In fact, any analytic equ<strong>at</strong>ion of st<strong>at</strong>e model of a phase transition<br />
gives the same mean field exponents, since in rel<strong>at</strong>ive units the form of the<br />
equ<strong>at</strong>ion near the critical point is always like equ<strong>at</strong>ion 5.34!
208 CHAPTER 5. CRITICAL EXPONENTS.<br />
5.3 Model free energy near a critical point.<br />
Fluctu<strong>at</strong>ions.<br />
By now it will be clear th<strong>at</strong> any realistic description of a phase transition<br />
near a critical point will have to include the effects of fluctu<strong>at</strong>ions. Without<br />
th<strong>at</strong>, we only have analytical theories with discontinuities <strong>at</strong> the transition,<br />
which lead to mean field values of the critical exponents. In order to illustr<strong>at</strong>e<br />
the approach to the problem of including fluctu<strong>at</strong>ions, we will again study a<br />
magnetic phase transition, but the procedure can be generalized to arbitrary<br />
systems.<br />
Scaled variables.<br />
The important variables which we can use to change the st<strong>at</strong>e of our model<br />
system are the temper<strong>at</strong>ure T and the magnetic field H. The critical point is<br />
given by the values Tc and Hc = 0. In order to focus on the behavior near the<br />
critical point we used scaled values of these variables:<br />
and<br />
T − Tc<br />
θ =<br />
Tc<br />
(5.66)<br />
h = H (5.67)<br />
where the scaling in the last case is simple, since we cannot divide by zero.<br />
Model energy.<br />
The important free energy in this case is a analogue to the Gibbs free energy<br />
(the magnetic Gibbs free energy) defined by<br />
g(θ, h) = 1<br />
(U − T S − MH) (5.68)<br />
N<br />
where we also divided by the total amount of m<strong>at</strong>erial to get an energy density.<br />
One important contribution to this free energy is the normal equilibrium thermodynamical<br />
form we have discussed before. This term only leads to continuous<br />
behavior, without any divergent behavior in the energy. The previous section<br />
showed th<strong>at</strong> the free energy for a van der Waals model is smooth, for example.<br />
This term is the background energy. The free energy, however, also contains a<br />
term describing the fluctu<strong>at</strong>ions in the system. This is the term we do not know<br />
how to write down formally, but we can separ<strong>at</strong>e the effects:<br />
g(θ, h) = gbackground(θ, h) + gfluctu<strong>at</strong>ions(θ, h) (5.69)
5.3. MODEL FREE ENERGY NEAR A CRITICAL POINT. 209<br />
Analysis of fluctu<strong>at</strong>ion term.<br />
When we get closer to the critical temper<strong>at</strong>ure, the effect of fluctu<strong>at</strong>ions<br />
will domin<strong>at</strong>e the behavior of the system. These fluctu<strong>at</strong>ions will cause all<br />
kinds of divergencies in the response functions. We therefore assume th<strong>at</strong> near<br />
the critical point only the second term in the last equ<strong>at</strong>ion is important. The<br />
analysis of critical exponents is therefore determined by the second term, and<br />
from now on we ignore the first term and omit the subscript fluctu<strong>at</strong>ions. The<br />
main question is, wh<strong>at</strong> kind of general st<strong>at</strong>ements can we make about this energy<br />
term!<br />
Scaling.<br />
Near a critical point the correl<strong>at</strong>ion length becomes very large. Any other<br />
length scale we have in the system is not important anymore, and the correl<strong>at</strong>ion<br />
length determines the behavior of the system. Suppose we have a picture of a<br />
system close to a critical point. We see bubbles of one phase in another, and the<br />
average size of these bubbles is rel<strong>at</strong>ed to the correl<strong>at</strong>ion length. If we change<br />
the values of the temper<strong>at</strong>ure or field slightly, we obtain a different picture with<br />
a different average size of the bubbles. If we have an absolute measure of the<br />
length scale we can identify these pictures. But imagine th<strong>at</strong> both pictures are<br />
scaled and th<strong>at</strong> one unit of length in each picture corresponds to the appropri<strong>at</strong>e<br />
correl<strong>at</strong>ion length. In this case the average size of the bubbles will be the same<br />
in both pictures. Can we still identify the pictures? Can we tell them apart?<br />
The experimental answer seems to be no. Therefore, the free energy due to<br />
the fluctu<strong>at</strong>ions in both cases must be rel<strong>at</strong>ed, and the density is changed by<br />
simply accounting for the extra amount of m<strong>at</strong>erial in a scaled unit volume.<br />
M<strong>at</strong>hem<strong>at</strong>ically this is expressed by saying th<strong>at</strong> the free energy g (due to the<br />
fluctu<strong>at</strong>ions) near the critical point should be a homogeneous function of its<br />
variables.<br />
Homogeneous functions.<br />
Homogeneous functions are easiest to discuss when only one variable is involved.<br />
By definition, a function f(x) is called homogeneous if<br />
f(λx) = S(λ)f(x) (5.70)<br />
where S is called a scaling function. This rel<strong>at</strong>ion has to hold for all values of<br />
λ and x. In this simple case it is easy to derive wh<strong>at</strong> the possible homogeneous<br />
functions are. When we apply to successive transform<strong>at</strong>ions we get<br />
but also<br />
f((µλ)x) = S(µλ)f(x) (5.71)
210 CHAPTER 5. CRITICAL EXPONENTS.<br />
or<br />
f((µλ)x) = f(µ(λx)) = S(µ)S(λ)f(x) (5.72)<br />
S(µλ) = S(µ)S(λ) (5.73)<br />
for all values of λ and µ. If we now take the deriv<strong>at</strong>ive with respect to λ we<br />
obtain<br />
and similarly for µ:<br />
from which follows:<br />
or<br />
which means th<strong>at</strong><br />
S ′ (µλ)µ = S(µ)S ′ (λ) (5.74)<br />
S ′ (µλ)λ = S ′ (µ)S(λ) (5.75)<br />
S(µ)S ′ (λ)λ = µS ′ (µ)S(λ) (5.76)<br />
S ′ (λ)λ<br />
S(λ) = µS′ (µ)<br />
S(µ)<br />
S ′ (λ)λ<br />
S(λ)<br />
(5.77)<br />
= p (5.78)<br />
where p is a constant. Since S(1)=1 we can solve this immedi<strong>at</strong>ely and we get<br />
Form of the scaling functions.<br />
S(λ) = λ p<br />
(5.79)<br />
We arrive <strong>at</strong> the important conclusion th<strong>at</strong> scaling functions are simple powers<br />
of the variable, where the constant p can take any real value. In equ<strong>at</strong>ion<br />
5.70 we can take x=1 and find<br />
f(λ) = λ p f(1) (5.80)<br />
which shows th<strong>at</strong> the homogeneous functions of one variable are simple powers.<br />
For example, for the volume V of a cube with side a we find th<strong>at</strong> V (λa) =<br />
λ 3 V (a).<br />
Homogeneous function in many variables.<br />
Homogeneous functions of many variables are defined in a similar manner,<br />
where we already include the fact th<strong>at</strong> the scaling functions are powers. The
5.4. CONSEQUENCES OF SCALING. 211<br />
only change we make is where we put the scaling function, and in this case it is<br />
easier to have the simple multiplier variable λ in front of the function:<br />
f(λ p x, λ q y) = λf(x, y) (5.81)<br />
Since the powers p and q can take any real value, we need to restrict ourselves<br />
to positive values of lambda only.<br />
Scaling applied to the free energy.<br />
The basic idea th<strong>at</strong> the correl<strong>at</strong>ion length is the only important length scale<br />
near the critical point transl<strong>at</strong>es in the following st<strong>at</strong>ement for the energy density.<br />
Near the critical point θ = 0 , h = 0 we have<br />
g(λ p θ, λ q h) = λg(θ, h) (5.82)<br />
The motiv<strong>at</strong>ion is more or less as follows. Nearer to the critical point the energy<br />
density becomes smaller, since the energy stored in a fluctu<strong>at</strong>ion is spread<br />
out over a larger volume. So if we make θ and h smaller, the energy density<br />
becomes smaller. Scaling theory simple assumes th<strong>at</strong> this reduction in temper<strong>at</strong>ure<br />
and field is homogeneous, like in equ<strong>at</strong>ion 5.82.<br />
5.4 Consequences of scaling.<br />
The scaling rel<strong>at</strong>ion in equ<strong>at</strong>ion 5.82 has some very important ) consequences.<br />
For example, if we calcul<strong>at</strong>e the magnetiz<strong>at</strong>ion m = we find th<strong>at</strong><br />
and therefore if we set θ = 1 and h = 0 we obtain<br />
or with t = λ p<br />
( ∂g<br />
∂h<br />
λ q m(λ p θ, λ q h) = λm(θ, h) (5.83)<br />
m(λ p , 0) = λ 1−q m(1, 0) (5.84)<br />
m(t, 0) = t 1−q<br />
p m(1, 0) (5.85)<br />
Since m(1, 0) = 0 we find therefore th<strong>at</strong> m(t, 0) = 0 for all values of t and th<strong>at</strong><br />
there is no magnetiz<strong>at</strong>ion. But here we have to be careful. Since λ > 0 this<br />
only holds for t > 0, and now our conclusion is indeed correct. Scaling tells us<br />
th<strong>at</strong> if the magnetiz<strong>at</strong>ion is zero <strong>at</strong> one point above the critical temper<strong>at</strong>ure, it<br />
is zero everywhere above the critical temper<strong>at</strong>ure.<br />
Rel<strong>at</strong>ion to critical exponent β.<br />
In order to discuss the behavior below the critical temper<strong>at</strong>ure we need to<br />
start with θ = −1. This gives:<br />
θ
212 CHAPTER 5. CRITICAL EXPONENTS.<br />
or with t = −λ p<br />
m(−λ p , 0) = λ 1−q m(−1, 0) (5.86)<br />
m(t, 0) = (−t) 1−q<br />
p m(−1, 0) (5.87)<br />
In this case m(−1, 0) is not zero and we find th<strong>at</strong> the magnetiz<strong>at</strong>ion near the<br />
critical temper<strong>at</strong>ure follows a simple power law as a function of the reduced temper<strong>at</strong>ure,<br />
as expected. A comparison with the definition of the critical exponent<br />
β gives<br />
β =<br />
1 − q<br />
p<br />
(5.88)<br />
and hence the value of the critical exponent follows immedi<strong>at</strong>ely from the scaling<br />
parameters p and q in the energy! This is true for all critical exponents! Since<br />
we only have two independent parameters in the energy scaling, we see th<strong>at</strong><br />
only two values of the critical exponents are independent, and th<strong>at</strong> the others<br />
should de dependent!<br />
Critical exponent δ.<br />
From equ<strong>at</strong>ion 5.83 we see immedi<strong>at</strong>ely th<strong>at</strong><br />
and with h = 1 we obtain<br />
or by defining |H| = λ q we find<br />
or<br />
Critical exponent γ.<br />
to<br />
λ q m(0, λ q h) = λm(0, h) (5.89)<br />
m(0, λ q ) = λ 1−q m(0, 1) (5.90)<br />
m(0, |H|) = |H| 1−q<br />
q m(0, 1) (5.91)<br />
δ = q<br />
1 − q<br />
(5.92)<br />
The susceptibility follows from χ = ( )<br />
∂m<br />
∂h and the scaling rel<strong>at</strong>ion 5.83 leads<br />
θ<br />
λ 2q χ(λ p θ, λ q h) = λχ(θ, h) (5.93)<br />
In this equ<strong>at</strong>ion we use again h = 0, and again distinguish between temper<strong>at</strong>ures<br />
above and below the critical temper<strong>at</strong>ure. We find
5.4. CONSEQUENCES OF SCALING. 213<br />
and<br />
t > 0 : χ(t, 0) = t 1−2q<br />
p χ(1, 0) (5.94)<br />
t < 0 : χ(t, 0) = (−t) 1−2q<br />
p χ(−1, 0) (5.95)<br />
which shows th<strong>at</strong> the critical exponents are indeed the same above and below<br />
the critical point:<br />
γ = γ ′ =<br />
but th<strong>at</strong> the pre-factors are different.<br />
Specific he<strong>at</strong>.<br />
2q − 1<br />
p<br />
(5.96)<br />
( 2<br />
∂ g<br />
The specific he<strong>at</strong> follows from ch = −T ∂θ2 )<br />
where the temper<strong>at</strong>ure in<br />
h<br />
front is the real temper<strong>at</strong>ure, not the scaled one. This gives<br />
which then leads to<br />
and<br />
λ 2p ch(λ p θ, λ q h) = λch(θ, h) (5.97)<br />
t > 0 : ch(t, 0) = t 1−2p<br />
p ch(1, 0) (5.98)<br />
t < 0 : ch(t, 0) = (−t) 1−2p<br />
p ch(−1, 0) (5.99)<br />
which shows again th<strong>at</strong> the critical exponents are indeed the same above and<br />
below the critical point:<br />
α = α ′ =<br />
and again th<strong>at</strong> the pre-factors are different.<br />
Scaling rel<strong>at</strong>ions.<br />
2p − 1<br />
p<br />
(5.100)<br />
Using the expressions for the critical exponents it is possible to derive rel<strong>at</strong>ions<br />
between these exponents by elimin<strong>at</strong>ing the variables p and q. For<br />
example, we have<br />
q<br />
(δ + 1)β = (<br />
1 − q<br />
+ 1)1 − q<br />
p<br />
1<br />
= = 2 − α (5.101)<br />
p<br />
and if we check this for the mean field exponents we see th<strong>at</strong> there indeed<br />
(δ + 1)β = 4 × 1<br />
2 = 2 − α indeed.<br />
Similarly we find th<strong>at</strong>
214 CHAPTER 5. CRITICAL EXPONENTS.<br />
q<br />
(δ − 1)β = (<br />
1 − q<br />
− 1)1 − q<br />
p<br />
= 2q − 1<br />
p<br />
= γ (5.102)<br />
which was already discovered by Widom before the formul<strong>at</strong>ion of scaling theory.<br />
It is easy to check th<strong>at</strong> the mean field exponents s<strong>at</strong>isfy this equality. Another<br />
equ<strong>at</strong>ion, named after Rushbrooke, is also easily derived:<br />
and<br />
α + 2β + γ = 2 (5.103)<br />
Note th<strong>at</strong> we can work backwards and find th<strong>at</strong> in mean field theory we have<br />
pMF = 1<br />
2<br />
qMF = 3<br />
4<br />
(5.104)<br />
(5.105)<br />
A final note, in this type of scaling we always find th<strong>at</strong> α = α ′ and γ = γ ′ ,<br />
because we assume th<strong>at</strong> the scaling is the same below and above the critical<br />
temper<strong>at</strong>ure.<br />
Comparison with experiment.<br />
We can look <strong>at</strong> the experimental d<strong>at</strong>a presented before, and construct the<br />
following table:<br />
type 1 − α+γ<br />
2 β 1 + γ<br />
β δ<br />
ferromagnetic Ni 0.39 0.33 5.0 4.2<br />
ferromagnetic F e 0.39 0.34<br />
anti-ferromagnetic RbMnF3 0.37 0.32<br />
liquid-gas Co2 0.34 0.34 4.5 4.2<br />
liquid-gas Xe 0.36 0.34 4.5 4.4<br />
mean field 0.5 0.5 3.0 3.0<br />
Ising, d=2 0.125 0.125 15 15<br />
Ising, d=3 0.31 0.31 5.0 5.0<br />
Heisenberg 0.38 0.38<br />
which is reasonable. One has to keep in mind th<strong>at</strong> critical exponents are very<br />
difficult to measure, and th<strong>at</strong> the error bars are large. If we take th<strong>at</strong> into<br />
account, we conclude th<strong>at</strong> the experimental results do not contradict scaling.<br />
Equalities or inequalities?<br />
The basis for the equalities can be discussed in a completely different manner.<br />
In chapter 2 we derived inequalities rel<strong>at</strong>ed to thermodynamic stability. For<br />
example, we found th<strong>at</strong>
5.4. CONSEQUENCES OF SCALING. 215<br />
CH − CM = T<br />
χT<br />
( )<br />
∂M<br />
2<br />
∂T H<br />
and since CM > 0 we can drop the term CM in the left hand side and find:<br />
CH ≥ T<br />
χT<br />
( )<br />
∂M<br />
2<br />
∂T H<br />
Near the critical point we have CH = c1|θ| −α , χT = c2|θ| −γ , and ( ∂M<br />
∂T<br />
c3|θ| β−1 , which gives<br />
c1|θ| −α ≥ T c23 |θ|<br />
c2<br />
2β+γ−2<br />
and this should be true even in the limit θ → 0, which leads to<br />
or<br />
as originally discovered by Rushbrooke.<br />
In a similar manner one can derive<br />
(5.106)<br />
(5.107)<br />
)<br />
H =<br />
(5.108)<br />
−α ≤ 2β + γ − 2 (5.109)<br />
α + 2β + γ ≥ 2 (5.110)<br />
γ ≥ β(δ − 1) (5.111)<br />
which was originally shown by Widom.<br />
The deriv<strong>at</strong>ion above is based on equilibrium thermodynamics. In scaling<br />
theory we describe a non-equilibrium st<strong>at</strong>e with fluctu<strong>at</strong>ions. But if we think<br />
of a given st<strong>at</strong>e with two phases present, where one phase is meta-stable, we<br />
can still apply the same kind of analysis to the meta-stable st<strong>at</strong>e, since it still<br />
corresponds to a minimum in the free energy.<br />
Explain this carefully.<br />
One can therefore argue th<strong>at</strong> the inequality 5.110 should also hold for the singular<br />
part of the free energy. Scaling theory confirms th<strong>at</strong>, but is also much<br />
more restrictive, because it forces the inequality to be an equality.<br />
Pair correl<strong>at</strong>ion function.<br />
In the previous chapter we discussed how the effects of a small perturb<strong>at</strong>ion<br />
are felt further away in the system. In Ginzburg-Landau theory it was<br />
possible to calcul<strong>at</strong>e the change in magnetiz<strong>at</strong>ion density caused by a perturb<strong>at</strong>ion<br />
<strong>at</strong> the origin. The function in equ<strong>at</strong>ion 4.56 is an example of a
216 CHAPTER 5. CRITICAL EXPONENTS.<br />
pair correl<strong>at</strong>ion function, i.e. a function describing how the magnetiz<strong>at</strong>ion <strong>at</strong><br />
two different sites is rel<strong>at</strong>ed. The formal definition of the pair correl<strong>at</strong>ion function<br />
Γ is<br />
Γ(T, ⃗r, ⃗r ′ ) = ⟨(m(⃗r) − m0)(m(⃗r ′ ) − m0)⟩ (5.112)<br />
where m0 is the average value of the magnetiz<strong>at</strong>ion. The pair correl<strong>at</strong>ion function<br />
will also depend on the temper<strong>at</strong>ure, of course, and on other thermodynamic<br />
variables. In a homogeneous system the pair correl<strong>at</strong>ion function will<br />
only depend on the distance |⃗r − ⃗r ′ | between the points. In Ginzburg-Landau<br />
theory we found th<strong>at</strong> the pair correl<strong>at</strong>ion is proportional to the function calcul<strong>at</strong>ed<br />
in equ<strong>at</strong>ion 4.56, since this function gives us the extra magnetiz<strong>at</strong>ion<br />
when the origin has a large magnetiz<strong>at</strong>ion (driven by a delta function applied<br />
field). Therefore, in th<strong>at</strong> case:<br />
Two critical exponents.<br />
Γ(T, r) ∝ 1 r<br />
e− ξ (5.113)<br />
r<br />
The critical exponents ν and ν ′ describe the behavior of the correl<strong>at</strong>ion<br />
length ξ close to the critical temper<strong>at</strong>ure:<br />
T > Tc : ξ(T ) ∝ (T − Tc) −ν<br />
T < Tc : ξ(T ) ∝ (Tc − T ) −ν′<br />
(5.114)<br />
(5.115)<br />
and again we expect th<strong>at</strong> scaling theory will give ν = ν ′ . Ginzburg-Landau<br />
theory gives ν = ν ′ = 1<br />
2 . At the critical temper<strong>at</strong>ure the correl<strong>at</strong>ion length is<br />
infinite, and the decay of the pair correl<strong>at</strong>ion function is given by the r dependence<br />
in front. Another critical exponent η is defined by:<br />
Γ(Tc, r) ∝ r 2−d−η<br />
(5.116)<br />
where d is the number of dimensions in the problem. Using the results of one<br />
of the problems in the previous chapter we see th<strong>at</strong> η = 0 in all dimensions for<br />
mean field (Ginzburg-Landau) theory.<br />
Momentum space.<br />
In many cases we need the Fourier transform of the pair correl<strong>at</strong>ion function,<br />
and calcul<strong>at</strong>e<br />
∫<br />
d d re ı⃗q·⃗r Γ(T, r) (5.117)
5.5. SCALING OF THE PAIR CORRELATION FUNCTION. 217<br />
which for small q is proportional to q η−2 . This is often used as the definition<br />
for η, but is of course completely equivalent to our definition. L<strong>at</strong>er we will<br />
discuss the exact definition of the structure factor, which is almost the same<br />
as the Fourier transform above. The difference is not essential for the current<br />
discussion.<br />
Rel<strong>at</strong>ion to the susceptibility.<br />
Since the pair correl<strong>at</strong>ion function describes the average of fluctu<strong>at</strong>ions, we<br />
can use the techniques from the end of Chapter 2 to rel<strong>at</strong>e this to a response<br />
function, which in this case is the susceptibility. In order to find the susceptibility<br />
we need to take an average over all space, and hence we find th<strong>at</strong><br />
∫<br />
χ(T ) ∝ d d rΓ(T, r) (5.118)<br />
or th<strong>at</strong> the susceptibility is proportional to the Fourier transform of the pair<br />
correl<strong>at</strong>ion function <strong>at</strong> q=0.<br />
Make sure th<strong>at</strong> you understand the previous argument based on chapter 2.<br />
5.5 Scaling of the pair correl<strong>at</strong>ion function.<br />
We assume th<strong>at</strong> near the critical point the correl<strong>at</strong>ion length is the only important<br />
length scale. As a result, the pair correl<strong>at</strong>ion function has to be a function<br />
and we can write<br />
of the r<strong>at</strong>io r<br />
ξ<br />
Γ(T, r) = f( r<br />
ξ(T ) )ξu (T ) = ˜ f( r<br />
ξ(T ) )ru<br />
(5.119)<br />
At the critical temper<strong>at</strong>ure the correl<strong>at</strong>ion length is infinite and therefore the<br />
second part of the equ<strong>at</strong>ion above gives<br />
Γ(Tc, r) = ˜ f(0)r u<br />
(5.120)<br />
which shows th<strong>at</strong> u = 2 − d − η. The Fourier transform of the pair distribution<br />
function is<br />
∫<br />
d d re ı⃗q·⃗r f( r<br />
ξ(T ) )ξu (T ) (5.121)<br />
which can be transformed using ⃗r = ξ(T )⃗x and ⃗s = ξ(T )⃗q into<br />
∫<br />
d d xe ı⃗s·⃗x f(x)ξ u+d (T ) (5.122)
218 CHAPTER 5. CRITICAL EXPONENTS.<br />
Rel<strong>at</strong>ion between the susceptibility and the correl<strong>at</strong>ion length.<br />
Since the susceptibility is proportional to the q=0 component of the Fourier<br />
transform of the pair correl<strong>at</strong>ion function, we see th<strong>at</strong> near the critical point<br />
the susceptibility is proportional to<br />
χ(T ) ∝ ξ u+d = ξ 2−η ∝ |T − Tc| −ν(2−η)<br />
which shows the important rel<strong>at</strong>ion between the critical exponents:<br />
5.6 Hyper-scaling.<br />
(5.123)<br />
γ = ν(2 − η) (5.124)<br />
Near the critical point the correl<strong>at</strong>ion length is the only important length scale,<br />
and this has one final consequence. The energy needed to cre<strong>at</strong>e a fluctu<strong>at</strong>ion,<br />
independent of its size, will be a constant proportional to Tc. L<strong>at</strong>er we can<br />
derive in a microscopic picture th<strong>at</strong> it will be on the order of kBTc, but for our<br />
present discussion it is sufficient to assume th<strong>at</strong> it is a constant per fluctu<strong>at</strong>ion.<br />
The volume of a fluctu<strong>at</strong>ion is proportional to ξ d , and therefore we find th<strong>at</strong><br />
near the critical point the energy density scales like<br />
g(θ, h = 0) ∝ ξ −d (T ) (5.125)<br />
This assumption is called hyper-scaling. Near the critical point we therefore<br />
have<br />
g(θ, h = 0) ∝ θ νd<br />
From the energy scaling rel<strong>at</strong>ion 5.82 we find:<br />
and with t = λ p we get<br />
which shows th<strong>at</strong> νd = 1<br />
p , or<br />
(5.126)<br />
g(λ p , 0) = λg(1, 0) (5.127)<br />
g(t, 0) = t 1<br />
p g(1, 0) (5.128)<br />
dν = 2 − α (5.129)<br />
which is NOT true for mean field theory combined with Ginzburg-Landau in<br />
three dimensions! The equ<strong>at</strong>ion above would give ν = 2<br />
3 , while Ginzburg-<br />
Landau gives ν = 1<br />
2 . Therefore, Ginzburg-Landau theory does not obey hyperscaling,<br />
and cannot be a complete theory <strong>at</strong> the transition. Note, however, th<strong>at</strong><br />
in four dimensions the equ<strong>at</strong>ion above is obeyed. Perhaps something is special<br />
in four dimensions!
5.7. VALIDITY OF GINZBURG-LANDAU THEORY. 219<br />
5.7 Validity of Ginzburg-Landau theory.<br />
Criterium for validity.<br />
The theory put forward by Ginzburg and Landau is only a start for a description<br />
of the role of fluctu<strong>at</strong>ions. The critical exponents in this theory have<br />
the mean field values, and in real experiments those values are different. But<br />
it is still a fair question to ask when this theory is a good approxim<strong>at</strong>ion. One<br />
criterium is the following. If we have fluctu<strong>at</strong>ions in a parameter, a theory th<strong>at</strong><br />
tre<strong>at</strong>s these as a perturb<strong>at</strong>ion needs to require th<strong>at</strong> the fluctu<strong>at</strong>ions are small,<br />
or th<strong>at</strong><br />
∆M ≪ M (5.130)<br />
which, obviously, can only be applied for temper<strong>at</strong>ures below the critical temper<strong>at</strong>ure.<br />
Note th<strong>at</strong> we use the total magnetiz<strong>at</strong>ion in this formula. In Ginzburg-<br />
Landau we have<br />
(∆M) 2 ∫<br />
= d 3 rd 3 r ′ ⟨(m(⃗r)−m0)(m(⃗r ′ ∫<br />
)−m0)⟩ = V d 3 ∫<br />
rΓ(T, r) ∝ V d 3 r 1<br />
r<br />
(5.131)<br />
which leads to something like<br />
(∆M) 2 = C1V ξ 2 (T ) (5.132)<br />
where C1 is a constant independent of the temper<strong>at</strong>ure.<br />
Inequality for validity.<br />
The total magnetiz<strong>at</strong>ion is simple, and we have M = V m0. Therefore we<br />
have a condition of the form<br />
C1V ξ 2 (T ) ≪ m 2 0V 2<br />
e− r<br />
ξ<br />
(5.133)<br />
which is always obeyed in the thermodynamic limit V → ∞. But th<strong>at</strong> is not<br />
correct. The thermodynamic limit should always be taken in the end,<br />
and we first need to consider the behavior as a function of temper<strong>at</strong>ure. If we<br />
approach the critical points the correl<strong>at</strong>ion length becomes infinite, signalling<br />
th<strong>at</strong> interchanging limits can be dangerous! Our criterium for validity should<br />
be th<strong>at</strong> the fluctu<strong>at</strong>ions measured in a relevant volume must be small. Since<br />
the range of fluctu<strong>at</strong>ions are given by the correl<strong>at</strong>ion length, the obvious choice<br />
is to consider a volume determined by this correl<strong>at</strong>ion length! Therefore, we<br />
take V = ξ 3 (T ).<br />
Since the magnetiz<strong>at</strong>ion itself is determined by the critical exponent β we<br />
have the following inequality th<strong>at</strong> gives a necessary criterium for the validity of<br />
Ginzburg-Landau theory:
220 CHAPTER 5. CRITICAL EXPONENTS.<br />
or<br />
C1 ≪ m 2 0ξ (5.134)<br />
C2 ≪ θ 2β−ν<br />
(5.135)<br />
where all constants are lumped together in C2. Using the appropri<strong>at</strong>e values<br />
β = ν = 1<br />
2 we get<br />
|θ| ≫ C 2 2<br />
(5.136)<br />
which shows th<strong>at</strong> close to the critical point Ginzburg-Landau theory is not<br />
self-consistent, i.e. it predicts fluctu<strong>at</strong>ions which are so large th<strong>at</strong> the basic<br />
assumptions of the theory are not valid anymore. Higher order gradients become<br />
important, and the series as an expansion in terms of gradients does not converge<br />
anymore.<br />
Necessary or sufficient.<br />
Strictly speaking the criterium above is only a necessary criterium for the<br />
validity of the theory. In general, however, it will also be sufficient. If |θ| ≫ C 2 2<br />
higher order gradients will be less and less important, and an expansion in terms<br />
of gradients is appropri<strong>at</strong>e. But this is only a st<strong>at</strong>ement about very high orders.<br />
Ginzburg-Landau theory assumes th<strong>at</strong> the lowest order gradient is the dominant<br />
term. This is not guaranteed, it could be possible th<strong>at</strong> all the action is in the<br />
next term. Therefore, the condition we have derives is not sufficient in general.<br />
In practice, however, one will often find th<strong>at</strong> the lowest order term is dominant,<br />
and th<strong>at</strong> the inequality derived above is also sufficient. An important exception<br />
does occur in all those cases where symmetry arguments make this term exactly<br />
zero! One should always keep th<strong>at</strong> in mind.<br />
Use of Ginzburg-Landau theory.<br />
Nevertheless, the theory of Ginzburg and Landau is an improvement, and<br />
will give a better description of the system as long as the inequality above<br />
is obeyed. Hence we have the following picture. Far away from the critical<br />
point mean field theory is a good description of the system under consider<strong>at</strong>ion.<br />
When we get closer to the critical point, we can improve by including gradient<br />
terms in the expansion of the free energy. Very close to the critical point we<br />
need a different theory, however. In this chapter we have given a model of the<br />
free energy th<strong>at</strong> describes the behavior <strong>at</strong> the critical point, but the critical<br />
exponents are parameters in th<strong>at</strong> model (following from the values of p and q<br />
in equ<strong>at</strong>ion 5.82). In order to derive the values of these critical exponents we<br />
need a different theory, however. This is usually done in st<strong>at</strong>istical mechanics,<br />
where we include the microscopic n<strong>at</strong>ure of the m<strong>at</strong>erial.
5.8. SCALING OF TRANSPORT PROPERTIES. 221<br />
Range of validity.<br />
The inequality |θ| ≫ C 2 2 gives us also an estim<strong>at</strong>e of the range of validity.<br />
In order to calcul<strong>at</strong>e numerical values we do need to know the constant C2, and<br />
hence we need the details of the model. It is possible th<strong>at</strong> Ginzburg-Landau<br />
theory is valid essentially everywhere for our measurement (apart from giving<br />
the values of the critical exponents), but there are also situ<strong>at</strong>ions where the<br />
excluded range is so large th<strong>at</strong> this theory is never applicable.<br />
Different dimensions.<br />
We can also apply the previous argument for a system in d dimensions. In<br />
th<strong>at</strong> case the pair correl<strong>at</strong>ion length is proportional to r 2−d (see problem in<br />
previous chapter), and the volume V is V = ξ d (T ). This leads to<br />
or<br />
C3 ≪ m 2 0ξ d−2<br />
C4 ≪ θ 2β−(d−2)ν<br />
which is valid even <strong>at</strong> the critical point as long as 2β − (d − 2)ν < 0 or<br />
Marginal dimension.<br />
(5.137)<br />
(5.138)<br />
d > d ∗ = 2(1 + β<br />
) (5.139)<br />
ν<br />
The constant d ∗ above is called the marginal dimension. In Ginzburg-<br />
Landau it has the value d ∗ = 4. The condition d > 4 can be interpreted as<br />
follows. If the number of dimensions is larger than four, mean field theory will<br />
be valid (as long as the conditions described in the necessary versus sufficient<br />
paragraph are met). This is the basis for the famous epsilon expansion, where<br />
all quantities are expanded in a power series in terms of d = d ∗ − ϵ and the limit<br />
d → 3 is taken to get results for the real world.<br />
If we are exactly <strong>at</strong> the marginal dimension, mean field theory does not<br />
necessarily describe the physics of the problem, but the difference can only be<br />
in terms th<strong>at</strong> diverge in a logarithmic fashion, i.e. slower than any arbitrary<br />
power of θ. Details can be quite messy in th<strong>at</strong> case.<br />
5.8 Scaling of transport properties.<br />
Different basis for scaling.
222 CHAPTER 5. CRITICAL EXPONENTS.<br />
We have introduced the scaling hypothesis by using the free energy. This<br />
seems an appropri<strong>at</strong>e choice, since the free energy is the most important quantity<br />
from which all other thermodynamic variables can be derived. Using two basic<br />
scaling parameters, p and q, we were able to derive all critical exponents. The<br />
important observ<strong>at</strong>ion is th<strong>at</strong> there are only two independent variables, and<br />
any set of two exponents can be used as a basic set. Of course, as soon as we<br />
introduce more st<strong>at</strong>e variables, we obtain more possibilities. For example, if<br />
we also allow pressure to vary in our free energy, more complex behavior can<br />
be observed. At first, increasing the pressure in a magnet will simply shift the<br />
critical temper<strong>at</strong>ure, but not change the critical exponents. In th<strong>at</strong> case the<br />
scaling of the pressure is irrelevant, and this can be expressed by setting the<br />
power r in λ r equal to zero. But we could also get things like tri-critical points,<br />
and now the story is different.<br />
The other question one can ask is if the free energy is the only possible<br />
starting point. The answer is no, any function of two variables can be used<br />
as a basis for scaling. One common choice is to start with the pair correl<strong>at</strong>ion<br />
function, and write:<br />
Γ(λ x r, λ y θ) = λΓ(r, θ) (5.140)<br />
and derive all critical exponents from the values of x and y. This is not hard to<br />
do, since we know th<strong>at</strong> the pair correl<strong>at</strong>ion function has the form<br />
with u = 2 − d − η, and therefore<br />
which yields:<br />
and<br />
with solutions<br />
and<br />
Γ(T, r) = f( r<br />
ξ(T ) )ξu (T ) (5.141)<br />
Γ(λ x r, λ y θ) = f(λ x+yν r<br />
ξ(T ) )λ−yuν ξ u (T ) (5.142)<br />
y = − 1<br />
uν =<br />
1<br />
dν − (2 − η)ν<br />
x + yν = 0 (5.143)<br />
−yuν = 1 (5.144)<br />
= 1<br />
2β =<br />
Show th<strong>at</strong> 2β = dν − (2 − η)ν.<br />
p<br />
2(1 − q)<br />
(5.145)
5.8. SCALING OF TRANSPORT PROPERTIES. 223<br />
x = −yν =<br />
2 − α<br />
2dβ =<br />
1<br />
2d(1 − q)<br />
which shows th<strong>at</strong> the two formul<strong>at</strong>ions are equivalent.<br />
Time dependent phenomena.<br />
(5.146)<br />
In Ginzburg-Landau theory we describe the thermodynamically stable st<strong>at</strong>e<br />
when we have applied a field <strong>at</strong> the origin. In a real process we can imagine th<strong>at</strong><br />
<strong>at</strong> time τ = 0 we apply a magnetiz<strong>at</strong>ion <strong>at</strong> the origin in a non-magnetic st<strong>at</strong>e.<br />
The effects of this seed <strong>at</strong> the origin will propag<strong>at</strong>e outwards, and regions away<br />
from the origin will become magnetized. The magnetized region will expand<br />
and finally in the limit τ → ∞ approach the Landau-Ginzburg solution. Since<br />
this process is a non-equilibrium process, it is hard to describe in terms of a free<br />
energy, but one can easily describe the pair correl<strong>at</strong>ion function as a function<br />
of time τ by Γ(r, τ, θ).<br />
Diffusion equ<strong>at</strong>ion.<br />
As we will see l<strong>at</strong>er, the change in magnetiz<strong>at</strong>ion over time is described by<br />
a diffusion equ<strong>at</strong>ion:<br />
( )<br />
∂m<br />
= D<br />
∂τ θ,⃗r<br />
⃗ ∇ 2 m (5.147)<br />
where D is the diffusion constant. In our example where we start with a delta<br />
function <strong>at</strong> the origin, m(r, τ = 0) = M0δ(⃗r) the solution is simple:<br />
m(⃗r, τ) =<br />
M0<br />
(4πDτ) 3<br />
r2 −<br />
e 4Dτ (5.148)<br />
2<br />
where we see th<strong>at</strong> the size of the magnetized area increases like √ τ as expected.<br />
This function m(⃗r, τ) is proportional to Γ(r, τ, θ).<br />
Time scaling.<br />
We now take into account the scaling of the time variable in a straightforward<br />
manner:<br />
Γ(λ x r, λ z τ, λ y θ) = λΓ(r, τ, θ) (5.149)<br />
which introduces a new parameter z and hence one new independent critical<br />
exponent.<br />
Scaling of the diffusion constant.
224 CHAPTER 5. CRITICAL EXPONENTS.<br />
The pair correl<strong>at</strong>ion function also obeys the diffusion equ<strong>at</strong>ion, given by:<br />
( )<br />
∂Γ<br />
= D<br />
∂τ θ,⃗r<br />
⃗ ∇ 2 Γ (5.150)<br />
where the diffusion constant depends on the temper<strong>at</strong>ure, and near the critical<br />
point is described by an exponent κ:<br />
Rel<strong>at</strong>ion between z and κ.<br />
and<br />
D(θ) ∝ θ −κ<br />
The deriv<strong>at</strong>ives in the diffusion equ<strong>at</strong>ion also scale with λ and we have<br />
which leads to<br />
which gives<br />
λ z<br />
(5.151)<br />
( )<br />
∂Γ<br />
(λ<br />
∂τ<br />
x r, λ z τ, λ y ( )<br />
∂Γ<br />
θ) = λ (r, τ, θ) (5.152)<br />
∂τ<br />
λ 2x ⃗ ∇ 2 Γ(λ x r, λ z τ, λ y θ) = λ ⃗ ∇ 2 Γ(r, τ, θ) (5.153)<br />
D(λ y θ) = D(θ)λ 2x−z<br />
κ =<br />
z − 2x<br />
y<br />
= z + 2yν<br />
y<br />
(5.154)<br />
(5.155)<br />
This shows th<strong>at</strong> we can obtain the value of z from a measurement of the diffusion<br />
constant and a knowledge of two other critical exponents.<br />
Consequences.<br />
If the time scaling in the pair correl<strong>at</strong>ion function is not important, and z=0,<br />
we find th<strong>at</strong> κ = 2ν and in Ginzburg-Landau theory this gives κ = 1. Therefore,<br />
we expect the diffusion constant to become infinitely large <strong>at</strong> the critical point.<br />
Any small perturb<strong>at</strong>ion propag<strong>at</strong>es with very large speed near the critical point.<br />
This is, of course, not realistic, and implies th<strong>at</strong> we need to use a rel<strong>at</strong>ivistic<br />
form of the equ<strong>at</strong>ions. Also, the diffusion equ<strong>at</strong>ion has a built-in assumption<br />
th<strong>at</strong> the diffusion is slow. So details near the critical point are again messy,<br />
but the fact th<strong>at</strong> the diffusion constant increases when we approach the critical<br />
point can easily be measured.
5.9. EXTRA QUESTIONS. 225<br />
5.9 Extra questions.<br />
1. Wh<strong>at</strong> is the physics determining the n<strong>at</strong>ure of the critical exponents?<br />
2. Wh<strong>at</strong> is universality?<br />
3. Wh<strong>at</strong> is the meaning of universality?<br />
4. Superconductivity example. Discuss the rel<strong>at</strong>ion between phase and vector<br />
potential.<br />
5. Wh<strong>at</strong> is mean field theory and why is it important?<br />
6. Can we find critical exponents in finite systems?<br />
7. How do we measure critical exponents?<br />
8. Wh<strong>at</strong> are scaled variables?<br />
9. Wh<strong>at</strong> is the meaning of the fluctu<strong>at</strong>ion term in the energy?<br />
10. Wh<strong>at</strong> does scaling of the free energy mean?<br />
11. Why do we only scale the fluctu<strong>at</strong>ion part?<br />
12. Why are there only two independent critical exponents?<br />
13. Wh<strong>at</strong> is the physical meaning of inequalities versus equalities?<br />
14. Wh<strong>at</strong> is the pair correl<strong>at</strong>ion function?<br />
15. Wh<strong>at</strong> does scaling of the pair correl<strong>at</strong>ion function mean?<br />
16. Wh<strong>at</strong> is hyper-scaling?<br />
17. Wh<strong>at</strong> is a marginal dimension?<br />
18. Could we start expansions starting <strong>at</strong> infinite dimensions?<br />
19. Wh<strong>at</strong> are consequences of mean field theory being correct in 4d?<br />
20. Explain time scaling.<br />
21. Can the diffusion constant be infinite?
226 CHAPTER 5. CRITICAL EXPONENTS.<br />
5.10 Problems for chapter 5<br />
Problem 1.<br />
Derive equ<strong>at</strong>ion 5.111.<br />
Problem 2.<br />
Show th<strong>at</strong> equ<strong>at</strong>ion 5.117 for small q is proportional to q η−2 .<br />
Problem 3.<br />
Derive equ<strong>at</strong>ion 5.154<br />
Problem 4.<br />
A sequence of critical exponents ∆l , ∆ ′ l<br />
is defined by<br />
( l ∂ G<br />
∂H l<br />
) ( ) −∆l<br />
T<br />
∝ − 1<br />
T<br />
Tc<br />
( ∂l−1G ∂H l−1<br />
)<br />
T<br />
(5.156)<br />
for l > 0 and T > Tc. The primed exponents are for T < Tc, where the pre-<br />
factor is<br />
(<br />
1 − T<br />
Tc<br />
) −∆ ′<br />
l<br />
. These exponents are called gap exponents. Show th<strong>at</strong><br />
∆ ′ 1 = 2 − α ′ − β and ∆ ′ 2 = β + γ ′ . Show th<strong>at</strong> in mean field theory ∆ ′ l<br />
all values of l.<br />
Problem 5.<br />
= 3<br />
2 for<br />
The scaling laws yield the following identity for the magnetiz<strong>at</strong>ion m as<br />
a function of scaled temper<strong>at</strong>ure θ and magnetic field h: λ q m(λ p θ, λ q h) =<br />
λm(θ, h). In terms of the critical temper<strong>at</strong>ure Tc we have θ =<br />
T −Tc<br />
Tc<br />
. The ex-<br />
ponents p and q are a property of the system under consider<strong>at</strong>ion. We measure<br />
m as a function of h for several temper<strong>at</strong>ures θ. If we plot these measurements<br />
in the m versus h plane we get a series of distinct curves. Suppose we now plot<br />
the d<strong>at</strong>a in a different manner. For a series <strong>at</strong> temper<strong>at</strong>ure θ we plot the d<strong>at</strong>a<br />
points <strong>at</strong> |θ| A h with value |θ| B m. How do we choose A and B in terms of p and<br />
q in such a manner th<strong>at</strong> all d<strong>at</strong>a points now fall on one universal curve?<br />
Problem 6.<br />
The scaling law yields the following identity for the Gibbs free energy per<br />
unit volume g as a function of scaled temper<strong>at</strong>ure θ and magnetic field h:
5.10. PROBLEMS FOR CHAPTER 5 227<br />
g(λpθ, λqh) = λg(θ, h). In terms of the critical temper<strong>at</strong>ure Tc we have θ =<br />
T −Tc . The exponents p and q are a property of the system under consider<strong>at</strong>ion.<br />
Tc<br />
Show th<strong>at</strong> we can write<br />
g(θ, h) = |θ| x G±( h<br />
)<br />
|θ| y<br />
where the plus or minus sign rel<strong>at</strong>es to being above or below the critical<br />
temper<strong>at</strong>ure. Rel<strong>at</strong>e the powers x and y to the powers p and q. Find an<br />
expression for G+ th<strong>at</strong> is consistent with Curies law for T > Tc, th<strong>at</strong> is m =<br />
, and wh<strong>at</strong> does th<strong>at</strong> tell you about the values of p and q.<br />
m0h<br />
T −Tc<br />
Problem 7.<br />
Near a liquid to gas critical point, <strong>at</strong> pressure pc and temper<strong>at</strong>ure Tc we can<br />
write the Gibbs free energy per unit volue, g, in terms of the reduced variables<br />
π = p−pc<br />
T −Tc<br />
and θ = . The dominant part of the free energy obeys the scaling<br />
pc Tc<br />
rel<strong>at</strong>ion:<br />
λg(π, θ) = g(λ z π, λ y θ)<br />
Find the value of the critical exponents γ and γ ′ in terms of z and y.<br />
Problem 8.<br />
Starting with equ<strong>at</strong>ion (5.83) show th<strong>at</strong> one can write<br />
m(θ, h) = |θ| β M±( h<br />
)<br />
|θ| βδ<br />
We use this to plot experimental d<strong>at</strong>a in the form<br />
versus<br />
m(θ, h)<br />
|θ| β<br />
h<br />
|θ| βδ<br />
which has the unknown parameters Tc, β, and δ. If these parameters are<br />
chosen correctly, the d<strong>at</strong>a points fall on two curves, one above Tc and one below.<br />
Using some spreadsheet applic<strong>at</strong>ion and the d<strong>at</strong>a in the table below, make<br />
this plot, and find your best values for the parameters.<br />
The following table (copied from Huang) shows for nickel the magnetiz<strong>at</strong>ion<br />
as a function of temper<strong>at</strong>ure and field:
228 CHAPTER 5. CRITICAL EXPONENTS.<br />
Temper<strong>at</strong>ure (Kelvin) Applied field (Oersted) Magnetiz<strong>at</strong>ion density (emu/g)<br />
678.86 4, 160 0.649<br />
678.86 10, 070 1.577<br />
678.86 17, 775 2.776<br />
653.31 2, 290 0.845<br />
653.31 4, 160 1.535<br />
653.31 6, 015 2.215<br />
653.31 10, 070 3.665<br />
653.31 14, 210 5.034<br />
653.31 17, 775 6.145<br />
653.31 21, 315 7.211<br />
629.53 1, 355 6.099<br />
629.53 3, 230 9.287<br />
629.53 6, 015 11.793<br />
629.53 10, 070 13.945<br />
629.53 14, 210 15.405<br />
629.53 17, 775 16.474<br />
614.41 1, 820 19.43<br />
614.41 3, 230 20.00<br />
614.41 6, 015 20.77<br />
614.41 10, 070 21.67<br />
614.41 14, 210 22.36<br />
614.41 17, 775 22.89<br />
601.14 3, 230 25.34<br />
601.14 6, 015 25.83<br />
601.14 10, 070 26.40<br />
601.14 14, 210 26.83<br />
601.14 17, 775 27.21<br />
Problem 9.<br />
Show th<strong>at</strong> 5.148 is indeed the solution of the diffusion equ<strong>at</strong>ion 5.147
Chapter 6<br />
Transport in<br />
<strong>Thermodynamics</strong>.<br />
Equilibrium thermodynamics demands the absence of macroscopic currents. It<br />
does not require th<strong>at</strong> the m<strong>at</strong>erial is exactly the same everywhere. Energy can<br />
flow back and forth, but we can describe a precise average value of the energy<br />
per unit volume <strong>at</strong> each point in space. In chapter two we have seen th<strong>at</strong> we<br />
can describe the probability of local fluctu<strong>at</strong>ions using the local entropy. We<br />
assumed th<strong>at</strong> measurement times were very long, and th<strong>at</strong> the net effect of<br />
the fluctu<strong>at</strong>ions is zero. We also assumed th<strong>at</strong> fluctu<strong>at</strong>ions <strong>at</strong> different points<br />
in space are not correl<strong>at</strong>ed. In chapter four we discussed how fluctu<strong>at</strong>ions are<br />
localized in space, and how the extent diverges near the critical point. In chapter<br />
five we used this correl<strong>at</strong>ion to describe its effects on the energy and on critical<br />
exponents.<br />
In this chapter we discuss another aspect of correl<strong>at</strong>ions between differences<br />
from average values. When we apply a small external field, we cre<strong>at</strong>e a continuous<br />
change in thermodynamic variables. This gives rise to net macroscopic<br />
currents. Such currents are difficult to describe in general, but here we take one<br />
limit in which we can make predictions. We assume th<strong>at</strong> the currents are so<br />
small, th<strong>at</strong> diffusion times are much larger than fluctu<strong>at</strong>ion times. Th<strong>at</strong> leads<br />
to the concept of local thermodynamic equilibrium.<br />
In this chapter we start with describing transport in general terms, and then<br />
giving examples of thermo-electric effects. In the next section we define the cases<br />
where we can discuss these effects based on thermodynamics. We develop the<br />
transport equ<strong>at</strong>ions, and the equ<strong>at</strong>ion for the gener<strong>at</strong>ion of entropy in such a<br />
non-equilibrium situ<strong>at</strong>ion. Next we follow Onsager, and show how the transport<br />
coefficients are rel<strong>at</strong>ed to the probabilities for local fluctu<strong>at</strong>ions. We then apply<br />
our formulas to cases where electronic transport takes place via diffusion, but<br />
now in the presence of a thermal gradient.<br />
229
230 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
6.1 Introduction.<br />
Wh<strong>at</strong> experimentalists do.<br />
Experimental sciences all have one thing in common. You are trying to<br />
discover something about a system. There are two ways of doing th<strong>at</strong>. You can<br />
sit down and observe the system. This is always the best first step, and one<br />
cannot underestim<strong>at</strong>e the importance of this step. But there comes a point in<br />
time where observing alone does not give much new inform<strong>at</strong>ion on a time scale<br />
of interest. For example, consider a system th<strong>at</strong> shows a very peculiar behavior<br />
one hour every three hundred years. This is longer th<strong>at</strong> our life time, and we<br />
will be imp<strong>at</strong>ient. It is possible for gener<strong>at</strong>ions of scientists to study this system<br />
and write down wh<strong>at</strong> they observe. After a thousand years or so we will have<br />
good inform<strong>at</strong>ion about th<strong>at</strong> system. The ”we” in this context is mankind, not<br />
an individual. Our culture, however, does not have th<strong>at</strong> p<strong>at</strong>ience, and we will<br />
do something about it.<br />
The second step in any experimental discovery is the following. You poke<br />
the system and see how it responds. Here we can make several choices. We<br />
can poke it gently or violently. Wh<strong>at</strong> is the difference? If we poke it gently,<br />
the system will return to its original st<strong>at</strong>e or to a nearby st<strong>at</strong>e which we still<br />
identify as being part of the original system. If we poke it violently, the system<br />
afterwards is completely different from the system before.<br />
Give examples of these two types of investig<strong>at</strong>ions.<br />
One of the problems with distinguishing gentle and violent is th<strong>at</strong> we do not<br />
know a priori where the border between these two is. Of course, we will have<br />
some idea from our careful observ<strong>at</strong>ions in step one. But even th<strong>at</strong> might not be<br />
sufficient. In th<strong>at</strong> case we start with the smallest poke we can do, and observe<br />
the reactions. We can only hope th<strong>at</strong> we did it gentle enough.<br />
How do these general remarks transl<strong>at</strong>e into the realm of physics? Especially<br />
in the context of thermodynamics. Th<strong>at</strong> is not too difficult. For example, we<br />
have a system and increase its temper<strong>at</strong>ure. We wait until the system is again<br />
in equilibrium.<br />
How long do we have to wait?<br />
After the waiting time is over we measure wh<strong>at</strong>ever we want to measure.<br />
Can we measure without perturbing the system?
6.1. INTRODUCTION. 231<br />
From experiments like the one sketched above we find equ<strong>at</strong>ions of st<strong>at</strong>e.<br />
For example, the previous experiment could have given us V (T, p) <strong>at</strong> standard<br />
pressure. We can then increase the pressure, too, and map out the whole curve.<br />
We will probably discover phase transitions in the m<strong>at</strong>erial. At th<strong>at</strong> point we<br />
need to make a judgement. For example, we see the change from ice to vapor.<br />
Is the vapor the same system as the ice or not? The answer depends on the<br />
context of our problem! Although the fact th<strong>at</strong> ice and vapor are different seems<br />
trivial to us, it assumes th<strong>at</strong> we can do a measurement to distinguish between<br />
the two phases! We ”see” the difference, literally. But if we only measure the<br />
volume, we cannot see the difference, apart from an observed jump in volume.<br />
If you are in the situ<strong>at</strong>ion th<strong>at</strong> ice and vapor are the same system, wh<strong>at</strong> if you<br />
further increase the temper<strong>at</strong>ure and the molecules in the vapor break apart?<br />
Gradients.<br />
In the last case, are we still studying the same system?<br />
In the previous subsection we assumed th<strong>at</strong> the change in control variable<br />
(the poke) is uniform over the whole sample. Often th<strong>at</strong> is not true, and we all<br />
know th<strong>at</strong> keeping a temper<strong>at</strong>ure uniform is hard to do. Again, it is a m<strong>at</strong>ter<br />
of tolerance. If the devi<strong>at</strong>ions from uniformity are small enough according to<br />
a specific standard we can ignore them. But we can also turn this around.<br />
Suppose we make the control variable non-uniform on purpose. For example,<br />
we take a block of ice and he<strong>at</strong> one end with a torch. Wh<strong>at</strong> happens? Is it<br />
interesting?<br />
How large is large?<br />
The question th<strong>at</strong> always needs to be answered is how large is the applied<br />
difference of control variable. For example, when we apply a voltage across a<br />
m<strong>at</strong>erial we observe a current. If the voltage is too large, we see a spark. In<br />
between, there can be many interesting types of behavior, too. In non-linear<br />
dynamics you studied the transitions from uniform convection to convection<br />
rolls. There was period doubling en route to chaos. This is all very complic<strong>at</strong>ed,<br />
and not easy to describe. The only part where we have some hope of success is<br />
when the currents are uniform. But even there we need to be careful.<br />
Linear response only.<br />
We are all familiar with Ohm’s law. The current I in response to a voltage<br />
difference V is given by I = V<br />
R . The resistance R for a given piece of m<strong>at</strong>erial<br />
can be measured once and for all, and does not vary for th<strong>at</strong> piece of m<strong>at</strong>erial if
232 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
we change the applied voltage. Hence the resistance R is a m<strong>at</strong>erial’s property<br />
and only depends on the st<strong>at</strong>e of the m<strong>at</strong>erial. For example, it will depend<br />
on the temper<strong>at</strong>ure of the m<strong>at</strong>erial! It also depends on the geometry of the<br />
m<strong>at</strong>erial, however. Make a wire twice as long, and the resistance doubles. This<br />
dependence is easy to scale out, though. In stead of the current I we use the<br />
current density j, the amount of current passing through a unit area. In stead<br />
of the applied voltage we use the electric field E. When the field is constant we<br />
have V = EL where L is the length of the m<strong>at</strong>erial. If the cross sectional area<br />
is A, we arrive <strong>at</strong> j = I EL<br />
RA<br />
A = RA . The quantity L is called the resistivity of the<br />
m<strong>at</strong>erial, and is a property independent of the geometry. It is more useful to<br />
replace it by its inverse, the conductivity σ. Also, we have to realize th<strong>at</strong> both<br />
current and field are vector quantities and hence we write in general:<br />
⃗j = σ · ⃗ E (6.1)<br />
where in the general case σ is a tensor, a three by three m<strong>at</strong>rix in this case. In<br />
cubic m<strong>at</strong>erials, however, the conductivity is simply a number times the unit<br />
m<strong>at</strong>rix. But in layered m<strong>at</strong>erials we know th<strong>at</strong> the conductivity parallel to the<br />
layers is different from the conductivity perpendicular to the layers!<br />
Higher order terms.<br />
In general, the conductivity σ depends on the external field, and this dependence<br />
can be very complic<strong>at</strong>ed. We restrict ourselves to the low field limit. In<br />
th<strong>at</strong> case we can write:<br />
⃗j = σ1 · ⃗ E + σ2 · ⃗ E ⃗ E + · · · (6.2)<br />
and the coefficients in the Taylor expansion are independent of the field. The<br />
second order conductivity is also a m<strong>at</strong>erials property, but is not used very<br />
often. Hence we will assume th<strong>at</strong> this term is not important, and only use<br />
linear response.<br />
Discuss types of experiments th<strong>at</strong> measure higher order response terms.<br />
Typical transport situ<strong>at</strong>ions.<br />
There are a number of cases where transport is important. For example, we<br />
can study the flow of gas through pipes. From a technical point of view, this<br />
is a very important applic<strong>at</strong>ion. The important parameters are pressure and<br />
temper<strong>at</strong>ure. In this scenario one typically follows the fluid while it is flowing<br />
through the system, but th<strong>at</strong> is not necessary. I will not discuss this applic<strong>at</strong>ion<br />
here, since I know the basics only. Th<strong>at</strong> is my shortcoming, however, and does<br />
not reflect <strong>at</strong> all on the importance of the subject!
6.2. SOME THERMO-ELECTRIC PHENOMENA. 233<br />
The applic<strong>at</strong>ion I will use as an example is conductivity in a solid. If we<br />
apply an electric field we will induce a current. In solids the pressure is typically<br />
constant <strong>at</strong> standard pressure, and we will ignore pressure as a variable. This<br />
allows us to discuss the process in terms of densities as shown l<strong>at</strong>er. But temper<strong>at</strong>ure<br />
is an important control parameter, as is magnetic field. We will not<br />
include magnetic field effects extensively, however. This is a detail only, and is<br />
easy to incorpor<strong>at</strong>e, as long as one is aware of extra minus signs. These minus<br />
signs are due to the fact th<strong>at</strong> reversing a current is equivalent to time reversal,<br />
but th<strong>at</strong> only works when one also inverts the magnetic fields.<br />
Steady st<strong>at</strong>e.<br />
Another important point is th<strong>at</strong> we will only discuss transport in the steady<br />
st<strong>at</strong>e. We assume th<strong>at</strong> all transients due to changes in the control parameters<br />
have died out. Hence there are no time dependencies in the applied fields, and<br />
we can write ⃗ E(⃗r) and T (⃗r). Since there are currents, there is time dependence<br />
in corresponding extensive parameters. Something is moving! For example, in<br />
our case we follow the electrons in a solid. The electrons are the vehicle for the<br />
electric current. In a steady st<strong>at</strong>e, the number of electrons in a certain volume is<br />
constant, and equal numbers of electrons are moving in and out. We also assume<br />
th<strong>at</strong> there are enough collisions th<strong>at</strong> these electrons are in thermal equilibrium<br />
with the ions in the solid, and have the same temper<strong>at</strong>ure. Therefore, the length<br />
scale implead by the applied external field should be long compared to the mean<br />
free p<strong>at</strong>h of the electrons. Hence we do not discuss so-called ballistic transport,<br />
where electrons shoot through a medium without interactions. These kind of<br />
processes need a different framework of description.<br />
In our applic<strong>at</strong>ion he<strong>at</strong> is a transport of energy by both the electrons and by<br />
the ions in the solid, via l<strong>at</strong>tice vibr<strong>at</strong>ions. We will not discuss the l<strong>at</strong>ter part,<br />
but real experiments will always have to take th<strong>at</strong> contribution into account!<br />
6.2 Some thermo-electric phenomena.<br />
Electric conductivity.<br />
Electric conductivity is a very familiar example of a transport phenomenon.<br />
It is so familiar, th<strong>at</strong> we easily make mistakes in using conductivity! The naive<br />
approach is to apply a voltage difference to a sample and to measure the current.<br />
Why is this naive? Because we need to connect the sample to the voltage source<br />
and the current meter! These wires have their own resistance, but even more<br />
important, the contact between the wires and the sample have resistance.<br />
An improvement is to reverse the procedure. A known current is driven<br />
through a sample. In a steady st<strong>at</strong>e there is no accumul<strong>at</strong>ion of charge, and we<br />
know th<strong>at</strong> all the current leaving the current source will go through the sample.
234 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
A<br />
B<br />
T1 T2<br />
Figure 6.1: Essential geometry of a thermocouple.<br />
Well, <strong>at</strong> least as long as there are no leaks! We than take two different wires<br />
and connect these to the sample in between the current leads. We measure the<br />
voltage difference between these wires, and from th<strong>at</strong> derive the resistance. We<br />
do not worry about the resistivity of the contacts, since the driving current does<br />
not pass though them, and since the measuring current can be made extremely<br />
small. A very ingenious method indeed!<br />
Can there be a problem? Yes! Several, actually. Because there is resistance,<br />
there will be a production of thermal energy. In such a case we can either<br />
keep the temper<strong>at</strong>ure constant or do the measurement in an adiab<strong>at</strong>ic fashion.<br />
Would th<strong>at</strong> make a difference? It probably will! L<strong>at</strong>er we will show th<strong>at</strong> it<br />
indeed does make a difference, and th<strong>at</strong> the adiab<strong>at</strong>ic conductivity is different<br />
from the constant temper<strong>at</strong>ure conductivity.<br />
Wh<strong>at</strong> else could go wrong? The temper<strong>at</strong>ure in the steady st<strong>at</strong>e of the<br />
experiment might not be uniform! In th<strong>at</strong> case the contacts used for measuring<br />
the voltage difference could be <strong>at</strong> different temper<strong>at</strong>ures, which is known to have<br />
consequences!<br />
Thermocouples.<br />
Consider the setup in figure 6.1. Two different m<strong>at</strong>erials (A and B) are<br />
used. M<strong>at</strong>erial B is cut in the middle, and the ends are connected to input<br />
leads. At these input leads we can either apply a voltage or drive a current.<br />
The connections between m<strong>at</strong>erial A and B are <strong>at</strong> two different temper<strong>at</strong>ures<br />
T1 and T2. During the experiment there will also be he<strong>at</strong> going through both<br />
m<strong>at</strong>erials. Since thermal energy will also be produced everywhere there will be<br />
a he<strong>at</strong> flow out of the m<strong>at</strong>erials into the environment, represented by the arrows<br />
in figure 6.1. We will now use this device in several setups.
6.2. SOME THERMO-ELECTRIC PHENOMENA. 235<br />
Seebeck effect.<br />
We measure the voltage difference V between the input leads when there<br />
is a temper<strong>at</strong>ure difference between the endpoints. In this case there is no<br />
electric current through the thermocouple. The voltage difference depends on<br />
the m<strong>at</strong>erials A and B only, and for small temper<strong>at</strong>ure differences we have<br />
V = −αAB(T2 − T1) (6.3)<br />
where we define the signs in the following manner. If T1 > T2 and if the<br />
thermocouple were connected to a resistor, the coefficient αAB is positive if the<br />
resulting electric current goes from point 1 to point 2 through m<strong>at</strong>erial A and<br />
returns through m<strong>at</strong>erial B and the resistor. If we label the input leads by 1<br />
and 2 according to which end they are connected to, we have V = V2 − V1.<br />
The extra minus sign in the definition of α is due to the distinction between<br />
internal and external currents, which follow an opposite p<strong>at</strong>h. This effect, the<br />
Seebeck effect, was discovered in 1821! Note th<strong>at</strong> this is before the discovery of<br />
Ohm’s law!<br />
Peltier effect.<br />
How can we use the Seebeck effect in a practical manner?<br />
We can also do the opposite. We drive a current I through the thermocouple.<br />
The current is positive if we drive it similar to the previous subsection. The<br />
current is driven into lead 1 and extracted <strong>at</strong> lead 2, if positive. In this case<br />
a positive amount of he<strong>at</strong> Q is gener<strong>at</strong>ed <strong>at</strong> connection 1 between m<strong>at</strong>erials A<br />
and B and the same amount of he<strong>at</strong> is absorbed <strong>at</strong> the other end, connection<br />
2. In addition to this he<strong>at</strong> flow between 2 and 1 there is, of course, also extra<br />
he<strong>at</strong> gener<strong>at</strong>ed all along the p<strong>at</strong>h, because we have both electric and thermal<br />
currents. But the Peltier effect only rel<strong>at</strong>es the he<strong>at</strong> flow between 1 and 2 and<br />
we have<br />
Q = ΠABI (6.4)<br />
This effect was discovered only ten years or so after the Seebeck effect. It has<br />
one very important applic<strong>at</strong>ion. We can use the Peltier effect to cool a sample<br />
connected to lead 2!<br />
Kelvin found an empirical rel<strong>at</strong>ion between the Seebeck effect and the Peltier<br />
effect in 1854:<br />
ΠAB = T αAB<br />
(6.5)
236 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
Thomson he<strong>at</strong>.<br />
Suppose we have both a he<strong>at</strong> flow and an electric current through the thermocouple.<br />
In th<strong>at</strong> case he<strong>at</strong> is gener<strong>at</strong>ed <strong>at</strong> every point along the thermocouple.<br />
This is represented by the arrows in figure 6.1. Suppose the distance along m<strong>at</strong>erial<br />
A (or B) is measured by a parameter x. The he<strong>at</strong> gener<strong>at</strong>ed <strong>at</strong> position<br />
x in an interval ∆x is given by ∆Q, and Thomson found th<strong>at</strong><br />
He also derived (empirically!) th<strong>at</strong><br />
∆Q = τI dT<br />
∆x (6.6)<br />
dx<br />
τA − τB = T dαAB<br />
dT<br />
(6.7)<br />
Note th<strong>at</strong> the coefficients defined so far always depend on the difference between<br />
two m<strong>at</strong>erials. Nowadays we use a reference m<strong>at</strong>erial <strong>at</strong> all temper<strong>at</strong>ures,<br />
and define the absolute thermoelectric coefficients with respect to this standard.<br />
For the absolute values we also have<br />
Magnetic effects.<br />
τ = T dα<br />
dT<br />
Wh<strong>at</strong> is the rel<strong>at</strong>ion between Thomson and Kelvin?<br />
(6.8)<br />
When we apply a magnetic field perpendicular to the thermocouple, the<br />
Lorentz force gives a sideways displacement of the charge carriers. These are<br />
transverse effects. The following four combin<strong>at</strong>ions are useful:<br />
• Hall: Drive an electric current through the m<strong>at</strong>erial and measure the<br />
transverse electric field.<br />
• Nernst: Drive a he<strong>at</strong> flow through the m<strong>at</strong>erial and measure the transverse<br />
electric field.<br />
• Ettingshausen: Drive an electric current through the m<strong>at</strong>erial and measure<br />
the transverse temper<strong>at</strong>ure gradient.<br />
• Righi-Leduc: Drive a he<strong>at</strong> flow through the m<strong>at</strong>erial and measure the<br />
transverse temper<strong>at</strong>ure gradient.<br />
For detailed discussions see books on solid st<strong>at</strong>e physics.
6.3. NON-EQUILIBRIUM THERMODYNAMICS. 237<br />
6.3 Non-equilibrium <strong>Thermodynamics</strong>.<br />
Is it possible?<br />
The words non-equilibrium thermodynamics do not seem to make sense <strong>at</strong><br />
first glance. By definition, thermodynamics describes a m<strong>at</strong>erial in equilibrium,<br />
and it is only then th<strong>at</strong> we are able to have equ<strong>at</strong>ions of st<strong>at</strong>e. If we are not in<br />
equilibrium, some st<strong>at</strong>e functions do not have their correct values (the entropy<br />
is not maximal, or some free energy is not minimal), and there is no correct<br />
m<strong>at</strong>hem<strong>at</strong>ical description.<br />
In order to discuss the situ<strong>at</strong>ion presented above, we need to consider the<br />
microscopic details of the m<strong>at</strong>erial, and follow a system as a function of time by<br />
solving the equ<strong>at</strong>ions of motion. This leads to concepts like chaotic dynamics,<br />
and some very interesting physics. Boltzmann was probably one of the first to<br />
give a beginning of an answer to this problem.<br />
When devi<strong>at</strong>ions are small.<br />
Wh<strong>at</strong> happens when a system is almost in equilibrium? When the devi<strong>at</strong>ions<br />
from equilibrium are small? Perturb<strong>at</strong>ion theory should apply, and a complete<br />
knowledge of the equilibrium st<strong>at</strong>e should be sufficient to understand wh<strong>at</strong> is<br />
happening. This is the situ<strong>at</strong>ion in which we can still use the formulas from<br />
equilibrium thermodynamics, and discuss wh<strong>at</strong> happens when we are not quite<br />
in equilibrium.<br />
Local stability.<br />
Suppose we divide our m<strong>at</strong>erials in small cells, with volume ∆V each. In each<br />
cell, with position ⃗ri we have a an amount of m<strong>at</strong>erial ∆Ni and a temper<strong>at</strong>ure<br />
Ti. We assume th<strong>at</strong> the m<strong>at</strong>erial in each cell is internally in thermodynamic<br />
equilibrium. Therefore, the usual equ<strong>at</strong>ions of st<strong>at</strong>e will give us the values of the<br />
other st<strong>at</strong>e variables pi, µi, and Si, as well as the energy Ui. We assume th<strong>at</strong><br />
we deal with the standard pV T system, if not we need to add one variable for<br />
each additional pair to the set of variables th<strong>at</strong> we need to know. This scenario<br />
is called local stability.<br />
Limits.<br />
In order for local stability to exist, the volume ∆V needs to be large. Wh<strong>at</strong><br />
is large? In general, this volume needs to contain many molecules. On the other<br />
hand, we would like to evalu<strong>at</strong>e quantities like N = ∑<br />
∆Ni via integrals. If we<br />
define the density in cell i by ni = ∆Ni<br />
∆V<br />
we would like to replace<br />
i
238 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
∑<br />
∫<br />
ni∆V ⇒ n(⃗r)d 3 r (6.9)<br />
i<br />
and in order to do so we need to take the limit ∆V → 0, which obviously causes<br />
problems. The answer is again found by assuming th<strong>at</strong> the vari<strong>at</strong>ions in the<br />
density and temper<strong>at</strong>ure are small. This implies th<strong>at</strong>, if we follow the Riemann<br />
sum ∑<br />
ni∆V as a function of decreasing value of ∆V , this sum reaches a stable<br />
i<br />
(converged) value while the volume ∆V is still large on a microscopic scale. At<br />
th<strong>at</strong> point we stop and do not enter the region where ∆V becomes so small th<strong>at</strong><br />
we see the effects of the microscopic structure.<br />
Tests for validity.<br />
The ideas developed in the previous paragraph give us a definition when we<br />
can still use the equ<strong>at</strong>ions of thermodynamics, but also allow for fluctu<strong>at</strong>ions.<br />
It also gives a recipe for a numerical test of this condition. We divide space in<br />
cells and calcul<strong>at</strong>e all quantities with the given cell size. Next, we choose smaller<br />
cells, and recalcul<strong>at</strong>e all our quantities. This gives us a plot of the value of the<br />
quantity versus cell size. By extrapol<strong>at</strong>ing these values to those obtained with<br />
∆V = 0 we find the error in our calcul<strong>at</strong>ions for each cell size. On the other<br />
hand, we also have an error due to the fact th<strong>at</strong> our cells only contain a finite<br />
number of particles. This is an error not included in the calcul<strong>at</strong>ions above, and<br />
has to be added. This error is of the form 1<br />
√ N<br />
V<br />
√ ∆Ni<br />
, which can be approxim<strong>at</strong>ed by<br />
1<br />
∆V . Adding these terms together we get a minimum error <strong>at</strong> some optimal<br />
value of the cell size, and if this minimal error is acceptable, our model is useful.<br />
If not, the sp<strong>at</strong>ial vari<strong>at</strong>ions in the density are too fast. From now on we will<br />
assume th<strong>at</strong> our procedure gives acceptable results according to this test.<br />
Choice of fields.<br />
In a standard pV T system we need to specify three st<strong>at</strong>e variables to completely<br />
identify the st<strong>at</strong>e. At each position ⃗r we fix the volume d 3 r and hence we<br />
need two other st<strong>at</strong>e variables to completely specify wh<strong>at</strong> is happening. Conceptually<br />
the easiest are n(⃗r) and T (⃗r), because we can imagine th<strong>at</strong> these can be<br />
measured very well. But we could also choose quantities like the energy density<br />
u(⃗r), which can have advantages in the theoretical formul<strong>at</strong>ions.<br />
Different coordin<strong>at</strong>e frames.<br />
Another choice which is often made is to use cells th<strong>at</strong> contain a fixed amount<br />
of m<strong>at</strong>erial ∆N. In this case the cells have variable volume d 3 v = n(⃗r)d 3 r, but<br />
all formulas can be derived by simply adding this weight function. All densities<br />
in this case are per mole, and not per unit volume as in the previous case.
6.3. NON-EQUILIBRIUM THERMODYNAMICS. 239<br />
Sometimes we want to study liquids th<strong>at</strong> support macroscopic currents. At<br />
each point we find a velocity field ⃗v(⃗r) which tells us in which direction the liquid<br />
is flowing <strong>at</strong> th<strong>at</strong> point. It can be very useful to use a coordin<strong>at</strong>e frame th<strong>at</strong> is<br />
moving with the liquid and to follow the amount of m<strong>at</strong>erial in the original cell.<br />
Work between cells.<br />
We need to digress a bit on how we define volume. If we have nothing but<br />
the m<strong>at</strong>erial itself, as in the case of fluid flow, we can only refer to the m<strong>at</strong>erial<br />
itself to define volume. This then leads autom<strong>at</strong>ically to the idea of following<br />
a certain amount of m<strong>at</strong>erial in fluid flow, as described above. When we study<br />
electron transport in solids, though, we have the <strong>at</strong>omic l<strong>at</strong>tice as a reference.<br />
Hence in the cases we will study, volume is defined by the <strong>at</strong>omic l<strong>at</strong>tice. If the<br />
electrons would not interact with the l<strong>at</strong>tice, the gas of electrons could expand or<br />
contract with respect to the l<strong>at</strong>tice, and the amount of electrons in a cell defined<br />
by the <strong>at</strong>omic structure would change because of correl<strong>at</strong>ed motion (pressure<br />
driven) and diffusion (chemical potential driven). In the rest of this chapter we<br />
assume th<strong>at</strong> the electrons interact strongly with the <strong>at</strong>omic l<strong>at</strong>tice. This means<br />
th<strong>at</strong> pressure difference do not lead to correl<strong>at</strong>ed transport of electrons, but to<br />
changes in the <strong>at</strong>omic configur<strong>at</strong>ions. The <strong>at</strong>oms act like pistons. As long as<br />
pressure gradients are small, the <strong>at</strong>oms will not move far because inter<strong>at</strong>omic<br />
forces are strong. So we can assume th<strong>at</strong> the cells defined by the <strong>at</strong>oms have a<br />
fixed volume, and th<strong>at</strong> pressure differences are simply absorbed by the l<strong>at</strong>tice<br />
in reaction.<br />
In our formul<strong>at</strong>ion the size of the cells is fixed, and mechanical work of the<br />
pdV variety is therefore not possible. The only two ways the cells can exchange<br />
energy with each other is to do chemical work µdN, which is an exchange of<br />
m<strong>at</strong>erial between the cells, or use he<strong>at</strong> transfer T dS. Of course, if we would have<br />
chosen cells with a fixed amount of m<strong>at</strong>erial, the situ<strong>at</strong>ion would be different<br />
and we could use mechanical work as well as he<strong>at</strong>.<br />
This situ<strong>at</strong>ion is typical for electronic transport in a solid. The electrons<br />
diffuse through the m<strong>at</strong>erial due to gradients in the chemical potential and/or<br />
temper<strong>at</strong>ure. There are corresponding gradients in the pressure, but those are<br />
compens<strong>at</strong>ed by the pressure due to the l<strong>at</strong>tice. We assume th<strong>at</strong> the l<strong>at</strong>tice<br />
is stiff, and th<strong>at</strong> changes in the l<strong>at</strong>tice pressure have very small effects on the<br />
positions of the <strong>at</strong>oms, so we can assume th<strong>at</strong> we do have small volume elements<br />
th<strong>at</strong> do not change.<br />
As a result, the formulas we will derive in the next section do not apply to<br />
fluid flow, in which there are currents due to pressure gradients, as described<br />
by Bernouilli. But we know th<strong>at</strong> in thermodynamic equilibrium the pressure<br />
gradient is a function of the gradients in the other intensive st<strong>at</strong>e variables, and<br />
we can therefore always express our currents in terms of the other gradients, and<br />
retain a formalism similar to the one to be developed. We just have to be very<br />
careful in making connections between transport coefficients and applied fields.<br />
By restricting ourselves to electron diffusion we avoid those complic<strong>at</strong>ions. See
240 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
textbooks on fluid dynamics for more details.<br />
6.4 Transport equ<strong>at</strong>ions.<br />
Slow processes only.<br />
In the previous section we have developed a model in which we have local<br />
equilibrium, because densities are varying slowly in space. But because there<br />
is a change in density and temper<strong>at</strong>ure between cells, there will be a flow of<br />
m<strong>at</strong>erial and he<strong>at</strong> in order to reach equilibrium. We will assume th<strong>at</strong> this flow<br />
is very slow. In other words, the local densities are changing, but <strong>at</strong> a r<strong>at</strong>e<br />
which is much slower than the r<strong>at</strong>e in which these cells get back to equilibrium<br />
after a small amount of m<strong>at</strong>erial or he<strong>at</strong> has been exchanged. This is a n<strong>at</strong>ural<br />
consequence of the assumption th<strong>at</strong> the system is always in local equilibrium.<br />
If transport is too fast, we cannot maintain local equilibrium.<br />
Continuity equ<strong>at</strong>ion.<br />
If we consider an arbitrary small volume ∆V and follow the change of an<br />
extensive quantity X in th<strong>at</strong> volume, the change d∆X<br />
dt in th<strong>at</strong> quantity is the<br />
sum of the cre<strong>at</strong>ion and/or destruction of X and the flow of X through the<br />
surface ∆S of the volume element. The flow is given by a current density ⃗ JX<br />
by following the amount of X going through an area ∆A with surface normal<br />
ˆn in a time ∆t. The corresponding amount of X flowing through this area is<br />
Therefore, the r<strong>at</strong>e of change in X is<br />
∆X = ⃗ JX · ˆn∆A∆t (6.10)<br />
∫<br />
d<br />
xd<br />
dt ∆V<br />
3 ∫<br />
V = Px(⃗r, t)d<br />
∆V<br />
3 <br />
V − ⃗JX · ˆnd<br />
∆S<br />
2 S (6.11)<br />
where the first term on the right, Px, is the local cre<strong>at</strong>ion (which could be<br />
neg<strong>at</strong>ive) per unit volume and the second term the flow through the surface.<br />
The quantity x is the density of X, or the amount of X per unit volume.<br />
The surface normal in this definition points outwards from the volume element.<br />
Using Gauss’ theorem this gives<br />
∫<br />
d<br />
xd<br />
dt ∆V<br />
3 ∫<br />
V = Pxd<br />
∆V<br />
3 ∫<br />
V −<br />
∆V<br />
and since the volume is arbitrary we have <strong>at</strong> each point in space:<br />
∂x<br />
∂t = Px − ⃗ ∇ · ⃗ JX<br />
⃗∇ · ⃗ JXd 3 V (6.12)<br />
(6.13)
6.4. TRANSPORT EQUATIONS. 241<br />
We have replaced the total deriv<strong>at</strong>ive by a partial deriv<strong>at</strong>ive, since we look <strong>at</strong><br />
the change of x(⃗r, t) <strong>at</strong> each point. The production is sometimes written as a<br />
time deriv<strong>at</strong>ive, too, and in order to distinguish one uses a not<strong>at</strong>ion like<br />
Px = Dx<br />
(6.14)<br />
Dt<br />
This not<strong>at</strong>ion (or vari<strong>at</strong>ions of it), is widely used in physics, and it reflects the<br />
change in quantity X due to both cre<strong>at</strong>ion and transport. In a coordin<strong>at</strong>e frame<br />
moving with the liquid this is the only term th<strong>at</strong> survives, which is a main reason<br />
for choosing such a frame. The continuity equ<strong>at</strong>ion is in this not<strong>at</strong>ion:<br />
Dx<br />
Dt (⃗r, t) = Px(⃗r, t) = ∂x<br />
∂t (⃗r, t) + ⃗ ∇ · ⃗ JX<br />
Is Dx<br />
Dt<br />
a real deriv<strong>at</strong>ive?<br />
Extensive st<strong>at</strong>e properties are conserved.<br />
(6.15)<br />
Consider two adjacent cells. If m<strong>at</strong>erial flows from one cell to the other, the<br />
total amount of m<strong>at</strong>erial does not change. In other words, if a cell loses m<strong>at</strong>erial,<br />
this m<strong>at</strong>erial is not cre<strong>at</strong>ed or destroyed, but transported to neighboring cells.<br />
For example, the energy of one cell changes because energy is flowing in to or<br />
out of the cell. We assume th<strong>at</strong> we do not have any chemical reactions, and<br />
there is no cre<strong>at</strong>ion of energy or m<strong>at</strong>erial. Hence we assume th<strong>at</strong><br />
Since energy and m<strong>at</strong>erial are conserved, we now have<br />
and<br />
Pu(⃗r, t) = 0 (6.16)<br />
Pn(⃗r, t) = 0 (6.17)<br />
⃗∇ · ⃗ JU = − ∂u<br />
∂t<br />
(6.18)<br />
⃗∇ · ⃗ JN = − ∂n<br />
(6.19)<br />
∂t<br />
where u is the energy density and n is the m<strong>at</strong>ter density. Note th<strong>at</strong> the inclusion<br />
of chemical reactions is conceptually not difficult, just tedious, because now we<br />
need to distinguish between several types of m<strong>at</strong>erial and have to include partial<br />
time deriv<strong>at</strong>ives.<br />
Entropy production and fluxes.
242 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
Consider a cell which is divided in to two equal parts by an area ∆A with<br />
surface normal ˆn pointing from the left to the right. The width of each half is<br />
1<br />
2 ∆L. The left side of this cell changes its energy by an amount ∆Ul and its<br />
amount of m<strong>at</strong>erial by ∆Nl. Because we always have local equilibrium, we know<br />
th<strong>at</strong> ∆Ul = Tl∆Sl + µl∆Nl in this part of the cell. If we assume no contact<br />
outside the cell, the m<strong>at</strong>erial and energy need to come from the right part of<br />
the cell, and therefore we have<br />
∆Ul = −∆Ur<br />
∆Nl = −∆Nr<br />
(6.20)<br />
(6.21)<br />
But local equilibrium gives us the conditions th<strong>at</strong> ∆Ur = Tr∆Sr + µr∆Nr for<br />
th<strong>at</strong> half of the cell. Therefore we find th<strong>at</strong><br />
∆Sl + ∆Sr = 1<br />
∆Ul − µl<br />
∆Nl + 1<br />
∆Ur − µr<br />
∆Nr<br />
Tl<br />
or by using the conserv<strong>at</strong>ion laws we find th<strong>at</strong> the entropy production is<br />
∆Sl + ∆Sr = ( 1<br />
Tr<br />
Tl<br />
Tr<br />
− 1<br />
)∆Ur + (<br />
Tl<br />
µl<br />
Tl<br />
Tr<br />
− µr<br />
)∆Nr<br />
Tr<br />
(6.22)<br />
(6.23)<br />
The amount of energy flowing through the area ∆A is ∆Ur and the amount<br />
of m<strong>at</strong>erial ∆Nr, all in a time ∆t. If the local energy current density <strong>at</strong> the<br />
origin is ⃗ JU and the local m<strong>at</strong>erial current density ⃗ JN , then we have<br />
The entropy produced in this process is given by<br />
∆S = ∆Sl + ∆Sr =<br />
∆Ur = ⃗ JU · ˆn∆A∆t (6.24)<br />
∆Nr = ⃗ JN · ˆn∆A∆t (6.25)<br />
(<br />
( 1<br />
−<br />
Tr<br />
1<br />
)<br />
Tl<br />
⃗ JU · ˆn + ( µl<br />
Tl<br />
− µr<br />
)<br />
Tr<br />
⃗ )<br />
JN · ˆn ∆A∆t (6.26)<br />
This can be rel<strong>at</strong>ed to the gradients in the following manner:<br />
1<br />
−<br />
Tr<br />
1<br />
Tl<br />
= ⃗ ( )<br />
1<br />
∇ · ˆn<br />
T<br />
1<br />
∆L<br />
2<br />
(6.27)<br />
and similar for the second term. This gives:<br />
∆S = 1<br />
(<br />
⃗∇<br />
2<br />
( )<br />
1<br />
· ˆn<br />
T<br />
⃗ JU · ˆn − ⃗ ∇<br />
(<br />
µ<br />
)<br />
· ˆn<br />
T<br />
⃗ )<br />
JN · ˆn ∆A∆L∆t (6.28)<br />
and therefore the r<strong>at</strong>e of change of the production of the entropy density s<br />
(entropy per unit volume) for this process is given by:
6.4. TRANSPORT EQUATIONS. 243<br />
Ps = 1<br />
(<br />
⃗∇<br />
2<br />
( )<br />
1<br />
· ˆn<br />
T<br />
⃗ JU · ˆn − ⃗ ∇<br />
(<br />
µ<br />
)<br />
· ˆn<br />
T<br />
⃗ )<br />
JN · ˆn<br />
(6.29)<br />
In this equ<strong>at</strong>ion we have chosen an arbitrary direction of the dividing partition.<br />
Of course, we need to take into account the complete flow. Consider a small<br />
cubic cell with faces orthogonal to the x, y, and z direction. The local entropy<br />
density production for the flow through each of the faces of this cube is given by<br />
the expression above, with either ˆn = ±ˆx, ˆn = ±ˆy, or ˆn = ±ˆz. We can use the<br />
expression for one side of the partition only, since we use the entropy density,<br />
or entropy per unit volume! If we add the effects the flow through all sides of<br />
the cube together we get the total entropy density production in this cubic cell.<br />
It is easy to see th<strong>at</strong><br />
∑<br />
( ⃗ A · ˆn)( ⃗ B · ˆn) = 2AxBx + 2AyBy + 2AzBz = 2 ⃗ A · ⃗ B (6.30)<br />
ˆn<br />
and hence the expression for the total energy density production is<br />
( ( )<br />
Ps = ⃗∇<br />
1<br />
·<br />
T<br />
⃗ JU − ⃗ (<br />
µ<br />
)<br />
∇ ·<br />
T<br />
⃗ )<br />
JN<br />
(6.31)<br />
Note th<strong>at</strong> using a cubic cell is not essential, it just makes the m<strong>at</strong>hem<strong>at</strong>ics much<br />
easier. Also note th<strong>at</strong> we get a factor two, because <strong>at</strong> each point we need to<br />
consider the flow into th<strong>at</strong> point from one side and the flow out of th<strong>at</strong> point<br />
towards the other side!<br />
Entropy flow.<br />
Wh<strong>at</strong> does Ps exactly mean?<br />
Next we define the flow of entropy ⃗ JS by rel<strong>at</strong>ing it to the flow of energy<br />
and the flow of particles.<br />
( )<br />
1<br />
(<br />
µ<br />
)<br />
⃗JS = ⃗JU − ⃗JN (6.32)<br />
T T<br />
As a consequence we find th<strong>at</strong><br />
⃗∇ · ⃗ JS =<br />
( )<br />
1<br />
⃗∇ ·<br />
T<br />
⃗ JU −<br />
(<br />
µ<br />
)<br />
⃗∇ · JN<br />
⃗ +<br />
T<br />
⃗ ∇<br />
( )<br />
1<br />
·<br />
T<br />
⃗ JU − ⃗ ∇<br />
(<br />
µ<br />
)<br />
·<br />
T<br />
⃗ JN<br />
(6.33)<br />
and using the continuity equ<strong>at</strong>ions for the energy and m<strong>at</strong>erial current we obtain:<br />
⃗∇ · ⃗ JS = −<br />
( )<br />
1 ∂u<br />
T ∂t +<br />
(<br />
µ<br />
)<br />
∂n<br />
T ∂t + ⃗ ∇<br />
( )<br />
1<br />
·<br />
T<br />
⃗ JU − ⃗ ∇<br />
(<br />
µ<br />
)<br />
·<br />
T<br />
⃗ JN<br />
(6.34)
244 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
Since we always have local equilibrium, we can use the first law of thermodynamics:<br />
dU = T dS + µdN (6.35)<br />
and show the sum of the first two terms is equal to − ∂s<br />
∂t and hence we find th<strong>at</strong><br />
for the local production of entropy we get<br />
( ( )<br />
Ps = ⃗∇<br />
1<br />
·<br />
T<br />
⃗ JU − ⃗ (<br />
µ<br />
)<br />
∇ ·<br />
T<br />
⃗ )<br />
JN<br />
(6.36)<br />
as we derived before. This justifies th<strong>at</strong> we defined the entropy current correctly.<br />
Transport equ<strong>at</strong>ions.<br />
Why is − ∂s<br />
∂t = − ( )<br />
1 ∂u<br />
T ∂t + ( ) µ ∂n<br />
T ∂t ?<br />
In our example we assume th<strong>at</strong> locally we have a pV T system, and because<br />
we define subregions by constant volume elements, only energy and mass can<br />
flow. In a more general case, we could also include other quantities, like the<br />
flow of magnetiz<strong>at</strong>ion. In this case we have for the first law in each local cell<br />
(with dV = 0) an equ<strong>at</strong>ion of the form:<br />
T dS = ∑<br />
xidXi<br />
which after dividing by T takes the form:<br />
Therefore, we have in the general case:<br />
and<br />
dS = ∑<br />
ϕidXi<br />
i<br />
i<br />
⃗JS = ∑<br />
i<br />
ϕi ⃗ Ji<br />
Ps = ∑<br />
⃗∇ϕi · ⃗ Ji<br />
i<br />
(6.37)<br />
(6.38)<br />
(6.39)<br />
(6.40)<br />
In general, each current will depend on the values of all intensive parameters<br />
and their gradients. We have to keep in mind, however, th<strong>at</strong> we assume th<strong>at</strong><br />
we are close to equilibrium. If all intensive parameters are constant, we are<br />
in equilibrium, and no net macroscopic flows are present. When we are close<br />
to equilibrium, the gradients in the intensive parameters will be small, and as<br />
a consequence we can expand the formulas for the currents in terms of the
6.5. MACROSCOPIC VERSUS MICROSCOPIC. 245<br />
gradients of the intensive parameters. An important case in practice is the<br />
situ<strong>at</strong>ion where the gradients are so small th<strong>at</strong> we need only linear terms. We<br />
will restrict ourselves to th<strong>at</strong> case and only study linear processes. The basic<br />
equ<strong>at</strong>ions are now<br />
⃗Ji = ∑<br />
Lik · ⃗ ∇ϕk<br />
k<br />
(6.41)<br />
where the m<strong>at</strong>rix coefficients Lik are functions of the intensive parameters (or<br />
of the combin<strong>at</strong>ions ϕi) only. The currents we consider here are only currents in<br />
energy, m<strong>at</strong>ter, etc. and not in the entropy. Of course, we can always elimin<strong>at</strong>e<br />
one of these currents in favor of the entropy current. This has the advantage<br />
th<strong>at</strong> we now have he<strong>at</strong> transport included as a separ<strong>at</strong>e entity.<br />
Onsager rel<strong>at</strong>ions.<br />
Are the variables Lik st<strong>at</strong>e functions?<br />
The coefficients Lik in the last formula obey some symmetry rel<strong>at</strong>ions. This<br />
is directly rel<strong>at</strong>ed to the fact th<strong>at</strong> fluctu<strong>at</strong>ions are not completely arbitrary, but<br />
are rel<strong>at</strong>ed to the entropy as described in chapter two. The other ingredient is<br />
time reversal symmetry. The general form of the rel<strong>at</strong>ion in the presence of a<br />
magnetic field is:<br />
Lik(T, µ, ⃗ H) = L T ki(T, µ, − ⃗ H) (6.42)<br />
where the transpose indic<strong>at</strong>es a transpose of the three by three m<strong>at</strong>rix in coordin<strong>at</strong>e<br />
space. Because magnetic fields give velocity dependent forces, we need<br />
to change the sign of the field when we reverse the direction of time. This is<br />
the cause for the minus sign in the last equ<strong>at</strong>ion. There are also additional<br />
minus signs in front of the equ<strong>at</strong>ions for properties th<strong>at</strong> mix magnetic and<br />
non-magnetic transport.<br />
6.5 Macroscopic versus microscopic.<br />
Thermal averages.<br />
In each cell we can define the local values of the extensive variables Xi and<br />
corresponding intensive variables ϕi. As mentioned before, we assume local<br />
equilibrium, and these variables are connected by equ<strong>at</strong>ions of st<strong>at</strong>e. If there<br />
are no external fields, all these values are independent of position and time. If<br />
there is an external field, however, the values in different cells will not be the
246 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
same anymore. For example, we can imagine th<strong>at</strong> there is a temper<strong>at</strong>ure field<br />
T (⃗r, t) present, and we want to study its consequences. Because we assume local<br />
equilibrium, the sp<strong>at</strong>ial and temporal vari<strong>at</strong>ion of this field has to be small. This<br />
implies th<strong>at</strong> we can study only the long wavelength and low frequency response<br />
of a system.<br />
The results will break down when either the wavelength is too small or<br />
the frequency is too high, and microscopic effects start to play a role. In the<br />
next sub-sections we rel<strong>at</strong>e the microscopic and macroscopic views, which also<br />
provides a proof for Onsager’s rel<strong>at</strong>ions. We do this in a general manner, independent<br />
of the microscopic n<strong>at</strong>ure of the m<strong>at</strong>erial. We simply consider the<br />
fluctu<strong>at</strong>ions in each cell. Some of the parameters describing these fluctu<strong>at</strong>ions,<br />
however, can only be calcul<strong>at</strong>ed in a microscopic theory. This is similar to the<br />
situ<strong>at</strong>ion in the Landau-Ginzburg theory as discussed in Chapter 4.<br />
Fluctu<strong>at</strong>ions.<br />
Assume th<strong>at</strong> the equilibrium system is determined by local values X0 i (⃗r, t)<br />
of the extensive parameters in each cell. Then in our non-equilibrium st<strong>at</strong>e we<br />
have in a given cell <strong>at</strong> point ⃗r the values Xi(⃗r, t) = X0 i (⃗r, t) + δXi(⃗r, t). The<br />
fluctu<strong>at</strong>ions are small if our cells contain many <strong>at</strong>oms (which is only possible in<br />
the long wavelength limit of external fields). If we now take a thermal average<br />
in each cell we find:<br />
and<br />
⟨δXi(⃗r, t)⟩ = 0 (6.43)<br />
⟨Xi(⃗r, t)⟩ = X 0 i (⃗r, t) (6.44)<br />
Since a thermal average is equivalent to a time average, we assume th<strong>at</strong> the<br />
fluctu<strong>at</strong>ions are much more rapid than the vari<strong>at</strong>ions in the external fields. In<br />
th<strong>at</strong> case we are able to do a time average of the quantity δXi(⃗r, t) over a time<br />
long enough th<strong>at</strong> it represents the thermodynamic limit, but short on the scale<br />
of the external perturb<strong>at</strong>ions. This is the reason for the low frequency limit.<br />
The fluctu<strong>at</strong>ions describe devi<strong>at</strong>ions from thermal equilibrium, and hence the<br />
quantities Xi(⃗r, t) and corresponding fields ϕi(⃗r, t) do not obey the equ<strong>at</strong>ions<br />
of st<strong>at</strong>e. Only the average quantities X 0 i (⃗r, t) and ϕ0 i<br />
(⃗r, t) do. The transport<br />
equ<strong>at</strong>ions derived in the previous section use these local thermal equilibrium<br />
values, and are defined in th<strong>at</strong> context only.<br />
Correl<strong>at</strong>ion functions.<br />
The fluctu<strong>at</strong>ions mentioned above are correl<strong>at</strong>ed, as we described in chapter<br />
two. Since the decay of fluctu<strong>at</strong>ions follows the laws of physics on the microscopic<br />
level, this correl<strong>at</strong>ion continues in time and we can discuss correl<strong>at</strong>ion<br />
coefficients of the form:
6.5. MACROSCOPIC VERSUS MICROSCOPIC. 247<br />
⟨δXi(⃗r, 0)δXj(⃗r ′ , t)⟩ (6.45)<br />
We can discuss correl<strong>at</strong>ions between fluctu<strong>at</strong>ions <strong>at</strong> difference points, since any<br />
excess in some extensive parameter has to flow away through neighboring cells.<br />
Again, we take averages to avoid all short time thermal fluctu<strong>at</strong>ions, and we<br />
still assume th<strong>at</strong> the cells are large compared to the detailed structure in the<br />
m<strong>at</strong>ter. Since the laws of n<strong>at</strong>ure are time invariant (here we assume th<strong>at</strong> there<br />
are no magnetic fields!) we have:<br />
⟨δXi(⃗r, 0)δXj(⃗r, t)⟩ = ⟨δXi(⃗r, 0)δXj(⃗r, −t)⟩ (6.46)<br />
and since we have transl<strong>at</strong>ional symmetry in time:<br />
⟨δXi(⃗r, 0)δXj(⃗r, t)⟩ = ⟨δXi(⃗r, t)δXj(⃗r, 0)⟩ (6.47)<br />
From this equ<strong>at</strong>ion we subtract ⟨δXi(⃗r, 0)δXj(⃗r, 0)⟩ and in the limit of small t<br />
we find<br />
⟨δXi(⃗r, 0) ∂δXj<br />
(⃗r, 0)⟩ = ⟨<br />
∂t<br />
∂δXi<br />
∂t (⃗r, 0)δXj(⃗r, 0)⟩ (6.48)<br />
Next we consider the current density th<strong>at</strong> is rel<strong>at</strong>ed to the quantity δXi in<br />
our cell, and call this current −→<br />
δJi. By using the symbol δ we make explicitly<br />
clear th<strong>at</strong> this is a microscopic current, rel<strong>at</strong>ed to fluctu<strong>at</strong>ions. Because there<br />
is no local cre<strong>at</strong>ion one has:<br />
This leads to<br />
∂δXj<br />
∂t = −⃗ ∇ · −→<br />
δJj<br />
(6.49)<br />
⟨δXi( ⃗ ∇ · −→<br />
δJj)⟩ = ⟨( ⃗ ∇ · −→<br />
δJi)δXj⟩ (6.50)<br />
where we have dropped the space and time arguments, since all quantities are<br />
<strong>at</strong> the same position ⃗r and same time t. This is important. If we want to<br />
rel<strong>at</strong>e fluctu<strong>at</strong>ions in quantities <strong>at</strong> different positions and times, we need the<br />
probability functions for these correl<strong>at</strong>ions, and we have no idea wh<strong>at</strong> they look<br />
like. On the other hand, probabilities for fluctu<strong>at</strong>ions <strong>at</strong> a given place and time<br />
have been discussed in chapter 2, and hence we do know how to calcul<strong>at</strong>e them.<br />
Therefore, we do have the means to evalu<strong>at</strong>e expressions like the last one.<br />
Fluctu<strong>at</strong>ions in fields.<br />
The local st<strong>at</strong>e of the system is specified by the extensive variables Xi(⃗r, t) =<br />
X 0 i (⃗r, t) + δXi(⃗r, t) in each cell positioned <strong>at</strong> ⃗r. Again, remember th<strong>at</strong> these<br />
cells are large enough to contain enough sufficient m<strong>at</strong>erial to be able to define<br />
local thermal equilibrium, but th<strong>at</strong> they are small on the scale of the vari<strong>at</strong>ions<br />
in the externally applied fields. Because we have
248 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
we have locally th<strong>at</strong><br />
dS = ∑<br />
ϕidXi<br />
i<br />
( )<br />
∂S<br />
ϕi =<br />
∂Xi<br />
(6.51)<br />
(6.52)<br />
where the partial deriv<strong>at</strong>ives only vary one of the extensive variables <strong>at</strong> a time.<br />
We assume local equilibrium, and hence<br />
ϕ 0 ( )<br />
∂S<br />
i = (X<br />
∂Xi<br />
0 1 , X 0 2 , · · ·) (6.53)<br />
<strong>at</strong> each point ⃗r and each time t.<br />
In thermodynamics, intensive quantities like temper<strong>at</strong>ure and pressure are<br />
only defined in the thermodynamic limit, as average quantities. Fluctu<strong>at</strong>ions in<br />
extensive quantities are easy to measure, <strong>at</strong> least in theory. We simply count the<br />
amount of m<strong>at</strong>erial, measure the volume, and measure the energy. Of course,<br />
we often only measure changes, but th<strong>at</strong> does not change the picture. Intensive<br />
quantities like temper<strong>at</strong>ure and pressure are measured using thermal contact<br />
with reservoirs, implying the need of thermal equilibrium. But even though we<br />
cannot technically measure these quantities <strong>at</strong> the level of fluctu<strong>at</strong>ions, we can<br />
easily define them by:<br />
( )<br />
∂S<br />
ϕi(⃗r, t) = (X1(⃗r, t), X2(⃗r, t), · · ·) (6.54)<br />
∂Xi<br />
or<br />
For small fluctu<strong>at</strong>ions we have<br />
ϕi(⃗r, t) = ϕ 0 i (⃗r, t) + ∑<br />
( 2 ∂ S<br />
k<br />
∂Xi∂Xk<br />
)<br />
(X 0 1 , X 0 2 , · · ·)δXk(⃗r, t) (6.55)<br />
δϕi(⃗r, t) = ϕi(⃗r, t) − ϕ 0 i (⃗r, t) = ∑<br />
SikδXk(⃗r, t) (6.56)<br />
where the m<strong>at</strong>rix S is symmetric.<br />
This same m<strong>at</strong>rix also is used in the probability function for small fluctu<strong>at</strong>ions<br />
in the Gaussian approxim<strong>at</strong>ion (see Chapter 2), where we have<br />
W (δX1, δX2, · · ·) = Ae 1<br />
2R<br />
∑<br />
k<br />
ij SijδXiδXj<br />
(6.57)<br />
We divided the change in entropy by the molar gas constant in order to have a<br />
dimensionless quantity in the exponent. Note th<strong>at</strong> in books where the amount<br />
of m<strong>at</strong>erial is measured in particles in stead of moles, one needs to divide by<br />
the Boltzmann constant! This also gives
6.5. MACROSCOPIC VERSUS MICROSCOPIC. 249<br />
( )<br />
∂W<br />
= R<br />
∂δXk<br />
−1 W δϕk<br />
(6.58)<br />
According to the maximum entropy principle, the probability function W<br />
takes its maximal value when all fluctu<strong>at</strong>ions are zero. This implies th<strong>at</strong> the exponent<br />
in the probability function is always neg<strong>at</strong>ive for non-zero fluctu<strong>at</strong>ions.<br />
Since the m<strong>at</strong>rix Sij is real and symmetric, it has a complete set of real eigenvalues.<br />
All these eigenvalues are neg<strong>at</strong>ive, or else there would be fluctu<strong>at</strong>ions<br />
for which the probability did not decrease. When all eigenvalues are neg<strong>at</strong>ive,<br />
the inverse of the m<strong>at</strong>rix Sij exists. We call this m<strong>at</strong>rix Tij and we therefore<br />
find th<strong>at</strong>:<br />
δXj = ∑<br />
Tjiδϕi<br />
and because S is symmetric, T is symmetric too.<br />
Finally, we consider the thermal average of the following quantity:<br />
i<br />
(6.59)<br />
∫<br />
⟨δXiδϕk⟩ = δXiδϕk(δX1, δX2, · · ·)W (δX1, δX2, · · ·)dδX1dδX2 · · · (6.60)<br />
Because of the rel<strong>at</strong>ion derived in 6.58 we have<br />
∫<br />
⟨δXiδϕk⟩ = R<br />
δXi<br />
( ∂W<br />
∂δXk<br />
and integr<strong>at</strong>ion by parts then gives<br />
)<br />
(δX1, δX2, · · ·)dδX1dδX2 · · · (6.61)<br />
⟨δXiδϕk⟩ = −Rδik<br />
This also gives us a represent<strong>at</strong>ion for the m<strong>at</strong>rix T, because<br />
⟨δXiδXj⟩ = ∑<br />
∑<br />
Tjk⟨δXiδϕk⟩ = −kB<br />
k<br />
k<br />
Tjkδik = −kBTji<br />
(6.62)<br />
(6.63)<br />
This also shows the symmetry of the m<strong>at</strong>rix T in an explicit manner. This rel<strong>at</strong>ion<br />
follows, of course, directly from the definition of the probability distribution<br />
in 6.57 given above. Similarly we find th<strong>at</strong><br />
⟨δϕiδϕj⟩ = ∑<br />
Sjk⟨δϕiδXk⟩ = −R ∑<br />
Sjkδik = −RSji<br />
Need for interactions.<br />
k<br />
k<br />
(6.64)<br />
In the previous discussion there is one important element missing. The forces<br />
responsible for currents are the gradients of intensive st<strong>at</strong>e variables, ⃗ ∇ϕi. The
250 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
probabilities for fluctu<strong>at</strong>ions in a cell only depend on the values of the extensive<br />
st<strong>at</strong>e variables Xi. In order to find the value of the gradient in such a field, we<br />
need to consider fluctu<strong>at</strong>ions in adjacent cells. They are uncoupled, however, in<br />
our current approxim<strong>at</strong>ion. Therefore,<br />
⟨(δXi(⃗r) − δXi(⃗r ′ )) 2 ⟩ = ⟨δX 2 i (⃗r)⟩ + ⟨δX 2 i (⃗r)⟩ (6.65)<br />
and the magnitude of this fluctu<strong>at</strong>ion is proportional to the volume of the cell.<br />
If we now consider two adjacent cubic cells of dimension L3 we see th<strong>at</strong> the<br />
typical value of the difference in the density is proportional to L 3<br />
3 and hence<br />
a typical value of the gradient is proportional to √ L. Hence typical values of<br />
gradients of densities are proportional to L−1 and they become very large for<br />
small values of L!<br />
Two things are missing in this approxim<strong>at</strong>ion. First, extensive quantities<br />
are conserved, and if values increase in one cell, the amount has to come from<br />
somewhere, and therefore values in neighboring cells tend to decrease. Second,<br />
there are interactions between the m<strong>at</strong>erial in one cell and in another. There<br />
is always gravity, and there are forces like the van der Waals interaction. In<br />
ionic fluids direct Coulomb interactions are possible. These effects are easily<br />
introduced in a microscopic theory, but then the calcul<strong>at</strong>ions are quite difficult.<br />
In thermodynamics we can do the same as we did in chapter four, however, and<br />
assume th<strong>at</strong> fluctu<strong>at</strong>ions are small and th<strong>at</strong> we can expand in a power series.<br />
The probability of fluctu<strong>at</strong>ions in a cell is now described by:<br />
with<br />
S(δX1, δX2, · · · , ∂aδX1, · · ·) = 1<br />
2<br />
W (δX1, δX2, · · · , ∂aδX1, · · ·) = Ae 1<br />
k B S<br />
∑<br />
ij<br />
SijδXiδXj + 1<br />
2<br />
∑<br />
iajb<br />
Fiajb∂aδXi∂bδXj<br />
(6.66)<br />
(6.67)<br />
The summ<strong>at</strong>ions over a and b are summ<strong>at</strong>ions over Carthesian coordin<strong>at</strong>es.<br />
Because the tensor Fiajb is rel<strong>at</strong>ed to a second order functional deriv<strong>at</strong>ive, it<br />
is symmetric. Therefore, it has a complete spectrum of eigenvalues, and again<br />
all these eigenvalues have to be neg<strong>at</strong>ive in order to guarantee th<strong>at</strong> the st<strong>at</strong>e<br />
without fluctu<strong>at</strong>ions has the highest probability. Because none of the eigenvalues<br />
are zero, the tensor has an inverse.<br />
The definition above is the only one possible in second order, and any microscopic<br />
theory has to reduce to this form in lowest order. The values of Fiajb<br />
can only be calcul<strong>at</strong>ed in a microscopic theory. This is a fundamental difference<br />
with the values of Sij which follow from standard thermodynamics. Note th<strong>at</strong><br />
there are no terms in second order with only one gradient. They disappear<br />
because of symmetry. Also note th<strong>at</strong> this form of theory is in spirit completely<br />
equivalent to Ginzburg-Landau!<br />
Properties for gradients.
6.5. MACROSCOPIC VERSUS MICROSCOPIC. 251<br />
It is easy to derive th<strong>at</strong> we have the following rel<strong>at</strong>ion for the gradients:<br />
⟨∂aδXi<br />
∂∂bδXj<br />
∂t<br />
⟩ = ⟨ ∂∂aδXi<br />
∂bδXj⟩ (6.68)<br />
∂t<br />
using the same reasoning as before. We also define the equivalent of the conjug<strong>at</strong>e<br />
intensive fields in a similar manner as we derived for the fluctu<strong>at</strong>ions in<br />
the fields ϕi by<br />
and we have exactly as before<br />
Onsager’s theorem derived.<br />
Fia = ∑<br />
jb<br />
Fiajb∂bδXj<br />
⟨∂aδXiFjb⟩ = −Rδijδab<br />
(6.69)<br />
(6.70)<br />
We are almost ready to derive the Onsager rel<strong>at</strong>ions. First of all, the r<strong>at</strong>e<br />
of change in the gradient of an extensive quantity is proportional to the current<br />
density pertaining to th<strong>at</strong> quantity:<br />
∂∂aδXi<br />
∂t ∝ δJia (6.71)<br />
which is easy to see, since doubling the current density will double the flow of<br />
m<strong>at</strong>erial and hence double the change in gradient. Next, from our definition of<br />
the fields F we see th<strong>at</strong> they behave like gradients of the intensive st<strong>at</strong>e variables.<br />
Since this is the only way we can define the conjug<strong>at</strong>e fields for fluctu<strong>at</strong>ions,<br />
these are the fields th<strong>at</strong> drive the currents. In the linear approxim<strong>at</strong>ion we have<br />
therefore:<br />
δJia = ∑<br />
jb<br />
LiajbFjb<br />
in the same way as in the previous section. Taken together, we now have:<br />
and hence<br />
⟨(∂aδXi)( d∂bδXj<br />
dt<br />
Similarly we find<br />
∂∂aδXi<br />
∂t<br />
)⟩ = C ∑<br />
kc<br />
∑<br />
= C<br />
jb<br />
LiajbFjb<br />
Ljbkc⟨∂aδXiFkc⟩ = −RLjbia<br />
⟨ ∂∂aδXi<br />
∂bδXj⟩ = −RLiajb<br />
∂t<br />
(6.72)<br />
(6.73)<br />
(6.74)<br />
(6.75)
252 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
and hence<br />
or<br />
Liajb = Ljbia<br />
Lik = L T ki<br />
(6.76)<br />
(6.77)<br />
Therefore, the Onsager rel<strong>at</strong>ions are a general consequence of the microscopic<br />
n<strong>at</strong>ure of m<strong>at</strong>ter, but we do not need to know any details. It is sufficient to<br />
know th<strong>at</strong> the macroscopic laws are the limit of the microscopic description,<br />
and th<strong>at</strong> the microscopic laws have time reversal symmetry. The form 6.66 for<br />
the fluctu<strong>at</strong>ions is simply the most general from we can write down in lowest<br />
order, and this form is sufficient for the purpose of proving Onsager’s rel<strong>at</strong>ions.<br />
The inclusion of magnetic fields is easy, we simply have to change the sign of<br />
all magnetic properties under time reversal. This gives some extra minus signs.<br />
Entropy production.<br />
In the previous section we derived the entropy production when currents are<br />
present in 6.40. When the currents are due to fluctu<strong>at</strong>ions only, this takes the<br />
form<br />
or<br />
Ps = ∑<br />
ia<br />
Ps = ∑<br />
iajb<br />
Fia · δJia<br />
LiajbFiaFjb<br />
(6.78)<br />
(6.79)<br />
and this has to be positive, since entropy is always produced when there are<br />
fluctu<strong>at</strong>ions. Therefore, the eigenvalues of the tensor L have to be positive! Even<br />
though this st<strong>at</strong>ement follows from an analysis of fluctu<strong>at</strong>ions, if can now be<br />
generalized to all currents, since they have the same microscopic underpinning.<br />
For example, consider the case where we only have a gradient in the chemical<br />
potential and only a flow of m<strong>at</strong>erial. In th<strong>at</strong> case we have ⃗ JN = L⃗ ∇ −µ<br />
T =<br />
− L<br />
T ⃗ ∇µ and because of the analysis above, L is positive. Therefore, the mass<br />
current is in the opposite direction of the gradient of the chemical potential, as<br />
is observed.<br />
6.6 Thermo-electric effects.<br />
Transport equ<strong>at</strong>ions for NQ system.
6.6. THERMO-ELECTRIC EFFECTS. 253<br />
We now apply the transport theory developed in the second section to a<br />
case where we have gradients in both the temper<strong>at</strong>ure and chemical potential.<br />
We have transport of m<strong>at</strong>erial and energy, or in different words, transport of<br />
m<strong>at</strong>erial and he<strong>at</strong>. Therefore we call this a NQ system. We repe<strong>at</strong> the equ<strong>at</strong>ions<br />
for the currents and entropy production:<br />
Ps =<br />
( )<br />
1<br />
⃗JS = ⃗JU −<br />
T<br />
( ( )<br />
⃗∇<br />
1<br />
·<br />
T<br />
⃗ JU − ⃗ ∇<br />
The transport equ<strong>at</strong>ions are in this case:<br />
and<br />
(<br />
µ<br />
)<br />
⃗JN (6.80)<br />
T<br />
(<br />
µ<br />
)<br />
·<br />
T<br />
⃗ JN<br />
⃗JU = LUU · ⃗ ∇ 1<br />
T + LUN · ⃗ ∇ −µ<br />
T<br />
)<br />
(6.81)<br />
(6.82)<br />
⃗JN = LNU · ⃗ ∇ 1<br />
T + LNN · ⃗ ∇ −µ<br />
(6.83)<br />
T<br />
In many cases it is useful to replace the energy flow by the he<strong>at</strong> flow. because<br />
we have ¯ dQ = T dS we define<br />
⃗JQ = T ⃗ JS = ⃗ JU − µ ⃗ JN<br />
(6.84)<br />
which more or less st<strong>at</strong>es th<strong>at</strong> the he<strong>at</strong> flow is the total flow of energy minus<br />
the flow of potential energy, represented by the last term in the equ<strong>at</strong>ion above.<br />
Using the he<strong>at</strong> flow, we find:<br />
( ( )<br />
Ps = ⃗∇<br />
1<br />
·<br />
T<br />
⃗ JQ + 1<br />
T ⃗ ∇(−µ) · ⃗ )<br />
JN<br />
(6.85)<br />
and<br />
or<br />
⃗JQ = (LUU − µLNU ) · ⃗ ∇ 1<br />
T + (LUN − µLNN ) · ⃗ ∇ −µ<br />
T<br />
⃗JQ = (LUU − µLNU − µLUN + µ 2 LNN ) · ⃗ ∇ 1<br />
T<br />
Similarly we have<br />
⃗JN = (LNU − µLNN) · ⃗ ∇ 1<br />
T<br />
which can be written in the form<br />
1<br />
⃗JN = L11<br />
T · ⃗ ∇(−µ) + L12 · ⃗ ∇ 1<br />
T<br />
(6.86)<br />
− 1<br />
T (LUN − µLNN ) · ⃗ ∇µ (6.87)<br />
− 1<br />
T LNN · ⃗ ∇µ (6.88)<br />
(6.89)
254 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
with<br />
1<br />
⃗JQ = L21<br />
T · ⃗ ∇(−µ) + L22 · ⃗ ∇ 1<br />
T<br />
L11 = LNN = L T 11<br />
L12 = LNU − µLNN<br />
L21 = LUN − µLNN = L T 12<br />
L22 = LUU − µLNU − µLUN + µ 2 LNN = L T 22<br />
(6.90)<br />
(6.91)<br />
(6.92)<br />
(6.93)<br />
(6.94)<br />
These last equ<strong>at</strong>ions can be written in the form of a similarity transform<strong>at</strong>ion:<br />
( L11 L12<br />
L21 L22<br />
)<br />
=<br />
( 1 0<br />
−µ 1<br />
) ( LNN LNU<br />
LUN LUU<br />
) ( 1 −µ<br />
0 1<br />
which implies th<strong>at</strong> the transformed tensor Lij is also positive definite.<br />
Units.<br />
)<br />
(6.95)<br />
It is useful to take time to consider the dimensions of all quantities involved,<br />
and which units we are employing. In thermodynamics all m<strong>at</strong>erial is counted<br />
in moles, not per particle. The transform<strong>at</strong>ion is easy, though, and involves<br />
Avogadro’s number NA. The current density in equ<strong>at</strong>ion 6.82 is measured in<br />
Jm −2 s −1 , while the current density in 6.83 is measured in Mm −2 s −1 . The<br />
temper<strong>at</strong>ure is in K and the chemical potential is in JM −1 , where we use the<br />
symbol M for moles. From 6.82 we then find the following for the transport<br />
coefficients:<br />
and similarly:<br />
Coefficient Dimensions<br />
LUU<br />
LUN<br />
LNU<br />
LNN<br />
JKm−1 s−1 MKm −1s−1 MKm −1s−1 M 2J −1Km−1s−1 Coefficient Dimensions<br />
L22<br />
L21<br />
L12<br />
L11<br />
JKm−1 s−1 MKm −1s−1 MKm −1s−1 M 2J −1Km−1s−1
6.6. THERMO-ELECTRIC EFFECTS. 255<br />
Electrical and thermal conductivity.<br />
For the rest of this chapter we confine ourselves to a specific case. Suppose<br />
we are describing the behavior of electrons in a metal. In th<strong>at</strong> case the flow of<br />
m<strong>at</strong>erial is equivalent to a flow of electrical current, and the he<strong>at</strong> flow is just the<br />
he<strong>at</strong> flow due to the electrons. Note th<strong>at</strong> in any experiment we also have a he<strong>at</strong><br />
flow due to the ions. This part of the he<strong>at</strong> flow has to be analyzed separ<strong>at</strong>ely,<br />
and should not be included in the Onsager reciprocity rel<strong>at</strong>ions! Basic quantities<br />
one defines are the electrical conductivity σ, which rel<strong>at</strong>es the electrical current<br />
density ⃗j to the electric field ⃗ E, and the thermal conductivity κ which rel<strong>at</strong>es<br />
the he<strong>at</strong> current density ⃗ JQ to the temper<strong>at</strong>ure gradient. The only contribution<br />
of interest to the chemical potential is the Coulomb potential, and we have:<br />
⃗j = −NAe ⃗ JN<br />
with −e the charge of an electron. This gives<br />
(6.96)<br />
⃗E = 1<br />
NAe ⃗ ∇µ (6.97)<br />
⃗j = N 2 A e2<br />
T L11 · ⃗ E + NAe<br />
T 2 L12 · ⃗ ∇T (6.98)<br />
⃗JQ = − NAe<br />
T L21 · ⃗ E − 1<br />
T 2 L22 · ⃗ ∇T (6.99)<br />
where the parameters Lij only account for the effects of the electrons.<br />
The entropy production expressed in these quantities is:<br />
Ps = − 1<br />
T 2 ⃗ ∇T · ⃗ JQ + 1<br />
T ⃗ E · ⃗j (6.100)<br />
Since the he<strong>at</strong> production is T Ps we see immedi<strong>at</strong>ely th<strong>at</strong> the second term is<br />
the standard he<strong>at</strong> production as described in electricity theory. The first term<br />
is the he<strong>at</strong> produced when a he<strong>at</strong> flow goes to a lower temper<strong>at</strong>ure.<br />
The electrical conductivity tensor σ is defined <strong>at</strong> constant temper<strong>at</strong>ure by<br />
⃗j = σ · ⃗ E, ⃗ ∇T = 0, and this gives<br />
σ = N 2 A e2<br />
T L11 (6.101)<br />
and has dimensions C 2 N −1 m −2 s −1 as needed. Remember th<strong>at</strong> the electric<br />
field has dimensions NC −1 . Also, because of the Onsager rel<strong>at</strong>ion we have th<strong>at</strong><br />
σ = σ T .<br />
The thermal conductivity tensor κ for the electrons is defined by the he<strong>at</strong><br />
current due to the electrons as a function of the thermal gradient, when there is<br />
no electrical current, or ⃗ JQ = −κ· ⃗ ∇T , ⃗j = 0. Because there is no electric current<br />
there is an electric field counteracting the force due to the thermal gradient:
256 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
or<br />
and hence<br />
which gives<br />
0 = N 2 A e2<br />
T L11 · ⃗ E + NAe<br />
T 2 L12 · ⃗ ∇T (6.102)<br />
⃗E = − 1<br />
NAeT L−1<br />
11 L12 · ⃗ ∇T (6.103)<br />
⃗JQ = 1<br />
T 2 (L21 · L −1<br />
11 L12 · ⃗ ∇T − L22 · ⃗ ∇T ) (6.104)<br />
and again because of Onsager we have κ = κ T .<br />
Cubic m<strong>at</strong>erials are easy.<br />
κ = 1<br />
T 2 (L22 − L21L −1<br />
11 L12) (6.105)<br />
In cubic m<strong>at</strong>erials all tensors in Carthesian space are simple numbers times<br />
the unit m<strong>at</strong>rix. Therefore, we have only three numbers th<strong>at</strong> characterize the<br />
transport, L11, L12 = L21, and L22. In th<strong>at</strong> case the electrical and thermal<br />
conductivity are given by:<br />
and we have<br />
κ = 1<br />
σ = N 2 A e2<br />
T L11 (6.106)<br />
T 2 (L22 − L 2 21L −1<br />
11<br />
) (6.107)<br />
σκ = N 2 A e2<br />
T 3 (L11L22 − L 2 21) (6.108)<br />
and since the m<strong>at</strong>rix of transport coefficients is positive definite, we see th<strong>at</strong> in<br />
this case:<br />
σκ > 0 (6.109)<br />
which, of course, can be easily generalized, and holds in all cases, when we<br />
interpret the gre<strong>at</strong>er than to mean being positive definite. This rel<strong>at</strong>ion is no<br />
surprise, since electrical currents always follow the electric field, and thermal<br />
currents always follow the inverse temper<strong>at</strong>ure gradient. But here we have<br />
rel<strong>at</strong>ed these common observ<strong>at</strong>ions to the production of entropy, which gives the<br />
fact th<strong>at</strong> the m<strong>at</strong>rix of transport coefficients is positive definite. The electrical<br />
conductivity is positive because of Newton’s law. The thermal conductivity is<br />
positive because of the second law of thermodynamics.
6.6. THERMO-ELECTRIC EFFECTS. 257<br />
Thermopower.<br />
Equ<strong>at</strong>ion 6.102 has interesting consequences. If there is a temper<strong>at</strong>ure gradient<br />
across a conductor, and if the end points are not connected and no current<br />
can flow, there will be a particle flow inside the m<strong>at</strong>erial causing a buildup of<br />
charge <strong>at</strong> the endpoints. The particle flow will stop when the electric force balances<br />
the thermal force, and the resultant electric field follows from 6.102, as<br />
given before:<br />
⃗E = − 1<br />
NAeT L−1<br />
11 L12 · ⃗ ∇T (6.110)<br />
The constant of proportionality is the absolute thermoelectric power tensor ϵ,<br />
defined by ⃗ E = ϵ ⃗ ∇T , and it follows th<strong>at</strong><br />
ϵ = − 1<br />
NAeT L−1<br />
11 L12<br />
(6.111)<br />
Note th<strong>at</strong> the n<strong>at</strong>ure of the charge carriers is reflected in the sign of ϵ. The<br />
carrier concentr<strong>at</strong>ion will be less <strong>at</strong> higher temper<strong>at</strong>ures. Therefore, if the charge<br />
carriers are positive, the low temper<strong>at</strong>ure part will have a larger positive charge,<br />
and the resulting field will be towards the high temper<strong>at</strong>ure. For neg<strong>at</strong>ive<br />
charged carriers this is the opposite.<br />
Suppose we have a m<strong>at</strong>erial with end points <strong>at</strong> temper<strong>at</strong>ures T1 and T2. No<br />
current flows, and therefore the difference in potential between the end points<br />
is<br />
∫ 2<br />
V2 − V1 = − ⃗E · d⃗r = −<br />
1<br />
∫ 2<br />
1<br />
d⃗r · ϵ · ⃗ ∇T (6.112)<br />
If we now take two different m<strong>at</strong>erials and connect the end points <strong>at</strong> one end,<br />
there will be a potential difference between the unconnected end points <strong>at</strong> the<br />
other end. The larger the difference in thermoelectric power between the two<br />
m<strong>at</strong>erials, the larger the voltage difference.<br />
A very simple expression can be obtained if we assume th<strong>at</strong> the temper<strong>at</strong>ures<br />
are very close together, and the m<strong>at</strong>erial is cubic. Then we have<br />
V2 − V1 ≈ (T1 − T2)ϵ (6.113)<br />
where the tensor is now reduced to a number times the unit m<strong>at</strong>rix. If we<br />
measure the voltage difference <strong>at</strong> the disconnected end points <strong>at</strong> T1 we find:<br />
V x<br />
1 − V y<br />
1 ≈ (T1 − T2)(ϵ y − ϵ x ) (6.114)<br />
which shows how a thermocouple can be used to measure a temper<strong>at</strong>ure difference.<br />
If we compare with formula 6.3 we see th<strong>at</strong>:<br />
αAB = ϵ A − ϵ B<br />
(6.115)
258 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
Reformul<strong>at</strong>ion of the transport equ<strong>at</strong>ions.<br />
The equ<strong>at</strong>ions for the electric and he<strong>at</strong> current densities can now be modified<br />
by elimin<strong>at</strong>ing the transport coefficients in terms of the measurable quantities<br />
we used in the previous subsections. We obtain:<br />
L11 = T<br />
N 2 σ (6.116)<br />
Ae2 T 2<br />
L12 = −NAeT L11 · ϵ = − σϵ (6.117)<br />
NAe<br />
L21 = L T 12 = −<br />
which yields the following equ<strong>at</strong>ions:<br />
Other thermoelectric effects.<br />
2 T<br />
NAe ϵT σ T<br />
(6.118)<br />
L22 = T 2 κ + T 3 ϵ T σ T ϵ (6.119)<br />
⃗j = σ · ⃗ E − σϵ ⃗ ∇T (6.120)<br />
⃗JQ = T ϵ T σ ⃗ E − (κ + T ϵ T σϵ) ⃗ ∇T (6.121)<br />
When a current flows through a m<strong>at</strong>erial <strong>at</strong> constant temper<strong>at</strong>ure, it causes<br />
an electric field across the m<strong>at</strong>erial, which in turn causes a he<strong>at</strong> flow. In this<br />
scenario we have<br />
⃗JQ = T ϵ T⃗j (6.122)<br />
for ⃗ ∇T = 0. If we force a current through two different m<strong>at</strong>erials connected<br />
in series, the he<strong>at</strong> flow will be different in the two m<strong>at</strong>erials. Therefore, <strong>at</strong> the<br />
junction he<strong>at</strong> is produced or absorbed. The amount is<br />
∆Q = T ∆ϵ T⃗j (6.123)<br />
This is the Peltier effect. By comparing with equ<strong>at</strong>ion 6.4 we see th<strong>at</strong><br />
ΠAB = T ∆ϵ (6.124)<br />
for cubic m<strong>at</strong>erials, which is the rel<strong>at</strong>ion found by Kelvin in an empirical manner.<br />
He<strong>at</strong> gener<strong>at</strong>ion in resistors.<br />
We all know the formula for the gener<strong>at</strong>ion of he<strong>at</strong> in a resistor, where power<br />
production is proportional to ⃗j · ⃗ E. This formula, however, assumes constant
6.6. THERMO-ELECTRIC EFFECTS. 259<br />
temper<strong>at</strong>ure. Wh<strong>at</strong> if there is also a temper<strong>at</strong>ure gradient in the resistor?<br />
We can find the answer quite easily. Assume th<strong>at</strong> we drive a current density<br />
⃗j through a resistor, and th<strong>at</strong> there is a temper<strong>at</strong>ure gradient ⃗ ∇T . These two<br />
quantities determine completely the st<strong>at</strong>e of the system, because of the transport<br />
equ<strong>at</strong>ions. The electric field in the resistor follows from 6.120:<br />
⃗E = ρ · ⃗j + ϵ ⃗ ∇T (6.125)<br />
where ρ is the resistivity tensor, the inverse of the conductivity tensor. The<br />
electric field is caused by both a charge transport due to the applied current<br />
and a charge transport due to the thermal gradient. The he<strong>at</strong> current follows<br />
from 6.121 and is<br />
⃗JQ = T ϵ T (⃗j + σϵ ⃗ ∇T ) − (κ + T ϵ T σϵ) ⃗ ∇T = T ϵ T⃗j − κ ⃗ ∇T (6.126)<br />
The entropy production follows from 6.100:<br />
Ps = − 1<br />
T 2 ⃗ ∇T · (T ϵ T⃗j − κ ⃗ ∇T ) + 1<br />
T (ρ · ⃗j + ϵ ⃗ ∇T ) · ⃗j (6.127)<br />
which leads to production of thermal energy <strong>at</strong> a r<strong>at</strong>e<br />
P = T Ps = ⃗j · ρ · ⃗j + 1<br />
T ⃗ ∇T · κ · ⃗ ∇T (6.128)<br />
as expected. The first term is the Joule he<strong>at</strong>, the second term the he<strong>at</strong> gener<strong>at</strong>ed<br />
while keeping a temper<strong>at</strong>ure gradient.<br />
Now we will perform a different experiment. We set up a resistor and fix<br />
the temper<strong>at</strong>ures <strong>at</strong> the end points, under conditions where there is no electric<br />
current. This sets up a he<strong>at</strong> flow in the resistor, and sets up a temper<strong>at</strong>ure profile<br />
in the resistor. Since he<strong>at</strong> is flowing, and he<strong>at</strong> is gener<strong>at</strong>ed, the temper<strong>at</strong>ure<br />
profile would change unless we keep each point in contact with a he<strong>at</strong> reservoir.<br />
This is exactly wh<strong>at</strong> we do in the following experiment. We take a conductor,<br />
and <strong>at</strong> each point of the conductor we make thermal contact with a reservoir,<br />
in such a manner th<strong>at</strong> the temper<strong>at</strong>ure of the conductor <strong>at</strong> each point remains<br />
<strong>at</strong> a constant value. Because of the gener<strong>at</strong>ion of he<strong>at</strong> and the he<strong>at</strong> currents,<br />
this means th<strong>at</strong> <strong>at</strong> each point of the conductor there is a he<strong>at</strong> flow into or out<br />
of the reservoir, needed to keep the temper<strong>at</strong>ure <strong>at</strong> th<strong>at</strong> point constant.<br />
We now ask the following question. In the previous set-up, if we now add an<br />
electric current in the conductor, wh<strong>at</strong> is the additional he<strong>at</strong> flow into or out<br />
of the reservoir needed to keep the temper<strong>at</strong>ure the same? In order to calcul<strong>at</strong>e<br />
this, we need to calcul<strong>at</strong>e the change in internal energy <strong>at</strong> each point.<br />
The energy transport is given by ⃗ JU = ⃗ JQ + µ ⃗ JN and the local change in<br />
energy follows from<br />
∂u<br />
∂t = −⃗ ∇ · ⃗ JU<br />
(6.129)<br />
In a steady st<strong>at</strong>e there is a constant profile in the particle density (no accumul<strong>at</strong>ion<br />
of charge!), and we have
260 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
and hence<br />
0 = ∂n<br />
∂t = −⃗ ∇ · ⃗ JN<br />
(6.130)<br />
∂u<br />
∂t = −⃗ ∇ · ⃗ JU = − ⃗ ∇ · ⃗ JQ − ( ⃗ ∇µ) ⃗ JN = − ⃗ ∇ · ⃗ JQ + ⃗ E · ⃗j (6.131)<br />
Using equ<strong>at</strong>ions 6.125 and 6.126 we get<br />
and since ⃗ ∇ · ⃗j = 0 this leads to<br />
∂u<br />
∂t = −⃗ ∇ · (T ϵ T⃗j − κ ⃗ ∇T ) + (ρ · ⃗j + ϵ ⃗ ∇T ) · ⃗j (6.132)<br />
∂u<br />
∂t = ⃗j · ρ · ⃗j + ⃗ ∇ · (κ ⃗ ∇T ) − T ( ⃗ ∇ϵ T ) · ⃗j (6.133)<br />
and since the thermopower only changes because the temper<strong>at</strong>ure changes we<br />
write this in the form<br />
∂u<br />
∂t = ⃗j · ρ · ⃗j + ⃗ ∇ · (κ ⃗ ∇T ) − T ( ⃗ ∇T ) · dϵT<br />
dT · ⃗j (6.134)<br />
Suppose there is no current. Than the change in internal energy is given by<br />
∂u0<br />
∂t = ⃗ ∇ · (κ ⃗ ∇T ) (6.135)<br />
and this determines the he<strong>at</strong> flow with the reservoir <strong>at</strong> each point. When there<br />
is a current, the additional he<strong>at</strong> flow is therefore:<br />
∂uj<br />
∂t = ⃗j · ρ · ⃗j − T ( ⃗ ∇T ) · dϵT<br />
dT · ⃗j (6.136)<br />
The first term is easy, and is the Joule he<strong>at</strong> gener<strong>at</strong>ed in each point in the<br />
conductor. In order for the conductor to remain <strong>at</strong> a constant temper<strong>at</strong>ure<br />
profile this Joule he<strong>at</strong> needs to go into the reservoir. The second term is a new<br />
he<strong>at</strong> source, and this is called the Thomson he<strong>at</strong>. It is linear in both the thermal<br />
gradient and the electric current and for small currents domin<strong>at</strong>es the Joule he<strong>at</strong>!<br />
The strength of this he<strong>at</strong> source is directly rel<strong>at</strong>ed to the temper<strong>at</strong>ure deriv<strong>at</strong>ive<br />
of the thermo-power.<br />
Comparing with equ<strong>at</strong>ion 6.6 we see th<strong>at</strong><br />
τ = T dϵ<br />
dT<br />
(6.137)<br />
for cubic m<strong>at</strong>erials, again in agreement with wh<strong>at</strong> Kelvin found empirically.<br />
Wh<strong>at</strong> happened to the minus sign when we went to the equ<strong>at</strong>ion above?
6.6. THERMO-ELECTRIC EFFECTS. 261<br />
Adiab<strong>at</strong>ic conductivity.<br />
A recurring theme in thermodynamics is th<strong>at</strong> the values of measured properties<br />
depend on the conditions under which experiments are performed. If we<br />
measure the conductivity of a m<strong>at</strong>erial <strong>at</strong> constant temper<strong>at</strong>ure we find the<br />
property σ. It is also possible to perform the experiment under adiab<strong>at</strong>ic conditions,<br />
in this case demanding th<strong>at</strong> ⃗ JQ = 0. Equ<strong>at</strong>ion 6.121 then gives<br />
and equ<strong>at</strong>ion 6.120 then leads to<br />
T ϵ T σ ⃗ E = (κ + T ϵ T σϵ) ⃗ ∇T (6.138)<br />
⃗j = σ · ⃗ E − σϵ(κ + T ϵ T σϵ) −1 T ϵ T σ ⃗ E (6.139)<br />
and hence the adiab<strong>at</strong>ic conductivity is given by<br />
σad = σ(1 − ϵ(κ + T ϵ T σϵ) −1 T ϵ T σ) (6.140)<br />
As always, we need to define which are the st<strong>at</strong>e variables we have under control.<br />
Note th<strong>at</strong> for very large temper<strong>at</strong>ures the adiab<strong>at</strong>ic conductivity goes to zero!<br />
Wh<strong>at</strong> does it mean th<strong>at</strong> the adiab<strong>at</strong>ic conductivity is zero?<br />
External versus total chemical potential.<br />
We rel<strong>at</strong>ed the gradient in the chemical potential to the electric field by<br />
⃗E = 1<br />
NAe ⃗ ∇µ (6.141)<br />
but this is obviously only the gradient of the external chemical potential. We<br />
also know from thermodynamics th<strong>at</strong><br />
SdT − V dp + Ndµ = 0 (6.142)<br />
Now assume th<strong>at</strong> the pressure is always constant, in which case we set dp = 0.<br />
Therefore we have for the gradients:<br />
or<br />
S ⃗ ∇T + N ⃗ ∇µtot = 0 (6.143)<br />
S ⃗ ∇T + NNAe ⃗ E + N ⃗ ∇µint = 0 (6.144)<br />
Therefore, if we apply an electric field and a temper<strong>at</strong>ure gradient, we autom<strong>at</strong>ically<br />
gener<strong>at</strong>e a gradient in the internal chemical potential, and hence a<br />
gradient in the amount of m<strong>at</strong>erial! This is a warning sign when we want to
262 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.<br />
compare thermodynamics and microscopic calcul<strong>at</strong>ions. A similar situ<strong>at</strong>ion occurs<br />
in dielectrics, where on a microscopic level we calcul<strong>at</strong>e the response of the<br />
system to a change in the total field, including the induced dipole field, driven<br />
by an applied field. In our case, a microscopic calcul<strong>at</strong>ion would give results of<br />
the form<br />
⃗jN = Rint · ⃗ ∇µint + Rext · ⃗ ∇µext + RT ⃗ ∇T (6.145)<br />
We then use the rel<strong>at</strong>ion 6.144 to distribute the effects of the internal chemical<br />
potential over the gradient contributions for temper<strong>at</strong>ure and applied field in<br />
order to get the response functions needed in thermodynamics.<br />
6.7 Extra questions.<br />
1. How can we take l<strong>at</strong>tice changes as a result of pressure gradients into<br />
account?<br />
2. Is it possible to construct a theory in which pressure also drives electron<br />
transport?<br />
3. How can we expand transport theory to include spin transport?<br />
4. How do we include he<strong>at</strong> transport due to the l<strong>at</strong>tice?<br />
6.8 Problems for chapter 6<br />
Problem 1.<br />
Consider the he<strong>at</strong> current density jQ and electric current density j through<br />
a straight metal wire of length d. The end points of the wire are connected to<br />
electrically insul<strong>at</strong>ing contacts, which can be held <strong>at</strong> fixed temper<strong>at</strong>ures. We<br />
assume th<strong>at</strong> the potential and temper<strong>at</strong>ure gradients are constant in the wire.<br />
The electric current density j and he<strong>at</strong> current density jQ are given by j =<br />
σE − σϵ dT<br />
dx and jQ = T ϵσE − (κ + T σϵ2 ) dT<br />
dx , where all coefficients are numbers,<br />
not tensors, and functions of temper<strong>at</strong>ure. (a) Wh<strong>at</strong> is the difference in electric<br />
potential between the end points as a function of the temper<strong>at</strong>ure difference?<br />
(b) The he<strong>at</strong> current density can be written in the form jQ = K × (T2 − T1).<br />
Find K in terms of κ, σ, and ϵ.<br />
Problem 2.<br />
For a copper to iron thermocouple we have the following d<strong>at</strong>a. When one<br />
junction is held <strong>at</strong> the temper<strong>at</strong>ure of melting ice, the emf <strong>at</strong> the other junction<br />
as a function of temper<strong>at</strong>ure is
6.8. PROBLEMS FOR CHAPTER 6 263<br />
with<br />
E = α1t + 1<br />
2 α2t 2 + 1<br />
3<br />
α3t 3<br />
α1 = −13.403 × 10 −6 , α2 = +0.0275 × 10 −6 , α3 = +0.00026 × 10 −6<br />
and t in degrees centigrade.<br />
Find the Peltier he<strong>at</strong> transferred <strong>at</strong> a junction <strong>at</strong> the temper<strong>at</strong>ure of (a)<br />
20 ◦ C, (b) 200 ◦ C, by a current of 3 mA in 5 minutes.<br />
If the current is 1 µA, <strong>at</strong> wh<strong>at</strong> temper<strong>at</strong>ure is the Peltier he<strong>at</strong> equal to zero?<br />
Problem 3.<br />
Consider the he<strong>at</strong> current density jq and electric current density je through<br />
a straight metal wire of length L. We assume th<strong>at</strong> the potential and temper<strong>at</strong>ure<br />
gradients are constant in the wire. The electric current density je and he<strong>at</strong><br />
current density jq are given by je = σE − σϵ dT<br />
dx and jq = T ϵσE − (κ + T σϵ2 ) dT<br />
dx ,<br />
where all coefficients are scalars, not tensors, and are functions of tempera-<br />
ture. The he<strong>at</strong> production is given by Pq = σ−1j 2 + κ dT<br />
T ( dx )2 . Find the electric<br />
conductivity as measured when there is no he<strong>at</strong> flow in the m<strong>at</strong>erial (adiab<strong>at</strong>ic<br />
conductivity), in terms of the quantities given in the formulas, and find the limit<br />
of the value of the adiab<strong>at</strong>ic conductivity for large temper<strong>at</strong>ures. Calcul<strong>at</strong>e the<br />
r<strong>at</strong>io of the he<strong>at</strong> production for a given electric current density in the adiab<strong>at</strong>ic<br />
case and in the normal constant temper<strong>at</strong>ure case.<br />
Problem 4.<br />
Using equ<strong>at</strong>ion (6.36) estim<strong>at</strong>e the he<strong>at</strong> production in the Niagara falls,<br />
assuming the temper<strong>at</strong>ure is constant when the w<strong>at</strong>er is falling.
264 CHAPTER 6. TRANSPORT IN THERMODYNAMICS.
Chapter 7<br />
Correl<strong>at</strong>ion Functions.<br />
Correl<strong>at</strong>ion functions are normally introduced in st<strong>at</strong>istical mechanics. It is<br />
easy to define the instantaneous positions of single <strong>at</strong>oms. It is also easy to<br />
define a function th<strong>at</strong> measures the position of two <strong>at</strong>oms simultaneously. After<br />
performing the appropri<strong>at</strong>e averages one can then construct probability distributions<br />
for finding distances between <strong>at</strong>oms.<br />
In <strong>at</strong>omic calcul<strong>at</strong>ions appropri<strong>at</strong>e averages typically means taking both sp<strong>at</strong>ial<br />
and temporal averages. We assume th<strong>at</strong> such averages yield the thermodynamical<br />
averages we need, and th<strong>at</strong> temper<strong>at</strong>ure follows. This is often the<br />
case, which is why introducing correl<strong>at</strong>ion functions on the <strong>at</strong>omic level is very<br />
useful.<br />
The calcul<strong>at</strong>ions mentioned above are, by construction, short range in n<strong>at</strong>ure.<br />
Long range correl<strong>at</strong>ions are small, and we know th<strong>at</strong> far away events<br />
should be uncorrel<strong>at</strong>ed. The zeroth order approxim<strong>at</strong>ion for large distances is<br />
therefore easy. The question arises wh<strong>at</strong> we can say about the first order approxim<strong>at</strong>ion<br />
how the correl<strong>at</strong>ion functions approach this limit. If we assume th<strong>at</strong><br />
long range fluctu<strong>at</strong>ions exist, but th<strong>at</strong> we always have local equilibrium, we can<br />
make st<strong>at</strong>ements about the approach to the uncorrel<strong>at</strong>ed st<strong>at</strong>e. This sets limits<br />
on correl<strong>at</strong>ions functions, and can be used to connect correl<strong>at</strong>ion functions with<br />
thermodynamical properties. This is the focus of the current chapter.<br />
Long range behavior in real space corresponds to low frequency behavior<br />
in Fourier space. Therefore, we are asking questions about low frequency (and<br />
hence low energy) excit<strong>at</strong>ions of the system. If our system is in thermal equilibrium,<br />
there are limit<strong>at</strong>ions on these low energy excit<strong>at</strong>ions.<br />
7.1 Description of correl<strong>at</strong>ed fluctu<strong>at</strong>ions.<br />
Phenomenological theories.<br />
In the previous chapters we have already mentioned the importance of in-<br />
265
266 CHAPTER 7. CORRELATION FUNCTIONS.<br />
teractions between parts of a m<strong>at</strong>erial a number of times. Without interactions,<br />
all m<strong>at</strong>erials would obey the ideal gas equ<strong>at</strong>ion of st<strong>at</strong>e, and we know th<strong>at</strong> to<br />
be not true. When we discussed the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e, we showed<br />
th<strong>at</strong> the additional parameters rel<strong>at</strong>e to interactions, an <strong>at</strong>tractive long range<br />
part and a strongly repulsive hard core interaction. In fact, we can say th<strong>at</strong> all<br />
devi<strong>at</strong>ions from the ideal gas law give us inform<strong>at</strong>ion about the interactions in<br />
a m<strong>at</strong>erial. The question remains, how to extract this inform<strong>at</strong>ion and how to<br />
interpret it. Van der Waals is just one simple model. The ultim<strong>at</strong>e answer can<br />
only be found in st<strong>at</strong>istical mechanics, where we calcul<strong>at</strong>e the partition function<br />
on a microscopic level, and derive equ<strong>at</strong>ions of st<strong>at</strong>e. A discussion on this<br />
level pertains to single phases. Even though the van der Waals equ<strong>at</strong>ion gives a<br />
phase transition, for each set of st<strong>at</strong>e variables only one phase is present, with<br />
the exception of the transition itself, where there is a mixture of both phases.<br />
Details of this mixture are not specified.<br />
Landau-Ginzburg theory is another way of incorpor<strong>at</strong>ing the effects of interactions.<br />
Here we recognize th<strong>at</strong> fluctu<strong>at</strong>ions exist, and th<strong>at</strong> the origin of<br />
these fluctu<strong>at</strong>ions is beyond the standard equilibrium theory. We model the<br />
effects of the fluctu<strong>at</strong>ions by a gradient term, and investig<strong>at</strong>e the consequences.<br />
This leads to the idea of a correl<strong>at</strong>ion length, and the divergence of physical<br />
properties <strong>at</strong> the phase transition. The theory is an improvement, but still too<br />
simple because it is always equivalent to mean field theory. This was elabor<strong>at</strong>ed<br />
upon in the chapter on critical exponents, where the effects of the ”real” free<br />
energy associ<strong>at</strong>ed with fluctu<strong>at</strong>ions was analyzed and rel<strong>at</strong>ions between critical<br />
exponents were developed.<br />
Finally, in the chapter on transport theory we used the existence of fluctu<strong>at</strong>ions<br />
to show why Onsager’s rel<strong>at</strong>ions are valid. Again, we used the gradient of<br />
st<strong>at</strong>e variables as a characteristic of local equilibrium. Any microscopic theory<br />
will have such terms in the macroscopic limit, and therefore Onsager’s rel<strong>at</strong>ions<br />
have to be valid for all m<strong>at</strong>erials.<br />
Correl<strong>at</strong>ions between fluctu<strong>at</strong>ions.<br />
In this chapter we will discuss the idea of interactions from a different perspective.<br />
We again divide space into cells, which are small on the macroscopic<br />
scale, but are large enough for the thermodynamic limit to be obeyed in each<br />
cell. We know th<strong>at</strong> the density in each cell fluctu<strong>at</strong>es around its average value,<br />
and we have used the local entropy to find the probability of such fluctu<strong>at</strong>ions.<br />
In order to discuss this we need to distinguish between instantaneous values<br />
of the density in a cell and its thermal average. Assume th<strong>at</strong> in thermal<br />
equilibrium the density in a cell <strong>at</strong> ⃗r is given by n(⃗r), in mols per cubic meter.<br />
The equilibrium density can be position dependent if we have an external<br />
potential which depends on position. For a homogeneous system we have, of<br />
course, n(⃗r) = N<br />
V , where N is the total amount of m<strong>at</strong>erial in the system and<br />
V the total volume.<br />
If we do an instantaneous measurement we find a value ρ(⃗r) for the density
7.1. DESCRIPTION OF CORRELATED FLUCTUATIONS. 267<br />
the cell. Because the density can fluctu<strong>at</strong>e, this is not equal to the equilibrium<br />
value, but we have<br />
⟨ρ(⃗r)⟩ = n(⃗r) (7.1)<br />
where the brackets denote taking the thermal average, or local equilibrium value.<br />
Now consider a measurement of the density <strong>at</strong> two different places, done<br />
simultaneously. We find a value ρ(⃗r)ρ(⃗r ′ ) In a simple minded theory without<br />
long range interactions fluctu<strong>at</strong>ions in neighboring cells are not correl<strong>at</strong>ed, and<br />
we have for the equilibrium value<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩ = ⟨ρ(⃗r)⟩⟨ρ(⃗r ′ )⟩ , ⃗r ̸= ⃗r ′<br />
(7.2)<br />
for fluctu<strong>at</strong>ions in different cells. For a homogeneous system the value is ( )<br />
N 2.<br />
V<br />
For the rest of the discussion we will assume th<strong>at</strong> we have a homogeneous system,<br />
unless st<strong>at</strong>ed otherwise.<br />
When we consider fluctu<strong>at</strong>ions in the same cell, however, we have to use the<br />
results from a previous chapter. Equ<strong>at</strong>ion 2.148 shows th<strong>at</strong><br />
and with<br />
⟨(∆N) 2 ⟩ = −kB<br />
( )<br />
∂S<br />
∂N U,V<br />
( ( 2 ∂ S<br />
∂N 2<br />
) ) −1<br />
V,U<br />
= − µ<br />
T<br />
(7.3)<br />
(7.4)<br />
the equ<strong>at</strong>ion for the fluctu<strong>at</strong>ions in density is quite complex, because temper<strong>at</strong>ure<br />
fluctu<strong>at</strong>es too.<br />
Density fluctu<strong>at</strong>ions only.<br />
We can simplify our model by assuming th<strong>at</strong> the system is <strong>at</strong> constant temper<strong>at</strong>ure<br />
T. Hence the only fluctu<strong>at</strong>ions we consider are fluctu<strong>at</strong>ions in the<br />
density. This is quite reasonable. Energy fluctu<strong>at</strong>es between the cells, but<br />
the transport of energy is a transport of momentum. M<strong>at</strong>erial oscill<strong>at</strong>es back<br />
and forth and the collisions transport energy. The transport of m<strong>at</strong>erial itself,<br />
though, needs a transport of mass, and this is typically slower. Hence if we<br />
look <strong>at</strong> the low frequency limit the effects of density fluctu<strong>at</strong>ions are the<br />
last too disappear. In other words, on the time scales where our description is<br />
useful, energy fluctu<strong>at</strong>ions have already died out, and we can assume a constant<br />
temper<strong>at</strong>ure everywhere.<br />
This point is not obvious. We have seen in chapter two th<strong>at</strong> we can integr<strong>at</strong>e<br />
out the energy fluctu<strong>at</strong>ions, and obtain a probability distribution function for<br />
the density fluctu<strong>at</strong>ions, with a variance depending on both fluctu<strong>at</strong>ions. Here<br />
we follow a different recipe, and assume th<strong>at</strong> the energy fluctu<strong>at</strong>ions are coupled<br />
to the outside world though the temper<strong>at</strong>ure reservoir. This means th<strong>at</strong> for a
268 CHAPTER 7. CORRELATION FUNCTIONS.<br />
given density change the average change in energy is now biased in order to<br />
keep the temper<strong>at</strong>ure the same. This changes the distribution of the density<br />
fluctu<strong>at</strong>ions, and makes the variance the simple function discussed in the next<br />
paragraph. Note th<strong>at</strong> with volume and temper<strong>at</strong>ure specified, the value of the<br />
density and the condition of local equilibrium together determine the value of<br />
the pressure and chemical potential. This means th<strong>at</strong> the pressure and chem-<br />
ical potential ) also fluctu<strong>at</strong>e, but are rel<strong>at</strong>ed by simple partial deriv<strong>at</strong>ives, like<br />
. The entropy fluctu<strong>at</strong>es, too, in a dependent manner!<br />
( ∂p<br />
∂N<br />
T,V<br />
At constant temper<strong>at</strong>ure the fluctu<strong>at</strong>ions are now given by<br />
⟨(∆N) 2 ( )<br />
∂N<br />
⟩ = kBT<br />
∂µ<br />
Because changes in intensive st<strong>at</strong>e variables are rel<strong>at</strong>ed by<br />
we see th<strong>at</strong><br />
T,V<br />
(7.5)<br />
0 = SdT − V dp + Ndµ (7.6)<br />
( )<br />
∂N<br />
∂µ T,V<br />
= N<br />
V<br />
( )<br />
∂N<br />
∂p T,V<br />
and the triple partial deriv<strong>at</strong>ive product rel<strong>at</strong>ion then tells us<br />
( )<br />
∂N<br />
∂p<br />
( )<br />
∂N<br />
= −<br />
∂V<br />
( )<br />
∂V<br />
∂p<br />
( )<br />
∂N<br />
=<br />
∂V<br />
T,V<br />
T,p<br />
T,N<br />
V κT<br />
T,p<br />
(7.7)<br />
(7.8)<br />
The partial deriv<strong>at</strong>ive <strong>at</strong> constant T and p is easy to find. If we change the<br />
volume <strong>at</strong> constant T and p, the amount of m<strong>at</strong>erial has to change in the same<br />
way, since we know th<strong>at</strong> N(V, T, p) = V n(T, p). In other words, if all intensive<br />
st<strong>at</strong>e variables are constant, the density is constant, and hence<br />
Combining all this leads to<br />
( ∂N<br />
∂V<br />
⟨(∆N) 2 ⟩ =<br />
)<br />
T,p<br />
= N<br />
V<br />
( ) 2<br />
N<br />
VcellkBT κT<br />
V<br />
(7.9)<br />
(7.10)<br />
where the volume involved is the volume of the local cell, hence the subscript.<br />
and the fluctu<strong>at</strong>ions in amount of<br />
Note th<strong>at</strong> for an ideal gas κT = 1<br />
p<br />
m<strong>at</strong>erial in mols is equal to Ncell<br />
NA<br />
= V<br />
NRT<br />
, which is independent of temper<strong>at</strong>ure!<br />
We can also include time, and discuss the correl<strong>at</strong>ion between fluctu<strong>at</strong>ions<br />
<strong>at</strong> different times. The time scale is set by the time needed for a local cell to<br />
reach equilibrium, time is also divided into cells. In th<strong>at</strong> case we have<br />
⟨ρ(⃗r, t)ρ(⃗r ′ , t ′ )⟩ =<br />
( ) 2<br />
N<br />
, ⃗r ̸= ⃗r<br />
V<br />
′<br />
or t ̸= t ′<br />
(7.11)
7.1. DESCRIPTION OF CORRELATED FLUCTUATIONS. 269<br />
⟨ρ(⃗r, t)ρ(⃗r, t)⟩ =<br />
( ) 2 (<br />
N kBT κT<br />
V Vcell<br />
where the l<strong>at</strong>ter follows using ⟨n 2 ⟩ = ⟨(∆n) 2 ⟩ + (⟨n⟩) 2 .<br />
Typical distances and times.<br />
)<br />
+ 1<br />
(7.12)<br />
How small can we choose our cell? We need enough m<strong>at</strong>erial in a cell to allow<br />
the cell to behave like a macroscopic system. Wh<strong>at</strong> is enough? Th<strong>at</strong> depends on<br />
our required precision. For example, if we have a cubic cell with sides of 10 −6 m,<br />
such a cell in a solid contains about 10 10 <strong>at</strong>oms, which is fairly large and gives<br />
fluctu<strong>at</strong>ions in particle numbers of order 10 −5 . In a gas, there are fewer <strong>at</strong>oms,<br />
say 10 6 , but this still has fluctu<strong>at</strong>ions much smaller than one percent. So we<br />
can say th<strong>at</strong> we probably should have distances larger than 1 µm. In terms of<br />
wavelengths of perturb<strong>at</strong>ions, we can say th<strong>at</strong> we need wavelength larger than<br />
1 µm.<br />
Wh<strong>at</strong> about time scales? In order to go to equilibrium we need energy to<br />
move back and forth in a cell, and a typical speed would be the speed of sound.<br />
Say th<strong>at</strong> is about 1000 m/s. The time it takes to travel 1 µm is 10 −9 s. So if<br />
we restrict ourselves to times larger than 1 µs we are in a good regime. In a<br />
frequency domain, this means frequencies below the MHz scale.<br />
Note th<strong>at</strong> these are just estim<strong>at</strong>es, and experiment only can tell us if macroscopic<br />
theories are sufficient or if we need a microscopic description. But wh<strong>at</strong><br />
is sufficient? Errors less than one percent? In addition, <strong>at</strong> low temper<strong>at</strong>ures we<br />
have to consider the effect of quantum fluctu<strong>at</strong>ions as well as thermodynamic<br />
fluctu<strong>at</strong>ions, and the former can domin<strong>at</strong>e the system.<br />
Using Dirac delta functions<br />
The two equ<strong>at</strong>ions we used for correl<strong>at</strong>ions between fluctu<strong>at</strong>ions can be combines<br />
using Kronecker delta symbols:<br />
⟨ρ(⃗r, t)ρ(⃗r ′ , t ′ )⟩ =<br />
( ) 2 (<br />
N kBT κT<br />
δc(⃗r, ⃗r<br />
V Vcell<br />
′ )δc(t, t ′ )<br />
) + 1<br />
(7.13)<br />
In this equ<strong>at</strong>ion we assume th<strong>at</strong> we can divide space into cells and these cells<br />
are small on a macroscopic scale, but large enough th<strong>at</strong> we can assume local<br />
equilibrium. We will use this equ<strong>at</strong>ion on a macroscopic scale only, and in th<strong>at</strong><br />
case we can replace summ<strong>at</strong>ions over cells by integr<strong>at</strong>ions. Because ∫ δ(⃗r)d 3 r =<br />
1 we can replace the Kronecker delta by a delta function times the cell volume,<br />
since on a macroscopic scale d 3 r = Vcell.<br />
For the Kronecker delta in time we can do the same, but we need to decide<br />
on a cell in time. In our current formalism we assume th<strong>at</strong> density fluctu<strong>at</strong>ions<br />
are slow, because they involve transport of m<strong>at</strong>erial. If a cell changes N, going<br />
to equilibrium involves the transport of thermal energy. We assume th<strong>at</strong> energy
270 CHAPTER 7. CORRELATION FUNCTIONS.<br />
transport is fast. We associ<strong>at</strong>e th<strong>at</strong> with a relax<strong>at</strong>ion time, τQ, and replace<br />
the Kronecker delta in time by a delta function times this relax<strong>at</strong>ion time.<br />
Altern<strong>at</strong>ively, we can use a relax<strong>at</strong>ion frequency defined by ωQτQ = 1. Adding<br />
these elements together we arrive <strong>at</strong> the macroscopic equ<strong>at</strong>ion:<br />
⟨ρ(⃗r, t)ρ(⃗r ′ , t ′ )⟩ =<br />
( ) 2 (<br />
N kBT κT<br />
δ(⃗r − ⃗r<br />
V ωQ<br />
′ )δ(t − t ′ )<br />
) + 1<br />
(7.14)<br />
In case the density is independent of time, which happens when we have<br />
time-independent external potentials, we need to revert back to the Kronecker<br />
delta in time and get<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩ =<br />
( ) 2<br />
N<br />
(kBT κT δ(⃗r − ⃗r<br />
V<br />
′ ) + 1) (7.15)<br />
7.2 M<strong>at</strong>hem<strong>at</strong>ical functions for correl<strong>at</strong>ions.<br />
Pair distribution function.<br />
The discussion in the previous section assumes th<strong>at</strong> each cell goes to equilibrium<br />
by exchanging he<strong>at</strong> and m<strong>at</strong>ter with a reservoir. We assume th<strong>at</strong> we<br />
know th<strong>at</strong> the values of the temper<strong>at</strong>ure and the chemical potential in each cell<br />
are given, and the standard way to specify is to use external reservoirs. But<br />
th<strong>at</strong> is not the correct picture in this case. Each cell interacts only with its<br />
neighbors, and thermal energy and m<strong>at</strong>ter have to come from these neighboring<br />
cells. If the amount of m<strong>at</strong>erial in our cell increases, the amount of m<strong>at</strong>erial in<br />
the neighboring cell decreases! The same is true for thermal energy.<br />
We are looking <strong>at</strong> time scales for which thermal equilibrium is established.<br />
Hence the exchange of thermal energy is fast, and we have de facto local equilibrium<br />
in all cells, and we can assume th<strong>at</strong> all cells are <strong>at</strong> the same temper<strong>at</strong>ure<br />
T . Therefore, is we measure an instantaneous value of the density ρ(⃗r) the measurement<br />
is slow enough th<strong>at</strong> this represents a local equilibrium value, and th<strong>at</strong><br />
the values of the density, temper<strong>at</strong>ure, and volume of the cell then determine<br />
the value of the chemical potential in the cell. If we now compare neighboring<br />
cells, we find a difference in chemical potential, which will lead to currents and<br />
in the end to the establishment of global thermal equilibrium.<br />
This implies, however, th<strong>at</strong> there is correl<strong>at</strong>ion between the values of the<br />
density in neighboring cells, and hence we cannot assume th<strong>at</strong> the thermal<br />
averages over neighboring cells are independent. In addition, the presence of<br />
interactions changes the notion th<strong>at</strong> fluctu<strong>at</strong>ions are independent. Th<strong>at</strong> also<br />
leads to correl<strong>at</strong>ions between different cells.<br />
We define the pair distribution function g by:<br />
⟨ρ(⃗r, t)ρ(⃗r ′ , t ′ )⟩ =<br />
( ) 2<br />
N<br />
g(⃗r, t, ⃗r<br />
V<br />
′ , t ′ ) (7.16)
7.2. MATHEMATICAL FUNCTIONS FOR CORRELATIONS. 271<br />
where we require th<strong>at</strong> the position variables are NOT in the same cell. If the<br />
system were not homogeneous, we would change this to<br />
⟨ρ(⃗r, t)ρ(⃗r ′ , t ′ )⟩ = n(⃗r, t)n(⃗r ′ , t ′ )g(⃗r, t, ⃗r ′ , t ′ ) (7.17)<br />
In homogeneous systems this function is only a function of the distances. In<br />
th<strong>at</strong> case we have no time dependence of the variables. We will now restrict<br />
ourselves to th<strong>at</strong> case, and write:<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩ =<br />
( ) 2<br />
N<br />
g(⃗r − ⃗r<br />
V<br />
′ ) (7.18)<br />
and use the same symbol g, but now with two arguments. Note th<strong>at</strong> the pair<br />
distribution function is a dimensionless quantity.<br />
It is important to realize th<strong>at</strong> this defines the pair distribution function as a<br />
thermodynamic function. Its definition relies on local equilibrium, and its values<br />
will depend on the thermodynamic st<strong>at</strong>e of the whole system. In other words,<br />
its values will depend on how we measure it, in the same way as the value of<br />
the he<strong>at</strong> capacity depends on working <strong>at</strong> constant temper<strong>at</strong>ure or entropy. This<br />
point should not be overlooked when one makes connections with microscopic<br />
definitions in st<strong>at</strong>istical mechanics!<br />
If we now require th<strong>at</strong> the pair distribution function is continuous in the<br />
limit of zero distance, we need to modify the definition of this function when<br />
the sp<strong>at</strong>ial arguments are in the same cell. Correl<strong>at</strong>ions in one cell are taken into<br />
account by fluctu<strong>at</strong>ions, but the values are now different because neighboring<br />
cells are correl<strong>at</strong>ed. Hence we cannot use the same expression as we derived for<br />
the uncorrel<strong>at</strong>ed system and therefore we write:<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩ =<br />
( ) 2<br />
N<br />
(g(⃗r − ⃗r<br />
V<br />
′ ) + F δ(⃗r − ⃗r ′ )) (7.19)<br />
with F an unknown function of temper<strong>at</strong>ure.<br />
Because of our conditions of local equilibrium, this definition only makes<br />
sense for distances and time differences which are large enough so we have<br />
local equilibrium. In a Fourier picture, this means th<strong>at</strong> we can only describe<br />
the long wavelength and low frequency limit. A microscopic theory is needed<br />
to give the pair distribution functions for all distances and times, but it is<br />
important to realize th<strong>at</strong> in the limit mentioned above the microscopic results<br />
should agree with the concepts we develop here. Our goal for this chapter<br />
is therefore to derive some rel<strong>at</strong>ions th<strong>at</strong> the pair distribution functions have<br />
to obey, based on simple thermodynamic properties. This will again give us<br />
limits for a microscopic theory. Also, many experiments are performed under<br />
conditions of local equilibrium, and for such experiments the results derived in<br />
this chapter should be valid. The key to deciding the validity is the question<br />
wh<strong>at</strong> is a large wavelength and wh<strong>at</strong> is a low frequency.<br />
In a general thermodynamical approach, where we ignore details on the<br />
<strong>at</strong>omic structure, the pair distribution function 7.27 depends only on the difference<br />
vector ⃗r − ⃗r ′ . In a gas or liquid one can simplify this even further, because
272 CHAPTER 7. CORRELATION FUNCTIONS.<br />
such fluids are isotropic, and the pair distribution function depends only on the<br />
scalar distance |⃗r −⃗r ′ |. In a solid, however, this symmetry is always broken. For<br />
example, one can study layered m<strong>at</strong>erials with properties th<strong>at</strong> are very different<br />
in plane as compared to out of plane. Therefore, we will study the more general<br />
form. Additional simplific<strong>at</strong>ions due to the isotropic n<strong>at</strong>ure of fluids can be<br />
derives easily.<br />
It is also important to realize th<strong>at</strong> the averages we are taking are thermodynamical<br />
or macroscopic averages. Th<strong>at</strong> is quite different from the quantum<br />
mechanical averages one includes on a microscopic scale. For example, in a two<br />
particle system we can rel<strong>at</strong>e the joint probability of finding a particle <strong>at</strong> place<br />
one and the other <strong>at</strong> place two directly to the wave function. In thermodynamics,<br />
we talk about many particles and averages due to the fact th<strong>at</strong> all these<br />
particles interact. The road from the microscopic behavior to the macroscopic<br />
setting is not clearly defined, and is a very interesting topic of study. Again, for<br />
another course.<br />
Since we study the pair distribution function as a thermodynamic quantity,<br />
we need to be precise which other st<strong>at</strong>e variables are kept constant, and which<br />
are not. In our simple case we assume a constant temper<strong>at</strong>ure T. This means<br />
th<strong>at</strong> the pressure will fluctu<strong>at</strong>e, since the pressure follows from the equ<strong>at</strong>ion of<br />
st<strong>at</strong>e, given the density and temper<strong>at</strong>ure. Remember, we assume local equilibrium!<br />
If we also allow the energy to fluctu<strong>at</strong>e, then the temper<strong>at</strong>ure will<br />
fluctu<strong>at</strong>e, too. Th<strong>at</strong> is a more general case. One could restrict this by assuming<br />
a constant pressure, which might be appropri<strong>at</strong>e in certain situ<strong>at</strong>ions. But<br />
no m<strong>at</strong>ter wh<strong>at</strong> the restrictions are, like always in thermodynamics one has to<br />
know wh<strong>at</strong> they are, and be aware of the fact th<strong>at</strong> different external constraints<br />
yield different results for thermodynamic quantities like the pair distribution<br />
function.<br />
The value of F can be chosen as follows. Suppose we integr<strong>at</strong>e ⃗r ′ in the<br />
correl<strong>at</strong>ion function over a very small volume ν around ⃗r. Th<strong>at</strong> leads to<br />
∫<br />
ν<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩d 3 r ′ =<br />
( ) 2 (∫<br />
N<br />
g(⃗r − ⃗r<br />
V ν<br />
′ )d 3 r ′ )<br />
+ F<br />
(7.20)<br />
Now consider an ideal gas, where there are no interactions between particles, in<br />
which case we have g = 1 and hence<br />
∫<br />
ν<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩d 3 r ′ =<br />
Next we integr<strong>at</strong>e ⃗r over the same volume to get<br />
∫<br />
ν<br />
∫<br />
ν<br />
( ) 2<br />
N<br />
(ν + F ) (7.21)<br />
V<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩d 3 rd 3 r ′ ( ) 2 (<br />
Nν<br />
= 1 +<br />
V<br />
F<br />
)<br />
ν<br />
(7.22)<br />
This procedure is allowed, even for a very small volume ν, since we are<br />
working in a time independent theory. By definition, we can wait infinitely long<br />
for our measurement, and hence we can establish thermal equilibrium. But wh<strong>at</strong>
7.2. MATHEMATICAL FUNCTIONS FOR CORRELATIONS. 273<br />
happens if we make the volume ν so small th<strong>at</strong> it is comparable with particle<br />
size? In th<strong>at</strong> case we can have only zero or one particle in our little volume,<br />
and we have, using Avogadro’s number NA:<br />
∫<br />
ν<br />
∫<br />
⟨ρ(⃗r)NAρ(⃗r ′ )NA⟩d 3 rd 3 r ′ ∫<br />
= ⟨ρ(⃗r)NA⟩d 3 N<br />
r = NA ν (7.23)<br />
V<br />
ν<br />
since 0 2 = 0 and 1 2 = 1. Hence in this limit we have<br />
or<br />
N<br />
V NA<br />
ν<br />
( ) 2 (<br />
Nν<br />
ν = 1 +<br />
V<br />
F<br />
)<br />
ν<br />
V<br />
NNAν =<br />
(<br />
1 + F<br />
)<br />
ν<br />
(7.24)<br />
(7.25)<br />
F = V<br />
− ν (7.26)<br />
NNA<br />
In this last equ<strong>at</strong>ion the first term on the right hand side is the average<br />
volume for one particle in the gas, and the second term is the volume of a<br />
particle itself, which is much smaller, and can be ignored for point particles.<br />
Note th<strong>at</strong> this leaves some questions for real particles in solids! Hence we have<br />
found a value for F , based on the knowledge th<strong>at</strong> m<strong>at</strong>ter is indeed made of<br />
<strong>at</strong>oms, and th<strong>at</strong> two <strong>at</strong>oms cannot be <strong>at</strong> the same place.<br />
This finally leads to our definition of the pair distribution function:<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩ =<br />
Pair correl<strong>at</strong>ion function.<br />
( ) 2 (<br />
N<br />
g(⃗r − ⃗r<br />
V<br />
′ ) + V<br />
δ(⃗r − ⃗r<br />
NNA<br />
′ )<br />
)<br />
(7.27)<br />
In many cases we are interested in the correl<strong>at</strong>ions between fluctu<strong>at</strong>ions.<br />
This follows quite easily from equ<strong>at</strong>ion 7.27:<br />
⟨(ρ(⃗r) − N<br />
V )(ρ(⃗r′ ) − N<br />
)⟩ =<br />
V<br />
and the function<br />
( ) 2 (<br />
N<br />
g(⃗r − ⃗r<br />
V<br />
′ , t − t ′ ) − 1 + V<br />
δ(⃗r − ⃗r<br />
NNA<br />
′ )<br />
)<br />
(7.28)<br />
h(⃗r − ⃗r ′ , t − t ′ ) = g(⃗r − ⃗r ′ , t − t ′ ) − 1 (7.29)<br />
is commonly called the pair correl<strong>at</strong>ion function. Without correl<strong>at</strong>ions between<br />
fluctu<strong>at</strong>ions, this function is zero unless we consider the fluctu<strong>at</strong>ions in one cell<br />
<strong>at</strong> one time.
274 CHAPTER 7. CORRELATION FUNCTIONS.<br />
It is instructive to evalu<strong>at</strong>e the following integral<br />
∫<br />
H = d 3 rh(⃗r) (7.30)<br />
where we take the correl<strong>at</strong>ion function <strong>at</strong> equal time, and omit the time argument.<br />
If we go back to equ<strong>at</strong>ion 7.27 we can integr<strong>at</strong>e both ⃗r and ⃗r ′ over the<br />
total volume and get<br />
or<br />
∫<br />
V<br />
∫<br />
V<br />
⟨ρ(⃗r)ρ(⃗r ′ )⟩d 3 rd 3 r ′ =<br />
⟨N 2 ⟩ =<br />
( ) 2 ( ∫<br />
N<br />
V g(⃗r)d<br />
V<br />
V<br />
3 r +<br />
( ) 2 ( ∫<br />
N<br />
V g(⃗r)d<br />
V<br />
V<br />
3 r +<br />
V 2<br />
NNA<br />
)<br />
) 2 V<br />
N<br />
(7.31)<br />
(7.32)<br />
Assume we have a system <strong>at</strong> temper<strong>at</strong>ure T and chemical potential µ. In<br />
th<strong>at</strong> case we have<br />
and hence<br />
or<br />
which leads to<br />
⟨(∆N) 2 ⟩ =<br />
( ) 2<br />
N<br />
V kBT κT + N<br />
V<br />
2 =<br />
V kBT κT + V 2 ∫<br />
= V<br />
( ) 2<br />
N<br />
V kBT κT<br />
V<br />
( ) 2 ( ∫<br />
N<br />
V g(⃗r)d<br />
V<br />
V<br />
3 r +<br />
V<br />
V 2<br />
g(⃗r)d 3 r +<br />
NNA<br />
V 2<br />
NNA<br />
)<br />
(7.33)<br />
(7.34)<br />
(7.35)<br />
H = kBT κT − V<br />
NNA<br />
(7.36)<br />
This is consistent with the equ<strong>at</strong>ions of st<strong>at</strong>e for an ideal gas, which has<br />
κT = 1<br />
p and hence H = 0. For an ideal gas we have g = 1 and h = 0 indeed.<br />
The equ<strong>at</strong>ion above is called the compressibility equ<strong>at</strong>ion of st<strong>at</strong>e. We can take<br />
a deriv<strong>at</strong>ive to get a rel<strong>at</strong>ed form:<br />
∫<br />
NA d 3 ( )<br />
∂h<br />
r<br />
∂T<br />
( )<br />
∂RT κT<br />
(⃗r, T ) =<br />
∂T<br />
(7.37)<br />
N,V<br />
which is another useful form of this equ<strong>at</strong>ion of st<strong>at</strong>e! Note th<strong>at</strong> in the equ<strong>at</strong>ion<br />
7.37 we use h(T, µ, V ; ⃗r), but th<strong>at</strong> the deriv<strong>at</strong>ive is <strong>at</strong> constant N!<br />
There are two useful limits of the equ<strong>at</strong>ion of st<strong>at</strong>e in equ<strong>at</strong>ion 7.36. In<br />
liquids and solids the compressibility is much smaller than th<strong>at</strong> of an ideal gas,<br />
and we have:<br />
N,V
7.3. ENERGY CONSIDERATIONS. 275<br />
H ≈ − V<br />
NNA<br />
Liquids and solids (7.38)<br />
which means th<strong>at</strong> fluctu<strong>at</strong>ions are reduced from ideal gas values because <strong>at</strong>oms<br />
cannot be <strong>at</strong> the same place. The other case occurs near phase transitions,<br />
where κT → ∞, in which case we have<br />
H → ∞ Near phase transitions (7.39)<br />
consistent with the fact th<strong>at</strong> fluctu<strong>at</strong>ions now have a very long range.<br />
7.3 Energy consider<strong>at</strong>ions.<br />
Energy of fluctu<strong>at</strong>ions.<br />
The m<strong>at</strong>erial in different cells interact with each other. There is an energy<br />
associ<strong>at</strong>ed with this interaction, which we call Ueff (⃗r, ⃗r ′ ). It is important to<br />
realize th<strong>at</strong> this is an effective interaction, since we look <strong>at</strong> the interactions<br />
between cells which are in local equilibrium. This interaction is much smaller<br />
than the real <strong>at</strong>omic interaction between the <strong>at</strong>oms in the cell, since screening<br />
and similar many body effects are included in this interaction. If the fluctu<strong>at</strong>ions<br />
in different cells where uncorrel<strong>at</strong>ed, this interaction would be zero. But since<br />
there is correl<strong>at</strong>ion, some interaction survives. Since we study the interaction<br />
between cells, this is clearly a long range interaction. In addition, we can assume<br />
th<strong>at</strong> the interaction depends on the difference between the positions only.<br />
The energy associ<strong>at</strong>ed with the fluctu<strong>at</strong>ions can be easily evalu<strong>at</strong>ed. If the<br />
energy for the homogeneous system is Uh then the energy due to the fluctu<strong>at</strong>ions<br />
is given by (since this is equal time, we leave out the time variable):<br />
Uf (T, V, N)) = 1<br />
∫<br />
2<br />
d 3 ∫<br />
r<br />
d 3 r ′ Ueff (⃗r − ⃗r ′ )⟨(ρ(⃗r) − N<br />
V<br />
N<br />
)(ρ(⃗r) − )⟩ (7.40)<br />
V<br />
The factor one half is needed to avoid double counting. Also, we need to be<br />
careful how to tre<strong>at</strong> the integr<strong>at</strong>ion when both position variables describe the<br />
same cell. Part of th<strong>at</strong> energy is already included in the thermodynamic description<br />
of the system <strong>at</strong> constant density. This can be done by replacing the<br />
correl<strong>at</strong>ion function by the pair correl<strong>at</strong>ion function without the term depending<br />
on F . Another way of looking <strong>at</strong> this is to do the integr<strong>at</strong>ion, and observe th<strong>at</strong><br />
the term with F looks like F Ueff (0), which can be absorbed in Uh. Hence we<br />
have<br />
Uf (T, V, N)) =<br />
2 N<br />
2V 2<br />
∫<br />
d 3 ∫<br />
r<br />
d 3 r ′ Ueff (⃗r − ⃗r ′ )h(⃗r − ⃗r ′ ) =<br />
N 2<br />
2V<br />
∫<br />
d 3 rUeff (⃗r)h(⃗r)<br />
(7.41)
276 CHAPTER 7. CORRELATION FUNCTIONS.<br />
The energy is a function of the total volume, total number of particles, and<br />
temper<strong>at</strong>ure. This needs some explan<strong>at</strong>ion. We originally defined fluctu<strong>at</strong>ions<br />
in a large cell of given volume, amount of m<strong>at</strong>erial, and energy. But then we<br />
assumed th<strong>at</strong> energy fluctu<strong>at</strong>ions are fast and th<strong>at</strong> the energy is always the<br />
thermal equilibrium value for the fluctu<strong>at</strong>ing density <strong>at</strong> the temper<strong>at</strong>ure for the<br />
whole system. In order for th<strong>at</strong> to be meaningful, it means th<strong>at</strong> we replace our<br />
boundary condition for the large cell by a condition of constant temper<strong>at</strong>ure.<br />
But we still assume th<strong>at</strong> the total amount of m<strong>at</strong>erial in the large volume is<br />
constant. We could also allow th<strong>at</strong> to change, by working <strong>at</strong> a given chemical<br />
potential.<br />
The way we have defined the pair correl<strong>at</strong>ion function is as a thermodynamic<br />
quantity, and it will depend on the st<strong>at</strong>e variables of the whole system. So we<br />
either have hN(T, V, N, ⃗s) or hµ(T, V, µ, ⃗s) and as a function of position these<br />
are different functions! Again, for the compressibility equ<strong>at</strong>ion of st<strong>at</strong>e 7.36 we<br />
need to use the pair distribution function as a function of µ!<br />
It can be useful to rewrite this by subtracting and adding a typical energy:<br />
Uf (T, V, N)) = V<br />
2<br />
( ) 2 ∫<br />
N<br />
V<br />
1<br />
2 V<br />
d 3 s (Ueff (⃗s) − Uave) h(T, V, N; ⃗s)+<br />
( ) 2 ∫<br />
N<br />
Uave<br />
V<br />
and using the fluctu<strong>at</strong>ion equ<strong>at</strong>ion of st<strong>at</strong>e we get<br />
Uf (T, V, N)) = 1<br />
2 V<br />
1<br />
2 V<br />
( ) 2 ∫<br />
N<br />
V<br />
If we now define the average energy by<br />
∫<br />
we have<br />
d 3 sh(⃗s) (7.42)<br />
d 3 s (Ueff (⃗s) − Uave) h(⃗s)+<br />
( ) 2<br />
N<br />
(RT κT −<br />
V<br />
1<br />
n )Uave<br />
NA<br />
d 3 ∫<br />
sUeff (⃗s)hN (T, V, N, ⃗s) = Uave(T, V, N)<br />
Uf (T, V, N) = 1<br />
2 V<br />
( ) 2<br />
N<br />
(RT κT −<br />
V<br />
1<br />
n )Uave<br />
NA<br />
(7.43)<br />
d 3 shN (T, V, N, ⃗s) (7.44)<br />
(7.45)<br />
There are two interesting applic<strong>at</strong>ions of this equ<strong>at</strong>ion. For an ideal gas we<br />
have κT = 1 1<br />
p = nRT and hence Uf (T, V, N) ≡ 0. This is not correct, however.<br />
Because the integral of the pair correl<strong>at</strong>ion function is zero, we cannot define<br />
the average energy (it would be infinite), and we are left with the basic formula<br />
in terms of the integral over the effective interaction.
7.3. ENERGY CONSIDERATIONS. 277<br />
Uave<br />
2N NA .<br />
Finally, we consider the effects of making a change in volume by changing<br />
coordin<strong>at</strong>es according to ⃗r ′ = λ⃗r. This gives V ′ = λ3V and<br />
For a solid we have κT ≈ 0 and hence Uf (T, V, N) ≈ − 1<br />
Uf (T, V ′ , N)) =<br />
2 N<br />
2V ′<br />
∫<br />
V ′<br />
d 3 r ′ Ueff (⃗r ′ )h(⃗r ′ ) (7.46)<br />
The additional pressure due to this interaction follows from pf = − ( ∂U<br />
∂V<br />
which gives in the limit λ → 1<br />
pf = − Uf (T, V ′ , N)) − Uf (T, V, N))<br />
V ′ − V<br />
)<br />
S,N<br />
(7.47)<br />
The difference in internal energy is<br />
N 2<br />
2V ′<br />
∫<br />
V ′<br />
d 3 r ′ Ueff (⃗r ′ )h(⃗r ′ 2 ∫<br />
N<br />
) − d<br />
2V V<br />
3 rUeff (⃗r)h(⃗r) (7.48)<br />
Now we rewrite the integr<strong>at</strong>ion in the first term to an integr<strong>at</strong>ion over V<br />
N 2<br />
2V ′<br />
∫<br />
d<br />
V<br />
3 2 ∫<br />
N<br />
rUeff (λ⃗r)h(λ⃗r) − d<br />
2V V<br />
3 rUeff (⃗r)h(⃗r) (7.49)<br />
This is rearranged in the form<br />
N 2<br />
2V ′<br />
∫<br />
d<br />
V<br />
3 ( ) 2 2 ∫<br />
N N<br />
r {Ueff (λ⃗r)h(λ⃗r) − Ueff (⃗r)h(⃗r)} + − d<br />
2V ′ 2V V<br />
3 rUeff (⃗r)h(⃗r)<br />
(7.50)<br />
For the first part we need to recognize th<strong>at</strong> we require the difference <strong>at</strong> constant<br />
entropy. Th<strong>at</strong> requires th<strong>at</strong> all rel<strong>at</strong>ive distributions remain the same, and th<strong>at</strong><br />
in the first integral we can replace h(λ⃗r) by h(⃗r). But th<strong>at</strong> is not sufficient,<br />
since we need to keep the total amount of fluctu<strong>at</strong>ions constant as well. This<br />
means th<strong>at</strong> we need to replace h(λ⃗r) by λ3h(⃗r), because this keeps the density<br />
constant. Therefore we have<br />
N 2<br />
2V ′<br />
∫<br />
d<br />
V<br />
3 r { λ 3 Ueff (λ⃗r) − Ueff (⃗r) } ( ) 2 2 ∫<br />
N N<br />
h(⃗r) + − d<br />
2V ′ 2V V<br />
3 rUeff (⃗r)h(⃗r)<br />
(7.51)<br />
N 2<br />
2V ′<br />
∫<br />
d<br />
V<br />
3 rλ 3 [⃗r · ⃗ 2 N<br />
∇Ueff (⃗r)]∆λh(⃗r) +<br />
2V ′<br />
∫<br />
d<br />
V<br />
3 r(λ 3 − 1)Ueff (⃗r)h(⃗r)<br />
V ′<br />
V<br />
( ) 2 2 ∫<br />
N N<br />
+ − d<br />
2V ′ 2V V<br />
3 rUeff (⃗r)h(⃗r) (7.52)<br />
Because λ3 = we see th<strong>at</strong> the second and third term cancel. Next we use<br />
and get<br />
∆V<br />
V<br />
= 3 ∆λ<br />
λ
278 CHAPTER 7. CORRELATION FUNCTIONS.<br />
N 2 ∫<br />
∆V d<br />
6V 2<br />
V<br />
3 rλ[⃗r · ⃗ ∇Ueff (⃗r)]h(⃗r) (7.53)<br />
In the limit ∆V going to zero λ goes to one, and we see th<strong>at</strong> the additional<br />
pressure due to the fluctu<strong>at</strong>ions is given by<br />
2 N<br />
pf = −<br />
6V 2<br />
∫<br />
d 3 rλ[⃗r · ⃗ ∇Ueff (⃗r)]h(⃗r) (7.54)<br />
V<br />
This is a very useful formula and can be generalized to much smaller length<br />
scales and <strong>at</strong>omic interactions, as long as we keep in mind th<strong>at</strong> the formula is<br />
derived using thermal equilibrium conditions. On smaller length scales we have<br />
to be careful how to define such conditions!
PART III<br />
Additional M<strong>at</strong>erial<br />
279
280
Appendix A<br />
Questions submitted by<br />
students.<br />
Students were asked to submit 10 question they would like to see included in<br />
the notes, similar to the questions already included, shown in italics. Errors are<br />
due to my editing. Some questions were left out because of duplic<strong>at</strong>ion.<br />
A.1 Questions for chapter 1.<br />
• Wh<strong>at</strong> is the high temper<strong>at</strong>ure reservoir for an internal combustion engine?<br />
• Are the rel<strong>at</strong>ions T<br />
T0 = 1 − ηC assured, derived, or observed?<br />
• Is there a standard or limit to which real gases can follow the ideal gas<br />
approxim<strong>at</strong>ions?<br />
• The energy of an ideal gas is D<br />
2 NRT , where D is the number of dimensions?<br />
Why is there a factor one-half for each dimension?<br />
• Is there an extensive variable th<strong>at</strong> we have not yet figured out how to<br />
measure, and how will this ignorance affect existing laws?<br />
• Work can be done in many different ways. Do different mechanisms for<br />
he<strong>at</strong> transfer lead to different T dS values?<br />
• Our bodies have more or less the same temper<strong>at</strong>ure, but do give of different<br />
amounts of he<strong>at</strong>. Can this be used to calcul<strong>at</strong>e the disorder in our bodies<br />
and predict human defects?<br />
• In a system th<strong>at</strong> is not in equilibrium, wh<strong>at</strong> can we say about deterministic<br />
behavior?<br />
• Do we have the same thermodynamic laws on the moon?<br />
281
282 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Are the laws of thermodynamics valid in acceler<strong>at</strong>ed systems?<br />
• Does the second law of thermodynamics hold for our mind? Could we<br />
cre<strong>at</strong>e more than we have learned?<br />
• Is it always possible to find other reversible processes th<strong>at</strong> have the same<br />
initial and final st<strong>at</strong>e?<br />
• The third law st<strong>at</strong>es th<strong>at</strong> lim<br />
T →0 S(T ) = 0, but does this not lead to contradictions<br />
when we also have ∆S = ∆Q<br />
T ?<br />
• How can we explain the second law of thermodynamics when an electric<br />
field acts on a dielectric and rearranges the dipoles?<br />
• Is the third law derived from experimental observ<strong>at</strong>ions?<br />
• Wh<strong>at</strong> would our world look like <strong>at</strong> T=0?<br />
• Are st<strong>at</strong>e variables always dependent on other st<strong>at</strong>e variables?<br />
• Give an example of a system th<strong>at</strong> can be made to be in thermodynamic<br />
equilibrium<br />
• Are there systems th<strong>at</strong> are not in equilibrium but th<strong>at</strong> can be modelled<br />
using thermodynamic equilibrium assumptions?<br />
• Wh<strong>at</strong> would be a consequence of the zeroth law being wrong?<br />
• Are there easy rel<strong>at</strong>ions for exact differentials when we have more than<br />
two variables? E.g. when is ¯ dg = h(x, y, z)dx + k(x, y, z)dy + j(x, y, z)dz<br />
exact?<br />
• Wh<strong>at</strong> is the importance of reversibility?<br />
• Can the efficiency be exactly zero?<br />
• If entropy change is irreversible, wh<strong>at</strong> does th<strong>at</strong> imply for the f<strong>at</strong>e of the<br />
universe?<br />
• If in the equ<strong>at</strong>ion lim<br />
N→∞ lim S = 0 we could interchange the limits, wh<strong>at</strong><br />
T →0<br />
would the consequences be?<br />
• How would things change if entropy was reversible?<br />
• Give some examples of reversible processes<br />
• Give some examples of irreversible processes<br />
• Give some examples of other extensive thermodynamic variables<br />
• Give some examples of other intensive thermodynamic variables
A.1. QUESTIONS FOR CHAPTER 1. 283<br />
• Find the exact differential df for f(x, y) = 3x y − 6yx 2 and y = y(u, v)<br />
• Wh<strong>at</strong> are the consequences of having a Carnot engine with η > 1?<br />
• Wh<strong>at</strong> are the consequences of having a Carnot engine with η < 0?<br />
• Wh<strong>at</strong> happens to the efficiency of a Carnot engine when T1 is finite and<br />
T2 → 0?<br />
• If the entropy of the universe is always increasing, is the universe a reversible<br />
process?<br />
• Why can’t we measure entropy directly?<br />
• How do you know when you have enough st<strong>at</strong>e variables?<br />
• Why does a process have to be reversible in order to study it in thermodynamics?<br />
• Why are the laws of thermodynamics numbered 0 to 3 and not 1 to 4?<br />
• In defining he<strong>at</strong> and work flow, which sign convention do you prefer and<br />
why?<br />
• How would you st<strong>at</strong>e the second law?<br />
• Why does the cycle used in the Carnot cycle produce the most efficient<br />
engine?<br />
• Why would you ever use a less efficient engine?<br />
• Why will a completely isol<strong>at</strong>ed system end up in equilibrium?<br />
• Why can’t we reach zero temper<strong>at</strong>ure?<br />
• If you would be designing a temper<strong>at</strong>ure scale, wh<strong>at</strong> would you base it on<br />
and why?<br />
• The universe is cooling down; can it reach zero temper<strong>at</strong>ure?<br />
• Is the big bang a reversible process?<br />
• Why would someone want to study macroscopic systems?<br />
• Are there any st<strong>at</strong>e variables describing a piece of iron th<strong>at</strong> we have not<br />
yet mentioned?<br />
• Give examples of st<strong>at</strong>e functions<br />
• Give an example of a spontaneous process<br />
• Wh<strong>at</strong> is the difference between Kelvin’s and Clausius’ st<strong>at</strong>ements of the<br />
second law?
284 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Why is temper<strong>at</strong>ure considered a st<strong>at</strong>e variable, but he<strong>at</strong> not?<br />
• Why can’t we measure temper<strong>at</strong>ure directly?<br />
• Why does the error in the integr<strong>at</strong>ion when adding many Carnot cycles to<br />
describe an arbitrary process approach zero when ∆T → 0 for each cycle?<br />
• Why is the T dS term necessary in the first law?<br />
• Why does the entropy increase when a system reaches equilibrium in a<br />
spontaneous process?<br />
• Wh<strong>at</strong> is a system?<br />
• How can you tell the st<strong>at</strong>e a system is in?<br />
• Is the n<strong>at</strong>ural evapor<strong>at</strong>ion/rain cycle an engine?<br />
• Does he<strong>at</strong> flow?<br />
• Why is the combin<strong>at</strong>ion of an electric motor and gener<strong>at</strong>or not capable of<br />
being more efficient than a Carnot engine?<br />
• Wh<strong>at</strong> makes an engine less efficient than a Carnot engine?<br />
• Considering the example of combining two Carnot engines to get a Carnot<br />
engine oper<strong>at</strong>ing between arbitrary temper<strong>at</strong>ures, does this show th<strong>at</strong> temper<strong>at</strong>ure<br />
is a st<strong>at</strong>e variable?<br />
• Are there consequences to the universe from the inequality of irreversible<br />
processes?<br />
• Does the maximiz<strong>at</strong>ion of entropy have any universal consequences?<br />
( )<br />
∂U<br />
• How does the st<strong>at</strong>ement lim<br />
T →0<br />
∂T = 0 about the limit of the partial<br />
V,N<br />
deriv<strong>at</strong>ive agree/disagree with the quantum mechanical st<strong>at</strong>ement th<strong>at</strong><br />
the energy of an oscill<strong>at</strong>or is never <strong>at</strong> the bottom of the well?<br />
A.2 Questions for chapter 2.<br />
• Why is the minimum energy principle important?<br />
• Wh<strong>at</strong> is the usefulness of writing all of these free energies in differential<br />
form?<br />
• How many Maxwell rel<strong>at</strong>ions can there be?<br />
• Why are response functions used in experiment?<br />
• If your finger had a small he<strong>at</strong> capacity, does this mean your finger would<br />
burn less if you touched a hot stove?
A.2. QUESTIONS FOR CHAPTER 2. 285<br />
• Wh<strong>at</strong> is the usefulness of writing S as a function of T, V, and X?<br />
• If the compressibility <strong>at</strong> constant entropy was less than zero, wh<strong>at</strong> consequences<br />
would this have on engines?<br />
• Based on physical grounds, why is the product in 2.102 less than or equal<br />
to 0?<br />
• Wh<strong>at</strong> are the possible uses of the magnetic susceptibility <strong>at</strong> constant temper<strong>at</strong>ure<br />
and the magnetic susceptibility <strong>at</strong> constant entropy?<br />
• Wh<strong>at</strong> would be some of the effects on closed systems if the expect<strong>at</strong>ion<br />
value of the product of delta U and delta N was zero.<br />
• In the third case in the beginning of chapter 2,how can we be sure th<strong>at</strong><br />
the equilibrium th<strong>at</strong> has been reached by suspending the mass is the same<br />
equilibrium without it?<br />
• Why we can choose all the independent variables to be extensive?<br />
• Wh<strong>at</strong> is more fundamental in Maxwell rel<strong>at</strong>ions?Is it the m<strong>at</strong>hem<strong>at</strong>ical<br />
rules or the physical rel<strong>at</strong>ions?<br />
• Are equ<strong>at</strong>ions 2.31 and 2.32 valid for irreversible process?<br />
• Explain the three lines after 2.38 about compressibility?<br />
• Prove equ<strong>at</strong>ion 2.52?<br />
• How can we explain in physical words equ<strong>at</strong>ion 2.84 which rel<strong>at</strong>es different<br />
response functions?<br />
• How our analysis of a magnetic system will change if we have a nonlinear<br />
magnetic response (ferromagnetic m<strong>at</strong>erial)?<br />
• How can we include the higher order susceptibilities in our analysis?<br />
• Is it possible to determine ω in equ<strong>at</strong>ion 2.144 by macroscopic analysis<br />
only?<br />
• Is the correl<strong>at</strong>ion between fluctu<strong>at</strong>ion in U and N trivial (reasonable)?<br />
• Are all free energies a minimum of their internal degrees of freedom when<br />
in equilibrium?<br />
• Why are the extensive entropy second deriv<strong>at</strong>ives rel<strong>at</strong>ed to a no-root<br />
condition?<br />
• Does the value − V<br />
4p<br />
in 2.136 have any special significance?<br />
• Why are the quantities in 2.144 a Gaussian?<br />
• Why does the exponent in 2.148 contain a factor of 1<br />
2 ?
286 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Should the fluctu<strong>at</strong>ions going to zero in the thermodynamic limit surprise<br />
you?<br />
• Is the correl<strong>at</strong>ion of the fluctu<strong>at</strong>ions a consequence of the uncertainty<br />
principle or does it represent a more general uncertainty rel<strong>at</strong>ionship?<br />
• When will the partial deriv<strong>at</strong>ives cancel in a special way and why?<br />
• Why do the deriv<strong>at</strong>ive permut<strong>at</strong>ions change sign?<br />
• In a non-equilibrium condition is it possible to have a neg<strong>at</strong>ive he<strong>at</strong> capacity<br />
and wh<strong>at</strong> would it mean?<br />
• Since there are several different ways to write the energy in thermodynamics,<br />
is there something fundamentally missing from thermodynamics<br />
th<strong>at</strong> is not missing from classical mechanics?<br />
• We have these things called Maxwell Rel<strong>at</strong>ions for different thermodynamics<br />
variables. Can we formul<strong>at</strong>e such rel<strong>at</strong>ions for other disciplines?<br />
• When doing flow in magnetics, or other solid st<strong>at</strong>e phenomenological processes,<br />
we run into vortices. These seem to be a special form of fluctu<strong>at</strong>ions.<br />
Are they?<br />
• Wh<strong>at</strong> are the implic<strong>at</strong>ions to the theory if we were ever to find a neg<strong>at</strong>ive<br />
compressibility or he<strong>at</strong> capacity.<br />
• Can we really make a closed system, and why would we want to, since we<br />
can’t interact with it.<br />
• Most of the inconsistencies in thermodynamics can be ironed out with<br />
quantum mechanics, but the first uses of thermodynamics were well before<br />
quantum mechanics, how did those scientists work out the problems?<br />
• With the three systems th<strong>at</strong> we discussed early in the chapter, only one<br />
was experimental, but all three were different, why?<br />
• Can you derive a general form of the Maxwell rel<strong>at</strong>ions th<strong>at</strong> we can then<br />
just plug the different variables into?<br />
• Why do we rely so much on the ideal gas law when it only works in<br />
equilibrium, and most systems we study are not in equilibrium?<br />
• How difficult is it to find Maxwell rel<strong>at</strong>ions for more than two variables?<br />
• How can a system exchange he<strong>at</strong> with the outside world, yet remain <strong>at</strong><br />
constant entropy?<br />
• Why are the SdT − V dp + Ndµ + Xdx terms not st<strong>at</strong>ed when expressing<br />
dU?<br />
• In equ<strong>at</strong>ion 2.66 is ( )<br />
∂U<br />
∂S V,N,X =<br />
( )<br />
∂U<br />
∂Sl V,N,X = Tl ?
A.3. QUESTIONS FOR CHAPTER 3. 287<br />
• Wh<strong>at</strong> is the general rule for figuring out where the minus signs come in<br />
when taking products of partial deriv<strong>at</strong>ives of many variables.<br />
• Does the rel<strong>at</strong>ion Cp<br />
CV<br />
= κT<br />
κS<br />
hold <strong>at</strong> very low temper<strong>at</strong>ures?<br />
• Why are all the other intensive-extensive variable pairs called work yet<br />
the temper<strong>at</strong>ure - entropy pair referred to as he<strong>at</strong> exchange.<br />
• For the inequality derived before 2.75, of the form ax 2 + bx + c ≥ 0, would<br />
imaginary roots have any physical meaning?<br />
• Can we say anything about the value of µ in the limit N<br />
V<br />
• Wh<strong>at</strong> is energy?<br />
→ 0?<br />
• How can you tell which thermodynamic potential is appropri<strong>at</strong>e?<br />
• In the first case (no friction) on page 52, how long will the oscill<strong>at</strong>ions<br />
last? Why?<br />
• Why is he<strong>at</strong> always gener<strong>at</strong>ed in an irreversible process?<br />
• Why are Legendre transform<strong>at</strong>ions useful?<br />
• Prove th<strong>at</strong> the Gibbs potential is a st<strong>at</strong>e function and th<strong>at</strong> its differential<br />
is exact.<br />
• How did the grand potential get its name?<br />
A.3 Questions for chapter 3.<br />
• Is energy the only factor to determine the phase st<strong>at</strong>e, wh<strong>at</strong> about other<br />
quantum factors?<br />
• How many phase transitions do we have?<br />
• Can the special rel<strong>at</strong>ivity and time play any role here?<br />
• Why is figure 3.13 impossible?<br />
• Can we give the phase transition an analysis based on chaos?<br />
• For the van der Waals equ<strong>at</strong>ion, if b=0 and T=an/R then p=0, why?<br />
• Why does the integral 3.34 yield A1-A2 ?<br />
• How will the van der Waals equ<strong>at</strong>ion be changed if we include magnetiz<strong>at</strong>ion?<br />
• Do we have any universal constants associ<strong>at</strong>ed with the van der Waals<br />
equ<strong>at</strong>ion?
288 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Why is the volume of the ice bigger than th<strong>at</strong> of w<strong>at</strong>er?<br />
• Are there any m<strong>at</strong>erials where the phase boundary for solid to liquid has<br />
a neg<strong>at</strong>ive slope?<br />
• Are there other systems where the energy of interest is not the Gibbs<br />
energy?<br />
• Why does the phase transition in Gibbs free energy always jump down?<br />
• Using stability and the model phase diagram in p − V space, can you<br />
explain why the triple point is where it is?<br />
• If we made a three dimensional graph of G(p,T) for 2 phases wh<strong>at</strong> would<br />
we see?<br />
• Can the Van Der Waals equ<strong>at</strong>ion be derived from a virial expansion?<br />
• Why look <strong>at</strong> the asymptotic forms?<br />
• Wh<strong>at</strong> do neg<strong>at</strong>ive pressures in the Van Der Waals equ<strong>at</strong>ion mean?<br />
• How important is scaling?<br />
• Why does crossing a phase boundary cause a jump in the value of the<br />
volume?<br />
• Do you think th<strong>at</strong> the phase boundary between solid and liquid will vanish<br />
if we can reach a high enough pressure?<br />
• Why is it often better to go around the critical point instead of crossing<br />
through the phase boundary?<br />
• Why do inaccessible regions only show up in phase diagrams when extensive<br />
variables are involved?<br />
• Is there anything besides the indices of refraction th<strong>at</strong> we can use to<br />
distinguish a liquid from a gas?<br />
• How are the forbidden regions of a V-T diagram rel<strong>at</strong>ed to the corresponding<br />
p-T graph?<br />
• Why do systems in forbidden regions phase separ<strong>at</strong>e?<br />
• Why is the eutectic point important?<br />
• Why are phase transitions important to thermodynamics?<br />
• Why is it important to include entropy as one of the st<strong>at</strong>e variables in our<br />
model system?<br />
• Wh<strong>at</strong> importance does the triple point have in industry?
A.3. QUESTIONS FOR CHAPTER 3. 289<br />
• How would we find Vc and Tc for w<strong>at</strong>er?<br />
• Why do we need weights in the equ<strong>at</strong>ion ∑ wiNi = N on page 122?<br />
• Why are there regions in which the cBT diagram which are not allowed?<br />
• Why is the ideal gas law an approxim<strong>at</strong>ion?<br />
• Why are some of the intermedi<strong>at</strong>e st<strong>at</strong>es unstable?<br />
• Wh<strong>at</strong> is the difference between mechanically stable and thermodynamically<br />
stable st<strong>at</strong>es?<br />
• Can w<strong>at</strong>er be he<strong>at</strong>ed over 100 degrees Celsius?<br />
• For p-V-T systems not in equilibrium, why can equ<strong>at</strong>ions of st<strong>at</strong>e like<br />
p=f(V,T) not exist?<br />
• In the p-T diagram for solid Ce in 3.2 with axis interchanged, does the fact<br />
th<strong>at</strong> this curve is not concave downwards contradict the ’consequences for<br />
second order deriv<strong>at</strong>ives’ discussed in chapter 2?<br />
• Where would a plasma ’fit in’ on a phase diagram like fig 3.1?<br />
• If a system were prepared <strong>at</strong> a given Magnetiz<strong>at</strong>ion M0 <strong>at</strong> high temper<strong>at</strong>ure<br />
and then cooled to a temper<strong>at</strong>ure which will force the st<strong>at</strong>e of the<br />
system into an inaccessible region, will the system phase separ<strong>at</strong>e?<br />
• If so, will the magnetiz<strong>at</strong>ion be in different directions, have different magnitudes,<br />
or both.<br />
• In a system with 4 independent st<strong>at</strong>e variables, wh<strong>at</strong> is the maximum<br />
number of phases th<strong>at</strong> can coexist?<br />
• We see th<strong>at</strong> neg<strong>at</strong>ive pressures are allowed by the Van der Waals equ<strong>at</strong>ion<br />
of st<strong>at</strong>e. Are neg<strong>at</strong>ive temper<strong>at</strong>ures also allowed?<br />
• During the binodal decomposition, why are time scales important?<br />
• In trying to achieve stability, does a system necessarily become mechanically<br />
stable first, then thermodynamically stable, or can these two processes<br />
occur simultaneously, incrementally.<br />
• Various groups of m<strong>at</strong>erials have common molecular sizes, interaction energies<br />
etc. within their groups. Are there various, commonly used, virial<br />
expansions for each such group?<br />
• When mixing two m<strong>at</strong>erials, we get a binary phase diagram. Wh<strong>at</strong> would<br />
the diagram look like if we were to mix three or more m<strong>at</strong>erials?<br />
• Why use the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e when it is limited in the<br />
systems th<strong>at</strong> it can describe?
290 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Are there systems with only meta-stable st<strong>at</strong>es?<br />
• Theoretically, we can have neg<strong>at</strong>ive pressure, why can’t we measure it?<br />
• Can we make a m<strong>at</strong>erial th<strong>at</strong> exists in an unstable region, but will not<br />
easily break down?<br />
• Why doesn’t neg<strong>at</strong>ive pressure correspond to a physical st<strong>at</strong>e?<br />
• When are non-unique energy solutions better <strong>at</strong> describing a system?<br />
• Can pressure be multi-valued for a single temper<strong>at</strong>ure?<br />
• Can temper<strong>at</strong>ure be multi-valued for single values of volume?<br />
• Can volume be multi-valued for single values of temper<strong>at</strong>ure?<br />
A.4 Questions for chapter 4.<br />
• Is there really no long-range order in a liquid?<br />
• Can we have spontaneous broken symmetries in a non-degener<strong>at</strong>e system?<br />
• Landau’s theory does not converge <strong>at</strong> Tc, but can’t we modify the theory<br />
so th<strong>at</strong> it does converge?<br />
• Can there be third order phase transitions, or is th<strong>at</strong> forbidden by the free<br />
parameters?<br />
• Can both he<strong>at</strong> capacities be modelled by the same response?<br />
• Wh<strong>at</strong> happens when there is a positive term in the fourth order sign of<br />
the expansion?<br />
• Is there a place where all three roots will have the same value of f for the<br />
same m?<br />
• Can we make our expansions work to get rid of anisotropies in the system?<br />
• How important are inhomogeneities in a system?<br />
• If the correl<strong>at</strong>ion length diverges <strong>at</strong> Tc, why is it so important in the<br />
theory?<br />
• Why do we need a non-equilibrium theory to describe the effects of fluctu<strong>at</strong>ions<br />
in a system?<br />
• How is snow different from ice?<br />
• Why must an order parameter be extensive?<br />
• Why do we measure the values of an order parameter via the associ<strong>at</strong>ed<br />
generalized force instead of measuring it directly?
A.4. QUESTIONS FOR CHAPTER 4. 291<br />
• Why do we compare free energies to determine if the st<strong>at</strong>e is an equilibrium<br />
st<strong>at</strong>e?<br />
• Prove th<strong>at</strong> there are no singularities in st<strong>at</strong>e functions for finite systems.<br />
• Where do you see hysteresis in E-and-M?<br />
• Why are there large fluctu<strong>at</strong>ions in the order parameter near the phase<br />
transition?<br />
• Why is it so hard to describe the system <strong>at</strong> the critical temp?<br />
• Why do small changes affect the whole system <strong>at</strong> the critical temp?<br />
• Can we describe phase transition by complex functions?<br />
• Can we describe phase transition by group theory?<br />
• Wh<strong>at</strong> is the effect of pressure on Landau form of free energy?<br />
• Is Landau theory equivalent to a unified theory in particle physics?<br />
• Is there any theory th<strong>at</strong> describes exactly wh<strong>at</strong>’s happening <strong>at</strong> a phase<br />
transition?<br />
• Is it possible to define b(T) in such away to have the correct exponent in<br />
(4.12)?<br />
• Wh<strong>at</strong>’s the meaning of nucle<strong>at</strong>ion and growth?<br />
• Do you know any real system with first order transition?<br />
• Is the description for fluctu<strong>at</strong>ions for macroscopic systems rel<strong>at</strong>ed to the<br />
uncertainty principle for microscopic systems?<br />
• Is the correl<strong>at</strong>ion length rel<strong>at</strong>ed to the susceptibility?<br />
• By wh<strong>at</strong> mechanism does a tap on a beaker of supercooled w<strong>at</strong>er cause it<br />
to suddenly crystalize?<br />
• Is it possible to apply our external field over such a large time scale th<strong>at</strong><br />
we can ignore fluctu<strong>at</strong>ions even near a phase transition?<br />
• Can we use Cauchy’s 1st theorem to evalu<strong>at</strong>e our Landau function <strong>at</strong> the<br />
critical point?<br />
• Is our ”Divide and conquer” technique for considering fluctu<strong>at</strong>ions analogous<br />
to our ”equivalent surface and volume charges” for considering dielectric<br />
samples?<br />
• Wh<strong>at</strong> physical consequences would there be if our ”f” in eq’n 4.37 were<br />
neg<strong>at</strong>ive, ie. fluctu<strong>at</strong>ions lowered our energy st<strong>at</strong>e?
292 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Wh<strong>at</strong> is our sigma*dm equivalent to supercooling and superconductivity?<br />
• Would analytic continu<strong>at</strong>ion help the evalu<strong>at</strong>ion of our Landau expansion<br />
<strong>at</strong> the critical point?<br />
• For the change in local magnetic field causing a perturb<strong>at</strong>ion in the magnetiz<strong>at</strong>ion<br />
everywhere; does this perturb<strong>at</strong>ion happen instantaneously or<br />
does it travel <strong>at</strong> the speed of light?<br />
• Does causality play a role in this perturb<strong>at</strong>ion?<br />
• At r=0 it is reasonable to speak of E=-gradient(Potential), why then is<br />
an expansion of free energy in terms of gradients of the order parameter<br />
not valid near Tc.<br />
• Why do experiments give different results than the Van Der Waals model<br />
<strong>at</strong> the phase transition?<br />
• Wh<strong>at</strong> would be a suitable order parameter for describing the phase transition<br />
between snow and ice?<br />
• Why is it important to compare the idea of a critical point in thermodynamics<br />
to the idea of a critical point in m<strong>at</strong>hem<strong>at</strong>ics?<br />
• Wh<strong>at</strong> kind of expansion would be needed to describe f exactly?<br />
• Would this analysis be appropri<strong>at</strong>e for approxim<strong>at</strong>ing the n<strong>at</strong>ure of a bar<br />
magnet?<br />
• Is the transition from liquid w<strong>at</strong>er to ice a second or first order phase<br />
transition?<br />
• At which point on the hysteresis will a butterfly flapping its wings 1000<br />
miles away cause the system to go from a meta-stable st<strong>at</strong>e to a stable<br />
st<strong>at</strong>e?<br />
• Wh<strong>at</strong> m<strong>at</strong>hem<strong>at</strong>ical oper<strong>at</strong>ions were used to derive (4.38)?<br />
• Why does rot<strong>at</strong>ional symmetry cause the solutions to be of the form<br />
ul(r)Ylm(ω)/r?<br />
• Wh<strong>at</strong> parameters would tell us whether a system is ice or snow?<br />
A.5 Questions for chapter 5.<br />
• Why in deriving the pair correl<strong>at</strong>ion function we assume a spherical symmetry?<br />
• Can we replace h and n by one critical exponent?
A.5. QUESTIONS FOR CHAPTER 5. 293<br />
• Is the law described by the correl<strong>at</strong>ion function obeying the third Newton<br />
law?<br />
• Wh<strong>at</strong> is the physical meaning of the fact th<strong>at</strong> the susceptibility is proportional<br />
to the q=0 component of the Fourier transform of the pair correl<strong>at</strong>ion<br />
function?<br />
• Are in reality the dependence of the response functions with respect to T<br />
is the same above and below the critical temper<strong>at</strong>ure<br />
• Wh<strong>at</strong> is the critical field?<br />
• Is there a mean field theory associ<strong>at</strong>ed with the first phase transition?<br />
• How can we be sure th<strong>at</strong> it is enough to tre<strong>at</strong> the system near the critical<br />
temper<strong>at</strong>ure in a perturb<strong>at</strong>ion theory?<br />
• How can Ginzburg Landau theory give valid measurements while wrong<br />
critical exponents?<br />
• Wh<strong>at</strong> is the maximum number of critical exponents?<br />
• The critical exponents shown have simple rel<strong>at</strong>ions to T and Tc. How<br />
can this be possible, when we know a st<strong>at</strong>e is dependent on <strong>at</strong> least two<br />
variables.<br />
• I have seen the reduced p,T,V written as t=T/Tc, etc. We do not do th<strong>at</strong><br />
in this chapter. Which is the correct version.<br />
• Since st<strong>at</strong>istical mechanics can get us to the heart of phase transitions,<br />
why bother doing this with thermodynamics.<br />
• How many other critical exponents are there.<br />
• Is there a simpler way to find our descriptions other than scaling.<br />
• How do we rel<strong>at</strong>e the functionals th<strong>at</strong> come out of scaling to experimental<br />
d<strong>at</strong>a.<br />
• Critical exponents can be hard to get out of st<strong>at</strong>e equ<strong>at</strong>ions, and they do<br />
not seem to give any new inform<strong>at</strong>ion. Why use them.<br />
• How can critical exponents get you back to the phase diagram. I just<br />
don’t see the connection.<br />
• If the exponents are the same for T¿Tc or T¡Tc, why do we bother to<br />
define them separ<strong>at</strong>ely.<br />
• Which is the most powerful and useful critical exponent.<br />
• Why would gamma not equal gamma prime?
294 APPENDIX A. QUESTIONS SUBMITTED BY STUDENTS.<br />
• Wh<strong>at</strong> would happen if Tc and pc could be neg<strong>at</strong>ive?<br />
• Why must delta M ¡¡ M?<br />
• How can you have a 2 dimensional system?<br />
• Why must a realistic description of a phase transition near a critical point<br />
include the effects of fluctu<strong>at</strong>ions?<br />
• Why does the correl<strong>at</strong>ion length become very large near a critical point?<br />
• Where else do we see scaling and universality?<br />
• Why are critical exponents so difficult to measure?<br />
• How can you measure a pair correl<strong>at</strong>ion function?<br />
• Why can we use any function of two variables as a basis for scaling?<br />
• Why do fluctu<strong>at</strong>ions occur near the critical point?<br />
• Would an engineer want to know the critical exponents of a steam engine?<br />
• Why is it important to know the mean field approach?<br />
• Wh<strong>at</strong> is the usefulness of using rel<strong>at</strong>ive variables?<br />
• Is the scaling function a m<strong>at</strong>hem<strong>at</strong>ical convenience or does it have physical<br />
significance?<br />
• If critical exponents are so hard to measure, wh<strong>at</strong> is their usefulness?<br />
• Could we have described the change in magnetiz<strong>at</strong>ion over time by something<br />
other than the diffusion equ<strong>at</strong>ion?<br />
• Would it be plausible to use a rel<strong>at</strong>ivistic form of the equ<strong>at</strong>ions?<br />
• Why does the scaling theory give nu=nu’?<br />
• Wh<strong>at</strong> would happen if we used the compressibility as the response function?
Appendix B<br />
Summaries submitted by<br />
students.<br />
Students were asked to submit a paragraph describing each chapter, written<br />
in a way they would like to see it included in an introductory description of<br />
the notes. Therefore these paragraphs reflect both the understanding of the<br />
m<strong>at</strong>erial and the exposition of the m<strong>at</strong>erial.<br />
B.1 Summaries for chapter 1.<br />
In thermodynamics one describes a macroscopic system, with many degrees of<br />
freedom, by st<strong>at</strong>e variables (which are time-averaged). For st<strong>at</strong>e variables to be<br />
useful in predicting intermedi<strong>at</strong>e values, thermodynamics concerns itself only<br />
with reversible processes in thermodynamic equilibrium in the thermodynamic<br />
limit. St<strong>at</strong>e variables are governed by the laws of thermodynamics. Laws zero<br />
and three give us fixed references. Law one is the conserv<strong>at</strong>ion of energy and<br />
law two tells us th<strong>at</strong> if all the processes are reversible we have the best scenario,<br />
otherwise entropy increases.<br />
<strong>Thermodynamics</strong> is the study of macroscopic systems in equilibrium. We use<br />
thermodynamics to help define experimental variables such as temper<strong>at</strong>ure and<br />
pressure. Using thermodynamics we can set up simple experiments to test our<br />
theories, and we can use our experimental results to find simple rel<strong>at</strong>ionships<br />
for our variables.<br />
Chapter 1 introduces extensive and intensive thermodynamical variables, talks<br />
about wh<strong>at</strong> they mean, talks about reversibility and irreversibility, and how<br />
quasi-st<strong>at</strong>ic systems work in thermodynamics. Chapter 1 also goes over the four<br />
laws of thermodynamics, and introduces some m<strong>at</strong>hem<strong>at</strong>ical formalism. The<br />
end of chapter 1 talks about entropy in the context of the third law. There are<br />
295
296 APPENDIX B. SUMMARIES SUBMITTED BY STUDENTS.<br />
also definitions of wh<strong>at</strong> we mean by macroscopic systems, st<strong>at</strong>e, thermodynamic<br />
equilibrium, and other thermodynamic vocabulary terms.<br />
Rules, rules, rules. Everything in life seems to be governed by rules. The best<br />
way to understand any system, from a society to a steam engine, is by looking<br />
<strong>at</strong> the laws th<strong>at</strong> govern its behavior. In this chapter we will learn the language<br />
of thermodynamics and study the laws th<strong>at</strong> apply to thermodynamic systems.<br />
The first law included he<strong>at</strong> as a form of energy and st<strong>at</strong>es th<strong>at</strong> the total energy<br />
of a system is conserved. The second law puts restrictions on the transfer of<br />
he<strong>at</strong> or the exchange of he<strong>at</strong> into work. It defines a preferred direction of time<br />
and tells us th<strong>at</strong> no engine can be more efficient than a Carnot engine. The<br />
third law says th<strong>at</strong>, in the thermodynamic limit, when the temper<strong>at</strong>ure goes to<br />
zero so does the entropy. It is thus impossible to reach T=0K using a reversible<br />
process in a finite number of steps. Finally, the zeroth law, which is so trivial<br />
it is often overlooked, st<strong>at</strong>es th<strong>at</strong> if two systems are in equilibrium with the<br />
same system, they are also in equilibrium with each other. This simple fact is<br />
wh<strong>at</strong> allows us to define universal standards for temper<strong>at</strong>ure, pressure, and all<br />
the other descriptors of our system. With these laws in hand we can begin to<br />
understand the behavior of thermodynamic systems throughout the land.<br />
This chapter will quickly review the 4 basic laws of thermodynamics with an<br />
emphasis on how the laws affect the real wold measurements. It will use the basic<br />
concepts of the Carnot engine to illustr<strong>at</strong>e how one might go about developing<br />
such measurements. The significance, limits, and possibilities will be touched<br />
upon as the framework is laid down.<br />
Reversible processes are important. There are four laws of thermodynamics,<br />
numbered 0 through 3. These laws hold for all systems. The Carnot engine is<br />
the most efficient engine there is. Entropy is an important quantity which is<br />
rel<strong>at</strong>ed to he<strong>at</strong> flow. You can only define entropy well in the thermodynamic<br />
limit. A Carnot engine is a good way to do work and to measure temper<strong>at</strong>ure.<br />
In thermodynamics we consider a macroscopic system and our variables are<br />
global averages. The thermodynamic variables th<strong>at</strong> uniquely determine the<br />
thermodynamic st<strong>at</strong>e are called st<strong>at</strong>e variables. To rel<strong>at</strong>e the different variables<br />
to each other we use two kinds of rel<strong>at</strong>ions. The first are the thermodynamic<br />
laws, which hold for all systems. The zeroth law discusses the conditions for<br />
thermal equilibrium. The first law is a st<strong>at</strong>ement about the conserv<strong>at</strong>ion of<br />
energy. The second law is a st<strong>at</strong>ement about maximal entropy in equilibrium.<br />
The third law is putting the entropy to be zero <strong>at</strong> zero absolute temper<strong>at</strong>ure.<br />
The second type of rel<strong>at</strong>ions are the equ<strong>at</strong>ions of st<strong>at</strong>e, which depend on our<br />
specific system. The equ<strong>at</strong>ions of st<strong>at</strong>e can be used only when the system is<br />
in equilibrium. For this point it is important to define two types of processes,
B.2. SUMMARIES FOR CHAPTER 2. 297<br />
reversible and irreversible.<br />
B.2 Summaries for chapter 2.<br />
Energy is stored in a gas. There are many different free energies because there<br />
are many different variables involved in a process. Response functions such as<br />
he<strong>at</strong> capacities are the most important way to obtain inform<strong>at</strong>ion. Thermodynamic<br />
processes have many variables involved. Hence, the calculus of many<br />
variables is used. When two systems come into contact, they are able to exchange<br />
energy by exchanging their extensive variables. Fluctu<strong>at</strong>ions play a role<br />
in m<strong>at</strong>erials.<br />
Chapter two introduces the minimum energy principle and discusses the consequences<br />
of things changing. The internal energy of a closed system is a minimum<br />
as a function of all parameters th<strong>at</strong> are not fixed by external constraints. Internal<br />
energy also gives an upper bound for the amount of work th<strong>at</strong> can be<br />
done in an irreversible process. One way to see the consequences of changing<br />
variables is via free energy. There are many different constructions for free energies,<br />
such as the Helmholtz free energy, the Gibbs potential, enthalpy, and<br />
the grand potential. Each of these formul<strong>at</strong>ions is useful in different situ<strong>at</strong>ions,<br />
depending on the system we are describing and the set of n<strong>at</strong>ural variables used<br />
with th<strong>at</strong> system. Once you decide which set of variables to use, you can use<br />
response functions, such as he<strong>at</strong> capacities, compressibilities, and the coefficient<br />
of thermal expansion to describe the effect of allowing your variables to<br />
change. Response functions help us describe the rel<strong>at</strong>ionships between different<br />
thermodynamic variables and tell us a lot about how a system behaves.<br />
The amount of energy which can be released by a gas depends on the process by<br />
which the gas is doing work. When trying to measure this energy one needs to<br />
decide which variables can, and should be controlled and which ones should be<br />
kept constant (eg. Volume may be difficult to measure and Entropy impossible<br />
to measure directly, so we may want to keep these variables constant in our<br />
measurements). Various ”Free Energies” then help us to express the amount of<br />
work a system can do with our desired variables held constant. By definition,<br />
these Free Energy st<strong>at</strong>e functions have exact differentials, and as a result, lead<br />
to equivalence rel<strong>at</strong>ions between various pairs of partial deriv<strong>at</strong>ives of many<br />
st<strong>at</strong>e variables with respect to other st<strong>at</strong>e variables (Maxwell Rel<strong>at</strong>ions). At<br />
equilibrium there is minimum energy and maximum entropy with respect to the<br />
concerned variables. This fact also leads to conditions on st<strong>at</strong>e variables. All<br />
intensive variables are the same for an isol<strong>at</strong>ed system in thermal equilibrium.<br />
With Newtonian Mechanics we use the Lagrangian and Hamiltonian to get the<br />
equ<strong>at</strong>ions of motion. With the Equ<strong>at</strong>ions of motion we have a fundamental
298 APPENDIX B. SUMMARIES SUBMITTED BY STUDENTS.<br />
understanding of the system th<strong>at</strong> we are working with. In <strong>Thermodynamics</strong> it<br />
is the Energy th<strong>at</strong> we use to get a fundamental understanding of the system.<br />
The internal energy is the basic form of energy th<strong>at</strong> we use. If we begin with<br />
the internal energy U, we can do successive transforms, or use the Maxwell<br />
rel<strong>at</strong>ions to find properties of the parameters th<strong>at</strong> define the system. Using this<br />
method we can arrive <strong>at</strong> the equ<strong>at</strong>ion of st<strong>at</strong>e for the system, which will define<br />
the system. We can also use this to find other measurable parameters, such as<br />
the compressibilities and capacitances.<br />
We will start the chapter by introducing the minimum internal energy and maximum<br />
entropy principles. Next we will generalize the minimum energy principle<br />
to other kinds of thermodynamic potentials (free energies). We will see th<strong>at</strong> the<br />
thermodynamic potentials th<strong>at</strong> describe our process depend on the variables<br />
under control. For each type of process we define the appropri<strong>at</strong>e thermodynamic<br />
potential, among those we will define the Internal Energy (U),Helmholtz<br />
(F),Gibbs (G),Entropy (H) and Grand Potential (Ω). Applying a m<strong>at</strong>hem<strong>at</strong>ical<br />
rules to derive the Maxwell rel<strong>at</strong>ions which connect changes in different st<strong>at</strong>e<br />
variables in different processes. Response functions like he<strong>at</strong> capacity, compressibility,<br />
and coefficient of thermal expansion, help us to obtain inform<strong>at</strong>ion<br />
about the thermodynamic system. Again we will use m<strong>at</strong>hem<strong>at</strong>ical concepts<br />
(partial deriv<strong>at</strong>ives ) to find rel<strong>at</strong>ions between the response functions and with<br />
the help of the minimum energy condition of equilibrium we will derive the<br />
inequalities for the he<strong>at</strong> capacities and compressibilities.Then we will look <strong>at</strong><br />
the stability requirements on thermodynamic functions other than the internal<br />
energy. When it comes to magnetic system we must be careful in our analysis<br />
and include the work term comes from the field cre<strong>at</strong>ed in the m<strong>at</strong>erial. Finally<br />
we will briefly discuss a small fluctu<strong>at</strong>ion in a thermodynamic limit and derive<br />
a very important probability rule.<br />
This chapter begins by introducing the minimum energy principle. This is used<br />
to lead into the different free energies and their definitions. Now with the five<br />
free energies defined along with the corresponding differential equ<strong>at</strong>ions, their<br />
linear n<strong>at</strong>ures (Gibbs-Duhem rel<strong>at</strong>ions) are explored. These linear rel<strong>at</strong>ions lead<br />
to several important properties such as dependence of some intensive variables<br />
on other variables. Maxwell rel<strong>at</strong>ions are introduced. Next, the basic m<strong>at</strong>erial<br />
properties such as he<strong>at</strong> capacity or compressibility are defined. A m<strong>at</strong>h<br />
review of multi-variable calculus is added to the list of tools next. Finally, with<br />
these tools, various rel<strong>at</strong>ionships between m<strong>at</strong>erial properties and solutions to<br />
some problems are examined. During the applic<strong>at</strong>ions to solutions a perturb<strong>at</strong>ional<br />
approach involving the role of fluctu<strong>at</strong>ions in m<strong>at</strong>erials is introduced.<br />
A large number of tools are assumed understood, such as the Gaussian n<strong>at</strong>ure<br />
of Maxwellian functions. After this chapter, there are a large number of problems<br />
solvable in thermodynamics with the tools presented. The main body of<br />
thermodynamics is presented here.
B.3. SUMMARIES FOR CHAPTER 3. 299<br />
B.3 Summaries for chapter 3.<br />
Discontinuities in equ<strong>at</strong>ions of st<strong>at</strong>e manifest themselves in the real world as<br />
phase transitions. Phase transitions enable us to tell when a m<strong>at</strong>erial becomes<br />
superconductive, or when it’s magnetic properties suddenly change, or where<br />
general ”jumps” occur in values of other extensive st<strong>at</strong>e variables. In a simple<br />
p-V-T system, the ideal gas law (PV=nRT) doesn’t help us to explain phase<br />
transitions, so we resort to virial expansions which take interaction between<br />
molecules; molecules actually taking up volume etc., into consider<strong>at</strong>ion. Van<br />
der Waals postul<strong>at</strong>ed one such ”not-so-ideal gas” equ<strong>at</strong>ion, valid for both liquid<br />
and gas phase. Stability requirements (partial (p) w.r.t. (v) ¡=0) and the<br />
minimum energy principle, help us to determine which values of p, V and T<br />
are allowable, from our <strong>at</strong> first seemingly multi-valued Van der Waals equ<strong>at</strong>ion<br />
solution.<br />
Chapter 3 addresses the issue of phase transitions, investig<strong>at</strong>ing discontinuities<br />
in the equ<strong>at</strong>ions of st<strong>at</strong>e. This is best seen by examining phase diagrams. The<br />
lines marked on the p-V diagrams represent the coexistence curves where a<br />
system can be in more than one st<strong>at</strong>e <strong>at</strong> a given pressure and temper<strong>at</strong>ure.<br />
When two such lines meet, we have the triple point, the p and T values <strong>at</strong><br />
which all 3 phases coexist. For multi-phase diagrams, the maximum number of<br />
phases which can coexist is one more than the number of independent intensive<br />
st<strong>at</strong>e variables needed to describe the system. Crossing a phase boundary leads<br />
to a change in st<strong>at</strong>e. This change causes a jump in the volume and the other<br />
extensive variables, while the energy (but not its deriv<strong>at</strong>ive) remains continuous.<br />
If a system is in a forbidden region of a phase diagram involving an extensive<br />
variable, it will phase separ<strong>at</strong>e. The Clausius-Clapeyron rel<strong>at</strong>ion tells us the<br />
slope of the coexistence curve in a p-V diagram (which is always positive) and<br />
can be expressed in terms of l<strong>at</strong>ent he<strong>at</strong>. The jump in volume of a phase<br />
transition is also seen in the van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e in which V is not<br />
a unique function of p and T. However, the van der Waals equ<strong>at</strong>ion allows for<br />
st<strong>at</strong>es with neg<strong>at</strong>ive pressures, which are never stable, so this equ<strong>at</strong>ion is only<br />
a good model for a liquid to gas phase transition, ending in a critical point.<br />
In this chapter phase transitions are explored. The boundary is defined. The rel<strong>at</strong>ionship<br />
<strong>at</strong> the boundary is m<strong>at</strong>hem<strong>at</strong>ically modelled in the Clausius-Clapeyron<br />
rel<strong>at</strong>ionship. L<strong>at</strong>ent he<strong>at</strong>s are defined in the context of this rel<strong>at</strong>ion. Further,<br />
the limit<strong>at</strong>ions on the possible coexistent phases are informally defined. Some<br />
applic<strong>at</strong>ions are described during the rest of the chapter while introducing additional<br />
concepts. Binary systems with chemical reactions are explored. The<br />
Van Der Waals equ<strong>at</strong>ion is next. The Virial expansion is introduced. The rest<br />
of the chapter deals with the behavior and validity conditions of the Van Der<br />
Waals equ<strong>at</strong>ion. The asymptotic form is explored to achieve insight.
300 APPENDIX B. SUMMARIES SUBMITTED BY STUDENTS.<br />
Phase transitions are discontinuities in the equ<strong>at</strong>ions of st<strong>at</strong>e. In the p versus<br />
T diagram, all three phases(solid, liquid, and gas) can coexist <strong>at</strong> the triple<br />
point. Also, the phase boundary between liquid and gas vanishes <strong>at</strong> the critical<br />
point. Two distinct phases can coexist <strong>at</strong> the phase boundary. We find a phase<br />
boundary by looking <strong>at</strong> the Gibbs free energy for two phases. The number of<br />
phases th<strong>at</strong> can coexist depends on the number of independent st<strong>at</strong>e variables.<br />
The Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is a better approxim<strong>at</strong>ion than the ideal<br />
gas law. A line which separ<strong>at</strong>es mechanically stable st<strong>at</strong>es from mechanically<br />
unstable st<strong>at</strong>es is a spinodal. A line th<strong>at</strong> separ<strong>at</strong>es a thermodynamically metastable<br />
st<strong>at</strong>e with a stable st<strong>at</strong>e is a binodal.<br />
In this chapter we will analyze the phase transition phenomena from a macroscopic<br />
point of view. Looking <strong>at</strong> a P-V-T system, we will first analyze the P-T<br />
phase diagram and describe the triple point and the critical point. Then we<br />
will analyze the V-T diagram which (like P-V diagram) characterized by the<br />
forbidden regions. Gibbs free energy analysis yields the Clausius-Clapeyron rel<strong>at</strong>ion,<br />
which gives a quantit<strong>at</strong>ive description of the phase boundaries. Next we<br />
will generalize our conclusions to a multi phase boundaries and give the binary<br />
phase diagrams as an applic<strong>at</strong>ion. Here the point where all three phases can<br />
coexist is called eutectic point. To give a more quantit<strong>at</strong>ive description of the<br />
phase transition we need to know the equ<strong>at</strong>ion of st<strong>at</strong>e. As a first try we will<br />
express it in a virial expansion form. The Van Der Waals equ<strong>at</strong>ion of st<strong>at</strong>e is a<br />
model which is valid for the change from the gas to the liquid phase. Here we<br />
will give a quantit<strong>at</strong>ive analysis for such important model.<br />
In chapter 3 we looked <strong>at</strong> the different ways to represent the phases of m<strong>at</strong>erials.<br />
We saw th<strong>at</strong> simple m<strong>at</strong>erials can have complex even convoluted phase diagrams.<br />
Simple binary m<strong>at</strong>erials can have highly unstable regions, and there are places<br />
in all phases diagrams where m<strong>at</strong>erials can simply not exist. We saw th<strong>at</strong> there<br />
are co-existence curves for all m<strong>at</strong>erials where we can have equal amounts of<br />
both phases. There are triple points in all phase diagrams where we can have<br />
solid-liquid-gas co-existing <strong>at</strong> the same time. There are however, no points<br />
on any phase diagrams where we can get four phases coming together <strong>at</strong> the<br />
same time. Phase diagrams can be simple or complex, but they give a good<br />
description of the system.<br />
B.4 Summaries for chapter 4.<br />
We use order parameters, which are st<strong>at</strong>e variables, to distinguish among phases<br />
of a system. A useful way to study the free energy per unit volume of a magnetic<br />
system is to expand it in terms of the magnetiz<strong>at</strong>ion of the system. Landau studied<br />
one such expansion, which only included even powers of the magnetiz<strong>at</strong>ion.<br />
This expansion, however, is not valid for when a system is <strong>at</strong> a phase transition.<br />
Ginzberg and Landau modified this expansion to include fluctu<strong>at</strong>ions. This
B.4. SUMMARIES FOR CHAPTER 4. 301<br />
modified form, however, is still incorrect precisely <strong>at</strong> the phase transition.<br />
Wh<strong>at</strong> happens <strong>at</strong> a phase transition? All the techniques and theories discussed<br />
so far enable us to model simple equilibrium systems. But wh<strong>at</strong> about the<br />
fluctu<strong>at</strong>ions and discontinuities <strong>at</strong> and around phase transitions? Chapter 4<br />
presents a good framework to analyze first and second order transitions with the<br />
Landau-Ginzburg theory. In order to examine these transitions, we must first<br />
determine an order parameter, an extensive st<strong>at</strong>e variable th<strong>at</strong> uniquely defines<br />
which phase the system is in. To measure the value of the order parameter<br />
we need to look <strong>at</strong> the generalized force associ<strong>at</strong>ed with it. Landau’s method<br />
describes phase transitions by parameterizing the free energy as a function of<br />
the order parameter. Because of the essential singularity <strong>at</strong> Tc, Landau’s power<br />
series fails to converge so it can’t explain wh<strong>at</strong> happens <strong>at</strong> the phase transition,<br />
but it is good for classifying kinds of phase transitions by comparing the st<strong>at</strong>es <strong>at</strong><br />
both sides of the transition. The order of a phase transition can be defined by its<br />
continuity. If it is possible to find a continuous p<strong>at</strong>h around a critical point from<br />
one phase to another (much like the p<strong>at</strong>h around an essential singularity with<br />
a branch cut) it is a second order transition. If there is no continuous p<strong>at</strong>h,<br />
it’s a first order transition. In first order transitions, the value of the order<br />
parameter is discontinuous <strong>at</strong> the critical temper<strong>at</strong>ure. If we go continuously<br />
through a co-existence curve, we end up in a meta-stable st<strong>at</strong>e. Near a phase<br />
transition, fluctu<strong>at</strong>ions become an issue. Ginzburg adapted Landau’s theory to<br />
better account for the effects of fluctu<strong>at</strong>ions by including sp<strong>at</strong>ial vari<strong>at</strong>ions in<br />
the order parameter. Ginzburg and Landau further improved the theory with<br />
an expansion of the free energy in terms of gradients of the order parameter.<br />
This provides a good description of the st<strong>at</strong>e of the system much closer to the<br />
critical temper<strong>at</strong>ure.<br />
We will start this chapter by defining the order parameter and how to measure<br />
it. Using a continuity concepts we will define the second order phase transition.<br />
Then we will show the similarity between the coexistence curve and the branch<br />
cut in a complex function. Giving an overview of a spontaneously broken symmetry<br />
before introducing the Landau theory of phase transition. Unlike the<br />
van der Waals equ<strong>at</strong>ion, the Landau theory of phase transition is more general<br />
one. Here we will analysis the Landau form of the free energy for the second<br />
and first order phase transition, and describe the behavior of the entropy, he<strong>at</strong><br />
capacity and susceptibility near the critical temper<strong>at</strong>ure. Here we will give another<br />
definition of the phase transition order, based on the continuity of the<br />
order parameter <strong>at</strong> the critical temper<strong>at</strong>ure. Trying to give more accur<strong>at</strong>e description<br />
near the phase transition we will study the Ginzburg-Landau theory<br />
for fluctu<strong>at</strong>ions.<br />
In chapter 4 we began the analysis th<strong>at</strong> will lead us to critical phenomena. We<br />
noticed th<strong>at</strong> phase transitions could be modelled by order parameters. Order
302 APPENDIX B. SUMMARIES SUBMITTED BY STUDENTS.<br />
parameters are st<strong>at</strong>e variables th<strong>at</strong> give the critical phenomena of the system.<br />
In the case of chapter 4 we looked <strong>at</strong> magnetiz<strong>at</strong>ion using Landau-Ginzburg<br />
theory, but we know th<strong>at</strong> this model can be used for other spontaneous broken<br />
symmetries.<br />
To describe the phase of a system we need a parameter which: 1) differenti<strong>at</strong>es<br />
between st<strong>at</strong>es of a system, while 2) being measured does not change the system<br />
(requires zero field for measurement). Such a st<strong>at</strong>e variable is termed an order<br />
parameter. When an order parameter and its external field couple to zero,<br />
the free energy is usually symmetric as a function of order parameter; so when<br />
we have a nonzero order parameter, there are <strong>at</strong> least 2 st<strong>at</strong>es for the system<br />
with the same energy. The act of measurement breaks this symmetry. Landau<br />
devised a model to parameterize the free energy as a function of order parameter.<br />
With an expansion of the free energy w.r.t order parameter we can look <strong>at</strong><br />
minima and see which values correspond to stable st<strong>at</strong>es. At 1st order phase<br />
transitions, the Gibbs function is continuous, but deriv<strong>at</strong>ives (e.g. (g/T)p =-s<br />
and (g/p)T=v) are discontinuous. At 2nd order phase transitions, G and first<br />
deriv<strong>at</strong>ives are continuous, but second deriv<strong>at</strong>ives are discontinuous. We look<br />
for phase transitions <strong>at</strong> discontinuities. . Landau’s expansion doesn’t converge<br />
<strong>at</strong> the critical points (essential singularities). Fluctu<strong>at</strong>ions become important<br />
near phase transitions. Fluctu<strong>at</strong>ions in a system causes parts of a system to<br />
be stable and other parts meta-stable, causing phase separ<strong>at</strong>ion of the system.<br />
Ginzburg, tried to account for fluctu<strong>at</strong>ions by making the order parameter a<br />
function of position. His model is still not valid <strong>at</strong> the critical point, but allows<br />
us to analyze the system much closer to it.<br />
B.5 Summaries for chapter 5.<br />
Critical exponents show how fast response functions diverge near the critical<br />
point. The mean field approach ignores the effects arising from fluctu<strong>at</strong>ions in<br />
the magnetiz<strong>at</strong>ion. When a system is far away from the critical point, mean<br />
field theory is a good approxim<strong>at</strong>ion. It is not a good approxim<strong>at</strong>ion, however,<br />
close to the critical point. The theory of Ginzberg and Landau gives a better<br />
description of a system near the critical point. We use scaling, which has<br />
consequences.<br />
Wh<strong>at</strong> is happening near the critical point and how can we describe the thermodynamic<br />
st<strong>at</strong>e there? We start by defining the four common critical exponents<br />
and discuss the universality behind th<strong>at</strong>. Landau model for superconductivity<br />
will be a good example for second order phase transition. Then we discuss the<br />
mean field theory, derive the critical exponents for Landau theory, and show<br />
th<strong>at</strong> van der waals model is a mean field model. Near the critical point the<br />
systems’ behavior determined by the fluctu<strong>at</strong>ion term of the free energy, which<br />
should be homogenous. As a consequence of the scaling we reduce the number
B.5. SUMMARIES FOR CHAPTER 5. 303<br />
of independent critical exponents. Next we apply the scaling on the pair correl<strong>at</strong>ion<br />
function and define the hyper scaling. Then we explore the validity of<br />
Ginzburg Landau theory. We end the chapter by introducing the time scaling<br />
in the correl<strong>at</strong>ion function.<br />
Critical exponents help us describe different r<strong>at</strong>es of divergence of response<br />
functions, near the critical point. These critical exponents are a unique characteriz<strong>at</strong>ion<br />
of phase transitions - i.e. Similar critical exponents indic<strong>at</strong>e similar<br />
processes <strong>at</strong> phase transition. Mean field models produce critical exponents of<br />
- b=1/2; a=1; g=0. Van Der Waals used such a model in his equ<strong>at</strong>ion of st<strong>at</strong>e,<br />
and these critical exponents were verified. Mean field theory is most useful far<br />
away from the critical point. In scaling theory we describe a non-equilibrium<br />
st<strong>at</strong>e with fluctu<strong>at</strong>ions. Including gradient terms in the expansion of free energy,<br />
improves our model near the critical point. The pair correl<strong>at</strong>ion function<br />
describes the average of these fluctu<strong>at</strong>ions. Near the critical point only x is important.<br />
No m<strong>at</strong>ter wh<strong>at</strong> you do though, everything still diverges <strong>at</strong> the critical<br />
point and we generally need to throw this whole mean field idea out the window.<br />
Chapter 5 discusses the n<strong>at</strong>ure of phase transitions th<strong>at</strong> end in a critical point<br />
by focusing on the properties <strong>at</strong> the critical point (which also determine wh<strong>at</strong><br />
happens away from the critical point). In order to characterize the divergent<br />
behavior due to fluctu<strong>at</strong>ions near the critical point, critical exponents are introduced.<br />
They show the speed of the divergence. Critical exponents are a<br />
unique characteriz<strong>at</strong>ion of the physics of phase transitions and are used to classify<br />
groups of similar transitions with identical exponents. Mean field theory<br />
ignores the details of fluctu<strong>at</strong>ions by averaging them out. Any mean field theory<br />
will always yield the same critical exponents. MFT is a good first approxim<strong>at</strong>ion<br />
for many systems whose measured critical exponents are close to the universal<br />
exponents predicted by the theory. Once the critical temper<strong>at</strong>ure and pressure<br />
are known, the van der Waals equ<strong>at</strong>ion is unique. We can then use rel<strong>at</strong>ive<br />
variables to write a parameter-free version of this equ<strong>at</strong>ion to derive values of<br />
the critical exponents. If we change the values of the temper<strong>at</strong>ure of the field<br />
slightly, we see a scaled version of the original picture because the free energy due<br />
to fluctu<strong>at</strong>ions near a critical point is a homogeneous function of its variables.<br />
Near the critical point, the correl<strong>at</strong>ion length is the only important length scale,<br />
so the energy density exhibits hyper-scaling there. Mean field theory works far<br />
away from the critical point. Closer to this point, the model is improved by<br />
including gradient terms in the expansions of the free energy. Very close to the<br />
critical point, however, we need a different theory.
304 APPENDIX B. SUMMARIES SUBMITTED BY STUDENTS.
Appendix C<br />
Solutions to selected<br />
problems.<br />
C.1 Solutions for chapter 1.<br />
Problem 1.<br />
The general form is ¯ df = A(x, y)dx + B(x, y)dy and is an exact differential if<br />
and only if ( )<br />
∂A<br />
=<br />
∂y x<br />
Therefore we have<br />
Part 1: A(x, y) = x2 , B(x, y) = y2 and<br />
( )<br />
∂A<br />
∂y<br />
( )<br />
∂B<br />
∂x<br />
Hence an exact differential. Using<br />
we find<br />
f(x, y) = f(0, 0) +<br />
x<br />
y<br />
( )<br />
∂B<br />
∂x y<br />
= 0<br />
= 0<br />
∫ x<br />
A(x<br />
0<br />
′ , 0)dx ′ ∫ y<br />
+ B(x, y<br />
0<br />
′ )dy ′<br />
f(x, y) = f(0, 0) + 1<br />
3 x3 + 1<br />
3 y3<br />
Part 2: A(x, y) = y2 , B(x, y) = x2 and<br />
( )<br />
∂A<br />
∂y<br />
x<br />
305<br />
= 2y
306 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Hence NOT an exact differential.<br />
( )<br />
∂B<br />
= 2x<br />
∂x y<br />
Part 3: A(x, y) = 3x2y2 , B(x, y) = 2x3y + y4 and<br />
( )<br />
∂A<br />
= 6x<br />
∂y x<br />
2 y<br />
( )<br />
∂B<br />
= 6x<br />
∂x<br />
2 y<br />
Hence an exact differential. Using<br />
we find<br />
Problem 2.<br />
h(x, y) = h(0, 0) +<br />
y<br />
∫ x<br />
A(x<br />
0<br />
′ , 0)dx ′ ∫ y<br />
+ B(x, y<br />
0<br />
′ )dy ′<br />
h(x, y) = h(0, 0) + x 3 y 2 + 1<br />
5 y5<br />
¯dq = xydx + 1<br />
x dy<br />
With the test given in problem 1 we have A(x, y) = xy , B(x, y) = 1<br />
x and<br />
( )<br />
∂A<br />
= x<br />
∂y<br />
( ∂B<br />
∂x<br />
Hence NOT an exact differential. Now use:<br />
)<br />
y<br />
x<br />
= − 1<br />
x 2<br />
¯ds = f(x)xydx + f(x) 1<br />
x dy<br />
With the test given in problem 1 we have A(x, y) = f(x)xy , B(x, y) = f(x) 1<br />
x<br />
and ( )<br />
∂A<br />
= f(x)x<br />
∂y<br />
x<br />
( )<br />
∂B<br />
= −f(x)<br />
∂x y<br />
1<br />
x2 + f ′ (x) 1<br />
x<br />
Hence in order th<strong>at</strong> this is an exact differential we need<br />
f(x)x = −f(x) 1<br />
x 2 + f ′ (x) 1<br />
x
C.1. SOLUTIONS FOR CHAPTER 1. 307<br />
f ′ (x)<br />
f(x) = x2 + 1<br />
x<br />
ln(f(x)) = C + 1<br />
3 x3 + ln(x)<br />
Where C is a constant. This gives, with D a constant:<br />
To check this answer we note<br />
f(x) = Dxe 1<br />
3 x3<br />
ds = Dx 2 ye 1<br />
3 x3<br />
dx + De 1<br />
3 x3<br />
With the test given in problem 1 we have A(x, y) = Dx 2 ye 1<br />
3 x3<br />
and ( )<br />
∂A<br />
∂y<br />
( )<br />
∂B<br />
∂x<br />
which is an exact differential indeed.<br />
Problem 3.<br />
Part a.<br />
x<br />
y<br />
= Dx 2 e 1<br />
3 x3<br />
= Dx 2 e 1<br />
3 x3<br />
dU = sin(x + y)dx + sin(x + y)dy<br />
, B(x, y) = De 1<br />
3 x3<br />
With the test given in problem 1 we have A(x, y) = sin(x + y) , B(x, y) =<br />
sin(x + y) and<br />
( )<br />
∂A<br />
= cos(x + y)<br />
∂y x<br />
( )<br />
∂B<br />
= cos(x + y)<br />
∂x<br />
which shows th<strong>at</strong> dU is an exact differential. We have<br />
Part b.<br />
y<br />
U(x, y) = U0 − cos(x + y)<br />
¯dW = sin(x) cos(y)dx
308 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
With the test given in problem 1 we have A(x, y) = sin(x) cos(y) , B(x, y) = 0<br />
and ( )<br />
∂A<br />
= − sin(x) sin(y)<br />
∂y x<br />
( )<br />
∂B<br />
= 0<br />
∂x<br />
which shows th<strong>at</strong> ¯ dW is NOT an exact differential.<br />
Part c. From the first law dU = ¯ dQ − ¯ dW we get<br />
Part d.<br />
Step 1 → 2<br />
¯dQ = [sin(x + y) + sin(x) cos(y)] dx + sin(x + y)dy<br />
W1,2 =<br />
y<br />
U(0, 0) = U0 − 1<br />
U(απ, 0) = U0 − cos(απ)<br />
U(απ, απ) = U0 − cos(2απ)<br />
U(0, απ) = U0 − cos(απ)<br />
∫ απ<br />
Hence Q1,2 = 2 − 2 cos(απ)<br />
Step 2 → 3<br />
0<br />
U2 − U1 = 1 − cos(απ)<br />
sin(x) cos(0)dx = − cos(απ) + 1<br />
U3 − U2 = cos(απ) − cos(2απ)<br />
W2,3 = 0 if x constant no work done!<br />
Hence Q2,3 = cos(απ) − cos(2απ)<br />
Step 3 → 4<br />
U4 − U3 = cos(2απ) − cos(απ)<br />
W3,4 =<br />
∫ 0<br />
απ<br />
sin(x) cos(απ)dx = (−1 + cos(απ)) cos(απ)<br />
Hence Q3,4 = cos(2απ) − cos(απ) − (1 − cos(απ)) cos(απ)<br />
Step 4 → 1<br />
U1 − U4 = cos(απ) − 1<br />
Hence Q4,2 = cos(απ) − 1<br />
Adding all components:<br />
W4,1 = 0 if x constant no work done!<br />
Wtotal = cos 2 (απ) − 2 cos(απ) + 1 = (1 − cos(απ)) 2
C.1. SOLUTIONS FOR CHAPTER 1. 309<br />
Qtotal = cos 2 (απ) − 2 cos(απ) + 1 = (1 − cos(απ)) 2<br />
Indeed Wtotal − Qtotal = 0 as needed for a cyclic process. This means th<strong>at</strong> Q is<br />
always positive, work is always produced, it is a he<strong>at</strong> engine. The only exception<br />
is α = 2 where nothing happened, Q=0.<br />
Problem 4.<br />
Step 1: Initial st<strong>at</strong>e (p0, T0, V0), final st<strong>at</strong>e (p0, T1, 2V0)<br />
Step 2: Initial st<strong>at</strong>e (p0, T1, 2V0) , final st<strong>at</strong>e (p1, 1<br />
2T1, 2V0)<br />
Step 3: Initial st<strong>at</strong>e (p1, 1<br />
2T1, 2V0) , final st<strong>at</strong>e (p1, T2, V0)<br />
Step 4: Initial st<strong>at</strong>e (p1, T2, V0) , final st<strong>at</strong>e (p0, T0, V0)<br />
The equ<strong>at</strong>ions for the ideal gas are pV = NRT and U = 3pV<br />
. This gives<br />
2<br />
T1 = 2T0 for step 1, p1 = 1<br />
2p0 for step 2, T2 = 1<br />
4T1 = 1<br />
2T0 for step 3, and<br />
p1 p0<br />
= T2 T0<br />
for step 4. The last equ<strong>at</strong>ion is consistent with the previous three.<br />
Hence we have<br />
Step 1 ∆U = 3<br />
2 p0V0<br />
Step 2 ∆U = 3<br />
2 (p1 − p0)2V0 = − 3<br />
2 p0V0<br />
Step 3 ∆U = − 3<br />
2 p1V0 = − 3<br />
4 p0V0<br />
Step 4 ∆U = 3<br />
2 (p0 − p1)V0 = 3<br />
4 p0V0<br />
Indeed the total change in internal energy is zero. It is easy to calcul<strong>at</strong>e the<br />
work done in all steps.<br />
Step 1 W = p0V0, hence Q = 5<br />
2 p0V0.<br />
Step 2 W = 0 , hence Q = − 3<br />
2 p0V0<br />
Step 3 W = −p1V0 = − 1<br />
2 p0V0, hence Q = − 5<br />
4 p0V0.<br />
Step 4 W = 0 , hence Q = 3<br />
4 p0V0<br />
This gives Wtotal = 1<br />
2p0V0 , Qin = 13<br />
Wtotal = Qin + Qout.<br />
The efficiency is η = W 2 = Qin 13<br />
The Carnot efficiency would be ηC = 1 − T2<br />
T1<br />
Problem 5.<br />
Part a.<br />
U(S, V ) = S2<br />
V gives<br />
T =<br />
4 p0V0, Qout = − 11<br />
( )<br />
∂U<br />
∂S V<br />
4 p0V0 which indeed obeys<br />
3 = 4 , which is much larger.<br />
= 2 S<br />
V
310 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
p = −<br />
( )<br />
∂U<br />
=<br />
∂V S<br />
S2<br />
V 2<br />
which gives the equ<strong>at</strong>ion of st<strong>at</strong>e 4p = T 2 . Therefore, <strong>at</strong> constant T we have<br />
constant p. Also, <strong>at</strong> constant S we see th<strong>at</strong> T V = 2S, constant.<br />
Part b.<br />
Carnot cycle<br />
Step 1: (p1, V1) → (p2, V2) isothermal <strong>at</strong> Th<br />
Step 2: (p2, V2) → (p3, V3) adiab<strong>at</strong>ic Th → Tl<br />
Step 3: (p3, V3) → (p4, V4) isothermal <strong>at</strong> Tl<br />
Step 4: (p4, V4) → (p1, V1) adiab<strong>at</strong>ic Tl → Th<br />
1 → 2<br />
2 → 3<br />
3 → 4<br />
4 → 1<br />
This gives<br />
W1,2 =<br />
∫ V2<br />
V1<br />
pdV = 1<br />
4 T 2 h(V2 − V1)<br />
Q1,2 = Th(S2 − S1) = Th( 1<br />
2 ThV2 − 1<br />
2 ThV1) = 1<br />
2 T 2 h(V2 − V1)<br />
W2,3 =<br />
∫ V3<br />
V2<br />
pdV = S 2 2<br />
W3,4 =<br />
∫ V4<br />
V3<br />
∫ V3<br />
V2<br />
(<br />
1 1<br />
dV =<br />
V 2 2 ThV2<br />
) 2<br />
( 1<br />
−<br />
V2<br />
1<br />
)<br />
V3<br />
Q2,3 = 0<br />
pdV = 1 2<br />
Tl (V4 − V3)<br />
4<br />
Q3,4 = Tl(S4 − S3) = Tl( 1<br />
2 TlV4 − 1<br />
W4,1 =<br />
∫ V1<br />
V4<br />
Wtotal = 1<br />
4 T 2 h(V2 − V1) + 1<br />
pdV = S 2 4<br />
∫ V1<br />
V4<br />
4 T 2 hV2(1 − V2<br />
2 TlV3) = 1<br />
2<br />
T 2<br />
l (V4 − V3)<br />
(<br />
1 1<br />
dV =<br />
V 2 2 TlV4<br />
) 2<br />
( 1<br />
−<br />
V4<br />
1<br />
)<br />
V1<br />
Q4,1 = 0<br />
V3<br />
) + 1 2<br />
Tl (V4 − V3) +<br />
4 1<br />
4<br />
2<br />
Tl V4(1 − V4<br />
V1<br />
For the adiab<strong>at</strong>ic steps we have V2Th = V3Tl and V4Tl = V1Th which gives<br />
T 2 V2<br />
hV2 V3<br />
= T 2<br />
l V3. Therefore<br />
Wtotal = 1<br />
4 T 2 h(V2 − V1) + 1<br />
4 T 2 hV2 − 1<br />
4<br />
2<br />
Tl V3 + 1 2<br />
T<br />
1<br />
2 T 2 h(V2 − V1) + 1 2<br />
Tl (V4 − V3)<br />
2<br />
4<br />
l (V4 − V3) + 1 2<br />
Tl V4 −<br />
4 1<br />
4 T 2 hV1 =<br />
)
C.1. SOLUTIONS FOR CHAPTER 1. 311<br />
Qin = 1<br />
2 T 2 h(V2 − V1)<br />
Qout = 1 2<br />
Tl (V4 − V3)<br />
2<br />
and Wtotal = Qin + Qout indeed. The efficiency is<br />
η = W<br />
2 Tl = 1 +<br />
Qin T 2 V4 − V3<br />
h<br />
V2 − V1<br />
which can be simplified because Tl(V4 − V3) = Th(V1 − V2) and gives<br />
as expected.<br />
Problem 6.<br />
η = 1 − Tl<br />
This problem is along the same lines as the previous one, with dU = T dS+HdM<br />
and work done by changing the magnetic field. Given is th<strong>at</strong> U = f(T ) , or<br />
U(S, M) = f(T (S, M)) and also<br />
M = CH<br />
T . In principle, this gives us U by using<br />
( )<br />
∂U<br />
T = = f<br />
∂S<br />
′ (T )<br />
M<br />
Th<br />
( ∂T<br />
∂S<br />
and a similar equ<strong>at</strong>ion for H. This approach is not needed here, though.<br />
Carnot cycle<br />
Step 1: (H1, M1) → (H2, M2) isothermal <strong>at</strong> Th<br />
Step 2: (H2, M2) → (H3, M3) adiab<strong>at</strong>ic Th → Tl<br />
Step 3: (H3, M3) → (H4, M4) isothermal <strong>at</strong> Tl<br />
Step 4: (H4, M4) → (H1, M1) adiab<strong>at</strong>ic Tl → Th<br />
Note: assume M2 < M1, so in step 1 we do work on the outside and he<strong>at</strong> flows<br />
in.<br />
1 → 2<br />
2 → 3<br />
∫ M2<br />
W1,2 = −<br />
M1<br />
HdM = − Th<br />
C<br />
U1,2 = 0<br />
∫ M2<br />
M1<br />
)<br />
M<br />
MdM = − Th<br />
2C (M 2 2 − M 2 1 )<br />
Q1,2 = U1,2 + W1,2 = − Th<br />
2C (M 2 2 − M 2 1 )<br />
U2,3 = f(Tl) − f(Th)<br />
Q2,3 = 0<br />
W2,3 = −U2,3 + Q2,3 = f(Th) − f(Tl)
312 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
3 → 4<br />
4 → 1<br />
∫ M4<br />
W3,4 = −<br />
M3<br />
U3,4 = 0<br />
HdM = − Tl<br />
∫ M4<br />
MdM = −<br />
C M3<br />
Tl<br />
2C (M 2 4 − M 2 3 )<br />
Q3,4 = U3,4 + W3,4 = − Tl<br />
2C (M 2 4 − M 2 3 )<br />
U4,1 = f(Th) − f(Tl)<br />
Q4,1 = 0<br />
W4,1 = −U4,1 + Q4,1 = f(Tl) − f(Th)<br />
This shows th<strong>at</strong> the total change in U is indeed zero and th<strong>at</strong><br />
hence<br />
Wtotal = − Th<br />
2C (M 2 2 − M 2 1 ) − Tl<br />
2C (M 2 4 − M 2 3 )<br />
Qin = − Th<br />
2C (M 2 2 − M 2 1 )<br />
η = W<br />
= 1 +<br />
Qin<br />
Tl M<br />
Th<br />
2 4 − M 2 3<br />
M 2 2 − M 2 1<br />
Finally, we need to reduce the last factor. We need to connect the change in M<br />
along the adiab<strong>at</strong>ic line:<br />
C<br />
2<br />
M 2 3 − M 2 2 = 1<br />
∫<br />
M3MdM =<br />
2 M2<br />
C<br />
∫ 3<br />
2 2<br />
∫ 3<br />
2<br />
( 1<br />
∫ 3<br />
C<br />
dU − dS) =<br />
T 2 2<br />
∫ Tl<br />
1 C<br />
dU =<br />
T 2 Th<br />
H<br />
dM =<br />
T<br />
1<br />
T f ′ (T )dT<br />
a similar expression can be found for the other adiab<strong>at</strong>ic process and we get<br />
or<br />
This gives<br />
as expected.<br />
M 2 3 − M 2 2 = M 2 4 − M 2 1<br />
M 2 3 − M 2 4 = M 2 2 − M 2 1<br />
η = 1 − Tl<br />
Th
C.1. SOLUTIONS FOR CHAPTER 1. 313<br />
Problem 7.<br />
In step 1, adiab<strong>at</strong>ic expansion. V0 → 2V0 and T0 → Tint. VT constant, hence<br />
Tint = 1<br />
2 T0<br />
In step 2, isothermal compression. 2V0 → V0 <strong>at</strong> constant T.<br />
Hence after 1 step T1 = 1<br />
2 T0. This means th<strong>at</strong> after n steps<br />
and we never reach zero.<br />
Problem 8.<br />
Part a.<br />
Initially U = U1 + U2 = C1T1 + C2T2.<br />
Finally U = (C1 + C2)Tf .<br />
No energy exchange, hence<br />
Tn = 1<br />
T0<br />
2n Tf = C1T1 + C2T2<br />
C1 + C2<br />
Part b.<br />
∆S = ∫ 1<br />
T dU holds for each system. Therefore<br />
∫ Tf 1<br />
∆Stotal =<br />
T1 T C1dT<br />
∫ Tf 1<br />
+<br />
T2 T C2dT =<br />
This gives<br />
C1(ln(Tf ) − ln(T1)) + C2(ln(Tf ) − ln(T2))<br />
∆Stotal = (C1 + C2) ln(Tf ) − C1 ln(T1) − C2 ln(T2)<br />
We know th<strong>at</strong> (C1 + C2)Tf = C1T1 + C2T2 and we need to show th<strong>at</strong> (C1 +<br />
C2) ln(Tf ) > C1 ln(T1) + C2 ln(T2)<br />
This follows from the n<strong>at</strong>ure of the logarithm. It is a concave function. In other<br />
words, if z = cx + (1 − c)y with c between 0 and 1, meaning th<strong>at</strong> z is between<br />
x and y, we have ln(z) > c ln(x) + (1 − c) ln(y).<br />
Problem 16.<br />
Consider the following differential:<br />
¯df = g(x, y, z)dx + h(x, y, z)dy + k(x, y, z)dz<br />
Wh<strong>at</strong> are the necessary and sufficient conditions for ¯ df to be exact?<br />
If the function f is a st<strong>at</strong>e function f(x, y, z) we have
314 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
from which it follows th<strong>at</strong><br />
( )<br />
∂f<br />
g(x, y, z, ) =<br />
∂x<br />
( )<br />
∂f<br />
h(x, y, z, ) =<br />
∂y<br />
( )<br />
∂f<br />
k(x, y, z, ) =<br />
∂z<br />
( )<br />
∂g<br />
∂y<br />
( )<br />
∂g<br />
∂z<br />
( )<br />
∂h<br />
∂z<br />
x,z<br />
x,y<br />
x,y<br />
( )<br />
∂h<br />
=<br />
∂x<br />
( )<br />
∂k<br />
=<br />
∂x<br />
( )<br />
∂k<br />
=<br />
∂y<br />
since in the second order deriv<strong>at</strong>ives of f we can interchange the order of the<br />
deriv<strong>at</strong>ives. Once these rel<strong>at</strong>ions are obeyed, rel<strong>at</strong>ions for higher order deriv<strong>at</strong>ives<br />
follow. Hence the rel<strong>at</strong>ions given above are the primary rel<strong>at</strong>ions. They<br />
are necessary.<br />
Suppose th<strong>at</strong> these rel<strong>at</strong>ions are true, can we derive th<strong>at</strong> the differential is exact?<br />
Yes, follow the line from the notes and define<br />
which gives<br />
H(x, y, z) =<br />
( )<br />
∂H<br />
∂y<br />
∫ x<br />
0<br />
∫ x<br />
dx<br />
0<br />
′ g(x ′ ∫ y<br />
, y, z) + dy<br />
0<br />
′ h(0, y ′ ∫ z<br />
, z) + dz<br />
0<br />
′ k(0, 0, z ′ )<br />
x,z<br />
dx ′<br />
( ) ∫ x<br />
∂H<br />
= dx<br />
∂z x,y 0<br />
′<br />
∫ x<br />
dx ′<br />
( )<br />
∂k<br />
∂x<br />
0<br />
∫ x<br />
0<br />
y,z<br />
x,z<br />
x,y<br />
y,z<br />
y,z<br />
x,z<br />
( )<br />
∂H<br />
= g(x, y, z)<br />
∂x y,z<br />
∫ x<br />
= dx<br />
0<br />
′<br />
( )<br />
∂g<br />
∂y<br />
( )<br />
∂h<br />
∂x<br />
(x<br />
x,z<br />
′ , y, z) + h(0, y, z) =<br />
(x<br />
y,z<br />
′ , y, z) + h(0, y, z) = h(x, y, z)<br />
( )<br />
∂g<br />
(x<br />
∂z x,y<br />
′ , y, z) +<br />
(x<br />
y,z<br />
′ , y, z) +<br />
dx ′<br />
( ∂k<br />
∂x<br />
)<br />
∫ y<br />
0<br />
dy ′<br />
∫ y<br />
0<br />
dy ′<br />
( ∂k<br />
∂y<br />
)<br />
( )<br />
∂h<br />
(0, y<br />
∂z x,y<br />
′ , z) + k(0, 0, z) =<br />
(0, y<br />
x,z<br />
′ , z) + k(0, 0, z) =<br />
(x<br />
y,z<br />
′ , y, z) + k(0, y, z) = k(x, y, z)
C.1. SOLUTIONS FOR CHAPTER 1. 315<br />
Consider an arbitrary p<strong>at</strong>h (ϕ(t), ψ(t), λ(t)) from (x0, y0, z0) to (x1, y1, z1). The<br />
functions ϕ,ψ, and λ are arbitrary except th<strong>at</strong> ϕ(0) = x0, ϕ(1) = x1, etc.<br />
Following the differential along this p<strong>at</strong>h gives<br />
∆f =<br />
∫ 1<br />
0<br />
(<br />
g(ϕ(t), ψ(t), λ(t)) dϕ<br />
(t) + h(ϕ(t), ψ(t), λ(t))dψ<br />
dt dt (t)+<br />
k(ϕ(t), ψ(t), λ(t)) dλ<br />
dt (t)<br />
)<br />
dt<br />
We define H(t) = H(ϕ(t), ψ(t), λ(t)) and find similar to the notes th<strong>at</strong><br />
∆f =<br />
∫ 1<br />
0<br />
dH<br />
dt = H(1) − H(0)<br />
dt<br />
which is p<strong>at</strong>h independent. Note th<strong>at</strong> we used all three conditions. Conditions<br />
on higher order deriv<strong>at</strong>ives can also be used, since our conditions can be derived<br />
by integr<strong>at</strong>ion. We need to make the extra assumption, though, th<strong>at</strong> the higher<br />
order deriv<strong>at</strong>ives do exist.<br />
Problem 17.<br />
Consider the following examples:<br />
¯df = 3x 2 y 2 zdx + 2x 3 yzdy + 6x 3 y 2 dz<br />
¯dg = 2xy 2 z 2 dx + 2x 2 yz 2 dy + 2x 2 y 2 zdz<br />
Are these differentials exact? If so, find the corresponding st<strong>at</strong>e function.<br />
First case:<br />
( ) 2 2 ∂3x y z<br />
= 6x<br />
∂y x,z<br />
2 yz<br />
( ) 2 2 ∂3x y z<br />
= 3x<br />
∂z x,y<br />
2 y 2<br />
( ) 3 ∂2x yz<br />
= 6x<br />
∂x y,z<br />
2 yz<br />
( ) 3 ∂2x yz<br />
= 2x<br />
∂z x,y<br />
3 y<br />
( ) 3 2 ∂6x y<br />
= 18x<br />
∂x y,z<br />
2 y 2<br />
( ) 3 2 ∂6x y<br />
= 12x<br />
∂y<br />
3 y<br />
x,z
316 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
which shows th<strong>at</strong> only the first of the three conditions is obeyed, hence this is<br />
not an exact differential.<br />
Second case:<br />
( ) 2 2 ∂2xy z<br />
= 4xyz<br />
∂y x,z<br />
2<br />
( ) 2 2 ∂2xy z<br />
= 4xy<br />
∂z x,y<br />
2 z<br />
( ) 2 2 ∂2x yz<br />
= 4xyz<br />
∂x y,z<br />
2<br />
( ) 2 2 ∂2x yz<br />
= 4x<br />
∂z x,y<br />
2 yz<br />
( ) 2 2 ∂2x y z<br />
= 4xy<br />
∂x y,z<br />
2 z<br />
( ) 2 2 ∂2x y z<br />
= 4x<br />
∂y<br />
2 yz<br />
which shows th<strong>at</strong> all three conditions are obeyed, hence this is an exact differential.<br />
It is not hard to guess th<strong>at</strong> the st<strong>at</strong>e function is given by<br />
Problem 18.<br />
x,z<br />
g(x, y, z, ) = g0 + x 2 y 2 z 2<br />
A system is completely described by the four st<strong>at</strong>e variables T,S,p, and V.<br />
1. Show th<strong>at</strong> the general form of the energy equ<strong>at</strong>ion is U = V f( S<br />
V ).<br />
2. If U = S α V 1−α find the equ<strong>at</strong>ions of st<strong>at</strong>e p(S,V) and T (S,V).<br />
3. The entropy S is not always easy to deal with, and we often elimin<strong>at</strong>e this<br />
variable. In the previous case, solve for p(T,V). Wh<strong>at</strong> is special in this<br />
equ<strong>at</strong>ion, and why is th<strong>at</strong> the case?<br />
4. Calcul<strong>at</strong>e the efficiency of a Carnot engine using a m<strong>at</strong>erial with the energy<br />
equ<strong>at</strong>ion U = S α V 1−α .<br />
We can always write U in the form U = V f(S, V ). Since U, S, and V are both<br />
extensive we have when we multiply all extensive variables by an arbitrary scale<br />
variable x:<br />
xU = xV f(xS, xV )
C.1. SOLUTIONS FOR CHAPTER 1. 317<br />
Combined with the original expression this gives<br />
f(S, V ) = f(xS, xV )<br />
for all non-zero values of x. This means th<strong>at</strong> f can only depend on the r<strong>at</strong>ion<br />
S<br />
V .<br />
From the first law:<br />
we have<br />
which leads to<br />
Note: this gives T S − pV = U !!<br />
The first equ<strong>at</strong>ion gives<br />
dU = T dS − pdV<br />
( )<br />
∂U<br />
T =<br />
∂S<br />
( )<br />
∂U<br />
p = −<br />
∂V<br />
U =<br />
S<br />
V<br />
= α U<br />
S<br />
= (α − 1) U<br />
V<br />
T S<br />
α<br />
U = pV<br />
α − 1<br />
( ) α−1<br />
S<br />
T = α<br />
V<br />
( ) α<br />
S<br />
p = (α − 1)<br />
V<br />
Since the left side is intensive, the right side can only depend on the r<strong>at</strong>io of S<br />
and V.<br />
We find<br />
( ) α<br />
α−1 T<br />
p = (α − 1)<br />
α<br />
hence p(T,V) does not depend on V. This is not surprising, since by scaling V we<br />
need th<strong>at</strong> p(T, xV ) = p(T, V ), and using x = 0 (which is allowed here, different<br />
from the case above!) shows th<strong>at</strong> p(T, V ) = p(T, 0) which is independent of V.<br />
In a Carnot cycle:<br />
Step 1: Isothermal expansion (p1, V1) to (p2, V2) <strong>at</strong> Th.<br />
Since the temper<strong>at</strong>ure is constant, the pressure is constant, and we have<br />
W1,2 = ph(V2 − V1)
318 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
and hence<br />
U2 − U1 = ph(V2 − V1)<br />
α − 1<br />
Q1,2 = αph(V2 − V1)<br />
α − 1<br />
Step 2: Adiab<strong>at</strong>ic expansion (ph, V2) to (pl, V3).<br />
The entropy is constant, Sh. Obviously we have<br />
Step 3 (similar to step 1):<br />
Step 4 (similar to step 2):<br />
Hence the total he<strong>at</strong> in is<br />
and the total work done is<br />
Q2,3 = 0<br />
W2,3 = U2 − U3 = (Th − Tl)Sh<br />
α<br />
W3,4 = pl(V4 − V3)<br />
Q3,4 = αpl(V4 − V3)<br />
α − 1<br />
Q4,1 = 0<br />
W4,1 = U4 − U1 = (Tl − Th)Sl<br />
α<br />
Qin = αph(V2 − V1)<br />
α − 1<br />
W = ph(V2 − V1) + pl(V4 − V3) + (Th − Tl)(Sh − Sl)<br />
α<br />
Using (from the equ<strong>at</strong>ions for the energy)<br />
we find<br />
or<br />
W =<br />
pV =<br />
α − 1<br />
T S<br />
α<br />
α − 1<br />
α Th(Sh<br />
α − 1<br />
− Sl) +<br />
α Tl(Sl − Sh) + (Th − Tl)(Sh − Sl)<br />
α
C.1. SOLUTIONS FOR CHAPTER 1. 319<br />
(No surprise, total he<strong>at</strong> exchange!!!)<br />
and<br />
which gives<br />
Problem 19.<br />
W = (Th − Tl)(Sh − Sl)<br />
Qin = Th(Sh − Sl)<br />
η = W<br />
=<br />
Qin<br />
Th − Tl<br />
Th<br />
= ηCarnot<br />
Consider the following cyclic process (Brayton), with an ideal mono-<strong>at</strong>omic gas<br />
as the working m<strong>at</strong>erial :<br />
1. Adiab<strong>at</strong>ic expansion from pA, V1, T1 to pB, V2, T2.<br />
2. Isobaric expansion from pB, V2, T2 to pB, V3, T3.<br />
3. Adiab<strong>at</strong>ic contraction from pB, V3, T3 to pA, V4, T4.<br />
4. Isobaric contraction from pA, V4, T4 to pA, V1, T1.<br />
Calcul<strong>at</strong>e the efficiency in terms of pA and pB. Compare this with the Carnot<br />
efficiency (indic<strong>at</strong>e the temper<strong>at</strong>ures used, and why you choose these references).<br />
Step 1: S constant (see notes)<br />
Step 2: p constant<br />
Step 3:<br />
Step 4:<br />
W1,2 = U1 − U2 = 3<br />
2 (V1pA − V2pB)<br />
W2,3 = pB(V3 − V2)<br />
W3,4 = 3<br />
2 (V3pB − V4pA)<br />
W4,1 = pA(V1 − V4)<br />
W = 5<br />
2 (pA(V1 − V4) + pB(V3 − V2))<br />
In step 2 the change in energy follows from U = 3<br />
2pV and is
320 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
U3 − U2 = 3<br />
2 pB(V3 − V2)<br />
Hence the energy flow in this step is<br />
Therefore<br />
In the adiab<strong>at</strong>ic steps we have<br />
Qin = Q2,3 = 5<br />
2 pB(V3 − V2)<br />
η = W<br />
= 1 +<br />
Qin<br />
pA(V1 − V4)<br />
pB(V3 − V2)<br />
pAV γ<br />
1<br />
pBV γ<br />
3<br />
= pBV γ<br />
2<br />
= pAV γ<br />
4<br />
with γ = 5<br />
3 , and hence V1V3 = V2V4 as in the notes.<br />
Also<br />
and hence<br />
or<br />
V3 = V4<br />
V2 = V1<br />
η = 1 −<br />
η = 1 −<br />
( pA<br />
pB<br />
( pA<br />
pB<br />
( pA<br />
pB<br />
( pA<br />
During the constant entropy parts we have<br />
Hence we have<br />
pB<br />
3 −<br />
V ∝ T 2<br />
p ∝ T 5<br />
2<br />
) 1<br />
γ<br />
) 1<br />
γ<br />
) γ−1<br />
γ<br />
η = 1 − T1<br />
T2<br />
which is a Carnot-like efficiency. However, we have an even lower temper<strong>at</strong>ure,<br />
T3 available, and it would be more efficient to run a Carnot engine between T1<br />
and T3!<br />
) 2<br />
5
C.2. SOLUTIONS FOR CHAPTER 2. 321<br />
C.2 Solutions for chapter 2.<br />
Problem 1.<br />
Part a.<br />
The Helmholtz Free Energy<br />
Part b.<br />
Ftotal = F (T, V1, N1) + F (T, V − V1, N − N1)<br />
This should be a minimum as a function of the internal parameter V1<br />
( ∂Ftotal<br />
∂V1<br />
This is zero if<br />
) ( )<br />
( )<br />
( )<br />
∂F<br />
∂F<br />
∂(V − V1)<br />
= (T, V1, N1) + (T, V − V1, N − N1)<br />
∂V<br />
∂V<br />
∂V1<br />
p(T, V1, N1) = p(T, V − V1, N − N1)<br />
where we used ( )<br />
∂F<br />
∂V = p.<br />
T,N<br />
Hence the pressure left and right needs to be the same.<br />
For the second order deriv<strong>at</strong>ives we find<br />
( 2 ∂ Ftotal<br />
∂V 2<br />
) ( 2 ∂ F<br />
=<br />
1 ∂V 2<br />
)<br />
( 2 ∂ F<br />
(T, V1, N1) +<br />
∂V 2<br />
)<br />
(T, V − V1, N − N1)<br />
where the second term now has a plus sign again. This has to be positive! In<br />
general, we can take V1 to be half V and we find th<strong>at</strong> we need<br />
or<br />
1<br />
The left hand side is equal to<br />
in the notes.<br />
Problem 2.<br />
−V κT =<br />
( 2 ∂ F<br />
∂V 2<br />
)<br />
> 0<br />
T,N<br />
( )<br />
∂p<br />
−<br />
∂V T,V<br />
V κT<br />
( ) ( ) ( )<br />
∂V ∂V ∂V<br />
= +<br />
∂p T,N ∂p S,N ∂S p,N<br />
> 0<br />
which is positive indeed, as has been shown<br />
( )<br />
( ) ( )<br />
∂S<br />
∂V ∂V<br />
= −V κS−<br />
∂p T,N<br />
∂S p,N ∂T p,N
322 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
where we used a Maxwell rel<strong>at</strong>ion based on the Gibbs Free Energy G(T,p,N).<br />
Therefore<br />
( )<br />
∂V<br />
κT − κS = α<br />
∂S p,N<br />
Using a Maxwell rel<strong>at</strong>ion based on the Enthalpy H(S,p,N) we find<br />
( )<br />
∂V<br />
∂S<br />
( )<br />
∂T<br />
=<br />
∂p<br />
( )<br />
∂T<br />
= −<br />
∂S<br />
( )<br />
∂S<br />
∂p<br />
p,N<br />
S,N<br />
where we also used the identity connecting three partial deriv<strong>at</strong>ives. The first<br />
partial deriv<strong>at</strong>ive is<br />
( )<br />
∂T<br />
=<br />
∂S p,N<br />
T<br />
Cp<br />
and the second is<br />
p,N<br />
( ) ( )<br />
∂S<br />
∂V<br />
= − = V α<br />
∂p T,N ∂T p,N<br />
as we used before. Combining all of this:<br />
Problem 3.<br />
T V α2<br />
κT − κS =<br />
Part a.<br />
Basic variables T, V, µ hence use the Grand Potential Ω = U − T S − µN.<br />
Part b.<br />
for RT
C.2. SOLUTIONS FOR CHAPTER 2. 323<br />
Part d.<br />
If S is infinity for µ = 0 th<strong>at</strong> is certainly also true <strong>at</strong> T=0 and µ = 0. Wh<strong>at</strong> is<br />
wrong?<br />
The formula only is valid for RT 0<br />
= 2S2<br />
> 0<br />
V 3<br />
= − 2S<br />
< 0<br />
V 2<br />
( 2 ∂ U<br />
∂S2 )<br />
V<br />
( )<br />
∂U<br />
∂S V<br />
= 2S<br />
V<br />
( )<br />
∂U<br />
=<br />
∂V S<br />
S2<br />
V 2<br />
( 2 ∂ U<br />
∂V 2<br />
)<br />
S<br />
Which gives 4p = T 2 , not surprising since we have only two intensive variables<br />
and they cannot be independent.<br />
Therefore U(T, V ) = 1<br />
4T 2V and CV = 1<br />
2T V = S ( of course! ).<br />
Also Cp does not exist, since we cannot change T while keeping p constant.<br />
The volume depends on pressure according to V = S √<br />
p and κS = − 1<br />
( )<br />
∂V<br />
V ∂p<br />
S =<br />
1<br />
2p<br />
Also here κT does not exist since we cannot change p <strong>at</strong> constant T.<br />
Finally, α cannot be defined, since we cannot change T <strong>at</strong> constant p.
324 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Problem 6.<br />
Hence<br />
Problem 7.<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
=<br />
V N<br />
( 2 ∂ U<br />
∂S2 )<br />
( 2 ∂ S<br />
∂U 2<br />
)<br />
V N<br />
V N<br />
( ) −1 ∂T<br />
∂U<br />
( )<br />
∂T<br />
=<br />
∂S<br />
V N<br />
V N<br />
= − 1<br />
T 2 CV<br />
= T<br />
CV<br />
( 2 ∂ U<br />
∂S2 )<br />
= −<br />
V N<br />
1<br />
T C2 V<br />
< 0<br />
Part a.<br />
( )<br />
∂S<br />
CM = T<br />
∂T MN<br />
( )<br />
∂S<br />
CH = T<br />
∂T HN<br />
( ) ( ) ( )<br />
∂S<br />
∂S ∂M<br />
= T<br />
+ T<br />
∂T MN ∂M T N ∂T<br />
( )<br />
∂S<br />
CH − CM = T<br />
∂M<br />
( )<br />
∂M<br />
∂T<br />
( )<br />
∂H<br />
= −T<br />
∂T<br />
( )<br />
∂M<br />
∂T<br />
or<br />
Part b.<br />
CH − CM = T<br />
T N<br />
( )<br />
∂H<br />
∂M T N<br />
HN<br />
( )<br />
∂M<br />
∂T HN<br />
MN<br />
( )<br />
∂M<br />
=<br />
∂T HN<br />
T<br />
χT<br />
HN<br />
HN<br />
( ) 2<br />
∂M<br />
∂T HN<br />
( )<br />
∂M<br />
χS =<br />
∂H SN<br />
( )<br />
∂M<br />
χT =<br />
∂H T N<br />
( ) ( ) ( )<br />
∂M ∂M ∂S<br />
=<br />
+<br />
∂H SN ∂S HN ∂H T N<br />
( ) ( )<br />
∂M ∂S<br />
χT − χS =<br />
∂S HN ∂H T N<br />
( )<br />
∂M<br />
∂S HN<br />
( ) ( )<br />
∂M ∂T<br />
=<br />
∂T HN ∂S HN<br />
( )<br />
∂M<br />
=<br />
∂T HN<br />
T<br />
CH<br />
( )<br />
∂S<br />
∂H<br />
( )<br />
∂M<br />
=<br />
∂T<br />
T N<br />
HN
C.2. SOLUTIONS FOR CHAPTER 2. 325<br />
and hence<br />
χT − χS =<br />
( )<br />
∂M<br />
∂T HN<br />
T<br />
CH<br />
( )<br />
∂M<br />
∂T HN<br />
Part c.<br />
Using the internal energy U(S,M) we have for stability<br />
( 2 ∂ U<br />
0 <<br />
∂S2 ) ( )<br />
∂T<br />
=<br />
=<br />
∂S<br />
T<br />
CM<br />
or CM > 0. Also<br />
or χS > 0.<br />
From parts a and b:<br />
0 <<br />
MN<br />
MN<br />
( 2 ∂ U<br />
∂M 2<br />
) ( )<br />
∂H<br />
=<br />
=<br />
SN ∂M SN<br />
1<br />
χS<br />
χT (CH − CM ) = CH(χT − χS)<br />
which gives χT CM = CHχS or CH and χT have the same sign.<br />
Finally, we need the mixed inequality.<br />
The right hand side is T<br />
CM χS<br />
and<br />
( )2 ( 2 2 ∂ U ∂ U<br />
<<br />
∂M∂S N ∂M 2<br />
)<br />
SN<br />
( 2 ∂ U<br />
∂S2 )<br />
MN<br />
while the left hand side is<br />
( ) ( ) ( ) ( )<br />
2 ∂ U ∂T<br />
∂T ∂S<br />
=<br />
=> −<br />
= −<br />
∂M∂S N ∂M SN ∂S MN ∂M T N<br />
T<br />
CM<br />
gives<br />
or<br />
( ) ( )<br />
∂H<br />
∂H<br />
= −<br />
∂T MN ∂M T N<br />
T<br />
CMχS<br />
><br />
[ ( )<br />
∂M T ∂T HN<br />
CM χT<br />
CH − CM<br />
χT<br />
] 2<br />
< CM<br />
χS<br />
( )<br />
∂M<br />
∂T HN<br />
= T (CH − CM )<br />
C 2 M χT<br />
= CH<br />
χT<br />
( )<br />
∂H<br />
∂T MN<br />
from which it follows th<strong>at</strong> CM<br />
χT > 0 or χT > 0 and since CH has the same sign,<br />
we have CH > 0. Parts a and b then give χT > χS and CH > CM.<br />
Part d.
326 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
From M = CNH<br />
T we get χT = CN<br />
T<br />
From U = 3<br />
2NkT we get CM = 3<br />
2Nk With ( )<br />
∂M<br />
CNH<br />
∂T = − HN T 2 we get using parts a and b:<br />
( 2 T CNH<br />
CH − CM =<br />
CN T 2<br />
) 2<br />
= CNH2<br />
T 2<br />
or CH = 3 MH<br />
2Nk + T<br />
and also<br />
or<br />
Problem 10.<br />
Part A.<br />
Part B.<br />
Part C.<br />
Part D.<br />
Part E.<br />
This means<br />
χT − χS = T<br />
χS = CN<br />
T −<br />
( )<br />
∂S<br />
=<br />
∂V T N<br />
∆Q = T ∆S = T<br />
CH<br />
( CNH<br />
T 2<br />
C 2 NH 2<br />
) 2<br />
3<br />
2 kT 3 + CH 2 T<br />
( )<br />
∂p<br />
=<br />
∂T V N<br />
NR<br />
V<br />
∆W = p∆V = NRT<br />
V ∆V<br />
( )<br />
∂S<br />
∆V =<br />
∂V T N<br />
NRT<br />
V ∆V<br />
∆U = ∆Q − ∆W = 0<br />
( )<br />
∂U<br />
= 0<br />
∂V T N<br />
or U = g(T, N).<br />
Part F.<br />
is intensive and cannot depend on N, therefore U = Nf(T ).<br />
U<br />
N<br />
Problem 11.
C.2. SOLUTIONS FOR CHAPTER 2. 327<br />
The initial st<strong>at</strong>e with N moles of m<strong>at</strong>erial has pressure p, temper<strong>at</strong>ure T, and<br />
volume V. Hence we can calcul<strong>at</strong>e the energy U, the entropy S, and the chemical<br />
potential µ. The final st<strong>at</strong>e has volume V ′ , amount of m<strong>at</strong>erial N ′ = N<br />
moles. Because there is no energy exchange, the final st<strong>at</strong>e has energy U ′ = U.<br />
Because we know three st<strong>at</strong>e variables in the final st<strong>at</strong>e, all other variables can<br />
be calcul<strong>at</strong>ed. Note th<strong>at</strong> the entropy S will change, even though there is no he<strong>at</strong><br />
exchange.<br />
Part A.<br />
For the ideal gas we have U = 3<br />
2 NRT , and hence T ′ = T . Using the ideal gas<br />
law we get p ′ V ′ = NRT ′ = NRT = pV , or p ′ = p V<br />
V ′ . The temper<strong>at</strong>ure stayed<br />
the same, but the pressure went down.<br />
Part B.<br />
Assume th<strong>at</strong> we take a reversible p<strong>at</strong>h from the initial to the final st<strong>at</strong>e. Because<br />
the entropy changes, this will be a p<strong>at</strong>h along which he<strong>at</strong> is exchanged and work<br />
is done, but in such a manner th<strong>at</strong> these two cancel. We need to calcul<strong>at</strong>e:<br />
We have<br />
T ′ − T =<br />
∫ V ′<br />
( ) ( )<br />
∂T<br />
∂T<br />
= −<br />
∂V U,N ∂U V,N<br />
V<br />
( )<br />
∂T<br />
dV<br />
∂V U,N<br />
( )<br />
∂U<br />
= −<br />
∂V T,N<br />
1<br />
CV<br />
( )<br />
∂U<br />
∂V T,N<br />
which we use since CV was given. Since we know partial deriv<strong>at</strong>ives of U <strong>at</strong><br />
constant values of extensive st<strong>at</strong>e variables, we write<br />
( )<br />
∂U<br />
=<br />
∂V T,N<br />
( )<br />
∂U<br />
+<br />
∂V S,N<br />
( )<br />
∂U<br />
∂S V,N<br />
( )<br />
∂S<br />
= −p + T<br />
∂V T,N<br />
( )<br />
∂S<br />
∂V T,N<br />
Since we do not like the partial deriv<strong>at</strong>ives of S we use Maxwell to get:<br />
and hence<br />
Finally:<br />
T ′ − T =<br />
( ) ( ) 2 ∂S<br />
∂ F<br />
= −<br />
=<br />
∂V T,N ∂T ∂V N<br />
∫ V ′<br />
V<br />
( )<br />
∂p<br />
=<br />
∂T V,N<br />
NR<br />
V − Nb<br />
( )<br />
∂U<br />
= −p +<br />
∂V T,N<br />
NR<br />
= aN2<br />
V − Nb V 2<br />
( )<br />
∂T<br />
∂V U,N<br />
2 ∫ ′<br />
V<br />
2aN<br />
dV = −<br />
3NR V<br />
which is neg<strong>at</strong>ive indeed, assuming th<strong>at</strong> a > 0.<br />
( )<br />
1 2aN 1 1<br />
dV = −<br />
V 2 3R V ′ V
328 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Problem 16.<br />
Part A.<br />
Part B.<br />
since S(0,V)=0 and hence<br />
Part C.<br />
( )<br />
∂S<br />
∂T V<br />
= 1<br />
T<br />
S(T, V ) =<br />
S(T, V ) =<br />
∫ T<br />
0<br />
∫ T<br />
0<br />
( )<br />
∂U<br />
∂T V<br />
= 2cV<br />
( )<br />
∂S<br />
dT<br />
∂T V<br />
2cV dT = 2cV T<br />
U = T S − pV ⇒ pV = T S − U = cV T 2<br />
or p = cT 2 . Note this is the example we used before starting with U(S, V ) = S2<br />
V<br />
and c = 1<br />
4 .<br />
Problem 17.<br />
Part A.<br />
At constant volume and number of moles dU = T dS − pdV + µdN = T dS and<br />
hence CV = T ∂S ∂U 3<br />
∂T = ∂T = 2NR. Part B.<br />
S(T, V, N)−S(T0, V, N) =<br />
Part C.<br />
∫ T<br />
T0<br />
( ) ∫ T<br />
∂S<br />
CV<br />
dT =<br />
∂T V,N T0 T<br />
lim S(T, V, N) = −∞<br />
T →0<br />
3<br />
dT = NR(ln(T )−ln(T0))<br />
2<br />
which is against the third law. Hence an ideal gas cannot be the correct description<br />
of a gas <strong>at</strong> very low temper<strong>at</strong>ure.<br />
Part D.<br />
H = U + pV = 5<br />
2 NRT<br />
Part E.<br />
At constant pressure and number of moles dH = T dS + V dp + µdN = T dS and<br />
hence CV = T ∂S ∂H 5<br />
∂T = ∂T = 2NR.
C.2. SOLUTIONS FOR CHAPTER 2. 329<br />
Problem 18.<br />
The mass of a mole of gas is m, and hence <strong>at</strong> height h the gravit<strong>at</strong>ional potential<br />
energy is mgh per mole. The total chemical potential is<br />
µtotal = µintrinsic + mgh<br />
which is constant in equilibrium. Therefore,<br />
and<br />
which leads to<br />
Problem 19.<br />
RT ln( n<br />
n<br />
nG<br />
nG<br />
) + mgh = C<br />
= e C−mgh<br />
RT<br />
mgh<br />
−<br />
n(h) = n(0)e RT<br />
Part A.<br />
For an adiab<strong>at</strong>ic process we have pV γ = constant, and the work done per mole<br />
is therefore<br />
W =<br />
∫ 2<br />
1<br />
pdV =<br />
∫ 2<br />
1<br />
c c<br />
(<br />
dV = V<br />
V γ 1 − γ<br />
1−γ<br />
2<br />
which is (with the ideal gas law pV = NRT ):<br />
W = 1<br />
1 − γ (pV2 − pV1) = R<br />
1 − γ (T2 − T1)<br />
− V 1−γ<br />
)<br />
1<br />
Remember this is per mole! Using the value of γ for w<strong>at</strong>er we get<br />
and hence<br />
W = 3R (T1 − T2)<br />
f = W<br />
H<br />
= 300R<br />
H<br />
Part B.<br />
The gain in height is 100m. The gain in potential energy is<br />
The amount of he<strong>at</strong> needed is<br />
∆Ug = (M + Mfuel + Mw<strong>at</strong>er) 100g
330 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Q = 1<br />
f ∆Ug = Hg<br />
3R (M + Mfuel + Mw<strong>at</strong>er)<br />
and hence the mount of fuel mf used is:<br />
mf = Hg<br />
3JR (M + Mfuel + Mw<strong>at</strong>er)<br />
Part C.<br />
Similar, the amount of w<strong>at</strong>er mw used is:<br />
Problem 20.<br />
mw = g<br />
3R (M + Mfuel + Mw<strong>at</strong>er)<br />
Part A.<br />
At room temper<strong>at</strong>ure, air (di<strong>at</strong>omic molecules) has its rot<strong>at</strong>ional degrees of<br />
freedom activ<strong>at</strong>ed, but not its vibr<strong>at</strong>ional, and hence we have cV = 5<br />
2R and<br />
cp = 7<br />
2R. Part B.<br />
The he<strong>at</strong> needed to cool N moles of air is Q = Ncp∆T .<br />
The amount of air exchanged per day is Va = 4 × 8 × 0.75m3 , which is about<br />
0.28 liters per second, or about Na = 0.013 moles per second. This requires an<br />
energy Qa = Nacp∆T per second. Hence:<br />
Qa = 0.013 × 7<br />
× 8.31 × 18J<br />
2<br />
or Qa is about 7 Joules per second.<br />
The efficiency of the refriger<strong>at</strong>or is η = W<br />
18<br />
W and this is about +Qa 300 . Hence the<br />
refriger<strong>at</strong>or uses about 120 Joules per second to oper<strong>at</strong>e.<br />
Part C.<br />
Now we also need to cool and condense w<strong>at</strong>er. The gas going into the refriger<strong>at</strong>or<br />
is now 90% dry air and 10% w<strong>at</strong>er <strong>at</strong> room temper<strong>at</strong>ure. Therefore, we need to<br />
cool and condense about 0.0013 moles of w<strong>at</strong>er per second. For w<strong>at</strong>er cp = 4R<br />
and hence the energy needed to cool the w<strong>at</strong>er is 0.1 × 4 7 7 or about 0.8 J per<br />
2<br />
second. This is an additional 0.1 J/s compared to the dry case.<br />
In addition, we need to condense 0.0013 moles of w<strong>at</strong>er per second. Since one<br />
mole of w<strong>at</strong>er is 18g, this is about 0.023 g per second, which requires 2.26 kJ<br />
per gram, or 52 J per second. This is by far the largest extra cost. Using the<br />
efficiency we find th<strong>at</strong> in this case the refriger<strong>at</strong>or needs an extra 870 Joules per<br />
second!<br />
Problem 21.<br />
Part A.
C.2. SOLUTIONS FOR CHAPTER 2. 331<br />
An amount of he<strong>at</strong> Q flows from the solid into the Carnot engine, an amount<br />
Q’ flows from the Carnot engine into the room temper<strong>at</strong>ure reservoir, and an<br />
amount of work W is done on the Carnot refriger<strong>at</strong>or. The efficiency is η =<br />
−W<br />
−Q ′ = TR−T<br />
. This gives<br />
TR<br />
Q = Q ′ − W =<br />
( )<br />
1<br />
− 1 W =<br />
η<br />
T<br />
TR − T W<br />
The amount of he<strong>at</strong> one needs to extract to change the temper<strong>at</strong>ure from T to<br />
T − ∆T is C∆T = AT 3 ∆T and hence the amount of work th<strong>at</strong> needs to be<br />
done to do this is<br />
∆W = (TR − T ) AT 2 ∆T<br />
Because ∆T is defined in the neg<strong>at</strong>ive T direction we find therefore<br />
dW<br />
dT = − (TR − T ) AT 2<br />
and hence the amount of work needed to cool the solid to zero temper<strong>at</strong>ure is<br />
Part B.<br />
and hence<br />
Part C.<br />
Part D.<br />
W =<br />
∫ 0<br />
TR<br />
( )<br />
∂S<br />
∂T V<br />
∆U =<br />
∆S =<br />
∫ 0<br />
TR<br />
− (TR − T ) AT 2 dT = A<br />
12 T 4 R<br />
= 1<br />
T<br />
∫ 0<br />
TR<br />
CdT =<br />
( )<br />
∂U<br />
∂T V<br />
= C<br />
T<br />
= AT 2<br />
AT 2 dT = − 1<br />
3 AT 3 R<br />
∫ 0<br />
TR<br />
AT 3 dT = − 1<br />
4 AT 4 R<br />
∆UR = −∆U + W = 1<br />
3 AT 4 R<br />
∆SR = −∆S = 1<br />
3 AT 3 R<br />
Note th<strong>at</strong> indeed ∆UR = TR∆SR.<br />
Problem 22.<br />
Part A.
332 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Part B.<br />
( ) ( ) 2 ∂S<br />
∂ F<br />
= −<br />
=<br />
∂V T,N ∂T ∂V N<br />
and hence<br />
∆Q = T<br />
( )<br />
∂S<br />
∆V<br />
∂V T,N<br />
( )<br />
∂p<br />
=<br />
∂T V,N<br />
N<br />
V R<br />
(<br />
1 + B(T ) N<br />
)<br />
+<br />
V<br />
N<br />
V RT B′ (T ) N<br />
V<br />
∆Q = p∆V +<br />
N 2<br />
V 2 RT 2 B ′ (T )<br />
Part C.<br />
The first term is the normal work done on the environment, the second term<br />
must therefore represent a change in internal energy!<br />
Problem 23.<br />
Suppose th<strong>at</strong> during the process the pressure and temper<strong>at</strong>ure in the tank are<br />
p and T . A small amount of air, ∆N moles, goes from the line into the tank.<br />
This air expands to pressure p and temper<strong>at</strong>ure T” and then exchanges he<strong>at</strong> to<br />
get to a common temper<strong>at</strong>ure.<br />
The initial internal energy of the air in the tank is N1u1, the final internal energy<br />
of the air in the tank is N2u2, where we have p1V = N1RT1 and p2V = N2RT2.<br />
There is an amount of he<strong>at</strong> Q added to the gas, which is supplied by the wall.<br />
The amount of m<strong>at</strong>erial added to the tank is N2 − N1 and by adding this<br />
m<strong>at</strong>erial we added energy to the system. Because the air is taken out of the<br />
line <strong>at</strong> constant pressure the amount of energy available to add to the tank<br />
is (N2 − N2)h ′ where h ′ is the molar enthalpy h ′ = u ′ + p ′ v ′ and v ′ is the molar<br />
volume in the compressed air line, p ′ v ′ = RT ′ . Hence<br />
N2u2 = N1u1 + Q + (N2 − N1)h ′<br />
which gives the rel<strong>at</strong>ion in the problem.<br />
This gives<br />
and<br />
Using cp = cV + R we find<br />
N2(u2 − u ′ − RT ′ ) = N1(u1 − u ′ − RT ′ ) + Q<br />
u2 − u ′ = cV (T2 − T ′ )<br />
N2(cV T2 − cpT ′ ) = N1(cV T1 − cpT ′ ) + Q
C.2. SOLUTIONS FOR CHAPTER 2. 333<br />
The amount of he<strong>at</strong> is supplied by the wall, but since the temper<strong>at</strong>ure of the<br />
wall increases this is a neg<strong>at</strong>ive number, with<br />
Q = −mtcs(T2 − T1)<br />
Using the equ<strong>at</strong>ion of st<strong>at</strong>e we arrive <strong>at</strong> an equ<strong>at</strong>ion for T2:<br />
p2<br />
T2<br />
V (cV T2 − cpT ′ ) = p1<br />
V (cV T1 − cpT ′ ) − mtcsR(T2 − T1)<br />
T2<br />
T1<br />
′<br />
′<br />
T T<br />
V (cV p2 − cpp2) = V (cV p1 − cpp1) − mtcsRT ′ ( T2<br />
T<br />
define x1 = T1<br />
T ′ and x2 = T2<br />
T ′ and we get<br />
T1<br />
)<br />
T ′<br />
′ − T1<br />
V (cV p2 − 1<br />
cpp2) = V (cV p1 − 1<br />
cpp1) − mtcsRT ′ (x2 − x1)<br />
x2<br />
cV (p2 − p1) + cp( p1<br />
(cV − cp)(p2 − p1) + cp(p1( 1<br />
x1<br />
x1<br />
x1<br />
− p2 RT<br />
) + mtcs<br />
x2<br />
′<br />
V (x2 − x1) = 0<br />
− 1) − p2( 1<br />
x2<br />
RT<br />
− 1)) + mtcs<br />
′<br />
V (x2 − x1) = 0<br />
We will get an approxim<strong>at</strong>e solution (exact is possible, too, since this is a second<br />
order equ<strong>at</strong>ion). Assume th<strong>at</strong> x1 = 1 + δ1, x2 = 1 + δ2, are close to 1, and keep<br />
only linear terms.<br />
RT<br />
(cV − cp)(p2 − p1) + cp(−p1δ1 + p2δ2) + mtcs<br />
′<br />
V (δ2 − δ1) = 0<br />
T<br />
δ2(mtcs<br />
′ 7<br />
+<br />
V 2 p2)<br />
T<br />
≈ δ1(mtcs<br />
′ 7<br />
+<br />
V 2 p1) + (p2 − p1)<br />
Divide by p ′ and define y1 = p1<br />
p ′ , y2 = p2<br />
p ′ :<br />
T<br />
δ2(mtcs<br />
′<br />
p ′ 7<br />
+<br />
V 2 y2)<br />
T<br />
≈ δ1(mtcs<br />
′<br />
p ′ V<br />
+ 7<br />
2 y1) + (y2 − y1)<br />
This expresses everything in dimensionless quantities. The quantity mtcs p ′ V is<br />
about 12. Hence we have<br />
14δ2 ≈ δ113 + 0.4<br />
which gives T2 about 304K or 32C and N2 about three times N1.<br />
Problem 24.<br />
Part A.<br />
T ′
334 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
where m is the mass density, M=mV.<br />
Part B.<br />
and hence<br />
F (T, V, N) = Fideal(T, V, N) − V 2 3m2 G<br />
5R<br />
V = 4π<br />
3 R3<br />
F (T, V, N) = Fideal(T, V, N) − 3m2 G<br />
5<br />
( ) 1<br />
3 4π<br />
V<br />
3<br />
5<br />
3<br />
( )<br />
∂F<br />
p = −<br />
=<br />
∂V T,N<br />
NRT<br />
V + m2 ( ) 1<br />
3 4π<br />
G V<br />
3<br />
2<br />
3<br />
κT = − 1<br />
V<br />
( )<br />
∂V<br />
∂p T,N<br />
( )<br />
∂p<br />
= −<br />
∂V T,N<br />
NRT<br />
V 2 + 2m2G 3<br />
κT = −<br />
V<br />
( ∂p<br />
∂V<br />
1<br />
)<br />
T,N<br />
Part C.<br />
stability needs positive κ and hence<br />
=<br />
NRT ≥ 2m2 G<br />
3<br />
NRT<br />
V<br />
( ) 1<br />
3 4π<br />
1 −<br />
V 3<br />
3<br />
1<br />
− 2m2 G<br />
3<br />
( ) 1<br />
3 4π<br />
V<br />
3<br />
5<br />
3<br />
This is not obeyed <strong>at</strong> low temper<strong>at</strong>ure of large volume!<br />
Problem 25.<br />
We have per mole th<strong>at</strong><br />
which gives<br />
or<br />
s(T ) =<br />
∫ 300<br />
0<br />
γT<br />
T<br />
( ) 1<br />
4π 2<br />
3<br />
3 V 3<br />
hs<br />
dT +<br />
300 +<br />
∫ 600<br />
cp<br />
300 T dT<br />
s(600K) = 300γ + hs<br />
300 + cp ln(2)
C.2. SOLUTIONS FOR CHAPTER 2. 335<br />
5.5cal/degree/mole = 1.5cal/degree/mole + 0.5cal/degree/mole + cp/ln(2)<br />
cp ≈ 5cal/degree/mole<br />
The molar specific he<strong>at</strong> cp for a mono-<strong>at</strong>omic gas is 5<br />
2<br />
R and for a di<strong>at</strong>omic gas<br />
<strong>at</strong> least 7<br />
2R, since rot<strong>at</strong>ional degrees of freedom are active <strong>at</strong> 600K. Hence the<br />
gas is mono-<strong>at</strong>omic.<br />
Problem 26.<br />
A substance has the following properties:<br />
(I) The amount of m<strong>at</strong>erial is fixed <strong>at</strong> N moles.<br />
(II) At constant temper<strong>at</strong>ure T0 the work W done by it on expansion from a<br />
volume V0 to a volume V is<br />
(III) The entropy is given by<br />
W = NRT0 ln( V<br />
S = NR V0<br />
V<br />
( T<br />
T0<br />
V0<br />
)<br />
) x<br />
In these expressions T0, V0, and x are fixed constants.<br />
(A) Calcul<strong>at</strong>e the Helmholtz free energy of the system.<br />
dF = −SdT − pdV<br />
F (T0, V ) − F (T0, V0) = −W = −NRT0 ln( V<br />
∫ T<br />
F (T, V ) − F (T0, V ) = −<br />
F (T0, V ) = F0 − NRT0 ln( V<br />
T0<br />
SdT ′ ∫ T<br />
= −<br />
T0<br />
V0<br />
)<br />
NR V0<br />
V<br />
V0<br />
)<br />
( T ′<br />
T0<br />
) x<br />
dT ′<br />
( ( ) )<br />
x+1<br />
V0 1 T<br />
F (T, V ) = F (T0, V ) − NRT0<br />
− 1<br />
V x + 1 T0<br />
F (T, V ) = F0 − NRT0 ln( V<br />
V0<br />
( ( ) )<br />
x+1<br />
V0 1 T<br />
) − NRT0<br />
− 1<br />
V x + 1 T0
336 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
(B) Find the equ<strong>at</strong>ion of st<strong>at</strong>e for the pressure p in terms of V , T , and N.<br />
( )<br />
∂F<br />
p = − =<br />
∂V T<br />
NRT0<br />
V<br />
NRT0 V0<br />
−<br />
V V<br />
( ( ) )<br />
x+1<br />
1 T<br />
− 1<br />
x + 1 T0<br />
(C) Find the work done on expansion <strong>at</strong> constant T, where T is arbitrary.<br />
W =<br />
∫ V<br />
V0<br />
Problem 27.<br />
pdV = NRT0 ln( V<br />
V0<br />
)+NRT0<br />
( V0<br />
V<br />
The he<strong>at</strong> capacity <strong>at</strong> constant pressure is given by<br />
( )<br />
∂S<br />
Cp = T<br />
∂T<br />
and the coefficient of thermal expansion is given by<br />
α = 1<br />
( )<br />
∂V<br />
V ∂T<br />
The amount of m<strong>at</strong>erial N is always kept constant.<br />
If we consider S(T, p) we have<br />
( ) ( )<br />
∂S ∂S<br />
dS = dT + dp<br />
∂T p ∂p T<br />
(A) Show th<strong>at</strong> this leads to<br />
T dS = T<br />
T dS = CpdT − αT V dp<br />
p<br />
p<br />
( )<br />
∂S<br />
dT + T<br />
∂T p<br />
T dS = CpdT + T<br />
) ( ( ) )<br />
x+1<br />
1 T<br />
− 1<br />
− 1<br />
x + 1 T0<br />
( )<br />
∂S<br />
dp<br />
∂p T<br />
( )<br />
∂S<br />
dp<br />
∂p T<br />
( ) ( ) ( )<br />
2 ∂S ∂ G ∂V<br />
= −<br />
− = −αV<br />
∂p T ∂T ∂p = ∂T p<br />
T dS = CpdT − T αV dp
C.2. SOLUTIONS FOR CHAPTER 2. 337<br />
One can show th<strong>at</strong> (but you do NOT have to!)<br />
where X is a finite constant.<br />
αV<br />
lim<br />
T →0 Cp<br />
= X<br />
(B) Show th<strong>at</strong> this has the consequence th<strong>at</strong> in an adiab<strong>at</strong>ic process one can<br />
never reach T = 0 by a finite change in pressure.<br />
At very low temper<strong>at</strong>ures<br />
CpdT = αV T dp<br />
dT = αV<br />
T dp<br />
dT<br />
T<br />
Cp<br />
= Xdp<br />
ln(T ) − ln(T ′ ) = X(p − p ′ )<br />
Therefore, if T → 0 the pressure difference goes to infinity.<br />
Problem 28.<br />
A container of volume V0 is divided into two equal parts by a removable wall.<br />
There are N0 moles of gas in equilibrium <strong>at</strong> temper<strong>at</strong>ure Ti on the left, the right<br />
side is empty. The container is isol<strong>at</strong>ed, no he<strong>at</strong> goes through. The internal<br />
wall is suddenly removed and the gas can expand freely until it finally fills up<br />
the whole container.<br />
(A) Is the gas in equilibrium just after the wall has been removed?<br />
No, it is not, it wants to expand and fill up all space.<br />
(B) Is the free expansion a reversible process?<br />
No, the process is not slow, the beginning is not in equilibrium, and only<br />
the final st<strong>at</strong>e is in equilibrium.<br />
(C) Does the entropy of the gas increase, decrease, or stay the same?<br />
The entropy increases (maximum entropy principle)<br />
(D) Does the energy of the gas increase, decrease, or stay the same?<br />
Stay the same, because there is no he<strong>at</strong> exchange, and no work done on<br />
the outside world.
338 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Assume th<strong>at</strong> we have a different, reversible process th<strong>at</strong> brings a gas <strong>at</strong> constant<br />
internal energy from a temper<strong>at</strong>ure Ti and volume 1<br />
2V0 to a volume V0 and<br />
temper<strong>at</strong>ure Tf . At each stage of this reversible process the gas is described by<br />
an internal energy function of the form U = u(T, V, N).<br />
(E) Is this reversible process adiab<strong>at</strong>ic?<br />
No, it is not. The internal energy is constant, but now work is done and<br />
he<strong>at</strong> has to flow in.<br />
(F) Express the response function ( )<br />
∂T<br />
∂V in terms of partial deriv<strong>at</strong>ives of<br />
U,N<br />
the energy function u(T, V, N).<br />
There are several approaches, for example, use the product rule:<br />
( ) ( )<br />
∂T<br />
∂T<br />
= −<br />
∂V U,N ∂U V,N<br />
( )<br />
∂U<br />
= −<br />
∂V T,N<br />
[ (∂U )<br />
∂T<br />
V,N<br />
] −1 ( )<br />
∂U<br />
∂V T,N<br />
(G) A mono-<strong>at</strong>omic ideal gas has u(T, V, N) = 3<br />
2NRT . How much does the<br />
temper<strong>at</strong>ure change in a free expansion from volume 1<br />
2V0 to V0?<br />
For the equivalent reversible process we have ( )<br />
∂U<br />
∂V = 0, and hence<br />
( )<br />
T,N<br />
∂T<br />
∂V = 0, which means th<strong>at</strong> the temper<strong>at</strong>ure T does not change. Since<br />
U,N<br />
the equivalent process brings us to the same final st<strong>at</strong>e, the free expansion<br />
also has no temper<strong>at</strong>ure change.<br />
Problem 29.<br />
Under adiab<strong>at</strong>ic compression a gas changes its temper<strong>at</strong>ure when the pressure is<br />
changed, ( in) a reversible manner. The response function describing this situ<strong>at</strong>ion<br />
∂T is Ω = ∂p<br />
S,N .<br />
( )<br />
T V α<br />
1 ∂V<br />
Show th<strong>at</strong> Ω = with α = + Cp V ∂T p,N and Cp the he<strong>at</strong> capacity <strong>at</strong> constant<br />
pressure.<br />
Note the sign error in the problem, α does not have a minus sign. Using the<br />
product rule we find:<br />
( ) ( ) ( )<br />
∂T<br />
∂T ∂S<br />
Ω =<br />
= −<br />
∂p S,N ∂S p,N ∂p T,N<br />
( )<br />
∂S<br />
Cp = T<br />
∂T p,N<br />
( ) ( ) ( )<br />
2 ∂S<br />
∂ G<br />
∂V<br />
= −<br />
= − = −V α<br />
∂p T,N ∂T ∂p N ∂T p,N
C.2. SOLUTIONS FOR CHAPTER 2. 339<br />
Combining this gives the required result.<br />
For an ideal mono-<strong>at</strong>omic gas αT = 1 and Cp = 5<br />
2NR. Find the function f<br />
defined by T = f(p) in this adiab<strong>at</strong>ic process.<br />
( )<br />
∂T<br />
=<br />
∂p<br />
T V α<br />
= 2V<br />
5NR<br />
Using the ideal gas law:<br />
This can be integr<strong>at</strong>ed to give:<br />
S,N<br />
Cp<br />
pV = NRT<br />
( )<br />
∂T<br />
=<br />
∂p S,N<br />
2pV 2NRT<br />
=<br />
5pNR 5pNR<br />
5 log(T ) = 2 log(p) + C(S, N)<br />
(<br />
p<br />
f(p) =<br />
p0<br />
) 2<br />
5<br />
where we changed the constant of integr<strong>at</strong>ion to p0.<br />
Problem 30.<br />
The Jacobian J is defined by<br />
J(x, y; u, v) =<br />
Show th<strong>at</strong> J(T, S; p, V ) = 1.<br />
Three approaches:<br />
use<br />
which gives<br />
J(T, S; p, V ) =<br />
use<br />
J(T, S; p, V ) =<br />
∂(x, y)<br />
∂(u, v) =<br />
( )<br />
∂x<br />
∂u v<br />
∂(T, S)<br />
∂(p, V ) =<br />
( )<br />
∂T<br />
∂p V<br />
( )<br />
∂S<br />
=<br />
∂V T<br />
( )<br />
∂T<br />
∂p V<br />
( )<br />
∂S<br />
+<br />
∂V p<br />
( ) ( )<br />
∂S ∂T<br />
−<br />
∂V T ∂p V<br />
( )<br />
∂y<br />
−<br />
∂v u<br />
= 2T<br />
5p<br />
( )<br />
∂x<br />
∂v u<br />
( ) ( )<br />
∂S ∂T<br />
−<br />
∂V p ∂V p<br />
( )<br />
∂S<br />
( )<br />
∂p<br />
∂p V ∂V T<br />
( )<br />
∂y<br />
∂u v<br />
( )<br />
∂S<br />
∂p V<br />
( ) ( ) ( )<br />
∂S ∂p ∂T<br />
−<br />
∂p V ∂V T ∂V p<br />
( )<br />
∂S<br />
∂p V
340 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
and get<br />
Mr. Maxwell gives:<br />
and hence<br />
A second route:<br />
which gives<br />
( ) ( ) ( )<br />
∂T ∂p ∂T<br />
= −<br />
∂p V ∂V T ∂V p<br />
J(T, S; p, V ) =<br />
( )<br />
∂T<br />
∂p V<br />
( ) ( ) 2 ∂S<br />
∂ F<br />
= − =<br />
∂V T ∂T ∂V<br />
J(T, S; p, V ) =<br />
J(T, S; p, V ) =<br />
J(T, S; p, V ) =<br />
( )<br />
∂T<br />
∂p V<br />
( )<br />
∂S<br />
∂V T<br />
( )<br />
∂p<br />
∂T V<br />
( )<br />
∂p<br />
∂T V<br />
= 1<br />
∂(T, S) ∂(T, S) ∂(T, V )<br />
=<br />
∂(p, V ) ∂(T, V ) ∂(p, V )<br />
( )<br />
∂S<br />
( )<br />
∂T<br />
∂V T ∂p V<br />
where we note th<strong>at</strong> the following partial deriv<strong>at</strong>ives are zero:<br />
( )<br />
∂T<br />
= 0<br />
∂V T<br />
Mr. Maxwell applies again, and the solution follows the end of the previous one.<br />
Third route:<br />
Take a single area in the p-V plane enclosed by a line L.<br />
∫ ∫<br />
dpdV = W<br />
A<br />
which is the work done in a cyclic process along L.<br />
Each point in A corresponds to a point (T, S) and hence maps A into A’<br />
∫ ∫<br />
A ′<br />
dT dS = Q<br />
is the he<strong>at</strong> into the system in the same process. Since the energy did not change<br />
we have W = Q, or<br />
∫ ∫ ∫ ∫<br />
dpdV =<br />
A<br />
A ′<br />
dT dS
C.2. SOLUTIONS FOR CHAPTER 2. 341<br />
for arbitrary areas A and corresponding areas A’. But the definition of the<br />
Jacobian tells us<br />
∫ ∫<br />
∫ ∫<br />
J(T, S; p, V )dpdV =<br />
A<br />
A ′<br />
dT dS<br />
and hence we have<br />
∫ ∫<br />
∫ ∫<br />
J(T, S; p, V )dpdV = dpdV<br />
A<br />
A<br />
for arbitrary areas A. This can only be true if the Jacobian is one.<br />
Problem 31.<br />
A fluid undergoes an adiab<strong>at</strong>ic throttling process. This means th<strong>at</strong> the fluid is<br />
pressed through a porous plug, with initial pressure p1 and final pressure p2.<br />
(A) Show th<strong>at</strong> the enthalpy of the fluid is constant in this process.<br />
As a result of this process, the temper<strong>at</strong>ure changes (remember when you pump<br />
air in your bicycle tire, or when you let the air escape?). The amount of change<br />
is measured by the Joule-Thomson coefficient ηJT , defined by<br />
( )<br />
∂T<br />
ηJT =<br />
∂p<br />
(B) Show th<strong>at</strong> ηJT = V (T α − 1)<br />
Cp<br />
(C) Wh<strong>at</strong> is the value for an ideal gas?<br />
Consider an amount of m<strong>at</strong>erial N taken from region 1 to region 2.<br />
The initial volume is V1, the final volume V2. The equipment th<strong>at</strong> keeps the<br />
pressure constant has to do work when these volumes change. The work done<br />
by this equipment on region 1 is p1V1, while the work done in region 2 is −p2V2<br />
(neg<strong>at</strong>ive!). There is no he<strong>at</strong> exchange, and the energy of the amount of m<strong>at</strong>erial<br />
changes from U1 to U2. Energy conserv<strong>at</strong>ion yields:<br />
H<br />
U2 − U1 = p1V1 − p2V2<br />
from which we get H1 = U1 + p1V1 = H2.<br />
( )<br />
∂T<br />
ηJT =<br />
∂p<br />
We always keep N constant, and leave out th<strong>at</strong> variable for the remainder of the<br />
problem. There are many ways to proceed. For example, we use the fact th<strong>at</strong><br />
deriv<strong>at</strong>ives <strong>at</strong> constant H are a problem, and write<br />
H,N
342 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
( ) ( )<br />
∂T ∂T<br />
= −<br />
∂p H ∂H p<br />
We have, with dH = T dS + V dp,<br />
which gives<br />
Cp = T<br />
( )<br />
∂H<br />
= −<br />
∂p T<br />
( )<br />
∂S<br />
=<br />
∂T p<br />
ηJT = − 1<br />
Again, from dH = T dS + V dp we find<br />
( )<br />
∂H<br />
= T<br />
∂p<br />
T<br />
Cp<br />
[ (∂H ) ] −1 (∂H )<br />
∂T<br />
( )<br />
∂H<br />
∂T p<br />
( )<br />
∂H<br />
∂p T<br />
( )<br />
∂S<br />
+ V<br />
∂p T<br />
Mr. Maxwell comes in handy again:<br />
( ) ( ) ( )<br />
2 ∂S ∂ G ∂V<br />
= − = − = −V α<br />
∂p T ∂T ∂p ∂T p<br />
and hence<br />
ηJT = − 1<br />
Cp<br />
(V − T V α)<br />
For an ideal gas we have αT = 1 and hence ηJT = 0. An ideal gas escaping<br />
from a tyre would not cause any cooling, or pumping an ideal gas into a tyre<br />
would not cause any he<strong>at</strong>ing!<br />
If we use<br />
(virial expansion), we find th<strong>at</strong><br />
pV α<br />
NRT<br />
pV = NRT (1 + N<br />
V B2(T ))<br />
pV<br />
NRT<br />
If we write α = 1<br />
T (1 + ∆α) we find<br />
= 1 + N<br />
V B2(T )<br />
pV<br />
− = −αN<br />
NRT 2 V B2(T ) + N<br />
V B′ 2(T )<br />
pV<br />
∆α = −(1 + ∆α)N<br />
NRT V B2(T ) + T N<br />
V B′ 2(T )<br />
(1 + N<br />
V B2(T ))∆α = −(1 + ∆α) N<br />
V B2(T ) + T N<br />
V B′ 2(T )<br />
p<br />
∂p<br />
T
C.3. SOLUTIONS FOR CHAPTER 3. 343<br />
In the low density limit we only need terms in first order:<br />
∆α ≈ − N<br />
V B2(T ) + T N<br />
V B′ 2(T ) = N<br />
V (T B′ 2(T ) − B2(T ))<br />
and by measuring the Joule-Thomson coefficient we can obtain the second virial<br />
coefficient<br />
C.3 Solutions for chapter 3.<br />
Problem 1.<br />
Note:<br />
p<br />
RT<br />
( )<br />
∂p<br />
∂T V N<br />
= N<br />
V<br />
= RN<br />
V<br />
(1 + R<br />
this can be neg<strong>at</strong>ive, if 2T < T0 − p0V<br />
NR .<br />
We have<br />
( )<br />
∂p<br />
∂V T N<br />
= RT (− N<br />
V<br />
p0<br />
(1 + R<br />
(T − T0) N<br />
V )<br />
p0<br />
p0<br />
(2T − T0) N<br />
V )<br />
2<br />
R 2N<br />
− (T − T0) )<br />
2 V 3<br />
for mechanical stability this has to be neg<strong>at</strong>ive, which gives<br />
or<br />
or<br />
(− N R<br />
−<br />
V 2<br />
(1 + R<br />
p0<br />
p0<br />
(T − T0)<br />
2 2N<br />
) < 0<br />
V 3<br />
(T − T0) 2N<br />
) > 0<br />
V<br />
(T0 − T )2NR < V p0<br />
which is always true if T > T0 . In the stable region we also have<br />
T0 − p0 V<br />
R N < T0 − 2(T0 − T ) = 2T − T0 < 2T<br />
and hence the isotherms are not crossing in the region of mechanical stability.<br />
The line separ<strong>at</strong>ing the stable and unstable region follows from<br />
NR<br />
p0V =<br />
1<br />
, T < T0<br />
2(T0 − T )
344 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
and hence<br />
T<br />
p = p0 , T < T0<br />
4(T0 − T )<br />
This shows th<strong>at</strong> the instability region starts <strong>at</strong> V → 0, p → ∞, T ↑ T0 and runs<br />
to V = 2NRT0<br />
, p = 0, forT → 0.<br />
p0<br />
There are no meta-stable st<strong>at</strong>es.<br />
Problem 2.<br />
Part a.<br />
or<br />
p = NRT aN<br />
e− V RT<br />
V − Nb<br />
pV<br />
NRT =<br />
V<br />
aN<br />
e− V RT<br />
V − NB<br />
in the limit V → ∞ the right-hand side goes to one.<br />
Part b.<br />
or<br />
which is zero if<br />
( )<br />
∂p<br />
= −<br />
∂V T N<br />
NRT<br />
aN<br />
e− V RT +<br />
(V − Nb) 2 NRT aN<br />
e− V RT<br />
V − Nb<br />
( )<br />
∂p<br />
∂V T N<br />
1<br />
= p(−<br />
(V − Nb)<br />
aN<br />
V 2RT =<br />
1<br />
(V − Nb)<br />
For the second order deriv<strong>at</strong>ive we have<br />
+ aN<br />
V 2 RT )<br />
aN<br />
V 2 RT<br />
([A])<br />
( 2 ∂ p<br />
∂V 2<br />
) ( ) (<br />
∂p<br />
1 aN<br />
=<br />
− +<br />
T N ∂V T N (V − Nb) V 2 ) (<br />
1 aN<br />
+ p<br />
− 2<br />
RT (V − Nb) 2 V 3 )<br />
RT<br />
which gives <strong>at</strong> the point where the first deriv<strong>at</strong>ive is zero th<strong>at</strong> we also need<br />
Dividing A by B gives<br />
V − Nb = V<br />
2<br />
or Vc = 2Nb. Inserting this in A gives<br />
1 aN<br />
(<br />
= 2<br />
(V − Nb) 2 V 3 ) ([B])<br />
kT
C.3. SOLUTIONS FOR CHAPTER 3. 345<br />
aN<br />
4N 2 b 2 RT<br />
= 1<br />
Nb<br />
or RTc = a<br />
4b .<br />
Finally, from the equ<strong>at</strong>ion of st<strong>at</strong>e we get pc = a<br />
4b2 e−2 .<br />
Part c.<br />
Similar to van der Waals.<br />
Problem 3.<br />
part a.<br />
Spherical drops V = 4π<br />
3 R3 , A = 4πR2 , and hence dV = 4πR2dR, dA = 8πRdR<br />
and dA = 2<br />
RdV .<br />
This gives<br />
dU = T dS − (p − σ 2<br />
)dV + µdN<br />
R<br />
and peff = p − 2σ<br />
R .<br />
part b.<br />
Equilibrium gives pout = peff = pin − 2σ<br />
R .<br />
part c.<br />
µg(p, T ) = µl(p + 2σ<br />
, T )<br />
R<br />
where p is the pressure in the gas.<br />
part d.<br />
Left hand side:<br />
Right hand side<br />
Reducing this to the volume<br />
or<br />
N∆µg = ∆G = Vgdp − SgdT<br />
N∆µl = ∆G = Vl(1 − 2σ<br />
R 2<br />
N∆µl = ∆G = Vl(1 − 6σ<br />
VlR<br />
( )<br />
∂R<br />
)dp − SldT<br />
∂p T N<br />
( )<br />
∂Vl<br />
)dp − SldT<br />
∂p T N<br />
N∆µl = ∆G = Vl(1 + 6σκT<br />
)dp − SldT<br />
R<br />
where we introduced the isothermal compressibility of the liquid. Along the<br />
coexistence line the changes are the same and hence<br />
Vgdp − SgdT = Vl(1 + 6σκT<br />
R<br />
)dp − SldT
346 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
or<br />
which gives<br />
(Sg − Sl)dT = (Vg − Vl(1 + 6σκT<br />
R ))dp<br />
dpco<br />
dT =<br />
Sg − Sl<br />
Vg − Vl(1 + 6σκT<br />
R )<br />
part e.<br />
With changing p and R INDEPENDENTLY we have for the left hand side<br />
and for the right hand side<br />
which gives<br />
or pco increases when R decreases.<br />
Problem 4.<br />
N∆µg = ∆G = Vgdp<br />
N∆µl = ∆G = Vl(dp − 2σ<br />
dR)<br />
R2 2σ<br />
(Vg − Vl)dp = −Vl dR<br />
R2 part a.<br />
The variables th<strong>at</strong> can change are the temper<strong>at</strong>ure and the Helium partial pressure,<br />
two in total, and hence three phases can coexist.<br />
part b.<br />
The total pressure is p. Therefore we have<br />
µice(p, T ) = µw<strong>at</strong>er(p, T ) = µgas(p − pHe, T )<br />
If the first two are independent of p, this fixes T and hence the temper<strong>at</strong>ure of<br />
the triple point does not change. From the last we find th<strong>at</strong> the pressure <strong>at</strong> the<br />
triple point is simple increased by the Helium partial pressure.<br />
Problem 10.<br />
Part A.<br />
In case Tg ̸= Tl he<strong>at</strong> will flow from one to the other until thermal equilibrium is<br />
reached and the temper<strong>at</strong>ure is the same everywhere. If pg ̸= pl one phase will<br />
do work on the other by changing the volumes until equilibrium is reached for<br />
which the pressure is the same everywhere.<br />
Part B.
C.3. SOLUTIONS FOR CHAPTER 3. 347<br />
Vl = ∂Gl<br />
∂p<br />
Nl + Ng = N0<br />
Vl + Vg = V0<br />
pVg = NgRT<br />
= Nl<br />
∂µl<br />
(p, T )<br />
∂p<br />
Part C.<br />
Multiply the fourth equ<strong>at</strong>ion by p and substitute the third and fourth equ<strong>at</strong>ion<br />
in the second, leads to<br />
or after elimin<strong>at</strong>ing Ng<br />
Nl + Ng = N0<br />
Nlp ∂µl<br />
∂p + NgRT = pV0<br />
Nl(p ∂µl<br />
∂p − RT ) = pV0 − N0RT<br />
Part D.<br />
In order for the solution to be valid we need 0 ≤ Nl ≤ N0 and 0 ≤ Vl ≤ V0.<br />
Problem 11.<br />
In the vicinity of the triple point of a m<strong>at</strong>erial the vapor pressure of the liquid<br />
is given by log( p<br />
20T0<br />
) = 10 − p0 T . The vapor pressure of the solid is given by<br />
log( p<br />
25T0<br />
) = 15 − p0 T .<br />
(A) Wh<strong>at</strong> are the temper<strong>at</strong>ure and the pressure of the triple point in terms of<br />
T0 and p0?<br />
The two equ<strong>at</strong>ions above give the coexistence curves gas-liquid and gassolid.<br />
At the triple point they meet. Hence<br />
10 − 20T0<br />
Tt<br />
= 15 − 25T0<br />
Tt<br />
Tt = T0<br />
pt = p0e −10
348 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
(B) Using the Clausius-Clapeyron equ<strong>at</strong>ion, show th<strong>at</strong><br />
dpco<br />
dT<br />
d log(pco)<br />
dT<br />
= l<br />
T ∆v<br />
= dpco<br />
dT<br />
1<br />
pco<br />
d log(pco)<br />
dT = l<br />
pcoT ∆v .<br />
(C) Is the volume per mole of the gas larger or smaller than th<strong>at</strong> of the liquid,<br />
assuming th<strong>at</strong> T0 and p0 are both positive?<br />
For the gas liquid curve we have:<br />
dpco<br />
dT<br />
= 20T0<br />
T 2<br />
which is positive. Since we need he<strong>at</strong> to go from liquid to gas, this shows<br />
th<strong>at</strong> the molar volume of the gas is larger.<br />
(D) Assume th<strong>at</strong> the values of the l<strong>at</strong>ent he<strong>at</strong> are independent of temper<strong>at</strong>ure<br />
near the triple point, and th<strong>at</strong> the molar volume of the gas is much larger<br />
than the molar volume of the liquid and of the solid. Calcul<strong>at</strong>e the r<strong>at</strong>ios<br />
lSL : lLG : lSG, where S,L, and G stand for solid, liquid, and gas.<br />
At the triple point we have:<br />
But we also have:<br />
which gives:<br />
and hence the r<strong>at</strong>io is:<br />
Problem 12.<br />
lLG = Tt∆v dpco<br />
dT = 20vG t<br />
lSG = Tt∆v dpco<br />
dT = 25vG t<br />
lSG = lSL + lLG<br />
lSL = 5v G t<br />
T0<br />
Tt<br />
lSL : lLG : lSG = 1 : 4 : 5<br />
T0<br />
Tt<br />
T0<br />
Tt
C.3. SOLUTIONS FOR CHAPTER 3. 349<br />
Grüneisen parameter.<br />
In many solids we can approxim<strong>at</strong>e the entropy by S = Nf(T/TD) where the<br />
Debye temper<strong>at</strong>ure TD is a function of the molar volume only, TD = g( V<br />
N ). The<br />
Grüneisen parameter γg is defined by<br />
Show th<strong>at</strong><br />
γg = − V<br />
NTD<br />
γg =<br />
V α<br />
CV κT<br />
dTD<br />
d V<br />
N<br />
For many solids the Grüneisen parameter is almost constant. If we assume th<strong>at</strong><br />
it is a constant, show th<strong>at</strong> we have the equ<strong>at</strong>ion of st<strong>at</strong>e<br />
P V = γgU + Nh( V<br />
N )<br />
This is the Mie-Grüneisen equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
α = 1<br />
V<br />
Combining everything<br />
or<br />
( )<br />
∂S<br />
CV = T<br />
∂T V,N<br />
= NT<br />
f<br />
TD<br />
′ ( T<br />
)<br />
TD<br />
κT = − 1<br />
( )<br />
∂V<br />
V ∂p T,N<br />
( )<br />
∂V<br />
∂T<br />
= 1<br />
( ) 2 ∂ G<br />
V ∂p∂T<br />
= − 1<br />
(<br />
∂S<br />
V ∂p<br />
p,N<br />
( ∂S<br />
∂V<br />
)<br />
( ∂V<br />
∂p<br />
α = − 1<br />
V<br />
T,N<br />
( )<br />
∂S<br />
∂V T,N<br />
= − NT<br />
T 2 f<br />
D<br />
′ ( T<br />
TD<br />
CV = NT<br />
α = κT<br />
TD<br />
NT<br />
V TD<br />
)<br />
T,N<br />
) dTD<br />
d V<br />
N<br />
N<br />
1<br />
N<br />
γgf ′ ( T<br />
= κT<br />
TD<br />
( ∂S<br />
∂V<br />
)<br />
)<br />
T,N<br />
T,N<br />
NT<br />
= γgf<br />
V TD<br />
′ ( T<br />
)<br />
TD<br />
f ′ ( T<br />
) =<br />
TD<br />
α NT V TD<br />
κT γg TD NT<br />
γg =<br />
V α<br />
CV κT<br />
)
350 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
From these equ<strong>at</strong>ions we find<br />
Integr<strong>at</strong>ion gives<br />
V α<br />
= γgCV<br />
κT<br />
( ) ( )<br />
∂S<br />
∂S<br />
V<br />
= γgT<br />
∂V T,N ∂T V,N<br />
( ) ( )<br />
∂S ∂U<br />
T<br />
=<br />
∂T V,N ∂T V,N<br />
( ) ( ) ( )<br />
2 ∂S<br />
∂ F ∂p<br />
= −<br />
=<br />
∂V T,N ∂T ∂V N ∂T<br />
( ) ( )<br />
∂p<br />
∂U<br />
V<br />
= γg<br />
∂T<br />
∂T<br />
V,N<br />
V p = γgU + I(V, N)<br />
V,N<br />
V,N<br />
But since I is extensive we can write it in the form Nh( V<br />
N ).<br />
Problem 13.<br />
Murnaghan equ<strong>at</strong>ion.<br />
The isothermal bulk modulus BT is given by BT = −V<br />
( )<br />
∂p<br />
∂V<br />
T<br />
= 1 . The bulk<br />
κT<br />
modulus can often be approxim<strong>at</strong>ed by BT (p, T ) = BT (0, T ) + βT p where βT is<br />
independent of pressure.<br />
Show th<strong>at</strong> this leads to the equ<strong>at</strong>ion of st<strong>at</strong>e<br />
p(V, T ) = BT<br />
( [V0<br />
(0, T )<br />
V<br />
βT<br />
] βT<br />
where V0 is the volume <strong>at</strong> p = 0 and temper<strong>at</strong>ure T.<br />
− 1<br />
( )<br />
∂p<br />
−V<br />
∂V T<br />
= BT (0, T ) + βT p<br />
The solution of the homogeneous equ<strong>at</strong>ion<br />
( )<br />
∂p<br />
−V<br />
∂V<br />
= βT p<br />
T<br />
)
C.3. SOLUTIONS FOR CHAPTER 3. 351<br />
is<br />
The general solution is therefore<br />
which leads after substitution to<br />
or<br />
After defining C = BT (0,T )<br />
V βT<br />
βT<br />
0<br />
Problem 14.<br />
p = CV −βT<br />
p = p0 + CV −βT<br />
βT (p − p0) = BT (0, T ) + βT p<br />
p0 = − BT (0, T )<br />
βT<br />
we get the required answer.<br />
Derive the virial coefficients Bj(T ) for the Van der Waals equ<strong>at</strong>ion of st<strong>at</strong>e.<br />
Comment on the sign of the second virial coefficient.<br />
Compare with<br />
and get<br />
p<br />
RT<br />
p<br />
RT<br />
(p + a<br />
p<br />
RT<br />
N 2<br />
= − a<br />
RT<br />
)(V − Nb) = NRT<br />
V 2<br />
= − a<br />
RT<br />
= − a<br />
RT<br />
p<br />
RT =<br />
N 2<br />
+<br />
V 2 N<br />
N 2 N<br />
+<br />
V 2 V<br />
N 2 N<br />
+<br />
V 2 V<br />
j=1<br />
V − Nb<br />
(<br />
1 − N<br />
V b<br />
∞∑<br />
j=0<br />
∞∑<br />
( ) j<br />
N<br />
Bj(T )<br />
V<br />
B1(T ) = 1<br />
B2(T ) = b − a<br />
RT<br />
) −1<br />
( ) j<br />
N<br />
b<br />
V<br />
j
352 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Bj(T ) = b j−1<br />
for j > 2<br />
The sign of the second coefficient is positive for RT > a 27<br />
b = 8 RTc. This indic<strong>at</strong>es<br />
th<strong>at</strong> <strong>at</strong> high temper<strong>at</strong>ure the pressure is higher than in an ideal gas, which is<br />
caused by the excluded volume term. If the temper<strong>at</strong>ure is low, the pressure<br />
is lower than in an ideal gas, due to the <strong>at</strong>tractive interactions. At very high<br />
temper<strong>at</strong>ures the thermal motion is so strong th<strong>at</strong> the <strong>at</strong>tractive interactions<br />
do not play a role anymore.<br />
Remember the Joule-Thomson coefficient?<br />
ηJT = N<br />
(T B ′ 2 − B2) = N<br />
Cp<br />
Cp<br />
( 2a<br />
RT<br />
− b)<br />
This is positive for T < 27<br />
4 Tc. Hence for low temper<strong>at</strong>ures the gas he<strong>at</strong>s up when<br />
compressed, but for high temper<strong>at</strong>ures actually cools down when compressed!<br />
Problem 15.<br />
The Gibbs free energy of pure m<strong>at</strong>erials A and B is given by NAµA(p, T ) and<br />
NBµB(p, T ). The entropy of mixing for m<strong>at</strong>erials A and B is −NAR log( NA<br />
NA+NB )−<br />
NBR log( NB<br />
NA+NB<br />
), while the interaction energy between A and B is given by<br />
A(T, p) NANB<br />
NA+NB . Show th<strong>at</strong> a system where we mix NA moles of A and NB<br />
moles of B is stable for 2RT > A and th<strong>at</strong> there are forbidden regions when<br />
2RT < A.<br />
G(p, T, NA, NB) = NAµA(p, T ) + NBµB(p, T ) + Gmix<br />
Gmix = A(T, p) NANB<br />
+<br />
NA + NB<br />
NART log( NA<br />
) + NBRT log(<br />
NA + NB<br />
NB<br />
)<br />
NA + NB<br />
The Gibbs energy per mole (divide by NA + NB) in terms of the concentr<strong>at</strong>ion<br />
c of B <strong>at</strong>oms ( c = NB ) is<br />
NA+NB<br />
g(p, T, c) = (1 − c)µA(p, T ) + cµB(p, T ) + A(T, p)c(1 − c)+<br />
cRT log(c) + (1 − c)RT log(1 − c)<br />
This has to be convex as a function of c.
C.4. SOLUTIONS FOR CHAPTER 4. 353<br />
This is positive if<br />
( 2 ∂ g<br />
∂c2 )<br />
= −2A(p, T ) + RT<br />
T,p<br />
1 1<br />
+ RT<br />
c 1 − c<br />
−2A(p, T ) + RT 1 1<br />
+ RT > 0<br />
c 1 − c<br />
2A(p, T )c(1 − c) < RT (1 − c) + RT C = RT<br />
Hence phase separ<strong>at</strong>ion occurs when<br />
RT < 2Ac(1 − c)<br />
The value of the right hand side is maximal for c = 1<br />
2 and unstable regions first<br />
form in the middle, when RT < 1<br />
2A. C.4 Solutions for chapter 4.<br />
Problem 1.<br />
∂f<br />
∂m = m(b(T ) + cm + dm2 ) with b(T ) = b0(T − T0). Extrema <strong>at</strong> m = 0 and<br />
m = m± = 1<br />
2d (−c ± √ c2 − 4db(T )).<br />
If T > T0 then b(T ) > 0 and m+ > m− > 0, as long as 4db(T ) < c2 , and then<br />
we have minima <strong>at</strong> m+ and 0. If T = T0 the extrema <strong>at</strong> m− and 0 coincide, and<br />
for T < T0 we have m+ > 0 > m− and now m=0 is a maximum. The energy<br />
differences are:<br />
f(m+, T ) − f(0, T ) = m 2 +( 1 1<br />
b(T ) +<br />
2 3 cm+ + 1<br />
4 dm2 +)<br />
Using b(T ) + cm+ + dm 2 + = 0 to elimin<strong>at</strong>e the linear term we get<br />
f(m+, T ) − f(0, T ) = m 2 +( 1 1<br />
b(T ) −<br />
6 12 dm2 +)<br />
which is neg<strong>at</strong>ive if 2b(T ) < dm2 + = −b(T ) − cm+, or 3b(T ) < −c 1<br />
2d (−c +<br />
√<br />
c2 − 4db(T ). This gives<br />
These two values are equal if<br />
6db(T )<br />
−c + c < √ c 2 − 4db(T )<br />
( 6db(T )<br />
−c + c)2 = c 2 − 4db(T )
354 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
and hence<br />
36d 2 b 2 (T )<br />
c 2<br />
− 12db(T ) = −4db(T )<br />
36d 2 b 2 (T ) = 8c 2 db(T )<br />
b(T ) = 2c2<br />
9d<br />
Tc = T0 + 2c2<br />
9db0<br />
This is larger than T0 and hence the phase transition is first order, the minimum<br />
<strong>at</strong> m+ becomes lower th<strong>at</strong> the minimum <strong>at</strong> 0.<br />
We also have<br />
which gives<br />
or<br />
f(m−, T ) − f(0, T ) = m 2 −( 1 1<br />
b(T ) −<br />
6 12 dm2−) f(m−, T ) − f(m+, T ) = 1<br />
6 b(T )(m2 − − m 2 +) − 1<br />
12 d(m4 − − m 4 +)<br />
f(m−, T ) − f(m+, T ) = 1<br />
12 (m2 − − m 2 +)(2b(T ) − d(m 2 − + m 2 +))<br />
Since 2b(T ) < dm 2 + we have<br />
2b(T ) − d(m 2 − + m 2 +) < −dm 2 − < 0<br />
and since m− − m+ = − 1<br />
√<br />
d c2 2 − 4db(T )0 we also have m− − m2 + = (m− −<br />
m+)(m− + m+) < 0. This gives f(m−, T ) − f(m+, T ) > 0 and hence if there is<br />
a minimum <strong>at</strong> m− it is a local minimum, and there is no other phase transition.<br />
Finally, for b(T ) = 2c2<br />
9d we have m+ = 1<br />
√<br />
2d (−c + c2 − 4 2c2<br />
9 ) or<br />
since<br />
√ c 2<br />
9<br />
Problem 2.<br />
= − 1<br />
3 c.<br />
m+(Tc) = −2c<br />
3d<br />
∂f<br />
∂m = b(p, T )m + c(p, T )m3 + dm 5 = 0<br />
gives the loc<strong>at</strong>ion of the extrema, b(p, T ) + c(p, T )m 2 0 + dm 4 0 = 0.
C.4. SOLUTIONS FOR CHAPTER 4. 355<br />
In case one, for c(p, T ) = c > 0, the d-term does not change wh<strong>at</strong> we had before,<br />
and the critical temper<strong>at</strong>ure is defined by b(p, Tc) = 0.<br />
In case two, for c(p, T ) = c < 0, we have the same situ<strong>at</strong>ion as in the first order<br />
transition discussed in the notes, with equ<strong>at</strong>ions<br />
This leads to<br />
and finally:<br />
f(m0, p, T ) − f(0, p, T ) = 0<br />
b(p, T ) + c(p, T )m 2 0 + dm 4 0 = 0<br />
1 1<br />
b(p, T ) +<br />
2 4 cm20 + 1<br />
6 dm40 = 0<br />
1 1<br />
b(p, T ) +<br />
2 2 cm20 + 1<br />
2 dm40 = 0<br />
m 2 0 = − 3c<br />
4d<br />
For part c we get th<strong>at</strong> for p > p0 the second order phase transition is always <strong>at</strong><br />
Tc = T0. For p < p0 we have m2 0 = 1<br />
2d (−c + √ c2 − 4db), or − c<br />
√<br />
1<br />
4d = 2d c2 − 4db,<br />
which gives<br />
−c = 2 √ c 2 − 4bd<br />
16bd = 3c 2<br />
16b0p(T − T0) = 3c0T 2 (p − p0) 2<br />
This clearly shows th<strong>at</strong> for p = p0 we have T = T0 and hence the curve Tc(p)<br />
for p < p0 connects smoothly to the curve Tc(p) = T0 for p > p0.<br />
In addition, when ∆p = p − p0 is small we have in lowest order<br />
or<br />
16b0p0∆T = 3c0T 2 0 (∆p) 2<br />
∆T<br />
∆p = 3c0T 2 0<br />
∆p<br />
16b0p0<br />
which shows th<strong>at</strong> the slope of the Tc(p) curve for p < p0 approaches 0 <strong>at</strong> the<br />
connection point, and hence also is continuous.<br />
Problem 3.<br />
We need to find the extrema and solve
356 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
and<br />
0 = ∂f<br />
= b(T )mx +<br />
∂mx<br />
1<br />
2 cmxm 2 y + dm 3 x<br />
0 = ∂f<br />
∂my<br />
= b(T )my + 1<br />
2 cmym 2 x + dm 3 y<br />
Clearly for b(T ) > 0 the only solution is mx = my = 0. For b(T ) < 0 the<br />
minimum follows from<br />
and<br />
b(T ) + 1<br />
2 cm2 y + dm 2 x = 0<br />
b(T ) + 1<br />
2 cm2 x + dm 2 y = 0<br />
Subtracting these two equ<strong>at</strong>ions yields:<br />
( 1<br />
2 c − d)(m2 x − m 2 y) = 0<br />
which shows th<strong>at</strong> for c ̸= 2d we get mx = my. These quantities then follow<br />
from<br />
b(T ) + ( 1<br />
2 c + d)m2 x = 0<br />
which is just wh<strong>at</strong> we had for a second order phase transition.<br />
Using the fact th<strong>at</strong> m 2 = m 2 x + m 2 y we find<br />
f(mx, my, T ) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
4 dm4 + 1<br />
4 (c − 2d)m2 xm 2 y<br />
If c = 2d the last term disappears, and we can find m as before, but do not<br />
know the r<strong>at</strong>io of mx and my. In other words, in this case the two equ<strong>at</strong>ions<br />
we had before became dependent.<br />
Problem 4.<br />
( )<br />
∂f<br />
= m(b0(T − T0) + cm<br />
∂m T<br />
2 + dm 4 ) = 0<br />
has a solution m=0, which is a minimum for T > T0.<br />
b0(T − T0) + cm 2 + dm 4 = 0<br />
has a solution if c2 ≥ b0(T − T0)d or T ≤ T0 + c2<br />
b0d . The energy <strong>at</strong> this solution<br />
is the lowest if T < Tc with
C.4. SOLUTIONS FOR CHAPTER 4. 357<br />
b0(Tc − T0) = 3c2<br />
16d<br />
(see text, equ<strong>at</strong>ion 4.32 ). Therefore we have<br />
(a) T > T0 + c2<br />
b0d , m=0 only solution<br />
(b) T0 + c2<br />
b0d > T > T0 + 3c2<br />
16b0d<br />
two solutions, m=0 is stable.<br />
(c) T0 + 3c2<br />
16b0d > T > T0, two solutions, non-zero m stable.<br />
(d) T0 > T , one solution, non-zero.<br />
Hence this looks like the first order transition in the van der Waals model, where<br />
the local minima only exist for a short range of values.<br />
Problem 5.<br />
( )<br />
∂f<br />
= m<br />
∂m T<br />
3 (c0(T − T0) + dm 2 ) = 0<br />
has a solution m=0, which is a minimum for T > T0.<br />
c0(T − T0) + dm 2 = 0<br />
has a solution for T < T0 given by<br />
√<br />
c0(T0 − T )<br />
m =<br />
d<br />
This is again a second order phase transition. The magnetiz<strong>at</strong>ion is again proportional<br />
to the square root of the temper<strong>at</strong>ure difference.<br />
The specific he<strong>at</strong> per unit volume follows from c = −T d2f dT 2 and is<br />
For T < T0 we have<br />
or<br />
or<br />
f(T ) = a(T ) + m 4<br />
f(T ) = a(T ) + c2 0(T0 − T ) 2<br />
T > T0, c = −T d2 a<br />
dT 2<br />
d 2<br />
(<br />
1<br />
4 c0(T − T0) + 1<br />
6 dm2<br />
)<br />
(<br />
1<br />
4 c0(T − T0) + 1<br />
6 dc0(T0 )<br />
− T )<br />
d<br />
f(T ) = a(T ) + c3 0(T − T0) 3<br />
12d 2
358 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
which gives<br />
c = −T d2a dT 2 − T c30(T − T0)<br />
4d2 and in this case the specific he<strong>at</strong> is continuous <strong>at</strong> the phase transition (but its<br />
deriv<strong>at</strong>ive is not). Also, again in this case the specific he<strong>at</strong> has an extra positive<br />
term below the phase transition.<br />
The susceptibility follows from<br />
χ −1 =<br />
( 2 ∂ f<br />
∂m2 )<br />
= 3c0(T − T0)m<br />
T<br />
2 + 5dm 4<br />
which is zero for T > T0. This means th<strong>at</strong> for these temper<strong>at</strong>ures a very small<br />
field will cause a large magnetiz<strong>at</strong>ion!<br />
For T < T0 we have<br />
χ −1 = 2c2 0(T − T0) 2<br />
d<br />
and hence the susceptibility diverges strongly <strong>at</strong> the transition temper<strong>at</strong>ure.<br />
Problem 6.<br />
The equ<strong>at</strong>ion to solve is again<br />
We solve this by doing a Fourier expansion<br />
∫<br />
ϕ(r) =<br />
∆ϕ + αϕ = − h0<br />
δ(⃗r) = 0 (C.1)<br />
f<br />
d d kψ(k)e ı⃗ k·⃗r<br />
where we used the fact th<strong>at</strong> we have a spherical solution. The delta function is<br />
given by<br />
δ(⃗r) = 1<br />
(2π) d<br />
∫<br />
d d ke ı⃗k·⃗r and hence we get<br />
and<br />
ψ(k)(−| ⃗ k| 2 + α) = − h0<br />
f<br />
ϕ(r) = h0<br />
f(2π) d<br />
∫<br />
d d k<br />
1<br />
1<br />
(2π) d<br />
k 2 − α eı⃗ k·⃗r<br />
where we have to keep in mind th<strong>at</strong> α < 0. We set again α = −ξ −2 and get
C.4. SOLUTIONS FOR CHAPTER 4. 359<br />
ϕ(r) =<br />
2 h0ξ<br />
f(2π) d<br />
∫<br />
d d 1<br />
k<br />
ξ2k2 + 1 eı⃗ k·⃗r<br />
Introducing a new variable ⃗q = r ⃗ k we get (the integr<strong>at</strong>ions are always over the<br />
whole infinite space, hence the domain of the integr<strong>at</strong>ion does not change):<br />
or<br />
ϕ(r) =<br />
2 h0ξ<br />
f(2π) drd ∫<br />
2<br />
2−d h0ξ<br />
ϕ(r) = r<br />
f(2π) d<br />
∫<br />
d d 1<br />
q<br />
ξ2q2r −2 + 1 eı⃗q·ˆr<br />
d d 1<br />
q<br />
ξ2q2 eı⃗q·ˆr<br />
+ r2 This shows th<strong>at</strong> in the limit r → 0 ϕ is proportional to r2−d , because the r<br />
dependence in the integral disappears.<br />
In general, we get the solution by doing a contour integr<strong>at</strong>ion in the complex k<br />
plane, and we pick up contributions from the poles <strong>at</strong> q = ±ırξ−1 . This shows<br />
r −<br />
th<strong>at</strong> the exponential behavior is of the form e ξ indeed. The correl<strong>at</strong>ion length<br />
has the same form as in 3 dimensions.<br />
Problem 7.<br />
From the equ<strong>at</strong>ions<br />
we see th<strong>at</strong><br />
T > Tc : S(T ) = − da<br />
dT<br />
T < Tc : S(T ) = − da<br />
dT<br />
(C.2)<br />
1 db<br />
−<br />
2 dT m2 (T ) (C.3)<br />
∆S = 1 db<br />
2 dT (Tc)m 2 (Tc)<br />
and hence the l<strong>at</strong>ent he<strong>at</strong> follows immedi<strong>at</strong>ely. Note th<strong>at</strong> a measurement of the<br />
l<strong>at</strong>ent he<strong>at</strong> together with a measurement of the magnetic moment gives us the<br />
slope of the second order coefficient in this model!<br />
Problem 8.<br />
The angle between ⃗ S and h<strong>at</strong>n is θ and hence<br />
f(T ; S, θ) = a(T ) + 1<br />
2 b0(T − T0)S 2 + 1<br />
2 c0(T − 1<br />
2 T0)S 2 cos 2 (θ) + 1<br />
4 dS4<br />
Therefore, if T > 1<br />
2T0 the angle dependent term is minimal for θ = 1<br />
2π and ⃗ S<br />
will be orthogonal to ˆn. In th<strong>at</strong> case we have
360 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
f(T ; S, θ = 1<br />
1<br />
π) = a(T ) +<br />
2 2 b0(T − T0)S 2 + 1<br />
4 dS4<br />
which is as studied before. For T > T0 we have S=0, and for T < T0 we have a<br />
non-zero solution, and a second order phase transition.<br />
On the other hand, if T < 1<br />
2T0 the angle dependent term will be minimal if ⃗ S<br />
is parallel to ⃗n ( or anti-parallel), and the energy is<br />
(<br />
1<br />
f(T ; S, θ = 0, π) = a(T ) +<br />
2 b0(T − T0) + 1<br />
2 c0(T − 1<br />
2 T0)<br />
)<br />
S 2 + 1<br />
4 dS4<br />
which gives an even larger non-zero solution for S. Hence this system shows two<br />
second order phase transitions. At T0 from zero S to non-zero S orthogonal to<br />
ˆn, and <strong>at</strong> 1<br />
2 T0 the non-zero S switches to parallel to ˆn.<br />
Problem 9.<br />
( )<br />
∂f<br />
=<br />
∂m T<br />
2<br />
R(T +<br />
m0<br />
1<br />
8 T0) sinh( 2m<br />
Using sinh(2x) = 2 sinh(x) cosh(x) we get<br />
which is zero if<br />
or<br />
m0<br />
) − 1<br />
RT0 sinh(<br />
m0<br />
m<br />
)<br />
m0<br />
( )<br />
∂f<br />
=<br />
∂m T<br />
R<br />
(<br />
4(T +<br />
m0<br />
1<br />
8 T0) cosh( m<br />
)<br />
) − T0 sinh(<br />
m0<br />
m<br />
)<br />
m0<br />
The first condition gives m=0.<br />
The second condition gives<br />
sinh( m<br />
) = 0<br />
m0<br />
4(T + 1<br />
8 T0) cosh( m<br />
cosh( m<br />
which has a solution when T0 ≥ 8T .<br />
The second order deriv<strong>at</strong>ive is<br />
and for m=0 this is<br />
( 2 ∂ f<br />
∂m2 )<br />
=<br />
T<br />
4<br />
m2 R(T +<br />
0<br />
1<br />
m0<br />
m0<br />
) − T0 = 0<br />
) = 2T0<br />
T0 + 8T<br />
8 T0) cosh( 2m<br />
m0<br />
) − 1<br />
m2 RT0 cosh(<br />
0<br />
m<br />
)<br />
m0
C.4. SOLUTIONS FOR CHAPTER 4. 361<br />
( 2 ∂ f<br />
∂m2 )<br />
(m = 0) =<br />
T<br />
4<br />
m2 R(T +<br />
0<br />
1<br />
8 T0) − 1<br />
m2 RT0 =<br />
0<br />
4R<br />
m2 (T −<br />
0<br />
1<br />
8 T0)<br />
which is a maximum when 8T ≤ T0.<br />
Therefore, we have a critical temper<strong>at</strong>ure 1<br />
8 T0 and a second order phase transition.<br />
Problem 10.<br />
0 =<br />
( )<br />
∂f<br />
=<br />
∂m T<br />
3<br />
2 b0(T − T0) √ m + cm 3<br />
which has solutions m=0 or 3<br />
2b0(T − T0) + cm2√m = 0.<br />
The l<strong>at</strong>ter only occurs for T < T0.<br />
The second order deriv<strong>at</strong>ive is<br />
( 2 ∂ f<br />
∂m2 )<br />
=<br />
T<br />
3<br />
4 b0(T − T0) 1<br />
√ + 3cm<br />
m 2<br />
and diverges <strong>at</strong> m=0. The form near m=0 is simple, however. For T > T0 there<br />
is a minimum ( positive infinite slope ) and for T < T0 it is a maximum. At the<br />
transition both solutions are degener<strong>at</strong>e. Hence we have again a second order<br />
phase transition <strong>at</strong> T0.<br />
The magnetic moment is<br />
.<br />
Problem 11.<br />
m =<br />
(<br />
3b0<br />
2c (T0<br />
) 2<br />
5<br />
− T )<br />
Consider a model for a second order phase transition in a magnetic field. The<br />
Helmholtz free energy per unit volume is given by<br />
f(T ; m) = a(T ) + 1<br />
2 b(T )m2 + 1<br />
c(T )m4<br />
4<br />
Assume c(T ) = c > 0, a positive constant, and b(T ) = b0(T − Tc) with b0 > 0.<br />
The magnetiz<strong>at</strong>ion density is still considered to be an internal parameter, which<br />
should be chosen to minimize the Helmholtz free energy. The condition is now<br />
( )<br />
∂f<br />
= H<br />
∂m T<br />
Calcul<strong>at</strong>e m(T,H) (hint: use Maple or equivalent) and plot m(H) for various<br />
temper<strong>at</strong>ures and m(T) for various fields.
362 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
has one or three real solutions.<br />
General case:<br />
H = b(T )m + cm 3<br />
z 3 + a2z 2 + a1z + a0 = 0<br />
q = 1<br />
3 a1 − 1<br />
9 a2 2<br />
r = 1<br />
6 (a1a2 − 3a0) − 1<br />
27 a3 2<br />
One real root if q3 + r2 > 0, three if q3 + r2 < 0<br />
b(T )<br />
In our case a2 = 0, a1 = c , a0 = − H<br />
c .<br />
Hence one root if<br />
It is possible to find the roots:<br />
q =<br />
b(T )<br />
3c<br />
r = H<br />
2c<br />
( ) 3<br />
b(T )<br />
+<br />
3c<br />
( ) 2<br />
H<br />
> 0<br />
2c<br />
−(b(T )) 3 < 27<br />
4 H2 c<br />
(Tc − T ) 3 < 27<br />
4b3 H<br />
0<br />
2 c<br />
(<br />
s± = r ± (q 3 + r 2 ) 1<br />
2<br />
) 1<br />
3<br />
z1 = (s+ + s−) − 1<br />
3 a2<br />
z2 = − 1<br />
2 (s+ + s−) − 1<br />
3 a2<br />
√<br />
3<br />
+ ı<br />
2 (s+ − s−)<br />
z3 = − 1<br />
2 (s+ + s−) − 1<br />
3 a2<br />
√<br />
3<br />
− ı<br />
2 (s+ − s−)<br />
Hence when we have only one root, we get (with a2 = 0):
C.4. SOLUTIONS FOR CHAPTER 4. 363<br />
m =<br />
⎛<br />
⎝ H<br />
2c +<br />
√ (b(T )<br />
3c<br />
) 3<br />
1.8<br />
1.6<br />
1.4<br />
1.2<br />
0.8<br />
0.6<br />
0.4<br />
1<br />
0.2<br />
0<br />
0<br />
1<br />
2<br />
T 3<br />
4<br />
5<br />
1<br />
2<br />
H<br />
Figure C.1: m versus T-H<br />
+<br />
( H<br />
2c<br />
When we have three roots we have<br />
and<br />
s± =<br />
⎛<br />
⎝ H<br />
2c<br />
This can be plotted via Maple.<br />
Problem 12.<br />
) 2<br />
⎞<br />
⎠<br />
1<br />
3<br />
+<br />
<br />
<br />
(<br />
± ı<br />
b(T )<br />
<br />
3c<br />
⎛<br />
3<br />
⎝ H<br />
2c −<br />
) 3<br />
z1 = 2ℜs+<br />
z2 = −ℜs+ + √ 3ℑs+<br />
z3 = −ℜs+ − √ 3ℑs+<br />
Consider the following model of a tri-critical point:<br />
4<br />
√ (b(T )<br />
3c<br />
( ) ⎞ 2<br />
H<br />
<br />
+ ⎠<br />
2c<br />
f(m, T, p) = 1<br />
2 a(T, p)m2 + 1<br />
4 b(T, p)m4 + 1<br />
6 cm6<br />
where c is a positive constant. The functions a and b are given by<br />
1<br />
3<br />
) 3<br />
+<br />
( H<br />
2c<br />
) 2<br />
⎞<br />
⎠<br />
1<br />
3
364 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
T − Tc p − pc<br />
a(T, p) = A + B<br />
Tc pc<br />
T − Tc p − pc<br />
b(T, p) = C + D<br />
Tc pc<br />
Depending on the p<strong>at</strong>h in p-T space, along p = f(T ) with pc = f(Tc), the<br />
behavior of the response functions and the order parameter takes on a different<br />
form. For example, the behavior of the magnetiz<strong>at</strong>ion m(p, T ) always varies<br />
like (Tc − T ) β , but the value of β depends on the p<strong>at</strong>h taken. The two distinct<br />
regions with different values of β are called critical and tri-critical.<br />
(A) Calcul<strong>at</strong>e the β this system.<br />
(B) Draw the phase diagram in p-T space.<br />
(C) Draw the critical and tri-critical regions in this phase diagram.<br />
(D) Give an explan<strong>at</strong>ion of these results, write it in such a way th<strong>at</strong> a new,<br />
incoming gradu<strong>at</strong>e student could understand it.<br />
( )<br />
∂f<br />
(m, T, p) = m<br />
∂m T,p<br />
( a(T, p) + b(T, p)m 2 + cm 4) = 0<br />
leads to either m = 0 or m 2 = 1<br />
2c<br />
(<br />
−b(T, p) ± √ b2 )<br />
(T, p) − 4a(T, p)c . This<br />
requires th<strong>at</strong> b2 (T, p) − 4a(T, p)c > 0.<br />
( 2 ∂ f<br />
∂m2 )<br />
(m, T, p) = a(T, p) + 3b(T, p)m<br />
T,p<br />
2 + 5cm 4<br />
indic<strong>at</strong>es th<strong>at</strong> m = 0 is a minimum when a(T, p) > 0. If this is the case both<br />
non-zero solutions for m 2 (if they exist) are positive only if b(T, p) < 0, and<br />
starting <strong>at</strong> the origin m = 0 we have a sequence minimum, maximum, minimum<br />
in f(m). If a(T, p) > 0 and b(T, p) > 0 the only solution is m = 0. If, on the<br />
other hand , a(T, p) < 0 then one of the solutions for m 2 is neg<strong>at</strong>ive, hence only<br />
one remains, and the sequence is maximum, minimum in f(m).<br />
The energy <strong>at</strong> the value m 2 = 1<br />
2c<br />
where we used<br />
(<br />
−b(T, p) + √ b 2 (T, p) − 4a(T, p)c<br />
f(m, T, p) = 1<br />
12 m2 ( 4a(T, p) + b(T, p)m 2)<br />
1<br />
6 a(T, p)m2 + 1<br />
6 b(T, p)m4 + 1<br />
6 cm6 = 0<br />
from the zero of the first deriv<strong>at</strong>ive. This is neg<strong>at</strong>ive when<br />
4a(T, p) + b(T, p)m 2 < 0<br />
)<br />
is
C.4. SOLUTIONS FOR CHAPTER 4. 365<br />
(when m 2 > 0 or b(T, p) < 0 ), which gives<br />
or<br />
8a(T, p)c − b 2 (T, p) + b(T, p) √ b 2 (T, p) − 4a(T, p)c < 0<br />
b(T, p) √ b 2 (T, p) − 4a(T, p)c < b 2 (T, p) − 8a(T, p)c<br />
Since b(T, p) < 0 this is always true if b 2 (T, p) − 8a(T, p)c > 0, which is always<br />
true if a(T, p) < 0. On the other hand, if b 2 (T, p) − 8a(T, p)c < 0 we need (note<br />
the change from < to > !!)<br />
or<br />
or<br />
b 2 (T, p) ( b 2 (T, p) − 4a(T, p)c ) > ( b 2 (T, p) − 8a(T, p)c ) 2<br />
12b 2 (T, p)a(T, p)c > 64a 2 (T, p)c 2<br />
3b 2 (T, p)a(T, p) > 16a 2 (T, p)c<br />
which is never true when a(T, p) < 0, as expected. If a(T, p) > 0 this leads to<br />
3b 2 (T, p) > 16a(T, p)c<br />
If this is true, certainly 4b 2 (T, p) > 16a(T, p)c and the minimum does exist.<br />
Also, this extends smoothly into the region b 2 (T, p) > 8a(T, p)c.<br />
In summary, if a(T, p) < 0 we always have a non-zero magnetiz<strong>at</strong>ion, and no<br />
local minimum <strong>at</strong> m = 0. If a(T, p) > 0 and b(T, p) > 0 the only solution is<br />
m = 0. If b(T, p) > 0 the line a(T, p) = 0 gives the line of second order phase<br />
transitions. If b(T, p) < 0 the line 3b 2 (T, p) = 16a(T, p)c gives a line of first<br />
order phase transitions.<br />
The analysis is now easiest in reduced coordin<strong>at</strong>es, π = p−pc<br />
pc<br />
which gives<br />
a(T, p) = Aθ + Bπ<br />
b(T, p) = Cθ + Dπ<br />
But we can make it even easier, if we use as coordin<strong>at</strong>es<br />
x = Aθ + Bπ<br />
y = Cθ + Dπ<br />
and θ =<br />
T −Tc<br />
, Tc<br />
In these coordin<strong>at</strong>es we have for x < 0 a non-zero magnetiz<strong>at</strong>ion with m=0 a<br />
maximum. For x > 0 and y > 0 we have m=0, and hence the line x = 0, y > 0<br />
gives a line of second order phase transitions. For x > 0 and y < 0 we have a
366 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Figure C.2: Figure 1.<br />
first order phase transition <strong>at</strong> x = 3<br />
16cy2 . In the region y < 0 and 0 < x < 3<br />
m=0 is a local minimum. See figure 1.<br />
The non-zero magnetiz<strong>at</strong>ion is given by the solutions of<br />
m 2 = 1<br />
(<br />
−b(T, p) +<br />
2c<br />
√ b2 )<br />
(T, p) − 4a(T, p)c<br />
16c y2<br />
If we approach the tri-critical point along b = 0 and a ↑ 0 we have m2 (<br />
=<br />
1 + √ )<br />
4|a(T, p)|c . Along b = 0 we have Dπ = −Cθ and hence |a| = |A −<br />
2c<br />
BC<br />
D ||θ| which gives m2 ∝ √ |θ|, or β = 1<br />
4 . On the other hand, if we approach<br />
the tri-critical point along a = 0 , b ↑ 0 we have m2 = |b|<br />
1<br />
c . This gives β = 2 .<br />
It is easy to show th<strong>at</strong> only along this straight line the critical exponent is 1<br />
2 ,<br />
any other line will have the a-term in the square root domin<strong>at</strong>ing near the tricritical<br />
point. If we approach the critical point along a parabolic line of the<br />
form b2 = 4acQ, with Q < 0, then the moment will look like<br />
m 2 = 1<br />
(<br />
√ )<br />
2c<br />
−b(T, p) + |b|<br />
Q − 1<br />
Q<br />
and now the exponent is always 1<br />
2 . This is equivalent to approaching the tricritical<br />
point along a = 0 and hence the critical region is really the line a = 0<br />
for b < 0.<br />
Problem 13.<br />
Consider the following model Helmholtz free energy per unit volume:<br />
f(T ; m) = a(T ) + e T 0 −T<br />
T 0 e −m2<br />
+ ce m2
C.4. SOLUTIONS FOR CHAPTER 4. 367<br />
where a(T ) is an arbitrary function of temper<strong>at</strong>ure, and T0 and c are positive<br />
constants. Show th<strong>at</strong> this system undergoes a second order phase transition,<br />
calcul<strong>at</strong>e the critical temper<strong>at</strong>ure Tc, and find the critical exponent β.<br />
Which is zero if m = 0 or<br />
( )<br />
∂f<br />
(<br />
= 2m −e<br />
∂m T<br />
T0−T T0 e −m2<br />
+ ce m2)<br />
T0 − T<br />
T0<br />
e T 0 −T<br />
T 0 e −m2<br />
= ce m2<br />
− m 2 = ln(c) + m 2<br />
2m 2 = T0 − T0 ln(c) − T<br />
T0<br />
Hence if T > T0(1 − ln(c)) there is only one solution and the system has zero<br />
magnetiz<strong>at</strong>ion. If T < T0(1 − ln(c)) there are three solutions, and because the<br />
energy goes to infinity for infinite m the solutions now are minimum, maximum,<br />
minimum. Since m = 0 is the middle solutions, it is now a maximum. The<br />
magnetiz<strong>at</strong>ion is therefore in th<strong>at</strong> case:<br />
m = ±<br />
√<br />
T0 − T0 ln(c) − T<br />
2T0<br />
which goes continuously to zero. The critical temper<strong>at</strong>ure is<br />
and the critical exponent is β = 1<br />
2 .<br />
Problem 14.<br />
Tc = T0(1 − ln(c))<br />
A magnetic system has a spontaneous magnetiz<strong>at</strong>ion below a critical temper<strong>at</strong>ure<br />
Tc, given by m2 = m2 0 Tc−T<br />
. The system is in a st<strong>at</strong>e <strong>at</strong> temper<strong>at</strong>ure<br />
Tc<br />
Ti = 2Tc and magnetiz<strong>at</strong>ion mi = 1<br />
2m0. This system is cooled down to very low<br />
temper<strong>at</strong>ure, but <strong>at</strong> constant magnetiz<strong>at</strong>ion. Describe the st<strong>at</strong>e of the system<br />
<strong>at</strong> this very low temper<strong>at</strong>ure. Calcul<strong>at</strong>e relevant quantities as a function of<br />
temper<strong>at</strong>ure.<br />
Define a temper<strong>at</strong>ure Ts by<br />
m(Ts) = mi<br />
m 2 Tc − Ts<br />
0 =<br />
Tc<br />
1<br />
4 m20
368 APPENDIX C. SOLUTIONS TO SELECTED PROBLEMS.<br />
Tc − Ts = 1<br />
4 Tc<br />
Ts = 3<br />
4 Tc<br />
For T > Ts nothing special is happening, the system is uniform, and a magnetic<br />
field is required to produce the magnetiz<strong>at</strong>ion. At T = Ts the required magnetic<br />
field is zero. When T < Ts the system will phase separ<strong>at</strong>e, part of it will<br />
have a neg<strong>at</strong>ive magnetiz<strong>at</strong>ion. No field is required. If the fraction of neg<strong>at</strong>ive<br />
magnetiz<strong>at</strong>ion is c, we have<br />
√ √<br />
Tc − T<br />
Tc − T<br />
mi = c(−m0 ) + (1 − c)(m0 )<br />
Tc<br />
√<br />
1<br />
Tc − T<br />
= (1 − 2c)<br />
2 Tc<br />
√ √<br />
Tc − Ts<br />
Tc − T<br />
= (1 − 2c)<br />
Tc<br />
1 − 2c =<br />
√ Tc − Ts<br />
Tc − T<br />
Tc<br />
Tc