Statistical Mechanics - Physics at Oregon State University

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58 CHAPTER 3. VARIABLE NUMBER OF PARTICLES variables U = T S + µN (Gibbs-Duhem), and hence Ω = 0. Also, T = ∂U ∂S N which is zero according to the formula U = Nω. What went wrong? There are two problems. First of all, in a T,S,µ ,N system we cannot choose T and µ as independent variables in thermodynamics. We have to introduce an extensive parameter. Second, we have to take the thermodynamic limit. The formula for N can be inverted to yield a formula for µ: µ = ω − kBT log(1 + 1 ) (3.82) N which tells us that for a given value of N the chemical potential and the temperature are related. In the thermodynamic limit N → ∞ we find µ = ω. For this value of the chemical potential the series for the grand partition function does not converge anymore, hence we have to perform all calculations for finite values of N and take the limit only in the end. The grand potential per particle is Ω(T, N) N = −kBT log(N + 1) N (3.83) and this quantity is zero in the thermodynamic limit. Hence we retrieved the result anticipated from thermodynamics. We also find S(T, N) = kB log(N + 1) + NkB log(1 + 1 ) (3.84) N which shows that in the thermodynamic limit S N = 0. In other words, temperature is not well-defined for this system. This problem is independent of the thermodynamic limit, but is inherent to our simple model where we use only one quantum number for both energy and number of particles. For a given number of particles the energy is fixed! Hence we really only have two independent variables in our system and thermodynamically our system is completely characterized once we know the value of N. In realistic physical models the parameters N and U have to be independent. An example is found in the next section. 3.7 Ideal gas in first approximation. The ideal gas is the first realistic and complete model we encounter. There are three sets of thermodynamic variables. We have volume and pressure, entropy and temperature, and chemical potential and number of particles. Therefore, we can choose T and µ independently, after which the value of p is fixed. In the original experiments with a gas as a function of pressure, volume, and temperature it was noted that the relations between these thermodynamic quantities were very similar for many different gases. Therefore, one postulated the so-called ideal gas laws and tried to explain why a real gas shows small (or large) differences from these ideal gas laws. Of course, we now know that the , and that most important parameter for ideal gas behavior is the density n = N V
3.7. IDEAL GAS IN FIRST APPROXIMATION. 59 the ideal gas laws are simply the limits of the correct functional dependencies to zero density. It is just fortunate that almost every gas is close to ideal, otherwise the development of thermal physics would have been completely different! The quantum mechanical model of an ideal gas is simple. Take a box of dimensions L × L × L, and put one particle in that box. If L is large, this is certainly a low density approximation. The potential for this particle is defined as 0 inside the box and ∞ outside. The energy levels of this particle are given by ɛ(nx, ny, nz) = 2 π 2 (n 2M L 2 x + n 2 y + n 2 z) (3.85) with nx ,ny, and nz taking all values 1,2,3,..... The partition function is easy to calculate in this case nx 2 − 2Mk e B T ( π L) 2 n 2 x Z1 = ny nxnynz e −βɛ(nxnynz) = 2 − 2Mk e B T ( π L) 2 n 2 y nz 2 − 2Mk e B T ( π L) 2 n 2 z (3.86) Define α2 = 2 π 2 (2ML2kBT ) and realize that the three summations are independent. This gives ∞ Z1 = e −α2n 2 n=1 3 (3.87) We cannot evaluate this sum, but if the values of α are very small, we can use a standard trick to approximate the sum. Hence, what are the values of α? If we take ≈ 10 −34 Js, M ≈ 5 × 10 −26 kg (for A around 50), and L = 1m, we find that α 2 ≈ 10−38 J kBT , and α ≪ 1 translates into kBT ≫ 10 −38 J. With kB ≈ 10 −23 JK −1 this translates into T ≫ 10 −15 K, which from an experimental point of view means always. Keep in mind, though, that one particle per m 3 is an impossibly low density! In order to show how we calculate the sum, we define xn = αn and ∆x = α and we obtain ∞ e −α2n 2 n=1 = 1 α ∞ n=1 e −x2n∆x (3.88) In the limit that ∆x is very small, numerical analysis shows that ∞ n=1 e −x2n∆x ≈ ∞ e −x2 0 dx − 1 1 − ∆x + O e ∆x 2 (3.89) where the integral is equal to 1√ 2 π. The term linear in ∆x is due to the fact that in a trapezoidal sum rule we need to include the first (n=0) term with a factor
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Statistical Mechanics Henri J.F. Ja
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IV CONTENTS 4.3 Gas of poly-atomic
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VI CONTENTS
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VIII INTRODUCTION Introduction. The
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X INTRODUCTION In chapter nine we d
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2 CHAPTER 1. FOUNDATION OF STATISTI
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108 CHAPTER 5. FERMIONS AND BOSONS
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120 CHAPTER 6. DENSITY MATRIX FORMA
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140 CHAPTER 7. CLASSICAL STATISTICA
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158 CHAPTER 8. MEAN FIELD THEORY: C
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192 CHAPTER 9. GENERAL METHODS: CRI
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230 APPENDIX A. SOLUTIONS TO SELECT
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