Statistical Mechanics - Physics at Oregon State University

A.2. SOLUTIONS FOR CHAPTER 2. 237
A.2 Solutions for chapter 2.
Problem 4.
Assume that the protons are labelled by i, with i running from 1 to N, and that
the spin of the proton is siµ. The energy of such a state is
i (−siµH). The
partition function is
which is
Z(T ) =
s1,···,sN
e 1
kT i siµH
Z(T ) = (Z1(T )) N
µH
µH
−
Z1(T ) = e kT +
+ e kT
The internal energy of the sample is U = −
i siµH, which is proportional to
the difference in population. Therefore the power absorbed is proportional to
U. We also have:
and hence
2 ∂
U = kT log(Z(T ))
∂T
U = kT 2 µH
µH
e+ kT − − e kT
N µH
µH
e + kT + e− kT
−µH
kT 2
When µH ≪ kT we replace the exponents by e x = 1 + x and get
µH µH
+ kT − (− kT U ≈ −NµH )
= −N
2
µ2H 2
kT
and the power is inversely proportional to T.
Problem 5.
(a) At low temperature Ē ≈ Nɛ1, all particles in ground state. At high temper-
N
ature all states are equally probable, and Ē ≈ 2 (ɛ1 + ɛ2). In thermodynamics
we have seen that
∂U
∂T → 0 for T → 0, because the entropy becomes zero.
V
Therefore the curve Ē(T ) starts with horizontal slope at T = 0 bends upwards,
bends down again and approaches the high temperature limit in an asymptotic
fashion. We expect the change from low to high temperature behavior to take
place near a temperature T0 given by ɛ2−ɛ1 = kT0, because now we have enough
thermal energy to make the transition from state 1 to state 2.