Statistical Mechanics - Physics at Oregon State University

A.5. SOLUTIONS FOR CHAPTER 5. 253
M = V −3
µ0
M
N
s
d 3 pse 1 p2
kB T (µ− 2m +sµ0B)
sinh(
= µ0
µ0B
kBT )
cosh( µ0B
kBT )
For T → ∞ this is zero. That makes sense, all states are equally probable,
independent of B, and hence there is no M as a function of B. Therefore χ = 0.
The next order term is:
Problem 5.
M
N
µ0B
= µ0
kBT
χ = Nµ2 0
kBT
The virial expansion is given by p
kT = ∞ j=1 Bj(T )
N j
V with B1(T ) = 1. Find
B2(T ) for non-interacting Fermions in a box.
Using
Ω = −2V kBT λ −3
T f 5
2 (λ)
N = 2V λ −3
T f 3
2 (λ)
and the fact that low density corresponds to small λ we get:
The pressure is still simple:
Ω ≈ −2V kBT λ −3
T (λ − λ2 5 −
2 2 )
N ≈ 2V λ −3
T (λ − λ2 3 −
2 2 )
p = − Ω
V ≈ 2kBT λ −3
T (λ − λ2 5 −
2 2 )
We need to solve for λ as a function of N. Using the fact that the density n = N
V
is small:
n ≈ 2λ −3
3
T λ(1 − λ2− 2 )
n(1 + λ2 − 3
2 ) ≈ 2λ −3
T λ