Statistical Mechanics - Physics at Oregon State University

252 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
Using the fact that B is very small we find up to first order:
N = 2I(T, µ)
M = 2µ 2
∂I
0B (T, µ)
∂µ T
χ = 2µ 2
∂I
0 (T, µ)
∂µ T
∂ log I
(T, µ)
∂µ
χ = Nµ 2 0
At T = 0 the integral is easy, since µ(T = 0) = ɛF :
with ɛF = p2F
2m . Hence
−3 4π
I(T = 0, ɛF ) = V
3 p3 −3 4π
F = V
3 (2mɛF ) 3
2
For small T we use µ = ɛF − π2
12
χ(T = 0) = Nµ 2 0
k 2 2
BT ɛF
N = 2I(T, ɛF − π2
12
Therefore, we have at low temperature:
I(T, ɛF − π2
12
k 2 B
ɛF
T 2
and taking the derivative with ɛF :
∂I
(T, ɛF −
∂µ
π2 k
12
2 2
BT )(1 + π2
12
which gives
or
T
ɛF
χ(T )(1 + π2
12
T
3 1
2
ɛF
which was obtained from:
k 2 B
2 T
)
ɛF
−3 4π
) = I(T = 0, ɛF ) = V
3 (2mɛF ) 3
2
k2 B
ɛ2 F
k2 B
ɛ2 F
2 T −3 4π
) = V
3 (2mɛF ) 3 3 1
2
2
2 T
) = χ(T = 0)
χ(T ) = χ(T = 0)(1 − π2
12
Large T, looks like Maxwell-Boltzmann
−3
N = V
s
k2 B
ɛ2 F
2 T
)
d 3 pe 1 p2
kB T (µ− 2m +sµ0B)
ɛF