Statistical Mechanics - Physics at Oregon State University

256 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
this can be written in the form
G = −kBT
o
1 − k log(1 + λe B T ( p2z eB 1
2m + mc (j+ 2 )) )
If we have the free energy, we also have the partition function, of course.
or
Using N = −
G = −kBT 2 eBL2
2πc
G = −kBT 2 eBL2
2πc
G = −kBT V
∂G
∂µ
T,V,B
N = V
pz,j
∞
L
2π −∞
eB
2π22 ∞
c −∞
eB
2π 2 2 c
1 − k log(1 + λe B T ( p2 z eB 1
2m + mc (j+ 2 )) )
dp
j
dp
and M = − ∂G
∂B
j
∞
dp
−∞
j
1 p2 eB 1
− k log(1 + λe B T ( 2m + mc (j+ 2 )) )
1 p2 eB 1
− k log(1 + λe B T ( 2m + mc (j+ 2 )) )
T,µ,V
we find
1
λ−1e 1 p2 eB 1
kB T ( 2m + mc (j+ 2 )) + 1
which gives in the limit λ → 0 (for high temperature):
eB
N ≈ λV
2π22 ∞
dp
c
1 p2
− k e B T ( 2m
or
N ≈ λV
N ≈ λV
Check: If B goes to zero we get
−∞
eB
2π22 ∞
eB
p2
2mck e− B T
− 2mk dpe B T
c −∞
eB
2π22 eB
∞
2mck e− B T 2mkBT
c −∞
N ≈ λV
j
eB 1
+ mc (j+ 2 ))
j
dxe −x2
1
4π23
3 ∞
2mkBT dxe
−∞
−x2
which is equal to the old result.
We can also take the small λ limit in the free energy and get
eB
G = −kBT V
2π22 ∞
dp
c
1 p2
− k λe B T ( 2m
−∞
j
e −j eB
mck B T
1
eB − mck 1 − e B T
eB 1
+ mc (j+ 2 ))
Comparing with the formula for N we see G = −NkBT , which should not come
as a surprise, since G = −pV and we are in the limit of an ideal gas!! Therefore: