Statistical Mechanics - Physics at Oregon State University

5.1. FERMIONS IN A BOX. 99
d
dz
∞
n=1
z n
n α (−1)n+1 =
∞
n=1
and hence we have for |z| < 1 :
n zn−1
nα (−1)n+1 = 1
z
∞
n=1
zn (−1)n+1
nα−1 (5.45)
d
dz fα(z) = 1
z fα−1(z) (5.46)
Analytical continuation then leads to the general result that this relation is valid
everywhere in the complex plane. As a result, we find:
N(T, µ, V ) = (2S + 1)
spin
degeneracy
V
simple
volume
dependence
or, using the density and the quantum density:
n
nQ(T )
λ −3
T
volume
scale
f 3
2 (λ)
density
effects
(5.47)
= (2S + 1)f 3 (λ) (5.48)
2
which clearly shows that the effects of the chemical potential only show up
through our special function f 3
2 (λ).
We can now use this equation to find µ(T, V, N). That follows the general
procedure we have outlined before. Can we always invert the equation? In thermodynamics
we found that > 0. This allows us to find the chemical
∂N
∂µ
T,V
potential as a function of N. Does our current result obey the same relation?
It should, of course, if we did not make any mistakes. We have
1
nQ(T )
∂n
∂µ T,V
=
1
nQ(T )
∂n
∂λ T,V
λ
kBT
= 2S + 1
kBT
f 1 (λ) (5.49)
2
Is this a positive quantity? It is better to answer this question by looking at
the integral forms. We have (using fα−1 = z d
dz fα: :
or
4
f 5 (λ) = √
2 π
∞
0
4
f 3 (λ) = √
2 π
4
f 3 (λ) = √
2 π
∞
0
x 2 dx log(1 + λe −x2
) (5.50)
∞
x 2 dx
0
x 2 dx λe−x2
1 + λe −x2
e −x2
λ −1 + e −x2
(5.51)
(5.52)
The last form already makes it clear that in λ increases the denominator decreases
and hence the integral increases. We also find from this expression: