Statistical Mechanics - Physics at Oregon State University

112 CHAPTER 5. FERMIONS AND BOSONS
From the formula for the chemical potential we obtain:
N
λ(T, N, V ) ≈
N + 2S + 1 e
2
π
2MkB T ( L )23 (5.125)
The last formula shows that limits cannot be interchanged without penalties:
and hence
lim λ(T, N, V ) = ∞ (5.126)
T →0
lim λ(T, N, V ) =
V →∞
N
→ 1 (5.127)
N + 2S + 1
lim
T →0 lim λ(T, N, V ) = lim
V →∞ V →∞ lim λ(T, N, V ) (5.128)
T →0
But we need to be careful. Since we required that 2
2M
take the limit V → ∞ first without violating this condition!
Grand energy at low temperatures.
( π
L )2 3 ≫ kBT we cannot
Since we know that the first term in the series for N, and hence in the series
for Ω, can be the dominant term and be almost equal to the total sum, we now
isolate that term. We could improve on this procedure by isolating more terms,
but that will not change the qualitative picture. Hence we write
Ω(T, µ, V )
V
= (2S + 1)kBT
(2S + 1)kBT
V
V
′
nx,ny,nz
log 1 − λe
ɛ(1,1,1)
− kB T
+
log(Znx,ny,nz(T, µ, V )) (5.129)
Now we replace the second term by the integral as before. The lower limit is
not strictly zero, but if V is large enough we can replace it by zero. Hence we
find
Ω(T, µ, V )
V
= (2S + 1)kBT
V
log 1 − λe − 32π 2
2MkB T L2
−
(2S + 1)kBT nQ(T )g 5
2 (λ) + error(T, µ, V ) (5.130)
where the error term is due to replacing the sum by the integral. If the temperature
is large and the volume is large, the limit of the first term in Ω is
zero, the second term dominates, and the error is small compared to the second
term. If the temperature is very small, and the volume is large but finite, the
error term is large compared to the second term, but small compared to the first
term! The error is largest in the transition region between these two cases, and
a description could be improved by adding more terms. In the thermodynamic