Statistical Mechanics - Physics at Oregon State University

62 CHAPTER 3. VARIABLE NUMBER OF PARTICLES
and the grand potential is
Ω(T, µ, V ) = −kBT zZ1 = −kBT e βµ V nQ(T )
(3.103)
The average number of particles is − , leading to
∂Ω
∂µ
T,V
N = e βµ V nQ(T ) (3.104)
which is consistent with 3.101.
We can also check the Gibbs-Duhem relation. We find
T S − pV + µN = NkBT
log( nQ(T
) 5
+ −
n 2
NkBT
V
n
V + kBT log(
nQ(T ) )N
(3.105)
T S − pV + µN = 3
2 NkBT (3.106)
which is equal to U indeed.
In the previous formulas we have assumed that we can replace the summations
by integrations. Is that still valid for N particles? Could anything go
wrong? In replacing the sum by an integral we essentially wrote
Z1 = I(1 + ε) (3.107)
where the constant I represents the result obtained using the integral and ε
is the relative error. This error is proportional to α, which is proportional to
1 − (V nQ(T )) 3 . We can replace the summation by an integral is V nQ(T ) ≫ 1.
For N particles we need to calculate
Z N 1 = I N (1 + ε) N ≈ I N (1 + Nε) (3.108)
as long as the error is small, or as long as Nɛ ≪ 1. This implies
or
N 3 ≪ V nQ(T ) (3.109)
n ≪ nQ(T )N −2
(3.110)
and in the thermodynamic limit this is never obeyed!! That is strange. The
formulas that resulted from our partition function are clearly correct, but the
error in the partition function is not small! What went wrong? The answer is
that we are not interested in the error in the partition function itself, but in the
error in the Helmholtz free energy, because that is the relevant quantity for all
physics. In the Helmholtz free energy we have:
F = Fideal − NkBT log(1 + ε) (3.111)