276 Grand canonical ensemble
ζ∂
∂ζζ∂
∂ζlnZ(ζ,V,T). (6.6.2)Using eqn. (6.4.25), this becomes
ζ∂
∂ζζ∂
∂ζlnZ(ζ,V,T) =ζ∂
∂ζ1
Z
∑∞
N=0NζNQ(N,V,T)=
1
Z
∑∞
N=0N^2 ζNQ(N,V,T)−1
Z^2
[∞
∑
N=0NζNQ(N,V,T)] 2
=〈N^2 〉−〈N〉^2. (6.6.3)
Thus, we have
(∆N)^2 =ζ∂
∂ζζ∂
∂ζlnZ(ζ,V,T). (6.6.4)Expressing eqn. (6.6.4) as derivatives ofZ(μ,V,T) with respect toμ, we obtain
(∆N)
2
= (kT)^2∂^2
∂μ^2
lnZ(μ,V,T) = (kT)^2∂^2
∂μ^2PV
kT. (6.6.5)
Sinceμ,V, andTare the independent variables in the ensemble, only the pressure in
the above expression depends onμ, and we can write
(∆N)^2 =kTV∂^2 P
∂μ^2. (6.6.6)
Therefore, computing the particle number fluctuations amounts to computing the
second derivative of the pressure with respect to chemical potential. This is a rather
nontrivial bit of thermodynamics, which can be carried out in a variety of ways. One
approach is the following: LetA(N,V,T) be the canonical Helmholtz free energy at a
particular value ofN. Recall that the pressure can be obtained fromA(N,V,T) via
P=−
(
∂A
∂V
)
. (6.6.7)
SinceA(N,V,T) is an extensive quantity, and we want to make theNdependence
in the analysis as explicit as possible, we define an intensive Helmholtz free energy
a(v,T) by
a(v,T) =1
N
A
(
N,
V
N
,T
)
, (6.6.8)
wherev=V/Nis the volume per particle anda(v,T) is clearly the Helmholtz free
energy per particle. Then,
P=−N
∂a
∂v∂v
∂V=−N
∂a
∂v1
N
=−
∂a
∂v