Particle number fluctuations 277From eqn. (6.6.9), it follows that
∂P
∂μ=
∂P
∂v∂v
∂μ=−
∂^2 a
∂v^2∂v
∂μ. (6.6.10)
We can obtain an expression for∂μ/∂vby
μ=∂A
∂N
=a(v,T) +N∂a
∂v∂v
∂N=a(v,T)−v∂a
∂v, (6.6.11)
so that
∂μ
∂v=
∂a
∂v−
∂a
∂v−v
∂^2 a
∂v^2=−v
∂^2 a
∂v^2. (6.6.12)
Substituting this result into eqn. (6.6.10) gives
∂P
∂μ=−
∂^2 a
∂v^2[
∂μ
∂v]− 1
=
∂^2 a
∂v^2[
v∂^2 a
∂v^2]− 1
=
1
v. (6.6.13)
Differentiating eqn. (6.6.13) once again with respect toμgives
∂^2 P
∂μ^2=−
1
v^2∂v
∂μ=
1
v^2[
v∂^2 a
∂v^2]− 1
=−
1
v^3 ∂P/∂v. (6.6.14)
Now, recall that the isothermal compressibility is given by
κT=−1
V
∂V
∂P
=−
1
v∂v
∂P=−
1
v∂P/∂v(6.6.15)
and is an intensive quantity. It is clear from eqn. (6.6.14) that∂^2 P/∂μ^2 can be ex-
pressed in terms ofκTas
∂^2 P
∂μ^2
=
1
v^2κT, (6.6.16)so that
(∆N)^2 =kT〈N〉v1
v^2κT=〈N〉kTκT
v, (6.6.17)
where the specific value ofNhas been replaced by its average value〈N〉in the grand
canonical ensemble. The relative fluctuations in particle number cannow be computed
from