# 6.4: Particle Number Fluctuations


In the grand canonical ensemble, the particle number $$N$$ is not constant. It is, therefore, instructive to calculate the fluctuation in this quantity. As usual, this is defined to be

$\Delta N = \sqrt{\langle N^2 \rangle - \langle N \rangle^2} \nonumber$

Note that

\begin{align*} \zeta{\partial \over \partial \zeta}\zeta {\partial \over \partial \zeta}\ln {\cal Z}(\zeta,V,T) &= {1 \over {\cal Z}}\sum_{N=0}^{\infty}N^2 \zeta^N Q(N,V,T) -{1 \over {\cal Z}^2} \left[\sum_{N=0}^{\infty} N \zeta^N Q(N,V,T)\right]^2 \\[4pt] &= \langle N^2 \rangle - \langle N \rangle^2 \end{align*}

Thus,

\begin{align*} \left(\Delta N\right)^2 &=\zeta{\partial \over \partial \zeta} \zeta {\partial \over \partial \zeta} \ln {\cal Z} (\zeta, V, T) \\[4pt] &= ({KT}^2){\partial^2 \over \partial \mu^2}\ln {\cal Z}(\mu,V,T) \\[4pt] &= kTV{\partial^2 P \over \partial \mu^2} \end{align*}

In order to calculate this derivative, it is useful to introduce the Helmholtz free energy per particle defined as follows:

$a(v,T) = {1 \over N}A(N,V,T) \nonumber$

where $$v={V \over N} = {1 \over \rho}$$ is the volume per particle. The chemical potential is defined by

\begin{align*} \mu &= {\partial A \over \partial N} \\[4pt] &=a(v,T) + N{\partial a \over \partial v}{\partial v \over \partial N} \\[4pt] &= a(v,T) - v{\partial a \over \partial v} \end{align*}

Similarly, the pressure is given by

$P = -{\partial A \over \partial V} = -N{\partial a \over \partial v}{\partial v \over \partial V} = -{\partial a \over \partial v} \nonumber$

Also,

${\partial \mu \over \partial v} = -v{\partial^2 a \over \partial v^2} \nonumber$

Therefore,

\begin{align*} \dfrac{\partial P}{\partial \mu} &= {\partial P \over \partial v}{\partial v \over \partial \mu} \\[4pt] &= {\partial^2 a \over \partial v^2} \left[v{\partial^2 a \over \partial v^2}\right]^{-1} \\[4pt] &= {1 \over v} \end{align*}

and

\begin{align*} \dfrac{\partial^2 P}{\partial \mu^2} &= {\partial \over \partial v}{\partial P \over \partial \mu}{\partial v \over \partial \mu} \\[4pt] &= {1 \over v^2} \left[ v {\partial^2 a \over \partial v^2}\right]^{-1} \\[4pt] &= -{1 \over v^3 \partial P/\partial v} \end{align*}

But recall the definition of the isothermal compressibility:

$\kappa_T = -{1 \over V}{\partial V \over \partial P}=-{1 \over v \partial p/\partial v} \nonumber$

Thus,

${\partial^2 P \over \partial \mu^2} = {1 \over v^2}\kappa_T \nonumber$

and

$\Delta N = \sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \nonumber$

and the relative fluctuation is given by

${\Delta N \over N} = {1 \over \langle N \rangle}\sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \sim {1 \over \sqrt {\langle N \rangle }}\rightarrow 0\;\;{as}\langle N \rangle \rightarrow \infty \nonumber$

Therefore, in the thermodynamic limit, the particle number fluctuations vanish, and the grand canonical ensemble is equivalent to the canonical ensemble.

This page titled 6.4: Particle Number Fluctuations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.