# Fluctuations in the grand canonical ensemble

As we have anticipated we expect that, similarly to what happens in the canonical ensemble, the fluctuations of the energy and number of particles around their mean values vanish in the thermodynamic limit so that the grand canonical ensemble is indeed equivalent to the canonical one.

The fluctuations of the energy around ${\displaystyle \left\langle {\mathcal {H}}\right\rangle }$ are computed exactly in the same way as we have done in The canonical and the microcanonical ensemble, and we get to the same result. We therefore now focus on the fluctuations of the number of particles around ${\displaystyle \left\langle N\right\rangle }$:

${\displaystyle \sigma _{N}^{2}=\left\langle N^{2}\right\rangle -\left\langle N\right\rangle ^{2}={\frac {1}{\beta ^{2}}}{\frac {\partial ^{2}\ln {\mathcal {Z}}}{\partial \mu ^{2}}}={\frac {1}{\beta ^{2}}}{\frac {\partial ^{2}}{\partial \mu ^{2}}}[-\beta \Phi (T,V,\mu )]}$
Now, from the definition ${\displaystyle \Phi ={\overline {E}}-TS({\overline {E}},V,{\overline {N}})-\mu {\overline {N}}}$ of grand potential we have (considering that ${\displaystyle \partial S/\partial V=P/T}$ and that, again, there are some derivatives that vanish because ${\displaystyle {\overline {E}}}$ and ${\displaystyle {\overline {N}}}$ define a minimum) that:
${\displaystyle {\frac {\partial \Phi }{\partial V}}=-P}$
Furthermore, ${\displaystyle \Phi }$ is extensive:
${\displaystyle \Phi (T,\lambda V,\mu )=\lambda \Phi (T,V,\mu )}$
and setting ${\displaystyle \lambda =1/V}$ we get ${\displaystyle \Phi (T,V,\mu )=V\Phi (T,1,\mu )}$: the grand potential is proportional to the volume. Therefore, ${\displaystyle \partial \Phi (T,V,\mu )/\partial V=\Phi (T,1,\mu )=-P}$ and so:
${\displaystyle \Phi (T,V,\mu )=V\Phi (T,1,\mu )=-VP}$
Thus:
${\displaystyle \sigma _{N}^{2}=-{\frac {1}{\beta }}{\frac {\partial ^{2}}{\partial \mu ^{2}}}\Phi (T,V,\mu )={\frac {V}{\beta }}{\frac {\partial ^{2}}{\partial \mu ^{2}}}P(T,\mu )}$
We must therefore find a way to express ${\displaystyle \partial ^{2}P/\partial \mu ^{2}}$. We define ${\displaystyle v(T,\mu )=V/{\overline {N}}}$ and ${\displaystyle f(T,v)=F/{\overline {N}}}$, and then write:
${\displaystyle {\frac {\partial P}{\partial \mu }}={\frac {\partial P/\partial v}{\partial \mu /\partial v}}}$
Since ${\displaystyle \Phi =-PV=F-\mu {\overline {N}}}$, we have:
${\displaystyle f{\overline {N}}-\mu {\overline {N}}=-PV\quad \Rightarrow \quad f(T,v)=\mu -vP(T,\mu )}$
and since:
${\displaystyle P=-{\frac {\partial F}{\partial V}}=-{\frac {\partial f}{\partial v}}}$
then:
${\displaystyle f(T,v)=\mu +v{\frac {\partial f}{\partial v}}\quad \Rightarrow \quad \mu =f-v{\frac {\partial f}{\partial v}}}$
We therefore have:
${\displaystyle {\frac {\partial P}{\partial v}}=-{\frac {\partial ^{2}f}{\partial v^{2}}}\quad \qquad {\frac {\partial \mu }{\partial v}}=-v{\frac {\partial ^{2}f}{\partial v^{2}}}}$
and substituting in ${\textstyle \partial P/\partial \mu }$, this leads to:
${\displaystyle {\frac {\partial P}{\partial \mu }}={\frac {1}{v}}}$
If we now further derive ${\displaystyle P}$ with respect to ${\displaystyle \mu }$:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}=-{\frac {1}{v^{2}}}{\frac {\partial v}{\partial \mu }}=-{\frac {1}{v^{2}}}{\frac {1}{\partial \mu /\partial v}}}$
and from the previous expression of ${\textstyle \partial \mu /\partial v}$ we have:
${\displaystyle {\frac {\partial \mu }{\partial v}}=-v{\frac {\partial ^{2}f}{\partial v^{2}}}=v{\frac {\partial P}{\partial v}}}$
so that:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}=-{\frac {1}{v^{2}}}\cdot {\frac {1}{v\cdot \partial P/\partial v}}=-\left(v^{3}{\frac {\partial P}{\partial v}}\right)^{-1}}$
We have done all these intricate computations because now we can use the definition of isothermal compressibility (see Response functions)[1]:
${\displaystyle K_{T}=-{\frac {1}{V}}{\frac {\partial V}{\partial P}}=-{\frac {1}{V}}{\frac {1}{\partial P/\partial V}}}$
(which as we can see is an intensive quantity) to write:
${\displaystyle {\frac {\partial ^{2}P}{\partial \mu ^{2}}}={\frac {K_{T}}{v^{2}}}}$
This way the variance of the number of particles can be written as:
${\displaystyle \sigma _{N}^{2}={\frac {V}{\beta }}\cdot {\frac {\partial ^{2}P}{\partial \mu ^{2}}}={\frac {V}{\beta }}\cdot {\frac {K_{T}}{v^{2}}}=N{\frac {K_{T}}{\beta v}}}$
(which is of course positive since ${\displaystyle K_{T}>0}$). We therefore see that:
${\displaystyle \sigma _{N}^{2}\propto N\quad \Rightarrow \quad \sigma _{N}\propto {\sqrt {N}}}$
and so the relative fluctuation of the number of particles is:
${\displaystyle {\frac {\sigma _{N}}{N}}\propto {\frac {1}{\sqrt {N}}}}$
which is a result analogous to what we have seen for the energy.

Therefore, as we expected, the fluctuations of the number of particles of a system in the grand canonical ensemble are negligible in the thermodynamic limit; this ultimately means that the grand canonical ensemble is equivalent to the canonical one.

1. This is of course equal to what we have found, because ${\displaystyle v=V/{\overline {N}}}$.