# How to calculate the particle number fluctuation in the Grand Canonical Ensemble

I'm trying to show that, in the Grand Canonical Ensemble, the particle number fluctuation is given by

$$\frac{(\Delta N)^2}{\langle N\rangle^{2}} = \frac{\kappa_{T}}{\beta V},$$

where $\Delta N$ is the particle number dispersion and $\langle N \rangle$ is the average number pf particles in the system. $\beta$ is the thermodynamic beta (which is just a constant in this problem), $V$ is the system's volume and

$$\kappa_{T} = -\frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_{T,\langle N \rangle}$$

is the isothermic compressibility, being $P$ the pressure.

Up to now I've managed to show that

\begin{eqnarray} (\Delta N)^{2} &=& \frac{1}{\beta}\left( \frac{\partial \langle N\rangle}{\partial \mu} \right)_{T,V}\\ \end{eqnarray}

where $\mu$ is the chemical potential. This is what I did next:

\begin{eqnarray} \left(\frac{\partial \langle N \rangle}{\partial \mu}\right)_{T,V} &=& \left(\frac{\partial V}{\partial \mu}\right)_{T,\langle N \rangle}\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu}\\ &=& \left(\frac{\partial V}{\partial P}\right)_{T,\langle N \rangle}\left(\frac{\partial P}{\partial \mu}\right)_{T,\langle N \rangle}\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu}\\ &=&-\kappa_{T} V\left(\frac{\partial P}{\partial \mu}\right)_{T,\langle N \rangle}\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu}. \end{eqnarray}

Now, using the Maxwell relations from the Grand potential we can show that

$$\left(\frac{\partial P}{\partial \mu}\right)_{T,\langle N \rangle}=\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu},$$ and assuming a constant density,

$$\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu} = \frac{\langle N \rangle}{V},$$

we finally arrive at

$$\frac{(\Delta N)^2}{\langle N\rangle^{2}} = -\frac{\kappa_{T}}{\beta V},$$

which, as you can see, has the wrong sign. I've found a similar derivation online where the author says that the chain rule I used is actually

\begin{eqnarray} \left(\frac{\partial \langle N \rangle}{\partial \mu}\right)_{T,V} &=& -\left(\frac{\partial V}{\partial \mu}\right)_{T,\langle N \rangle}\left(\frac{\partial \langle N \rangle}{\partial V}\right)_{T,\mu} \end{eqnarray}

and I cannot understand where did she get this minus sign from in the chain rule.

Do you know what I'm doing wrong?

Thank you very much.

Let's look at your formulae. Once one gets $$(\partial{\left}/\partial{\mu})_{T,V}$$, $$\left$$ should be understood as a function of $$(T,V,\mu)$$. In order to connect this derivative to a derivative at fixed $$\left$$, the best way to avoid confusion is to start from the differential of $$\left$$: $$d\left = \left(\frac{\partial{\left}}{\partial{\mu}}\right)_{T,V}d\mu + \left(\frac{\partial{\left}}{\partial{V}}\right)_{T,\mu}dV + \left(\frac{\partial{\left}}{\partial{T}}\right)_{\mu,V}dT.$$ So we get: $$\left(\frac{\partial{\left}}{\partial{\mu}}\right)_{T,V}d\mu + \left(\frac{\partial{\left}}{\partial{V}}\right)_{T,\mu}dV = 0,$$ valid for any finite variation of $$\mu$$ and $$V$$ at fixed $$\left$$ and $$T$$. Thus, we can get easily $$\left(\frac{\partial{V}}{\partial{\mu}}\right)_{T,\left} = -\left(\frac{\partial{\left}}{\partial{\mu}}\right)_{T,V} \left(\frac{\partial{V}}{\partial{\left}}\right)_{\mu,T},$$ which looks like your "chain rule", but has the right sign.
A different way to get the same result follows a more "thermodynamic" path. Let's start with the inverse of the quantity you were manipulating, i.e. $$(\partial{\mu}/\partial{N})_{V,T}$$. By using the Gibbs-Duhem relation: $$d\mu = -\frac{S}{N}dT + \frac{V}{N}dP,$$ at fixed T we have: $$\left(\frac{S}{N}\right)_{V,T}= \frac{V}{N}\left(\frac{\partial{P}}{\partial{N}}\right)_{V,T}= -\frac{V}{N}\left(\frac{\partial{\mu}}{\partial{V}}\right)_{N,T}= -\left(\frac{V}{N}\right)^2\left(\frac{\partial{P}}{\partial{V}}\right)_{V,T}$$ (Maxwell relations have been used twice to go from the second to the third and from the third to the forth formula) from which one can easily extract the relation with the isothermal compressibility.