Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is

$\stackrel{—}{N}=\frac{kT}{Z}\frac{\partial Z}{\partial \mu }$

where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is

$\overline{N^2}=\frac{{\left(kT\right)}^2}{Z}\frac{{\partial }^{2}Z}{\partial {\mu }^2}$

Use these results to show that the standard deviation of $N$is

${\sigma }_N=\sqrt{kT\left(\partial \stackrel{—}{N}/\partial \mu \right)},$

in analogy with Problem$6.18$Finally, apply this formula to an ideal gas, to obtain a simple expression for${\sigma }_N$in terms of$\overline{N}$Discuss your result briefly.

We have to given the average number of particles in the system is$\stackrel{—}{N}=\frac{kT}{Z}\frac{\partial Z}{\partial \mu }$, the mean square number of particles is $\overline{N^2}=\frac{{\left(kT\right)}^2}{Z}\frac{{\partial }^{2}Z}{\partial {\mu }^2}$and the standard deviation of $N$is${\sigma }_N=\sqrt{kT\left(\partial \stackrel{—}{N}/\partial \mu \right)}$.

The grand partition function equals the sum over the Gibbs factors, that is:

$Z=\sum _s{e}^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}$

take the partial derivative of the partition function with respect to , so :

$\begin{array}{c}\frac{\partial Z}{\partial \mu }=\frac{1}{kT}\sum _{s}N\left(s\right)e^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}\\ \sum _{s}N\left(s\right)e^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}=kT\left(\frac{\partial Z}{\partial \mu }\right)\end{array}$

dividing both sides by the grand partition function to get:

$\frac{1}{Z}\sum _{s}N\left(s\right)e^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}=\frac{kT}{Z}\left(\frac{\partial Z}{\partial \mu }\right)$

the LHS is just the average $N$, so:

localid="1650885600621" $\stackrel{—}{N}=\frac{kT}{Z}\left(\frac{\partial Z}{\partial \mu }\right)$ ...(1)

take the partial derivative again for the grand partition function with respect to$\mu$, to get:

localid="1650885614351" $$\begin{gathered} \frac{{\partial }^{2}Z}{\partial {\mu }^2}=\frac{1}{k^2{T}^2}\sum _s{\left(N\left(s\right)\right)}^2{e}^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT} \\ \sum _s{\left(N\left(s\right)\right)}^2{e}^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}=k^2{T}^2\frac{{\partial }^{2}Z}{\partial {\mu }^2} \end{gathered}$$

Dividing both sides by the grand partition function to get:

$\frac{1}{Z}\sum _s{\left(N\left(s\right)\right)}^2{e}^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}=\frac{k^2{T}^2}{Z}\frac{{\partial }^{2}Z}{\partial {\mu }^2}$

the LHS is just the average $N^2$, therefore:

localid="1650885704869" $\stackrel{—}{N^2}=\frac{k^2{T}^2}{Z}\frac{{\partial }^{2}Z}{\partial {\mu }^2}$ ...(2)

take the partial derivative for the average number of particles with respect to $\mu$to get:

localid="1650885695468" role="math" $\begin{array}{r}\frac{\partial \stackrel{—}{N}}{\partial \mu }=\frac{\partial }{\partial \mu }\left(\frac{1}{Z}\sum _{s}N\left(s\right)e^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}\right)\\ \frac{\partial \stackrel{—}{N}}{\partial \mu }=-\frac{1}{Z^2}\frac{\partial Z}{\partial \mu }\sum _{s}N\left(s\right)e^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}+\frac{1}{ZkT}\sum _s{\left(N\left(s\right)\right)}^2{e}^{-\left[E\left(s\right)-\mu N\left(s\right)\right]/kT}\end{array}$

substitute from (1) and (2) to get:

$$\begin{gathered} \frac{\partial \stackrel{—}{N}}{\partial \mu }=-\frac{\stackrel{—}{N}}{kT}\stackrel{—}{N}+\frac{\stackrel{—}{N^2}}{kT} \\ \frac{\partial \stackrel{—}{N}}{\partial \mu }=-\frac{{\stackrel{—}{N}}^2}{kT}+\frac{\stackrel{—}{N^2}}{kT} \\ kT\frac{\partial \stackrel{—}{N}}{\partial \mu }=\stackrel{—}{N^2}-{\stackrel{—}{N}}^2 \end{gathered}$$

the standard deviation is defined as:

${\sigma }_N^2=\stackrel{—}{N^2}-{\stackrel{—}{N}}^2$

combine this equation with the previous one to get:

$$\begin{gathered} {\sigma }_N^2=kT\frac{\partial \stackrel{—}{N}}{\partial \mu } \\ {\sigma }_N=\sqrt{kT\frac{\partial \stackrel{—}{N}}{\partial \mu }} \end{gathered}$$