Q. 7.6 - An Introduction to Thermal Physics

Question

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. 

Step-by-Step Solution

Answer

The simple expression for ${\sigma }_N$  in terms of $\overline{N}$ is ${\sigma }_N=\sqrt{kT\frac{\partial \stackrel{—}{N}}{\partial \mu }}$.

Step 1: Given Information

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

Step 2: Simplify

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:

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

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

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

Step 3: Calculation

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:

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

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

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

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:

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