91Ó°ÊÓ

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

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

$Z=\underset{s}{∑}{e}^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}$

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

$\begin{array}{c}\frac{∂Z}{∂\mathrm{μ}}=\frac{1}{kT}\underset{s}{∑}N\left(s\right)e^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}\\ \underset{s}{∑}N\left(s\right)e^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}=kT\left(\frac{∂Z}{∂\mathrm{μ}}\right)\end{array}$

dividing both sides by the grand partition function to get:

$\frac{1}{Z}\underset{s}{∑}N\left(s\right)e^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}=\frac{kT}{Z}\left(\frac{∂Z}{∂\mathrm{μ}}\right)$

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

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

take the partial derivative again for the grand partition function with respect to$\mathrm{μ}$, to get:

localid="1650885614351" $$\begin{gathered} \frac{{∂}^{2}Z}{∂{\mathrm{μ}}^2}=\frac{1}{k^2{T}^2}\underset{s}{∑}{\left(N\left(s\right)\right)}^2{e}^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT} \\ \underset{s}{∑}{\left(N\left(s\right)\right)}^2{e}^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}=k^2{T}^2\frac{{∂}^{2}Z}{∂{\mathrm{μ}}^2} \end{gathered}$$

Dividing both sides by the grand partition function to get:

$\frac{1}{Z}\underset{s}{∑}{\left(N\left(s\right)\right)}^2{e}^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}=\frac{k^2{T}^2}{Z}\frac{{∂}^{2}Z}{∂{\mathrm{μ}}^2}$

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

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

take the partial derivative for the average number of particles with respect to $\mathrm{μ}$to get:

localid="1650885695468" role="math" $\begin{array}{r}\frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}=\frac{∂}{∂\mathrm{μ}}\left(\frac{1}{Z}\underset{s}{∑}N\left(s\right)e^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}\right)\\ \frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}=−\frac{1}{Z^2}\frac{∂Z}{∂\mathrm{μ}}\underset{s}{∑}N\left(s\right)e^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}+\frac{1}{ZkT}\underset{s}{∑}{\left(N\left(s\right)\right)}^2{e}^{−\left[E\left(s\right)−\mathrm{μ}N\left(s\right)\right]/kT}\end{array}$

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

$$\begin{gathered} \frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}=−\frac{\stackrel{—}{N}}{kT}\stackrel{—}{N}+\frac{\stackrel{—}{N^2}}{kT} \\ \frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}=−\frac{{\stackrel{—}{N}}^2}{kT}+\frac{\stackrel{—}{N^2}}{kT} \\ kT\frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}=\stackrel{—}{N^2}−{\stackrel{—}{N}}^2 \end{gathered}$$

the standard deviation is defined as:

${\mathrm{σ}}_N^2=\stackrel{—}{N^2}−{\stackrel{—}{N}}^2$

combine this equation with the previous one to get:

$$\begin{gathered} {\mathrm{σ}}_N^2=kT\frac{∂\stackrel{—}{N}}{∂\mathrm{μ}} \\ {\mathrm{σ}}_N=\sqrt{kT\frac{∂\stackrel{—}{N}}{∂\mathrm{μ}}} \end{gathered}$$