Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of E² is

$\overline{E^2}=\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial {\beta }^2}$

Then use this result and the results of the previous two problems to derive a formula for ${\sigma }_E$in terms of the heat capacity, $C=\partial \overline{E}/\partial T$

You should find${\sigma }_E=kT\sqrt{C/k}$

For any system in equilibrium with a reservoir at temperature T, the average value of E² is

$\overline{E^2}=\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial {\beta }^2}$

The partition function is:

$Z=\sum _s{e}^{-\beta E\left(s\right)}$

Where $\beta$is Boltzmann's constant and is given as $\beta =\frac{1}{kT}$

By partial derivative of partition function with respect to $\beta$

$\frac{\partial Z}{\partial \beta }=\sum _s-E\left(s\right)e^{-\beta E\left(s\right)}$

By second partial derivative of partition function with respect to $\beta$

$\frac{{\partial }^{2}Z}{\partial {\beta }^2}=\sum _{s}E(s{)}^2{e}^{-\beta E\left(s\right)}$

Multiplying by $Z/Z$

$\frac{{\partial }^{2}Z}{\partial {\beta }^2}=Z\sum _{s}E(s{)}^2\frac{e^{-\beta E\left(s\right)}}{Z}$

The probability is:

$P=\frac{1}{Z}{e}^{-\beta E\left(s\right)}$

Therefore,

$\frac{{\partial }^{2}Z}{\partial {\beta }^2}=Z\sum _{s}E\left(s{)}^{2}P\right(s)$

The average energy is:

$\overline{E}=\sum _{s}E\left(s\right)P\left(s\right)$

The squared average energy:

$\overline{E^2}=\sum _{s}E\left(s{)}^{2}P\right(s)(2)$

Substitute into (1)

$\frac{{\partial }^{2}Z}{\partial {\beta }^2}=Z\overline{E^2}\left(3\right)$

From problem 6.16,

$\frac{\partial Z}{\partial \beta }=-Z\overline{E}\left(4\right)$

Substitute into (3)

$$\begin{gathered} \frac{\partial }{\partial \beta }\left(\frac{\partial Z}{\partial \beta }\right)=Z\overline{E^2} \\ \frac{\partial }{\partial \beta }(-Z\overline{E})=Z\overline{E^2} \\ -Z\frac{\partial \overline{E}}{\partial \beta }-\overline{E}\frac{\partial Z}{\partial \beta }=Z\overline{E^2} \\ \overline{E^2}=-\frac{\partial \overline{E}}{\partial \beta }-\frac{\overline{E}}{Z}\frac{\partial Z}{\partial \beta } \end{gathered}$$

Using (4) we get:

$$\begin{gathered} \overline{E^2}=-\frac{\partial \overline{E}}{\partial \beta }-\overline{E}(-\overline{E}) \\ \overline{E^2}=-\frac{\partial \overline{E}}{\partial \beta }+{\overline{E}}^2 \\ \overline{E^2}-{\overline{E}}^2=-\frac{\partial \overline{E}}{\partial \beta }\left(5\right) \end{gathered}$$

Using the chain rule of partial derivative:

$\frac{\partial \overline{E}}{\partial \beta }=\frac{\partial \overline{E}}{\partial T}\frac{\partial T}{\partial \beta }$

Therefore,

$\frac{\partial T}{\partial \beta }={\left(\frac{\partial \beta }{\partial T}\right)}^{-1}={\left(\frac{\partial }{\partial T}\frac{1}{kT}\right)}^{-1}={\left(-\frac{1}{k{T}^2}\right)}^{-1}=-k{T}^2$

Substitute into (5):

$\overline{E^2}-{\overline{E}}^2=C_{v}k{T}^2$

But ${\sigma }_E^2=\overline{E^2}-{\overline{E}}^2$

$$\begin{gathered} {\sigma }_E^2=C_{v}k{T}^2 \\ {\sigma }_E=T\sqrt{C_{v}k} \end{gathered}$$