91Ó°ÊÓ

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

$\stackrel{¯}{E^2}=\frac{1}{Z}\frac{{∂}^{2}Z}{∂{\mathrm{β}}^2}$

The partition function is:

$Z=\underset{s}{∑}{e}^{-\mathrm{β}E\left(s\right)}$

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

By partial derivative of partition function with respect to $\mathrm{β}$

$\frac{∂Z}{∂\mathrm{β}}=\underset{s}{∑}-E\left(s\right)e^{-\mathrm{β}E\left(s\right)}$

By second partial derivative of partition function with respect to $\mathrm{β}$

$\frac{{∂}^{2}Z}{∂{\mathrm{β}}^2}=\underset{s}{∑}E(s{)}^2{e}^{-\mathrm{β}E\left(s\right)}$

Multiplying by $Z/Z$

$\frac{{∂}^{2}Z}{∂{\mathrm{β}}^2}=Z\underset{s}{∑}E(s{)}^2\frac{e^{-\mathrm{β}E\left(s\right)}}{Z}$

The probability is:

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

Therefore,

$\frac{{∂}^{2}Z}{∂{\mathrm{β}}^2}=Z\underset{s}{∑}E\left(s{)}^{2}P\right(s)$

The average energy is:

$\stackrel{¯}{E}=\underset{s}{∑}E\left(s\right)P\left(s\right)$

The squared average energy:

$\stackrel{¯}{E^2}=\underset{s}{∑}E\left(s{)}^{2}P\right(s)(2)$

Substitute into (1)

$\frac{{∂}^{2}Z}{∂{\mathrm{β}}^2}=Z\stackrel{¯}{E^2}\left(3\right)$

From problem 6.16,

$\frac{∂Z}{∂\mathrm{β}}=-Z\stackrel{¯}{E}\left(4\right)$

Substitute into (3)

$$\begin{gathered} \frac{∂}{∂\mathrm{β}}\left(\frac{∂Z}{∂\mathrm{β}}\right)=Z\stackrel{¯}{E^2} \\ \frac{∂}{∂\mathrm{β}}(-Z\stackrel{¯}{E})=Z\stackrel{¯}{E^2} \\ -Z\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}-\stackrel{¯}{E}\frac{∂Z}{∂\mathrm{β}}=Z\stackrel{¯}{E^2} \\ \stackrel{¯}{E^2}=-\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}-\frac{\stackrel{¯}{E}}{Z}\frac{∂Z}{∂\mathrm{β}} \end{gathered}$$

Using (4) we get:

$$\begin{gathered} \stackrel{¯}{E^2}=-\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}-\stackrel{¯}{E}(-\stackrel{¯}{E}) \\ \stackrel{¯}{E^2}=-\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}+{\stackrel{¯}{E}}^2 \\ \stackrel{¯}{E^2}-{\stackrel{¯}{E}}^2=-\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}\left(5\right) \end{gathered}$$

Using the chain rule of partial derivative:

$\frac{∂\stackrel{¯}{E}}{∂\mathrm{β}}=\frac{∂\stackrel{¯}{E}}{∂T}\frac{∂T}{∂\mathrm{β}}$

Therefore,

$\frac{∂T}{∂\mathrm{β}}={\left(\frac{∂\mathrm{β}}{∂T}\right)}^{-1}={\left(\frac{∂}{∂T}\frac{1}{kT}\right)}^{-1}={\left(-\frac{1}{k{T}^2}\right)}^{-1}=-k{T}^2$

Substitute into (5):

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

But ${\mathrm{σ}}_E^2=\stackrel{¯}{E^2}-{\stackrel{¯}{E}}^2$

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