Question

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

$\overline{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }=-\frac{\partial }{\partial \beta }\ln Z$

where $\beta =1/kT$. These formulas can be extremely useful when you have an explicit formula for the partition function.

Short answer

Hence proved,

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

Step-by-step solution

Given information

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

$\overline{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }=-\frac{\partial }{\partial \beta }\ln Z$

Explanation

The partition function is given by:

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

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

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

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

Multiplying by a factor of $-1/Z$

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

The probability is given as:

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

Equation (1) becomes,

role="math" localid="1647371210348" $-\frac{1}{Z}\frac{\partial Z}{\partial \beta }=\sum _{s}E\left(s\right)P\left(s\right)$

The average energy is:

role="math" localid="1647371238971" $\overline{E}=\sum _{s}E\left(s\right)P\left(s\right)$

Therefore,

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