Consider a classical "degree of freedom" that is linear rather than quadratic $E=c\left|q\right|$for some constant $c$. (As example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat derivation of the equipartition theorem for this system, and show that the average energy isrole="math" localid="1646903677918" $\stackrel{-}{E}=kT$.

The classical degree of freedom is described by$E=c\left|q\right|$, where$c$is a constant.

The formula to calculate the partition function is given by

$Z=\sum _q{e}^{-\beta E\left(q\right)}..................\left(1\right)$

Here, $Z$is the partition function and $\beta =\frac{1}{kT}$is the Boltzmann factor, $k$being Boltzmann constant and $T$being the absolute temperature.

Substitute $c\left|q\right|$for $E$into equation (1) and change the summation by integral sign to obtain the partition function.

$Z=\frac{1}{∆q}{\int }_{-\infty }^{\infty }{e}^{-\beta c\left|q\right|}dq.................\left(2\right)$

Simplify equation (2) in step 2 to obtain an expression for the partition function.

$\begin{array}{rcl}Z& =& \frac{1}{∆q}\left[{\int }_{-\infty }^0{e}^{\beta cq}dq+{\int }_0^{\infty }{e}^{-\beta cq}dq\right]\\ & =& \frac{1}{∆q}\left[\frac{1}{\beta c}{\left[e^{\beta cq}\right]}_{-\infty }^0-\frac{1}{\beta c}{\left[e^{-\beta cq}\right]}_0^{\infty }\right]\\ & =& \frac{2}{∆q\beta c}..................\left(3\right)\end{array}$

The formula to calculate the average energy is given by

$\stackrel{-}{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }.................\left(4\right)$

Here, $\stackrel{-}{E}$is the average energy.

Substitute the value of $Z$from equation (3) into equation (4) and simplify to obtain the required average energy.

role="math" localid="1646905279668" $\begin{array}{rcl}\stackrel{-}{E}& =& -\frac{1}{\frac{2}{∆q\beta c}}\frac{\partial }{\partial \beta }\left(\frac{2}{∆q\beta c}\right)\\ & =& -\beta \frac{\partial }{\partial \beta }\left(\frac{1}{\beta }\right)\\ & =& \frac{1}{\beta }\\ & =& kT\end{array}$