I’ve realized that there are very few topics of physics covered on this site, so this post is meant to fix it a little bit. Let’s do a thermodynamics problem today then. So let’s start with some discussion of the equiparition theorem: what exactly is it and why should you really care? After all, equipartition is a pretty big word and you would have probably tuned out by now if a professor was lecturing about this. This is important some problems, like the one we will solve today! Let’s get started, shall we?

Problem

The energy of a photon is $E=pc$. Treating the photon as a classical object, compute the average energy of a photon at temperature $T$. In addition, find the adiabatic index $\gamma$ for a 3D photon gas.

Solution

We start by using the title of this post to help us out a lot. Using the Pythagorean theorem because we need to work in 3 dimensions,

\[E = c \sqrt{p_x^2 + p_y^2 + p_z^2}\]

Next, using the equipartition theorem, we can find the following expressions.

\[k_B T = \left \langle p_x \frac{\partial E}{\partial x} \right \rangle = c \left \langle \frac{p_x^2}{\sqrt{p_x^2 + p_y^2 + p_z^2}} \right \rangle\] \[k_B T = \left \langle p_y \frac{\partial E}{\partial x} \right \rangle = c \left \langle \frac{p_y^2}{\sqrt{p_x^2 + p_y^2 + p_z^2}} \right \rangle\] \[k_B T = \left \langle p_z \frac{\partial E}{\partial x} \right \rangle = c \left \langle \frac{p_z^2}{\sqrt{p_x^2 + p_y^2 + p_z^2}} \right \rangle\]

Using linearity after adding all three equations up, we have

\[3k_B T = c \left \langle \frac{p_x^2+p_y^2+p_z^2}{\sqrt{p_x^2+p_y^2+p_z^2}} \right \rangle = c \left \langle \sqrt{p_x^2 + p_y^2 + p_z^2} \right \rangle\]

This is a big yayyyyy because we recognize the expression on the far right: the energy of a photon. Therefore,

\[\boxed{\left \langle E \right \rangle = 3 k_B T}\]

Finally, we can use this to finish off the problem. Recall from statistical mechanics that the volumetric heat coefficient is

\[C_V = \frac{\partial \langle E \rangle}{\partial T} \bigg \vert_V = 3k_B\]

We can also use the fact that $C_P = C_V + k_B$ to find that

\[C_P = 4 k_B\]

Finally, since $\gamma = \frac{C_P}{C_V}$, so finally

\[\boxed{\gamma = \frac{4}{3}}\]