thermodynamics - Ultra-relativistic gas

What is the physical significance of the relation $E=3NkT$ for classical ultra-relativistic gas? Why is it greater than ideal gas for which $E=(3/2)NkT$?

Answer

Nice question. Sometimes we get used to a certain fact, such as equipartition with $(1/2)kT$ per degree of freedom, that we forget that it's not always true, or what assumptions are required in order to make it true. I had to refresh my memory on how equipartition works.

Basically the $(1/2)kT$ form of the equipartition theorem is a special case that only works if the energy consists of terms that are proportional to the squares of the coordinates and momenta. The 1/2 comes from the exponent in these squares.

The WP article on equipartition has a discussion of this. There is a general equipartition theorem that says that

$$\langle x \frac{\partial E}{\partial x} \rangle = kT,$$

where $x$ could be either a coordinate or a conjugate momentum. If $E$ has a term proportional to $x^m$, the partial derivative has a factor of $m$ in it. In the ultrarelativistic case, where $E\propto\sqrt{p_x^2+p_y^2+p_z^2}$, you don't actually have a dependence on the momenta (momentum components) that breaks down into terms proportional to a power of each momentum. However, I think it's pretty easy to see why we end up with the result we do, because in one dimension, we have $|\textbf{p}|=|p_x|$, which does have the right form, with an exponent of 1.