I was reading this Phys SE post and found it to be fairly unsatisfying.The most upvoted answer has a statement as follows:
And according to classical equipartition, each of these infinite possible oscillatory modes should contain, on average, an energy of $\frac{1}{2}k_B T.$
Another Phys SE post makes a statement which aligns with what I originally thought on the subject; that there was $\frac{1}{2} k_B T$ energy per degree of freedom. For a single mode of a wave we assume that there is only one position component and one momentum component, classically,
$$E=\frac{1}{2}kx^2 +\frac{1}{2m}p^2 $$
My understanding is that when we count the number of modes in a cavity, for example, we are already taking the geometry into account so we can't include degrees of freedom for the three spatial components in the above equation.
Then using the canonical distribution,
$$P(E)=\frac{1}{Z}\exp(-E/k_BT)$$
Secondary question: why is it appropriate to use this probability distribution? My understanding from thermodynamics/stat. mech is that this distribution applies to a system in a heat bath.
It's pretty straightforward to evaluate the integral for the average energy per mode using the Gaussian integrals,
$$\bar{E} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} P(E)E\ dxdp = k_B T$$
I think it's a bit interesting that the limits of the position extend from negative infinity to positive infinity. I would think that the requirement is that the distribution is well centered inside the range of position limits and this is just an approximation.
Is this method of evaluating the average energy per mode accurate?
