Unruh effect, temperature and energy density

Question

The Unruh effect basically states that an accelerated observer will see warm gas of particles following a blackbody distribution with some temperature T, where as an inertial observer would see none.

If I understand correctly, black-body radiation contains all particles, although almost every source I've been able to find only mentions the EM black-body curve.

My question is: how does one determine the energy density of per unit volume of space that an accelerated observer would expect to see (given a temperature T)? I know the relevant distributions are Fermi-Dirac (for fermions) and Bose-Einstein (for bosons). It's not at all clear to me how to use both, or if the Fermi-Dirac distribution should be used at all, given that the Unruh 'warm gas' is a gas of photons (primarily)?

Is a good approximation simply to use Planck's radiation law and the Stefan-Boltzmann law?

Answer 1

You can find energy density of a photon gas, which is what the Unruh Effect would generate, here:

https://en.wikipedia.org/wiki/Photon_gas

Pressure would be 1/3 times U/V

A photon gas does not conserve photons. It is described by two variables: temperature and volume.

Since you only need acceleration (or an equivalent gravitational field) to calculate the Unruh Effect Temperature, and all you need is Temperature and Volume to determine the internal energy of a photon gas, calculating the temperature per unit of volume should be relatively straightforward. Making sure that the units match (Lorentz–Heaviside, Gauss, etc.) might be the hardest part.