# Ultra-relativistic gas

## Main Question or Discussion Point

Hei

I need to know the ratio of specific heats, $$\gamma$$ for an ultra-relativistic gas, in which kT >> $$m_{p}c^{2}$$, assuming that it is satisfied the equation for a politropic gas $$\epsilon=\frac{P}{\gamma-1}$$, where $$\epsilon$$ is the internal energy density.
(What is the difference between relativistic and ultra-relativistic?)

It must be something very easy, I have already the solution for:

Ionized Non-relativistic gas: (kT<< $$m_{e}c^{2}$$)

$$\epsilon=\frac{3}{2}nkT+\frac{3}{2}nkT$$
$$P = nkT + nkT$$
So $$\gamma=5/3$$.

Ionized Relativistic gas: ($$m_{e}c^{2}$$ << kT << $$m_{p}c^{2}$$)

$$\epsilon=\frac{3}{2}nkT+3nkT$$
$$P = nkT + \frac{1}{3}3nkT$$
So $$\gamma=13/9$$.

But all this doesn't make much sense to me, could you shed some light over it, please?

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Astronuc
Staff Emeritus
Is one assuming hydrogen?

At these temperatures, one has a plasma, so there is basically an electron gas and an ion (nuclei) gas, which I guess one is assuming Te = Ti.

Also since 1 eV = 11605 K roughly, 1 keV is 11.605 MK which is pretty darn hot, and one is looking at 511 keV.

1 MeV = 11.605 GK (11 billion K)!

Yes, it's hydrogen. I forgot to tell, this is about high energy astrophysics, the kind of situation that one can find in a neutron star or in the accreting material around a black hole. Quite extreme locations.

It's not too hard to derive the statistical mechanics of a relativistic gas. Start with a confining box, and count the number of states in momentum space. Convert to state density in energy, but with the relation E=pc instead of E=p^2/2m. Calculate partition function by the usual integration, and the usual rules of statistical mechanics gives all the desired thermodynamic quantities...

Astronuc
Staff Emeritus
It's not too hard to derive the statistical mechanics of a relativistic gas. Start with a confining box, and count the number of states in momentum space. Convert to state density in energy, but with the relation E=pc instead of E=p^2/2m. Calculate partition function by the usual integration, and the usual rules of statistical mechanics gives all the desired thermodynamic quantities...
True, but bear in mind that this has atleast two components - an electron gas, will be treated relativistically at lower energies, than nuclei, which from the mp is hydrogen (i.e. protons). Realistically, there would be deuterons and alpha particles and possibly heavier nuclei. But certainly the electron mass/momentum would be treated relativisitically.

But how about the intense magnetic/electric fields which are not normally part of molecular kinetics models (of neutral gases)?

True, but bear in mind that this has at least two components - an electron gas, will be treated relativistically at lower energies, than nuclei, which from the mp is hydrogen (i.e. protons). Realistically, there would be deuterons and alpha particles and possibly heavier nuclei. But certainly the electron mass/momentum would be treated relativistically.

But how about the intense magnetic/electric fields which are not normally part of molecular kinetics models (of neutral gases)?
The multiple components shouldn't be a problem --- they are essentially non-interacting as far as the statistical mechanics go (i.e the interaction doesn't further constrain the available phase space of the entire system). For similar reasons, I don't think the magnetic fields or electric fields change things too much. If there were no externally applied field (which would then have to be incorporated in the energy of the states), internal field should average out to be zero --- a mean field approximation. As long as you weren't too interested in the non-equilibrium physics, the procedure above should give the correct thermodynamics.

Astronuc
Staff Emeritus