Calculate Gamma for Ultra-Relativistic Gas: kT >> m_{p}c^{2}

[Nuindacil]

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?

 

[Astronuc]

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 T_e = T_i.

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)!

 

[Nuindacil]

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.

 

[genneth]

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]

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 m_p 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)?

 

[genneth]

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 m_p 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]

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).

Well, I am not so sure about the 'non-interaction'. High speed electrons would repeatedly lose energy due to brehmstrahlung and cyclotron radiation, so in addition to the momentum/energy distribution of e's and p's, is one also considering photons?

 

[genneth]

Well, I am not so sure about the 'non-interaction'. High speed electrons would repeatedly lose energy due to brehmstrahlung and cyclotron radiation, so in addition to the momentum/energy distribution of e's and p's, is one also considering photons?

Perhaps I shouldn't have said non-interacting --- rather that the interaction does not add new degrees of freedom, or constrain the available state space any more --- energy and momentum conservation are already in there. Thus the results from the statistical mechanics will still be accurate.