Fermi gas in relativistic limit

[Mayan Fung]

TL;DR Summary

Why can we use the low-temperature limit to study Fermi gas in the ultra-relativistic limit?

In a statistical mechanics book, I learned about the degenerate pressure of a Fermi gas under the non-relativistic regime. By studying the low-temperature limit (T=0), we got degenerate pressure is $\propto n^{5/3}$ (n is the density).

And then I was told that in astrophysical objects, the fermions are in the relativistic regime so if we deal with an ultra-relativistic fermi gas: $\epsilon = chk$, and also the low-temperature limit(T=0), then we can arrive at degenerate pressure $\propto n^{4/3}$

My question is: If the fermions are in the ultra-relativistic limit, then it mush be very hot. How can we use the low-temperature limit to solve the problem?

 

[PeterDonis]

If the fermions are in the ultra-relativistic limit, then it must be very hot.

Not necessarily. If the fermions are being confined by very high pressure due to a strong gravitational field (such as in a neutron star), then they can be at zero temperature but still be relativistically degenerate.

 

[Mayan Fung]

Not necessarily. If the fermions are being confined by very high pressure due to a strong gravitational field (such as in a neutron star), then they can be at zero temperature but still be relativistically degenerate.

But we are taking the limit that $pc>>mc^2$ so that
$$\epsilon = \sqrt{p^2c^2+m^2c^4} = pc$$
I think a large momentum $p$ means a high speed? Or is it also related to gravity strength?

 

[PeterDonis]

I think a large momentum $p$ means a high speed?

Not if you are equating "high speed" with "high temperature". The momentum in this case is due to the confinement of the fermions to smaller and smaller regions of space, not due to high temperature.

 

[Mayan Fung]

Not if you are equating "high speed" with "high temperature". The momentum in this case is due to the confinement of the fermions to smaller and smaller regions of space, not due to high temperature.

Very clear! Thanks for that!

 

[Mayan Fung]

@PeterDonis

I just read an article https://www.nationalgeographic.com/science/space/universe/white-dwarfs/

It says that white dwarf has a temperature of around 100,000K. But I also saw this low-temperature approximation in calculating the degeneracy pressure of a white dwarf. It makes me confused again

 

[PeterDonis]

It says that white dwarf has a temperature of around 100,000K.

Yes.

But I also saw this low-temperature approximation in calculating the degeneracy pressure of a white dwarf.

The energy equivalent of 100,000 K is still much less than the rest mass of an electron. So kinetic pressure even at that temperature is still negligible compared to degeneracy pressure, and the zero temperature approximation, which neglects kinetic pressure, is still a good one.

 

[PeterDonis]

we are taking the limit that $pc>>mc^2$

Yes, and note that, even at a temperature of 100,000 K, the momentum of an electron due to temperature times $c$ is much less than $m c^2$, so it is negligible compared to the momentum due to degeneracy (gravity confining the electron to a small volume of space).