# White dwarf Ultra-relativistic electrons

• I

## Summary:

Ultra relativistic & degenerate electrons in a white dwarf

## Main Question or Discussion Point

Hi! I have read a calculation about a white dwarf star assuming the electrons are both ultra-relativistic and degenerate.
My questions is - how come the ultra-relativistic assumption doesn't contradict the "degeneracy" assumption.
Degenerate means the electrons are filling out all the lowest levels up to E_F (Fermi energy) with no excitation beyond that. But doesn't the fact that the electron gas is ultra-relativistic mean that electrons have a lot of energy, and some are excited beyond E_F (even though E_F is quite large for these objects) so that it's not degenerate anymore?

Thank you.

Related Astronomy and Astrophysics News on Phys.org
Drakkith
Staff Emeritus
Degenerate means the electrons are filling out all the lowest levels up to E_F (Fermi energy) with no excitation beyond that. But doesn't the fact that the electron gas is ultra-relativistic mean that electrons have a lot of energy, and some are excited beyond E_F (even though E_F is quite large for these objects) so that it's not degenerate anymore?
I don't think so. I'm no expert in this matter, but I believe that the energy levels can be very, very high and still be the lowest available for an electron. Besides, just because an electron is ultra-relativistic doesn't mean that it is 'excited'. If the available state requires that an electron have a very large amount of kinetic energy, then any electron in that state simply has to have that much kinetic energy. The electron wouldn't be in an excited state in this case as it would occupy the lowest possible state given the environmental conditions.

Drakkith - that makes sense. Thank you!

I would assume that we still need to have

$$m_e c^2 \ll K \ll E_F$$

where ##K## is the kinetic energy of the electron. That is, it cannot be "too ultra-relativistic" where the kinetic energy of the highest level electrons approaches ##E_f## and stay degenerate.

Drakkith
Staff Emeritus
Not sure. My understanding is that the kinetic energy of the electrons can be quite high in degenerate matter, with speeds being a substantial fraction of the speed of light for electrons in the higher energy levels. Note that the Fermi energy is defined as the difference between the largest and smallest occupied energy levels, so by definition some electrons must have energies very close to ##E_F## if the matter is degenerate.

But, like I said, I'm no expert here.

Staff Emeritus
2019 Award
You do realize that even at T = 0 white dwarf electrons are degenerate and relativistic. And that just means that <v> ~ c for a good fraction of the filled states.

So does degenerate just mean that ##kT## (thermal excitations) are too small to cause electrons near Fermi level to be excited to the next level?

Drakkith
Staff Emeritus
So does degenerate just mean that ##kT## (thermal excitations) are too small to cause electrons near Fermi level to be excited to the next level?
Not sure. My very limited understanding was that degenerate means that there are very few lower energy states available for the particles to occupy, forcing them to occupy higher energy states when forced into a smaller volume. That could be entirely incorrect on my part though.

So does degenerate just mean that ##kT## (thermal excitations) are too small to cause electrons near Fermi level to be excited to the next level?
In an ideal gas, particles can occupy the same state in momentum space; in a degenerate gas particles cannot occupy the same state and are forced to occupy different states in momentum space - this means the degenerate gas is not as compressible as an ideal gas, and the pressure depends on the density and not the temperature - in fact, since the particles are forced to inhabit higher energy levels, E > kT and the degenerate particles have energies exceeding the thermal energy; in a relativistic degenerate gas, the velocities of the particles approach the speed of light and so must be modified by a relativistic correction otherwise they would become greater than c. This means that the dependence of the pressure on the density in the relativistic case is not as great as in the non-relativistic case.

Thanks for the replies. I am trying to imagine how the degeneracy pressure arises. If the volume is getting smaller, then imagining the different states as standing waves, each wavelength is getting smaller and its energy is getting larger. But why does it mean that states are forced to overlap?

Thanks for the replies. I am trying to imagine how the degeneracy pressure arises. If the volume is getting smaller, then imagining the different states as standing waves, each wavelength is getting smaller and its energy is getting larger. But why does it mean that states are forced to overlap?
I don't think you appreciate the scale of the problem we're dealing with - a white dwarf star contains one and a half times the mass of the Sun in a volume the size of the Earth. The densities approach atomic scale, yet the material is still an ionized plasma. This is why the electron states begin to overlap and the Pauli exclusion principle needs to be applied, otherwise the entire star would just collapse to the size of a nucleus under the force of gravity.

But let's get back to your original question about electrons that are in the relativistic regime; If you have a star where the density is low but the temperature is exceedingly high, then it's possible that their motion is relativistic but not degenerate. For a star with a very high density it becomes degenerate and as you add more and more particles, their energies become greater and greater, necessitated by the Pauli exclusion principle, until they reach the point where their motions become relativistic. Even though these are exceedingly large energies, they're still the minimal energy possible provided by the exclusion principle and are below the Fermi energy.