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I am trying to understand why the cores of red giants are isothermal. They become supported by degenerate electrons due to extreme pressures relative to the thermal energy in the core which forces the electrons to occupy the lowest energy states up to close to ionization energy. I just don't understand what makes the cores nearly isothermal.

Does the lack of temperature dependence allow the thermal energy to conduct throughout the core?

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  • $\begingroup$ You have the right answer for why the gas is isothermal, but beware of one thing-- your wording sounded like you believe that degeneracy is making the pressure high, but actually it is the released kinetic energy that makes the pressure high. All degeneracy does is make the temperature low, so it drives down the thermal pressure. So you're right the thermal pressure is low compared to the total gas pressure, but degeneracy only acts on the former, the latter is just from kinetic energy. $\endgroup$
    – Ken G
    Commented Apr 7, 2017 at 1:15
  • $\begingroup$ There is no such thing as pressure = degeneracy pressure + thermal pressure. It is just pressure caused by the momentum distribution of the particles, which is a function of (number) density and temperature. The statement about ratio of pressure to thermal energy (density) is correct: Roughly 2/3 for a non-degenerate ideal gas, and of order $(E_F/kT)^2$ for a degenerate gas. $\endgroup$
    – ProfRob
    Commented Apr 7, 2017 at 7:49

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The thermal conductivity of a degenerate fermion gas is extremely high, because the mean free paths ofthe fermions are very large. The fermions can have lots of momentum and much higher speeds than would be the case for a non-degenerate gas of similar temperature. And, because there are few empty low energy states, then they cannot easily suffer dissipative scattering events unless the fermion involved is above the local Fermi energy.

As a result, heat in the form of fermion (electrons in the case of a helium core supported by electron degeneracy pressure) kinetic energy, can be transferred readily until the gas becomes isothermal.

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  • $\begingroup$ "And, because there are few empty low energy states, then they cannot easily suffer dissipative scattering events unless the fermion involved is above the local Fermi energy." This makes a lot of sense to me. Electrons are able to conduct heat from the core outward because they are not easily caught. Naturally this process will occur from hotter regions to cooler ones. does this prevent the core from heating up? Or is that some process related to neutrino loss? $\endgroup$
    – handroski
    Commented Apr 7, 2017 at 1:20
  • $\begingroup$ @handroski In principle a degenerate gas is easy to heat up (I assume you mean raise the temperature), because it has a very low heat capacity. It is more complicated in the He core, because the ions are non-degenerate. They can act as a sink of thermal energy. The core does heat up, that's how carbon is produced... $\endgroup$
    – ProfRob
    Commented Apr 7, 2017 at 7:54

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