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Degenerate gases are excellent conductors of heat. However, the fermions that compose the gas will not expand outwards due to heat, except in incredibly high temperatures.

  • Why is this? Does it have to do with the fact that the fermions occupy the lowest possible energy states up to the Fermi energy?

  • Would that make the gas require high amounts of energy to compress or expand?

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In a degenerate gas of fermions, the fermions fully occupy momentum states from zero up to a momentum corresponding to the Fermi energy. It is the momentum of the fermions that leads to degeneracy pressure.

As long as the kinetic energy of particles at the Fermi energy is much less than $kT$, then the fermions can be considered completely degenerate, so that the situation above applies and there are no fermions that occupy energy states higher than the Fermi energy. The Fermi energy only depends on the density of fermions.

The pressure is given by the following integral $$ P = \frac{1}{3}\int g(p) F(p) v\ \mathrm{d}p,$$ where $g(p) = 8\pi p^2/h^3$ is the density of available momentum states, $F(p)$ is the occupation number of those states, and $v$ is the particle velocity. For a degenerate gas, the integral is easy because $F(p)=1$ up to the Fermi momentum and zero thereafter. What this means is that temperature does not feature in the integrand or in its limits. Therefore the pressure is independent of temperature.

If the gas is heated (for instance, nuclear fusion reactions are present), then initially the temperature can rise with no increase in the pressure. It is not until $kT$ approaches the Fermi energy that a significant number of energy states above the Fermi energy become occupied and the pressure becomes temperature dependent again.

The work done to compress a gas is $PdV$, whether it is degenerate or not. For a given density of particles, the pressure of a degenerate gas is lower than that of a perfect gas. So from that point of view it is easier to compress. On the other hand if heat can escape from the gas and the compression can be done isothermally, then the pressure of a perfect gas increases with density, but the pressure of a (non-relativistic) gas increases as density to the power 5/3, so is harder to compress.

In compact stars, the significance is that a gas can collapse and then cool and settle into a degenerate state at high density and thereafter maintain a constant pressure. This means that white dwarfs and neutron stars can cool without shrinking. In the cores of low-mass stars, or in white dwarfs, these properties mean that fusion reactions can ignite in an explosive, runaway fashion, since the nuclear reaction rates are highly temperature dependent, but the degeneracy pressure does not respond to an increasing pressure.

EDIT: I feel I need to refine this answer in the light of comments and an answer contributed by Ken G in Why is the release of energy during the He-flash in stars almost explosive?

The answer to your headline question actually ought to be that in isolation, degenerate gases do expand if you add enough heat to them. However, the point is that by the time you have added enough heat to make them expand significantly then they can no longer be considered degenerate gases. This is because the heat capacity of a degenerate gas is very small, so for a given amount of added heat, the temperature can increase enormously, hence lifting the degeneracy as explained above.

In white dwarfs and the cores of low-mass stars, this is prevented from happening initially because the electrons that provide most of the pressure are not the only species present. Most of the heat energy from thermonuclear reactions is actually deposited in the non-degenerate ions. These however make only a small contribution to the total pressure and hence the electrons remain degenerate and the gas does not expand significantly·

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  • $\begingroup$ If I understand star evolution correctly, then during the helium flash, a star will have an electron degenerate core that doesn't expand from temperature. This allows the core to heat up until helium fusion is possible, and this will lead to a thermal runaway. Eventually, the degenerate core will expand from heat, causing it to cool and become stable. Am I wrong? $\endgroup$ – Sir Cumference May 11 '16 at 14:37
  • $\begingroup$ The core contracts until He fusion starts. If the core is quite degenerate at that point then you will get a He flash because the T dependence is very strong and the pressure does not increase at first. Eventually the degeneracy is lifted by high temperatures and the pressure begins to increase, the core expands, the H-burning shell is pushed out and extinguished. See physics.stackexchange.com/questions/174801/… physics.stackexchange.com/questions/154983/… $\endgroup$ – Rob Jeffries May 11 '16 at 15:08
  • $\begingroup$ So at high enough temperatures, even a degenerate core will expand? $\endgroup$ – Sir Cumference May 11 '16 at 15:24
  • $\begingroup$ @SirCumference Yes, once $kT$ rises to a few times the Fermi kinetic energy. $\endgroup$ – Rob Jeffries May 11 '16 at 15:40

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