Is a cold (3800K, or so) white dwarf still be considered/in degenerate state. Or, how can a gas be degenerate and cold?


1 Answer 1


I'm not entirely sure what your source of confusion is, so I'll just describe how temperature and degeneracy relate and hope that clears it up.


All particles which are fermions are subject to the Pauli Exclusion Principle. This means no two fermions can occupy the same quantum state. They will resist doing so. What that really means, doesn't matter too much so for now, just accept it as an axiom.

Electrons are fermions. A star, before it becomes a white dwarf, is filled with electrons just minding their own business. Then, as the star begins to form itself into a white dwarf, the crushing pressure of gravity squeezes all the electrons into a tight space. Gravity tries to push these electrons into the same quantum state, but the Pauli exclusion principle forbids this.

Fermi Energy

Note: The following is a gross over-simplication, but it gets the idea across without being too terribly wrong. Don't take it as gospel truth.

You can define a set of all the possible quantum states. For simplicity, I'm just going to call them $\mathrm{state}\ 1$, $\mathrm{state}\ 2$, ..., $\mathrm{state}\ 67$, ..., $\mathrm{state}\ 10^{10483}$, etc. (ordered by energy). From above, we know each state can be occupied by only one electron. In general, these electrons will occupy a variety of non-sequential states. Pre-white dwarf, we might order the electrons by state and say

  • $\mathrm{electron}\ 1$ is in $\mathrm{state}\ 43$,
  • $\mathrm{electron}\ 2$ is in $\mathrm{state}\ 12084$,
  • $\mathrm{electron}\ 3$ is in $\mathrm{state}\ 4.187\times10^{78}$
  • etc.

As the electrons are squeezed in the white dwarf, they'll try to get as close to the same quantum state as possible without violating the Pauli exclusion principle. A fully degenerate star would have electrons in the following states:

  • $\mathrm{electron}\ 1$ is in $\mathrm{state}\ 1$,
  • $\mathrm{electron}\ 2$ is in $\mathrm{state}\ 2$,
  • $\mathrm{electron}\ 3$ is in $\mathrm{state}\ 3$
  • etc.
  • $\mathrm{electron}\ n$ is in $\mathrm{state}\ n$

In this completely degenerate state, all electrons have been forced to occupy all the lowest energy states. The Fermi energy is the energy of the last electron in the highest state. If all electrons happen to be squeezed such that they're all below the Fermi energy, then we consider the object degenerate.

Temperature and Degeneracy

Gravity might try to squash all the electrons down into the lowest quantum states, but the ability of gravity to achieve this depends on the temperature. Electrons at a higher temperature have higher energies and it becomes harder for gravity to suppress them. With a certain set of assumptions1, one can say that for a star of a given temperature, $T$, and a given density, $\rho$, it will be degenerate if2:

$$\frac{T}{\rho^{2/3}} < 1261\ \mathrm{K\ m^2\ kg^{-2/3}}$$

From this equation, you can hopefully see that being colder actually helps a white dwarf be degenerate. The two ways to make a star (or anything really) become degenerate are to cool it down and increase the density. For white dwarf formation, the fact that it's degenerate is more a function of the high density than it is the cool temperature.

1Assumptions and preceding equations are in An Introduction to Modern Astrophysics, 2nd ed., pg. 566.

2As a preemptive measure, this equation does not apply universally. You cannot apply it to yourself and declare yourself degenerate!

  • $\begingroup$ Thank you for your explanation zephyr. I do understand that degeneracy is independent of temperature. I am trying to understand why it is that degenerate material takes so long to "cool" down. Is it just a good conductor, or insulator, or both? What would it feel like to touch such matter? $\endgroup$
    – rdinardo
    Commented Jul 28, 2017 at 22:20

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