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I am facing a dilemma. The fact that matter is ionized allows ions and electrons to be much closer together than they are in atoms (Bohr radius $a = 0.5 \cdot 10^{− 10} \mathrm{m}$), and the result is that some stars, including white dwarfs, have a very small radius and a very high density that could not be had if matter were not ionized. A white dwarf, however, is a star that has run out of nuclear fuel; therefore, it slowly cools down. At some point the temperature becomes so low that matter would tend to regroup into non-ionized atoms. But to do that it would have to increase its radius and decrease its density, and this it cannot do because the change in gravitational energy it would take is greater than all the energy it has left. So, in a sense, the star would get to a situation where it does not have enough energy to cool down again. In other words, the temperature would stop decreasing without the star being in equilibrium with the part of the universe that surrounds it, and that is obviously not possible. How to succeed in explaining this?

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3 Answers 3

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There is an answer to your question.

The point is that this answer cannot be given in terms of classical mechanics. It is highly dependent on Quantum Mechanics. You assume that when the star cools down enough, the free electrons will recombine with the ions to form non-ionised atoms. But this does not happen.

The matter is highly compressed because of the gravitational energy, and as you correctly assume, cannot expand because there is not enough energy to fight against gravitation. And it does keep cooling.

But here is where a purely quantum phenomenon takes place, due to the combination of two purely quantum laws of physics : the Pauli exclusion principle (two electrons cannot be in the same quantum state) and the Heisenberg uncertainty principle. When there are many electrons piled up in a very small volume - and thus, their position is rather well determined - by Heisenberg uncertainty principle there must be a rather large uncertainty in their momentum. So even the lowest possible quantum state, the one where an electron would "like to fall" at very low temperature, has a non-negligible kinetic energy. But because of the Pauli exclusion principle only one electron can fall in that energy level. Even at very low temperature, the second one can only fall down on the next energy level, and the third one just above and so on. So even when the white dwarf keeps cooling indefinitely, the electrons keep a rather large kinetic energy. This energy is much higher than the recombination energy, so the situation is indeed that of ionized matter. The electrons are compressed much closer to the ions than for non-ionized atoms but with all this kinetic energy they whizz by the ions, without binding to them as atoms.

When the star is hot, the classical thermal energy helps this purely quantum (Pauli principle on top of Heisenberg uncertainty) kinetic energy to resist gravitation. As the star cools, the classical thermal energy decreases, and the gravitation will tend to compress the star more and more. The more you compress matter, the more the zero-temperature quantum kinetic energy increases (Heisenberg uncertainty : less uncertainty on the position hence more uncertainty on the momentum $p$) and resists the compression. This zero-temperature kinetic energy is indeed the source of the electron degeneracy pressure mentioned by S.G. in his answer. I just wanted to explain it in simple terms in my answer rather than referring to a different website.

So an equilibrium is found even at zero temperature.

Now here is a beautiful twist. As long as the total mass of the white dwarf is not too high, the equilibrium can be reached, because the kinetic energy goes like the square of the momentum, usually called $p$ (Indeed $E_k=p^2/(2m))$ when the electrons are not relativistic, and this grows fast when gravitation tries to compress matter, fast enough to resist it. But if the mass is too large the electrons are compressed to the point they become relativistic. And then the kinetic energy grows only like the momentum $p$, $E_k=pc$, not as its square. And this growth is not fast enough to resist gravitation and the white dwarf keeps collapsing to become a neutron star or even a black hole. This is the mechanism of the Chandrasekhar limit which is about 1.4 solar masses.

Chandrasekhar found this limit by trying to answer your question, you know! So it is not at all a naive question, but one that needs a good understanding of quantum mechanics to solve for cold white dwarf, and of special relativity to see how it breaks down at too large a mass.


EDIT

After seeing the answer from ProfRob I want to add something. He mentions that the maximum energy per electron is of the order of a MeV, but the way he writes it, it is as if gravitational energy is unable to give more "degeneracy" energy to the electrons. But this is not so. What happens is that the MeV range is just the energy where the electrons become relativistic. If the mass of the star is small enough, the equilibrium happens below that mass and it stays a white dwarf. If it is too heavy, the energy reaches this value and keeps increasing. If one does not see white dwarfs where electrons have a higher energy, it is not because it cannot be reached, but because beyond that it does not stop.

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  • $\begingroup$ The electrons in a white dwarf are highly degenerate with $E_F/k_BT$ of several hundred, even at temperatures of tens of thousands of K. The only way to increase the Fermi energy is to add mass to the white dwarf. White dwarrfs of constant mass do not get significantly smaller or denser as they cool and so the Fermi energy does not significantly increase with time. $\endgroup$
    – ProfRob
    Commented Nov 8, 2023 at 15:40
  • $\begingroup$ A typical white dwarf (of mass $0.6M_\odot$) has an average density of a few $10^{9}$ kg/m$^3$, is made of ionised carbon and has a typical electron Fermi (total) energy of 0.8 MeV, whatever its temperature. More massive white dwarfs have higher Fermi energies, less massive white dwarfs have lower Fermi energies. They are "of order MeV". Actually, the stuff you've written about the Chandrasekhar mass has nothing to do with the question at all. I've reversed my upvote for your needless edit. $\endgroup$
    – ProfRob
    Commented Nov 8, 2023 at 15:43
  • $\begingroup$ @ProfRob Before becoming a white dwarf, a star is hot. Fuel exhausted, it goes through a path in the Hertzsprung-Russell diagram en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram but eventually it radiates away and shrinks. Once it is degenerate, only then does it deserves the name "white dwarf". But the process of cooling does happen, before it has this name. The fact that there is an upper limit to the Fermi energy is exactly what is the meaning of Chandrasekhar mass. Why else are there no white dwarfs with higher Fermi energy ? $\endgroup$
    – Alfred
    Commented Nov 8, 2023 at 17:06
  • $\begingroup$ The question is about white dwarfs and I don't understand your comment about maximum Fermi energies, as I said for a given WD it doesn;t change. The "Chandrasekhar mass" - as explained by you, is the mass at which the radius of the WD shrinks to zero and the Fermi energy does in fact become infnite. The actual upper limit, is set by General Relativity (not considered by Chandrasekhar) or by electron capture and both occur when the Fermi energy reaches around 10 MeV in the core of the white dwarf. None of which has anything to do with the OP's question. Your downvote has been reciprocated. $\endgroup$
    – ProfRob
    Commented Nov 8, 2023 at 18:10
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    $\begingroup$ There is a maximum Fermi energy - it is arounfd 10 MeV (for a carbon white dwarf) as I said above. However, that is irrelevant to the question and I did not write it in my answer, so I'm not sure what you are on about. My downvote is because your answer and your comments give the impression that you do not understand that the Fermi energy of a white dwarf is more-or-less fixed as it cools. $\endgroup$
    – ProfRob
    Commented Nov 8, 2023 at 18:24
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The electrons in a white dwarf form a degenerate gas. That is, the Pauli Exclusion Principle demands that no two electrons occupy the same quantum state. That means that electrons at high density must occupy a broad range of energy states from zero up to some very large value (of order 1-10 MeV, depending on the white dwarf mass). It is this (kinetic) energy that provides the electron degeneracy pressure that supports the star.

When the white dwarf cools, the electrons cannot fall to much lower energies, since these quantum states are already occupied. Indeed the white dwarf can continue to cool towards zero Kelvin (the coolest white dwarfs in the universe at the moment have surface temperatures of about 3000 K and interiors at around 50,000 K) and since the electrons maintain their MeV energies, the overall gas remains fully ionised.

In fact there is almost no thermal energy that can be extracted from the electrons. Most of the thermal energy is in the accompanying non-degenerate ions. These ions can cool into lower energy states and can lock into a crystal structure once their thermal energy drops below some multiple of the Coulomb energy between them. This has no effect on the ionisation state or overall structure of the white dwarf since, it is the degenerate, high energy electrons that dominate the pressure.

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  • $\begingroup$ This seems to me to go to the point: electrons are degenerate. Moreover, the idea that at lower temperatures there is de-ionization is wrong even from a classical point of view: the classical atom is unstable from the very beginning. $\endgroup$
    – Quillo
    Commented Nov 9, 2023 at 19:35
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OP's comment: The fact that matter is ionized allows ions and electrons to be much closer together than they are in atoms

This is not necessarily true. An ionized gas has to contains much more energy than a non ionized one to remain in that state. Ionization also means more number of particles available (compared to the non-ionized gas) to exert pressure. So if we keep pressure and temperate fixed than an ionized gas should occupy a larger volume by $PV = NkT$ (using the ideal gas law in an approx sense).

OP's comment: white dwarfs, have a very small radius and a very high density that could not be had if matter were not ionized

A typical white dwarf is mostly supported by the electron degeneracy pressure (against gravity). So the inner regions need not be ionized in that sense as the pressure is not coming from the motion of charged ions. The compactness of white dwarfs is due to non existence of nuclear fusion to support against gravity and not because it is ionized.

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