How much does thermal expansion affect neutron stars? Would the loss of temperature cause a neutron star to be more densely packed and thus collapse into a black hole?
[Physics] Can a neutron star become a black hole via cooling
astrophysicsblack-holesneutron-starsstellar-physicsthermodynamics
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Actually, before you got to the Schwarzschild radius, the pressure previously supporting the neutron star against its own gravity would no longer be able to do so, and the entire star would violently collapse. Some matter would be thrown out, while the rest would become a black hole. The singularity at the center would be among all the rest of the singularities at the centers of all other black holes for the densest objects in the Universe.
Terminology note: the Chandrasekhar limit $M_C \approx 1.4 M_\text{sun}$ is for electron-degenerate matter. The analogous limit for neutron-degenerate matter, $M_\text{TOV} \sim 2.5 M_\text{sun}$, is named for Tolman, Oppenheimer, and Volkoff. We have much less confidence in our estimate for the TOV limit than we do in the Chandrasekhar limit, because we know less about the equation of state for neutron-degenerate matter than we do for electron-degenerate matter.
We are aware of several stable neutron stars with masses $M_C < M_\text{object}$; there’s a partial list in Wikipedia article linked above. But I suspect you were asking about the stability of neutron stars with masses above $M_\text{TOV}$.
There is speculation in the literature about the possible existence of quark stars, in which the nucleon degrees of freedom dissolve and the star is supported by degeneracy pressure among the free quarks. It’s possible in principle that a neutron star which accumulated mass beyond $M_\text{TOV}$ could collapse to a quark star, analogous to the collapse of a white dwarf (or of an electron-degenerate stellar core) to a neutron star. But we know even less about the equation of state for quark matter than we know for neutron matter. I don’t think it’s known for certain that the mass limit for a quark star is any larger than the mass limit for a neutron star. It’s also unknown whether quark stars would consist of up and down quarks, like normal baryonic matter, or whether the phase transition would produce a substantial fraction of strange quarks.
The Wikipedia page lists a number of (unconfirmed) quark star candidates, and describes why confirmation is so difficult. It may well be the case that quark stars don’t exist, and that an overmassive neutron star is definitely doomed to become a black hole.
The neutron-star merger event GW170817 produced an object with final mass $2.74^{+0.04}_{-0.01}M_\text{sun}$. That gravitational-wave event suggested the new object collapsed to a black hole on a timescale of a few seconds (as opposed to milliseconds, or hours). If you’re interested in the nitty-gritty details of black hole formation from “supermassive neutron stars,” that would be a path into the literature.
Best Answer
No (or at least not much). One of the essential properties of stars that are largely supported by degeneracy pressure, is that this pressure is independent of temperature and that is because although a neutron star may be hot, it has such a small heat capacity, it contains very little thermal energy$^{*}$.
When a neutron star forms, it cools extremely rapidly by the emission of neutrinos, on timescales of seconds. During this phase, the neutron star does contract a little bit (tens of per cent), but by the time its interior has cooled to a billion Kelvin, the interior neutrons are degenerate and the contraction is basically halted. It is possible that a (massive) neutron star could make the transition to a black hole before this point.
If it does not do so, then from there, the neutron star continues to cool (but actually possesses very little thermal energy, despite its high temperature), but this makes almost no difference to its radius.
$^{*}$ In a highly degenerate gas the occupation index of quantum states is unity up the Fermi energy and zero beyond this. In this idealised case, the heat capacity would be zero - no kinetic energy can be extracted from the fermions, since there are no free lower energy states. In practice, and at finite temperatures, there are fermions $\sim kT$ above the Fermi energy that can fall into the few free states at $\sim kT$ below the Fermi energy. However, the fraction of fermions able to do so is only $\sim kT/E_F$, where $E_F$ is the kinetic energy of fermions at the Fermi energy. At typical neutron star densities, this fraction is of order $T/10^{12}\ {\rm K}$, so is very small once neutron stars cool (within seconds) below $10^{10}$ K.
What this means is that the heat capacity is extremely small and that whilst the neutrons in a neutron star contain an enormous reservoir of kinetic energy (thus providing a pressure), almost none of this can be extracted as heat.