In our everyday experience termperature is due to the motion of atoms, molecules, etc. A neutron star, where protons and electrons are fused together to form neutrons, is nothing but a huge nucleus made up of neutrons. So, how does the concept of temperature arise?
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6$\begingroup$ So you think the neutrons aren't moving? $\endgroup$– Jon CusterCommented Jul 31, 2014 at 18:26
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$\begingroup$ The neutrons are of course moving very fast indeed, but this has little to do with temperature (or vice versa). $\endgroup$– ProfRobCommented Aug 3, 2014 at 20:20
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2 Answers
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First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force.
Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere in the spectrum, whose peak wavelength can be converted to a temperature using Wien's displacement law:
$$T=\frac{b}{\lambda_{\rm max}}$$
with $b\sim2.9\times10^{-3}\rm mK^{-1}$. Neutron stars peak in the x-ray, and picking a wavelength of $1\;\rm nm$ (roughly in the middle of the logarithmic x-ray spectrum) gives a temperature of about $3$ million $\rm K$, which is in the ballpark of what is typically quoted for a neutron star.
More broadly than the motion of atoms or molecules, you can think of temperature as a measurement of the internal (not bulk) kinetic energy of a collection of particles, and energy is trivially related to temperature via Boltzmann's constant (though to get a more carefully defined concept of temperature requires a bit more work, see e.g. any derivation of Wien's displacement law).
answered Jul 31, 2014 at 18:27
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$\begingroup$ Interestingly, that temperature corresponds to an energy scale $kT≈250\,\mathrm{eV}$, which is quite cold from the perspective of typical nuclear excitations of a few mega-eV. $\endgroup$– rob ♦Commented Aug 1, 2014 at 0:32
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$\begingroup$ For a neutron star, I wouldn't be surprised if a significant portion of the entropy was accounted for by the spin alignment of the neutrons and the magnetic field, rather than mechanical kinetic energy. $\endgroup$ Commented Aug 1, 2014 at 0:58
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1$\begingroup$ @rob 50eV would be a cold temperature for the outer skin of an old neutron star. Neutron star interiors will be MeV to GeV. $\endgroup$– ProfRobCommented May 26, 2020 at 7:41
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1$\begingroup$ @rob I suppose my point is that this answer is speaking only about the surface temperature, where the concept of temperature apples in just the same way it would to any gas. Yes, the B-field can play an energetic role at the surface, but only near the surface. Indeed, once the interior cools, it becomes superconducting and the B-field is expelled. $\endgroup$– ProfRobCommented May 26, 2020 at 15:42
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The answer given by Kyle refers of course only to the surface or photospheric temperature of the neutron star - the temperature of the layer from which photons can escape to reach an observer. In these outer layers the relationship between temperatures and particle motions is more-or-less consistent with the "everyday" Maxwell-Boltzmann picture referred to by the OP.
However, the bulk of a neutron star is much hotter than this, probably by a factor of 100 or so. In thermal terms, a neutron star consists of an isothermal core (the vast bulk of the star) surrounded by a very thin (maybe a few metres) insulating blanket, across which there is a big temperature drop.
Inside the neutron star there is a shell of material containing neutron-rich nuclei and degenerate electrons, where the traditional concept of temperature (at least as applied to the nuclei) still has some merit in terms of the kinetic energies of the nuclei. In the inner regions, containing $>95$ percent of the mass, there are mainly neutrons with a small (1%-ish) fraction of protons and electrons. These are all degenerate gases.
In degenerate gases, the concept of temperature is a bit more slippery. The neutrons occupy energy states according to Fermi-Dirac statistics in the low temperature/high density limit; the kinetic energy of the neutrons becomes almost independent of temperature and entirely dependent on density. As a consequence the gas pressure is independent of temperature and these degenerate gases contain very little thermal energy, even when they are at extremely high temperatures.
This is why neutron stars cool extremely rapidly - their degenerate interiors contain much less thermal energy than a non-degenerate gas at similar temperatures.
