In what part of the spectrum is it radiating? In the infrared, in the microwave? Or is not radiating anymore at all?

In russian:

Чему сейчас равна температура поверхности и ядра нейтронной звезды, которая образовалась 12 миллиардов лет назад? В каком диапазоне она сейчас излучает? В инфракрасном, микроволновом? Или не излучает вообще?

  • $\begingroup$ Cute avatar, Voix. $\endgroup$ – Mark C Nov 10 '10 at 19:22
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    $\begingroup$ Translation into English: "What is the temperature of the surface and core of a neutron star formed 12 billion years ago now equal to? In what part of the spectrum is it radiating? In the infrared, in the microwave? Or is not radiating anymore at all? $\endgroup$ – Mark C Nov 10 '10 at 19:25
  • $\begingroup$ Another question is, what suppose to happen with neutron star on long run ? what makes it burn, what kind of fusion/decay is happening there ? What if it cools, then what degenerated matter looks like after it cools ? will the gravitational equilibrium be ruined after some burn time ? how it explodes if it can explode at all ? $\endgroup$ – user299 Nov 11 '10 at 6:06
  • $\begingroup$ @Mark, Thanks for the translation into English :) Avatar is the Russian version of Winnie the Pooh :) $\endgroup$ – voix Nov 11 '10 at 20:01
  • $\begingroup$ You're welcome. Yes, Param-taram-taram-param! $\endgroup$ – Mark C Nov 12 '10 at 0:31

This depends a lot on the heating and cooling mechanisms you have in your model. Certainly initially neutron stars cool when neutrinos created in nuclear reactions escape. However, this is highly sensitive to such effects as superconductivity of the nucleon species (which both suppresses nuclear reactions and reduces heat capacity) and the density profile (which is in turn determined by the endlessly debated equation of state for the matter). At some point, the star likely reaches a thermal equilibrium of sorts, cooling by blackbody radiation from the surface.

Of course cooling is not the only process. Just a few of the heating ideas that have been floated over the years include:

  • Friction between superfluid and non-superfluid layers sapping energy out of the star's rotation;
  • The latent heat of crystallization of the crust being released whenever the crust "cracks" due to its equilibrium obliquity being reduced as the star spins down;
  • Conversion of magnetic field energy into heat via, e.g., pair production;
  • Redistribution of species as the centrifugally-induced chemical potentials change with spindown;
  • Interaction with the interstellar medium; and even
  • Colliding and interacting with the occasional free-floating magnetic monopole, if you believe in such things.

A rather extensive review of all this and more can be found in this review by Tsuruta, if you have access. The paper is replete with cooling curves for all sorts of models - too many to summarize here. Extrapolating to $12~\mathrm{Gyr}$ leads to temperatures1 of anywhere from a few Kelvin (obviously the lower limit based on the temperature of the CMB reservoir in which the system sits) to $10^4~\mathrm{K}$ (blackbody peaks in the UV) and higher.

1 Often the axes are labeled by luminosity rather than temperature, the conversion being simple enough given the model's radius. Also note that all observables in the paper are properly gravitationally redshifted; if you want to know about the radiation field just above the surface, you have to blueshift the photons back into potential well.

  • $\begingroup$ The surface and core temperature will be quite different won't they? I would have thought the details of the atmosphere (which determine the ratio) would be very important. $\endgroup$ – Rob Jeffries Aug 3 '14 at 21:54

You'll have to wait for a real physicist to jump in for a numerical answer. As far as I know, it doesn't burn but is simply radiating the energy it started out with. I suspect the thermal conductivity is quite high. If you assume the thermal conductivity is so high that there is essentially no temperature difference between the photosphere and the core you should be able to use the Stefan Boltzmann law to determine the rate of heat loss. But even then you'd need to know the heat capacity of matter in that form, so any physicists willing to jump in with actual theory?

If I assume heat capacity is independent of temperature, the cooling equation (after throwing out all the constants) would look like: $$\frac{dT}{dt}=-T^4$$ That has a solution, (again throwing out constants) like $T=t^{-1/3}$.

There could be other sources of energy, initial magnetic field, rotation, gravitational (it would probably shrink at least somewhat as it cools), and it there is any matter nearby the potential for stuff to fall onto it. {That's about as far as a dare venture, again hopefully an astrophysicist will jump in}

  • $\begingroup$ This assume the heat capacity is independent of temperature, but this isn't the case for a ball of degenerate neutrons. $\endgroup$ – Rob Jeffries Jan 1 '15 at 19:50

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