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The following rough approximation seems to show a neutron star at its Tolman–Oppenheimer–Volkoff limit of 2.17 solar masses and 12km radius, has gravitational binding energy on the same order as its relativistic mass-energy:

$$\frac{3}{5} \frac{(2.17*1.989*10^{30}kg)^2G}{12km}=6.2*10^{48}J$$ $$2.17∗1.989∗10^{30}kg *c^2=3.88*10^{48}J$$

While numbers and formula I've used are very rough -- which would explain the apparent 60% excess -- it does seem as though once a TOV star has settled down through radiation, just about all of its mass has been effectively converted to energy. Is this correct? If not, how much of its primordial (H, He, etc.) mass has been converted to energy?

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  • $\begingroup$ The relativistic binding energy of a neutron star is well discussed in the book "Compact Stars" by Glendenning. Related: physics.stackexchange.com/q/2902/226902 $\endgroup$
    – Quillo
    Commented Nov 1, 2023 at 6:45
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    $\begingroup$ That is a Newtonian form of the gravitational potential energy (which is negative) for a uniform body. The gravitational binding energy is the sum of that and the internal energy of the star, which is also negative for a bound star, but smaller in magnitude. Are you asking what fraction of $M$ is in the form of rest mass? The gravitational mass $M$ includes all contributions. See physics.stackexchange.com/questions/133625/… $\endgroup$
    – ProfRob
    Commented Nov 1, 2023 at 8:10
  • $\begingroup$ I invoked "primordial" rather than "rest" mass because, for example, iron-56 has a rest mass that is lower than its primordial mass that arose during nucleosynthesis era in lambdaCDM. I'm trying to get rid of any residual loss of energy to mass for the most extreme conditions. $\endgroup$ Commented Nov 1, 2023 at 11:47
  • $\begingroup$ What does "residual loss of energy to mass" mean? $\endgroup$
    – PM 2Ring
    Commented Nov 1, 2023 at 12:10
  • $\begingroup$ Stellar synthesis of iron-56 is via exothermic fusion results in lower mass than its reactants. $\endgroup$ Commented Nov 1, 2023 at 20:20

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When a neutron star is at the TOV limit then the particles within it (predominantly neutrons, though there may be phase changes into heavier hadronic matter or quark matter) are becoming, or are, highly relativistic. As a result, their mass-energy is dominated by kinetic energy and not rest mass.

However, the mass that you are putting into your formula (for gravitational potential energy, not gravitational binding energy) is the gravitational mass. This is the mass that an external observer judges the neutron star to have on the basis of its gravitational effects. This is going to be (approximately) the sum of the rest mass of the particles that make up the neutron star plus the mass-equivalent of the gravitational binding energy (which is negative). i.e. It is lower than the rest mass of the particles that make up the neutron star.

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  • $\begingroup$ Are you saying that a neutron star that is not rotating is made up of neutrons that are moving close to light speed? $\endgroup$ Commented Nov 1, 2023 at 20:13
  • $\begingroup$ @JamesBowery yes, of course. That's what degeneracy pressure is. $\endgroup$
    – ProfRob
    Commented Nov 1, 2023 at 22:19
  • $\begingroup$ The loss of gravitational potential energy while descending to become part of the stellar mass may not make it into the ultimate kinetic energy of the neutrons due to radiation into the surrounding space during and even after descent is completed. So I don't see where this relativistic kinetic energy comes from. $\endgroup$ Commented Nov 2, 2023 at 14:50
  • $\begingroup$ @JamesBowery If you confine fermions they have kinetic energy because they become degenerate. In this case, the confining is done by gravity. Radiative losses from a neutron star are utterly negligible and degenerate neutrons (or degenerate fermions in general) cannot emit electromagnetic radiation since they cannot reduce their energy. Of course the difference between the energy of some bigger, non-equilibrium state must be got rid of before the neutron star can achieve equilibrium. I still am unsure what you are actually asking. $\endgroup$
    – ProfRob
    Commented Nov 2, 2023 at 17:38
  • $\begingroup$ Are you asking how much of the rest mass is lost as neutrinos during the collapse maybe? $\endgroup$
    – ProfRob
    Commented Nov 2, 2023 at 17:41

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