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Does the Schwarzschild radius exist at the exact tipping point where neutron degeneracy pressure fails?

For example, consider a hypothetical neutron star that is being held up just below the limit of neutron degeneracy pressure. If the neutron star gains enough mass (assume a constant radius), it will collapse into a black hole. Does neutron degeneracy pressure fail at the exact point where the neutron star gains enough mass appropriate for the resultant black hole's Schwarzschild radius (where the escape velocity = c)? If so, why do these seemingly unrelated limits match?

Or... does the existing neutron star's radius exceed the resultant black hole's radius? In other words, does neutron degeneracy pressure fail, and as the the neutron star's radius decreases when collapsing, does it eventually reach a smaller (than in the previous scenario) Schwarzschild radius? If this is the case, what is the "in-between" state of matter as the neutron star collapses into a black hole?

Thank you!

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    $\begingroup$ Theoretically there can be quarks stars between neutron stars and black holes and then it's the quark degeneracy that further resists the inward force of gravity. en.wikipedia.org/wiki/Quark_star $\endgroup$ – Triatticus Aug 31 '18 at 15:12
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The Schwarzschild radius of a mass is proportional to the mass. This means that you can have a very wide range of density for objects about to be at their S radius. For neutron stars, there is a range of masses where the neutron matter is well outside the S radius. For more massive objects the radius of neutronic matter would be inside the S radius. For very large masses, on the range of the mass of our galaxy, the density where the object would be just outside its S radius would be roughly that of air.

So, no, there is no special relationship between the S radius and the density of neutron stars.

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  • $\begingroup$ I don't see how this answers any part of the question. $\endgroup$ – Rob Jeffries Jan 17 at 21:56
  • $\begingroup$ @RobJeffries: It does seem to me like it answers it. The OP's question is a naive one, and it just needs a simple answer at a beginner's level. The argument presented here is sufficient to dispel the OP's misconception. $\endgroup$ – user4552 Jan 17 at 23:40
  • $\begingroup$ @BenCrowell it fails to answer the question in para. 1; it does not correctly state that neutron stars become unstable at radii larger than $R_s$; it does not dispel the notion that neutron degeneracy pressure "fails"; it claims that massive neutron stars have radii smaller than $R_s$; it does not discuss why instability sets in at around $R_s$ and does not discuss the state of matter at very high densities. $\endgroup$ – Rob Jeffries Jan 18 at 0:26
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The hydrostatic equilibrium between gravity and the internal kinetic energy (pressure) of a neutron star in General Relativity is governed by the Tolman-Oppenheimer-Volkoff equation. This is like the usual Newtonian hydrostatic equilibrium equation except that there are multiplicative terms on the right hand side that increase the required pressure gradient as the pressure of the fluid increases and as the curvature of space increases.

What this means is that as you increase the central pressure of the neutron star, the pressure gradient required to support it becomes even larger. That is because the kinetic energy of the neutrons (or whatever else is in the interior) is actually contributing to the mass-energy that is curving space. This eventually becomes self-defeating - an additional increase in the central pressure to provide an increased pressure gradient just increases the required pressure gradient by even more! The star becomes unstable and collapses.

This imposes a limit on $M/R$ and for a given equation of state implies a maximum mass for the neutron star -- the Tolman-Oppenheimer-Volkoff limit.

The maximum compactness of a neutron star can be estimated in various ways. e.g. on p.260-261 of "Black Holes, White Dwarfs and Neutron Stars" (Shapiro & Teukolsky) it is shown that if causality is satisfied and the speed of sound is less than the speed of light, then for a stable neutron star, $GM/Rc^2 < 0.405$, which means that $R > 1.23R_s$, where $R_s$ is the Schwarzschild radius. Even if you abandon causality, it is easy to show from the TOV equation itself that even with infinite central pressure, that $GM/Rc^2 < 8/9$ and hence $R>1.13 R_s$.

Thus the radii of stable neutron stars are always somewhat above the Schwarzschild radius of a black hole with the same mass. The reason that the radius at which they become unstable is similar to $R_s$ is just that this is the size scale at which the GR effects that cause the instability become very important. It is not that degeneracy pressure "fails", it is that no stable solution is possible in GR.

Note that this final conclusion about the minimum radius as a function of $R_s$ is almost independent of assumptions about what provides the equation of state, but the corresponding mass at which this compactness limit is reached is sensitive to the exact equation of state. Neutron stars are not supported by ideal neutron degeneracy pressure, they are mainly supported by repulsive strong nuclear forces between closely packed neutrons. This allows them to achieve maximum masses of at least $2.1M_{\odot}$ and still satisfy the compactness limits above. If neutron stars were supported by ideal neutron degeneracy pressure than no neutron stars would exist with $M>0.75 M_{\odot}$, as was shown by Oppenheimer & Volkoff in 1939.

What is the "in between" state of matter at even higher densities during the collapse to a black hole? Nobody knows. It is quite likely that there are new "hadronic phases", which basically means the kinetic energy of neutrons gets turned into the rest mass of heavy hadrons like $\Lambda$ and $\Sigma$ particles. It is also possible that at very high densities (maybe 10 times that of nuclear matter), the quarks attain "asymptotic freedom" and there is a new phase of matter consisting of effectively free up, down and strange quarks.

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  • $\begingroup$ Neutron stars are not supported by neutron degeneracy pressure, they are supported by repulsive strong nuclear forces between closely packed neutrons. I don't think this is quite right. It's not an either-or. The behavior of the strong nuclear force matters, and so does the exclusion principle. If you ignore the strong force, then you get an estimate that's off but in the right ballpark. If you ignore the exclusion principle, then I don't think you get anything sensible. $\endgroup$ – user4552 Jan 17 at 21:25
  • $\begingroup$ @BenCrowell at densities of $\sim 10^{18}$ kg/m$^3$, ideal NDP is a factor of 5-10 smaller than required to support a 2 solar mass neutron star. physics.stackexchange.com/questions/291503/… $\endgroup$ – Rob Jeffries Jan 17 at 21:36
  • $\begingroup$ Adding "mainly" seems like a good edit. But I still don't think it makes sense to talk as if x% is neutron degeneracy pressure and 100-x% is the strong force. We don't have any system we can even talk about where x is not present, and there is no reason AFAICT to think these are additive effects. $\endgroup$ – user4552 Jan 17 at 23:38
  • $\begingroup$ @BenCrowell there is no doubt as to the meaning of what I am saying. The strong force interaction is almost entirely responsible for the existence of neutron stars. $\endgroup$ – Rob Jeffries Jan 18 at 0:16

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