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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.m.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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