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Stars can only collaps and form black holes if their masses are above the Chandrasekhar limit, $M>M_{\rm Pl}^3/M_{\rm hydrogen}^2$. When the universe eventually cools down enough, the black holes can start to evaporate via emission of Hawking radiation.

Why isn't there a point in the evaporation of the black holes (corresponding to a lower limit of black hole size) where the electron degeneracy pressure wins over the gravitational interaction again, so that the black holes become "normal" matter states again without all the special black hole properties?

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    $\begingroup$ You assume there are still "electrons" inside that black hole horizon. But no one knows what actually happens to matter inside the horizon. $\endgroup$
    – ACuriousMind ♦
    Mar 24, 2016 at 18:20

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You need to be precise about what you mean by a black hole.

In the real world black holes don't exist. So a collapsing mass doesn't become a black hole then unbecome a black hole as it evaporates. It was never a black hole.

The Schwarzschild and Kerr metrics are idealised solutions that are time independent, so they have existed for an infinite time and continue to exist for an infinite time. And neither contains any mass, electrons or otherwise. They are both vacuum solutions with an ADM mass but a stress-energy tensor that is everywhere zero (except at the singularity where it is undefined).

So if you start with a Schwarzschild or Kerr geometry and introduce evaporation they won't suddenly stop being a black hole because they are entirely geometrical constructs.

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    $\begingroup$ That proposition by Hawking is quite speculative, isn't it? I'm not sure you should state it as scientific consensus. $\endgroup$
    – Rococo
    Mar 24, 2016 at 20:23
  • $\begingroup$ Rennie is always there for a solid answer. Thanks for making this sight fun. $\endgroup$
    – Muze
    Mar 25, 2016 at 2:41
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    $\begingroup$ @Rococo: that an event horizon takes an infinite coordinate time to form has been known since at least the discovery of the Oppenheimer-Snyder metric (published 1939). I suppose Hawking radiation is still unproven though I'd guess most of us regard it as very likely to exist. While there are speculative elements to Hawking's paper, the idea that only apparent horizons can exist is far from new. $\endgroup$
    – John Rennie
    Mar 25, 2016 at 6:04
  • $\begingroup$ Okay, thanks for the clarification. Is it nonetheless true that, if you jump into an Oppenheimer-Snyder black hole plus Hawking radiation, after a long time relative to the initial collapse but before significant evaporation, you will hit the singularity (or whatever replaces it in a full quantum gravity description) in finite proper time? $\endgroup$
    – Rococo
    Mar 25, 2016 at 15:16
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    $\begingroup$ @Rococo: see Does any particle ever reach any singularity inside the black hole?. 't Hooft (yes that's the Nobel prize winning physicist!) thinks you will hit the singularity. He has a Nobel prize and I don't so I'm inclined to believe what he says, but I'm not sure I fully understand this. $\endgroup$
    – John Rennie
    Mar 25, 2016 at 15:59
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The Chandrasekhar limit (as defined by Chandrasekhar) takes no account of General Relativity. It arises when an electron degenerate object in equilibrium tends towards an infinite density at a specific mass - the Chandrasekhar mass.

In GR, the "Chandrasekhar mass" for ideal electron degeneracy is lower, but more importantly, the point of instability and collapse occurs at finite density.

Therefore, if you compress an object sufficiently for it to collapse beyond the density at which electron degeneracy or indeed neutron degeneracy or any other equation of state can support it (it doesn't matter because a GR instabiity sets in at finite density for any proposed equation of state), then an astrophysical black hole will form.

Even were the black hole then to lose mass by evaporation, the density of the matter would always be such (in the frame of reference of the collapsing matter the density heads of course rapidly to infinity) that it could never again be supported in GR by any equation of state.

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