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This is a broad question but it's well documented that GR and QM are very well tested in their own domains but they conflict around black holes.

Picture a neutron star slowly accreting matter until it's mass is sufficient to bring about an event horizon. It resists gravity owing to the Pauli exclusion principle and it must surely be comprised of the same 'stuff' as the event horizon forms. Why do we then rely on GR and assume everything collapses to a singularity which seems illogical in nature when the most sensible ('we don't know yet') answer surely should be that there exists a 'black star' under the event horizon? It seems that QM is overshadowed by GR in this instance when GR seems to give more illogical answers.

As a thought experiment, if we had a very heavy neutron star and fired one photon at a time at it; I would imagine the surface begins to redshift more and more as time goes on. There would surely reach a point where the miniscule deviations mean that a 'ravine' can no longer emit anything to an observer but a 'mountain' could. It would seem to be on a tipping point of being both a black hole and a neutron star but the mountain is still supported from below from seemingly below the horizon that is forming.

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    $\begingroup$ 1) The equations work out such that we can know of neutron stars that are not as dense as the densest we can see, but is just much bigger i.e. bigger mass, can form an event horizon outside of it, i.e. collapse into a black hole. This shows that matter that is ostensibly still normal is able to form black holes. 2) From our understanding of how quantum degeneracy pressure varies as density increases, we deduce that for black holes, at some point, not even quantum degeneracy pressure can avoid the singularity. $\endgroup$ Commented Oct 17, 2023 at 19:49
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    $\begingroup$ Your question is based on a number of statements, which you assume to be correct, but virtually none of them is, so the entire question has no meaning. While the references to the event horizon, Pauli exclusion principle, and singularity are incorrect, what really stands out is “the mountains supported from below the horizon”. This is not at all how the Schwarzschild geometry works. However, based just on the title, the answer is trivial - there are no black holes (or even gravity at all) in Quantum Mechanics. $\endgroup$
    – safesphere
    Commented Oct 17, 2023 at 20:39

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When we say that general relativity and quantum mechanics are not compatible, we mean a number of different circumstances. For example, by linearizing the Einstein-Hilbert action, one can obtain a legal quantum theory, which, however, cannot be renormalized. This means that even if one treats the resulting quantum theory as an effective theory, in regimes where gravity is strong, such as inside black holes, no predictions can be made.

Another aspect is related to Einstein's equations, for example. On the left-hand side we have tensors, which therefore have functions as components, and on the right-hand side the stress-energy tensor. It is not clear what it means to have an operator-valued distribution (as in the quantum case) on one side, and geometry on the other. In semiclassical theory we get around the problem by considering the expectation value of the stress-energy tensor, but the resulting solution sits on a background and it is not clear how to analyze non-linearities.

The thought experiment you reported is therefore not quite correct, since we do not know how to couple quantum interactions with gravity in strong regimes.

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OP asked: Why do we then rely on GR and assume everything collapses to a singularity which seems illogical in nature when the most sensible ('we don't know yet') answer surely should be that there exists a 'black star' under the event horizon?

To properly answer this question, we would require a quantum theory of gravity. In classical GR, the statement that a sufficiently compact neutron star should collapse to a black hole is supported by the Penrose singularity theorem.

It states that the spacetime of a classical gravitational system can be shown to be geodesically incomplete if:

  1. A trapped surface forms
  2. The energy density of a region of space is not negative (weak energy condition). This implies that geodesics end in finite parameter time, hence reaching a singularity.

The Pauli degeneracy pressure of ordinary matter within the neutron star satisfies both the conditions laid above. Hence, a (singularity-possessing) black hole would eventually form.

One interesting phenomenological observation as to why a black hole cannot be supported by ordinary degeneracy pressure is:

  1. The radius of a degeneracy pressure supported neutron star goes as $R \sim M^{-\beta}; \beta>0$ (e.g. see this)
  2. The radius of a Schwarzschild black hole goes as $R = 2GM$, i.e., $R \sim M$

So we find that for black holes the radius goes up with increasing mass and vice-versa for degeneracy pressure supported neutron stars.

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  • $\begingroup$ This answer is incorrect. “the statement that a sufficiently compact neutron star should collapse to a black hole is supported by the Penrose singularity theorem” - No, it is not. As we observe from outside, the trapped surface never forms in the eternity of cosmological time, so the singularity theorem doesn’t apply. And since nothing in the universe is older than the current cosmological time, there are no singularities in real astrophysical black holes. Also the radius of the Schwarzschild black hole is timelike and therefore is zero meters, so the second part is also incorrect. $\endgroup$
    – safesphere
    Commented Oct 18, 2023 at 9:28
  • $\begingroup$ I never said that you have to use the coordinate w.r.t. an outside observer. Again I do not know what you are saying in the second part either. You seem to be using the coordinate system of an external observer. For the second bit you can refer to this paper for more context: arxiv.org/pdf/2307.06164.pdf $\endgroup$
    – S.G
    Commented Oct 18, 2023 at 9:40
  • $\begingroup$ It is globally hyperbolic because the singularity is enclosed by the horizon. The presence of a singularity is coordinate independent as you can see yourself by working out the Kretschmann scalar for Schwarzchild spacetime en.m.wikipedia.org/wiki/Kretschmann_scalar $\endgroup$
    – S.G
    Commented Oct 18, 2023 at 20:11
  • $\begingroup$ Then what's the point of the Cosmic censorship principle? en.wikipedia.org/wiki/Cosmic_censorship_hypothesis. Additionally, I do not think I can convince you on the singularity bit. It seems that you are again fixated on using the coordinate system of an outside observer. $\endgroup$
    – S.G
    Commented Oct 19, 2023 at 19:59
  • $\begingroup$ No, all my “light cone” and “causality” statements were coordinate independent. Cosmic censorship mostly prevents naked timelike singularities, which can be in the past light cone of other events and thus could affect things in unpredictable ways. Cosmic Censorship doesn’t apply to spacelike singularities (at least not to all). For example, it doesn’t apply to the Big Bang, Big Crunch, or to Schwarzschild black holes. $\endgroup$
    – safesphere
    Commented Oct 20, 2023 at 9:32

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