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This question already has an answer here:

Imagine a very heavy (tens of solar masses) star in its final moments before collapsing to form a black hole. The gravitational force exerted by the weight of the star overcomes the neutron degeneracy pressure and the star continues to collapse inward until the escape velocity at some radius from the star exceeds the speed of light. Now pause at the exact moment this happens.

Am I correct in thinking that at this point we have a black hole, since there is a radius at which light could not escape? If the answer is yes, then from what I understand this black hole must have a singularity. Assuming that a singularity can only exist behind an event horizon, this singularity must have come into existence at this exact moment.

My question is twofold:

  • Are all of the above assumptions / assertions correct?
  • How does a singularity instantly form?

Apologies if the question is naive - I am just a curious innocent rather than an expert!

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marked as duplicate by John Rennie black-holes Apr 12 '16 at 9:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate of: When does a singularity start to exist during a black hole formation? $\endgroup$ – John Rennie Apr 11 '16 at 17:27
  • $\begingroup$ Not really relevant to your question, but neutron degeneracy pressure is certainly not the thing that has to be overcome in order to form a black hole. The equation of state of dense hadronic matter is much harder than that. $\endgroup$ – Rob Jeffries Apr 11 '16 at 21:25
  • $\begingroup$ Thanks for that, I'll look into it. Am I still correct in assuming neutron degeneracy pressure is responsible for keeping a neutron star as a neutron star? $\endgroup$ – acernine Apr 12 '16 at 12:12
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According to the cosmic censorship hypothesis all singularities are assumed to be behind horizons. This is not the same as saying "all horizons contain a singularity".

The black hole solutions to Einstein's equations (which contain singularities) are all stationary, vacuum spacetimes. Stationary means that the spacetime does not explicitly depend on time. The spacetime remains the same for all time. Vacuum means there is no matter anywhere in the spacetime. As a star collapses to form a black hole, the spacetime is dynamic, and there is matter. If we wait long enough, the spacetime will settle into a stationary black hole solution (according to the no-hair theorem). Eventually all of the matter will collapse to a single point. This does not mean that the instant the horizon forms the singularity appears.

If we imagine a star collapsing to form a black hole, at some point the collapsing matter reaches a critical size where all of the matter is inside the Schwarzschild radius. The horizon forms, but where is the matter? It is all inside the horizon, very close to where it was just before the horizon formed. As time passes the matter continues to collapse, eventually forming a point of infinite density (if classical GR is correct) and infinite curvature, the singularity.

One could imagine a different theory of gravity where some quantum gravity effect provides a pressure force stronger than neutron degeneracy pressure and stops the the collapse before the singularity is formed. If all of the matter is inside the horizon when this happens, it would be very difficult (impossible for classical GR) to tell the difference between this state and a black hole. This would be an object with a horizon, but with no singularity at all.

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  • $\begingroup$ Thanks for the detailed answer! I'd up-vote it but I'm afraid I don't have enough rep. $\endgroup$ – acernine Apr 11 '16 at 17:34

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