Black hole without singularity?

  • My question is about the equivalence about having an event horizon and having a singularity.



    In one side the implication looks pretty obvious:




    • A singularity implies having an event horizon and therefore a black hole. Since the mass is compressed in a zero volume space, if you get close enough there will be a point where the escape velocity gets bigger than the speed of light so you will get a black hole by definition.



    But what about the opposite? Does having an event horizon imply the existence of a singularity?



    Could it be that you have a neutron star massive enough to reach a escape velocity equal to the speed of light but not strong enough to make the matter collapse?



    Even if such star can not exist because the strong force collapses before reaching an event horizon, this doesn't mean an equivalence.



    It just means that for some specific value of the maximum strong force this is not possible, but image now an imaginary exotic matter that has a way bigger strong force.



    For such "science fiction" matter, it would be possible to reach an event horizon without collapsing to a singularity, right?



    Or is it really an equivalence between this two concepts, such that no matter how resistant matter is to collapse it will never reach an event horizon?


    The existence of the Kerr Newmann metric suggests that it *might* be possible to have a singularity without an event horizon. I don't know what the latest consensus is amongst real physicists.

  • zephyr

    zephyr Correct answer

    5 years ago

    Does having an event horizon imply the existence of a singularity?




    An event horizon is not an inherent component of any given object. It's not like once a star turns into a black hole, it suddenly gets an event horizon. The event horizon is merely a mathematical boundary which defines the distance from a mass $M$ where the escape velocity equals the speed of light. I can calculate such a boundary for a black hole, for the Sun, the Earth, or even you. So I guess the answer here is no, having an event horizon does not imply the existence of a singularity.




    Could it be that you have a neutron star massive enough to reach a escape velocity equal to the speed of light but not strong enough to make the matter collapse?




    The answer here, technically, is no. The reason being that once it requires a speed greater than or equal to the speed of light to escape your object, it is necessarily a black hole. That is the definition of a black hole. So that means this neutron star you propose is actually a black hole. Another equivalent definition of a black hole is any object whose mass is concentrated inside that object's event horizon.



    But you might still ask, could you have a black hole where the mass inside the event horizon is not a singularity. This would require some sort of support to prevent the matter from collapsing down to the singularity. The answer to this is that it is currently unknown. The problem is that inside event horizons, suddenly you need to work with both GR and quantum field theory but those two theories don't play nice. Instead you should be using a Quantum Gravity theory but this theory hasn't been developed. So ultimately any answer to this would be a guess until this theory gets fully fleshed out.


    If you write an equation for the escape velocity equal to the speed of light and plug in all the numbers for the Earth, wouldn't you get no solution? (I say this assuming that you correctly compute the gravity *inside* the Earth, where the gravity continually decreases due to mass outside your radius having no net gravity. If I recall correctly, you could get a solution inside the Earth by just running the numbers for the entire mass of the Earth, but this is obviously an invalid solution at the surface and beyond.)

    @jpmc26 Yes you are correct. What I was proposing in my answer was that, to calculate any given mass' event horizon, you assume it is a point mass and use the standard $r=2GM/c^2$ equation. Of course, you'll find that if you do this for the Earth, the radius is much smaller than the Earth's actual radius, hence how we know the Earth is not a black hole (aside from some more obvious evidence). The whole point was that the fact that you do get an "invalid" solution, as you call it, tells you the Earth isn't a black hole. It does still technically have an event horizon though.

    Then the Earth doesn't have an event horizon. An Earth-mass black hole would. Or a human-mass one. You're not calculating the boundary for any of the objects you mention. Just for black holes with equivalent mass.

    This is the problem. No one has been able to peek under the skirts of a black hole, so all we can do is make guesses as to what's there. Current mathematical formulas for black holes tend to end it an asymptote or a division by zero. Maybe eventually a new math or a new equation will be able to describe what happens at that point.

    See @MarkFoskey's answer below. Everything inside an event horizon necessarily ends up in the same place (or a least so close to it that GR breaks down and something quantum happens).

    @SteveLinton Was that worth a downvote? I'm not wrong. There is currently no theory capable of accurately describing what happens inside the event horizon (including GR) so to assert that you know 100% for sure that everything inside an event horizon necessarily becomes a singularity is stating knowledge you can't be sure about.

    @zephyr point taken. I may have overreacted. I cannot apparently undo it unless you edit the answer in some way though. GR can describe what happens "almost everywhere" inside the event horizon, but that description always fails at some point on everyones future worldline.

    There is no theory capable of **testably** describing what happens inside the event horizon. GR does describe what happens inside the event horizon and your answer ought to have been based on that.

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Content dated before 7/24/2021 11:53 AM