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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?

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    $\begingroup$ 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. $\endgroup$
    – Harry Johnston
    Mar 10, 2017 at 0:49

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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.

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    $\begingroup$ 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.) $\endgroup$
    – jpmc26
    Mar 9, 2017 at 21:54
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    $\begingroup$ @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. $\endgroup$
    – zephyr
    Mar 10, 2017 at 13:31
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    $\begingroup$ 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. $\endgroup$
    – toniedzwiedz
    Mar 11, 2017 at 11:23
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    $\begingroup$ 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. $\endgroup$
    – Howard Miller
    Mar 11, 2017 at 21:14
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    $\begingroup$ 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). $\endgroup$
    – Steve Linton
    Mar 14, 2018 at 17:04
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Singularity means "my theory doesn't work here". In other words, GR is unable to predict what happens at the point, so it calls this point a singularity.

The most important thing is not to mistake map for the territory. GR is the map, a real black hole is the territory. GR is the map which allows us to predict what we will find in the territory.

If the map says "don't really know about this point" you really shouldn't expect that when you go into the territory you will see an Unmeasurable Infinite Thing there. It's very much against our historical experience. To date, time after time we observed normal finite things in the territory, but we never have seen Unmeasurable Infinite Things. In every case when an old map said we would see infinity, we found that measurements of a territory were finite and so falsified that map (that theory).

So it seems we should expect singularity as a word to only refer to the map. You really shouldn't expect to observe a singularity (a map thing) when your actual ship enters a black hole (a territory thing).

It might turn out that GR is approximately right about an event horizon, but we already know it is not good enough to describe what is at the center.

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    $\begingroup$ A lot of us amateurs don't realize that what we think about the universe is really just a set of equations that tend to predict what we observe, and sometimes what we observe doesn't exactly match what the equations predict. And some people actually love that when it happens. $\endgroup$
    – Howard Miller
    Mar 11, 2017 at 23:07
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    $\begingroup$ Since the question asks about singularities, which are a prediction/problem with GR, then an answer in terms of GR would be better. $\endgroup$
    – ProfRob
    Mar 15, 2018 at 9:20
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A real neutron star would start to collapse when the strength of its gravity exceeds the strength of the neutron degeneracy pressure, before it has an event horizon.

As you approach the event horizon, the force required to stop a stationary mass from falling in approaches infinity. So I don't think any finite force, fictional or otherwise, can keep a star star-shaped after it reaches the critical density at which it has an event horizon.

That being said, predicting exactly what happens at or inside the event horizon would likely require a quantum theory of gravity, and I don't have one of those.

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    $\begingroup$ All observed neutron stars have masses too high to be supported by "neutron degeneracy pressure". As was established by Oppenheimer and Volkhoff in the late 1930s. $\endgroup$
    – ProfRob
    Mar 15, 2018 at 8:53
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The answer specifically about the question of a neutron star disappearing inside an event horizon but remaining in some sort of equilibrium is no. At least, it is no according to General Relativity, which is the only respectable game in town at present.

It is no for two reasons. Firstly, in GR, the pressure that supports a star is also a source of gravity (or space-time curvature). The increasing pressure required to support a neutron star of increasing $M/R$ ultimately becomes self-defeating, irrespective of what mechanism or particle provides that pressure. This limit is reached at about 1.2 to 1.4 Schwarzschild radii (depending on how the pressure and density of the material are related), and stable objects cannot be smaller.

Second, the mathematics of GR clearly shows that inside the event horizon an object cannot be stationary and that its radial coordinate must decrease and a singularity (or a breakdown in GR as $r \rightarrow 0$ if you prefer) will be formed on a timescale of $\sim r_s/c$, where $r_s$ is the Schwarzschild radius. This is as inevitable as the increase of time is outside the event horizon.

The details might be slightly different for a spinning (Kerr) BH. The formation of a singularity is still expected, but an isolated Kerr BH may form a ring-like singularity. This does not alter the impossibility of having a stable/static object inside the event horizon (in GR) and a "singularity" is expected to form.

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zephyr is right that you would need quantum gravity to really understand what happens inside event horizons. But the traditional description of what would happen inside the event horizon of a black hole (more or less ignoring quantum mechanics) is that there is no force that can stop matter from forming a singularity. The coordinate system inside the event horizon is such that, speaking crudely, the future direction points towards the center. So you can't just have a pile of matter dense enough to fit inside an event horizon, and at the same time strong enough not to collapse to a singularity.

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  • $\begingroup$ Is there a calculation of the maximum proper (subjective) time to reach the centre? IE once you're inside a (Schwarzchild for the sake of simplicity) event horizon of radius R, how long can your future experience be before you necessarily experience a singularity (from the inside!). $\endgroup$
    – Steve Linton
    Mar 14, 2018 at 17:01
  • $\begingroup$ @SteveLinton Yes there is. For a Schwarzschild BH it is $\pi r_s/2c$, where $r_s$ is the Schwarzschild radius. $\endgroup$
    – ProfRob
    Mar 15, 2018 at 8:55
  • $\begingroup$ So for the largest known supermassive black-hole (about 10^10 solar masses) you have about 1.8 days after you cross the event horizon before you must hit the singularity, no matter what you do. So much for the Heechee! $\endgroup$
    – Steve Linton
    Mar 15, 2018 at 11:18
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A black hole could be any size. It could be the size of a planetary system. Given a specific distribution of matter, I think you don't necessarily need a singularity to be present.

But given the fact that matter inside the black hole cannot escape it, it will eventually all move closer and closer and end up creating that singularity.

Even if we imagine a rigid body black hole with structures to keep it from imploding, internal entropy would eventually make those structures collapse and you'd still get a singularity pretty fast.

That's all based on my understanding on black holes, I am not a scientist. I'd love if someone could tell me if I understand correctly.

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  • $\begingroup$ "Internal entropy would eventually make those structures collapse" this puts you firmly on the quantum scale, so please explain this theory of internal entropy in quantum terms. You won't win Nobel for your current formulation, and OP's question is in fact at Nobel prize level and it deserves a serious answer. $\endgroup$
    – kubanczyk
    Mar 18, 2018 at 21:31
  • $\begingroup$ It is theorized that black holes lose mass via Hawking radiation. And it is not necessarily true that the matter inside would move closer and closer together. IMO, since singularities involve infinity, they are not physically real or possible - they are just a mathematical model. Infinity cannot exist in the physical world, as one would need a crossing point from the finite to the infinite, which by the nature of infinity is a paradox. Infinity is a mathematical concept, not a number. If a physics model gives you infinity, then you made a math error, and/or your model is incomplete. $\endgroup$
    – Tristan
    May 18, 2018 at 22:45

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