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Why does there have to be a singularity in a black hole, and not just a very dense lump of matter of finite size? If there's any such thing as granularity of space, couldn't the "singularity" be just the smallest possible size?

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The "singularity" in a black hole doesn't have any matter in it, necessarily. A "singularity" doesn't mean a point of infinite density, it means a place you can reach be geodesics where space-time is not a manifold. – Ron Maimon Jan 1 '12 at 17:40
+1, because the answers you got are very interesting – JollyJoker Jan 10 '12 at 5:38
dulicated or nearly duplicated by – Ben Crowell Aug 29 '13 at 22:30

5 Answers 5

up vote 11 down vote accepted

It's important to understand the context in which statements like "there must be a singularity in a black hole" are made. This context is provided by the model used to derive the results. In this case, it was classical (meaning "non quantum") general relativity theory that was used to predict the existence of singularities in spacetime. Hawking and Penrose proved that, under certain reasonable assumptions, there would be curves in spacetime that represented the paths of bodies freely falling under gravity that just "came to an end". For these curves, spacetime behaved like it had a boundary or an "edge". This was the singularity the theory predicted. The results were proved rigorously mathematically, using certain properties of differential equations and topology.

Now in this framework, spacetime is assumed to be smooth - it's a manifold - it doesn't have any granularity or minimum length. As soon as you start to include the possibilities of granular spacetime, you've moved outside the framework for which the original Hawking Penrose theorems apply, and you have to come up with new proofs for or against the existence of singularities.

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So, then it could be said that any way you look at it, GR breaks down (at the singularity) - singularity or not? – Per Jan 1 '12 at 16:12
Yes, I've often heard it phrased as "GR predicts its own downfall". This is maybe a bit harsh - it's more a case that GR is making a statement about its domain of applicability. – twistor59 Jan 1 '12 at 16:19
@Ron Maimon, well we don't know that yet, there may be naked singularities, the question is still open. But I agree with you strange properties of spacetime doesn't mean that somethng is wrong with the theory, it could be that space time has strange properties. It is not the first in physics. – MBN Jan 10 '12 at 6:30
@pabouk: Not clear. The singularity is timelike if the black hole is spinning or charged, and in this case, we don't have exact collapse solutions. It should be possible to figure out the behavior of exact charged collapsing dust, but it hasn't been done. The singularity theorem has been misinterpreted to mean that there is a spacelike singularity which swallows all matter, and this is only true for spherically symmetric collapse, it fails when the singularity is timelike. My own opinion about what happens in this case is that there is a partial explosion, but this is not what others think. – Ron Maimon Oct 9 '14 at 0:55
@pabouk: Regarding 'settled state', the end point for the black hole is always a Kerr-Neumann static solution, the major issue is what happens to the interior matter in the case that it misses the singularity, i.e. when you have a charged/rotating collapse. My feeling on this is that it is partially ejected, and the black hole is only produced to the extent that mass-energy is destroyed at the singularity, and this is only massless stuff, for highly charged or highly rotating black holes. I suspect the full answer requires string theory, the classical theory has nonsense extra universes. – Ron Maimon Oct 9 '14 at 0:58

Actually there are no singularities inside the black hole. It is just a mathematical special point on the coordinate system, which does not correspond to any real-world singularities.

The event horizon of a black hole is just an impenetrable barrier on which the time is frosen so nothing can pass it. In another model the black hole as a whole behaves as a viscous liquid with quite limited density (the density decreases as the BH's mass rises).

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This is only part of the story, from the exterior perspective. The infalling observer can see falling through, and this is the essence of "black hole complementarity". This is not my downvote, but please fix this answer--- it is well known to be only half the story. – Ron Maimon Jan 9 '12 at 23:14
The infalling observer will not reach the horizon before the BH evaporates. Your claim is derived from a theory that does not account for the fact that there cannot be eternal, non-changing, non-evaporating black holes. But you are free to believe in anything. Even if you were right and infalling observer indeed would reach the horizon, it would not be a subject of scientific research because impossibility to report any observations from there would make application of scientific method impossible. Anything outside of reach of scientific method is not science. – Anixx Jan 10 '12 at 1:52
We have had the exact same discussion before: the exterior of the quantum black hole is complementary to the infalling description. If the black hole is evaporating, to the extent that it is evaporating, it is nonclassical. A quantum black hole still allows objects to fall through from their own perspective, and measurements on the horizon would allow you to reconstruct what is happening in the interior. There is no demonstration that black hole infalling is one-way in a quantum theory, what you do inside affects the Hawking emission, and there could be non-thermal stuff, you could come out. – Ron Maimon Jan 10 '12 at 4:51
"A quantum black hole still allows objects to fall through from their own perspective, and measurements on the horizon would allow you to reconstruct what is happening in the interior." Oh well. Just bear in mind that all information about infalling observer will always be outside the black hole (to the extent which allows to reconstruct the falling observer at any time) and no direct observation of what is inside is possible. You are free to believe that there is a copy of the same information inside it, this is a matter of interpretation, or rather, belief. It is outside of reach of science – Anixx Jan 10 '12 at 14:55
While you are right that you can think that it never went in, and this is a consistent point of view, it is the external point of view, the picture you get on your camera will contain images of the interior when the camera comes out again. The relative position of stuff in the camera will feel that it is falling through, and this is because the notion of "which side of the horizon am I at" is a quantum indefinite notion, which depends on your basis for the Hilbert space. These insights are old, they are due to Susskind, and they come from reconciling unitarity with Hawking radiation. – Ron Maimon Jan 10 '12 at 22:55

See Carter 1968 for why rotating black holes that have incoming disturbances may not have a singularity at all.

A stationary non - rotating hole will have a singularity. But no one thinks that these exist in nature. But with rotation that singularity 'shrinks' to a ring. The set of paths that hit the singularity is shrunk to a mathematical 2D plane from 'all directions' with the Swarzschild Soln. Then with incoming 'noise' it may be that there are no paths - geodesics - that lead to a singularity.

All exact solutions of General Relativity are done with asymptotically flat space, which does not exist in the real world. So while the theory of GR admits singularities, in a real classical GR world they likely don't exist.

Carter actually always talks about a singularity, but one with no paths to it. No ouchy at the end of a path. With no paths to a singularity - is it really there? I would think not, and as Carter points out, others do too. (Lifshitz and Khalatnikov).

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There must be null geodesics which hit the singularity by the Penrose theorem--- the outgoing null rays have to defocus somewhere, and they can't do it at any ordinary point. There is no requirement that any non-null geodesic must hit the singularity, and I don't think they do for rotating/charged black holes, although this is a minority position. – Ron Maimon Jan 2 '12 at 9:06
-1: I have to downvote, because you haven't fixed your wrong statement that Carter says that there is no singularity after perturbation. Carter is NOT saying thi: it contradicts the most basic implications of Penrose's theorem. He is only noting that a generic rotating+charged perturbation leads to a time-like singularity, so that only null rays hit the singularity. This is the content of the paper you linked, it doesn't say what you say here. – Ron Maimon Jan 3 '12 at 6:03
See the paper, in the 'Implications' section at the end. "Thus as the symmetry is progressively reduced... the extent of the class of geodesics reaching the singularity is steadily reduced likewise, until in the case with both charge and rotation there are almost none at all, which suggests that after further reduction of symmetry, incomplete geodesics might cease to exist altogether." As I said Carter thinks that real black holes in nature don't have any geodesics that end on the singularity. – Tom Andersen Jan 9 '12 at 21:10
Another quote, earlier in the paper: "Thus we conclude that when the solution is charged, no timelike geodesics can reach the singularity". I think I see what you are saying - that Carter says the singularity is there, its just that there are no paths to it. I don't see the difference myself - if there are no paths to an 'object' - does the object exist? – Tom Andersen Jan 9 '12 at 21:16
The key word is "almost", and I agree with Carter that nearly all paths miss the singularity. Only a measure zero set of all geodesics, null geodesics, hit the singularity and it doesn't have to be all null geodesics either. Carter is saying "the singularity is there, and there are very few paths to it", namely null geodesics directed at the ring. The idea that the singularity can be removed is speculation by Carter, which it is not really polite to recall, because it only shows he hadn't fully internalized the singularity theorems in 1968 (I am sure he did subsequently). – Ron Maimon Jan 9 '12 at 22:59

For any experiments a spherical Black Hole behaves the same way as if its mass was uniformly distributed over its surface or uniformly distributed over its volume or concentrated in its center. These variants are indistinguishable.

It is impossible to find exact distribution of mass inside a black hole because it has no internal structure, due to holographic principle (if it had, it would be possible to transfer information out of black hole via gravitation waves).

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Please comment what do you not like with this answer. Note that for non-spherical state the BH behaves like if its mass was uniformly distributed over its surface. – Anixx Jan 4 '12 at 2:27
this is a consistent point of view, and I would upvote, but the statement that the interior solution does not make sense is no good. The interior can be found from Einstein's equation uniquely, it is just reconstructed from information available the exterior in the quantum theory. But the exterior is also a reconstruction, at least sufficiently near the black hole, and all space-time is reconstructed in string theory. – Ron Maimon Mar 24 '12 at 5:18
What is "interior solution"? Can its existence be proven by a scientific experiment or falsified? If not, it is not science. – Anixx Mar 24 '12 at 8:22
it is possible to make a black hole so big, that the entire Earth will fall through and nothing will happen. Do you seriously believe that the Earth will not continue as it always has through the transition? I have explained before that I believe you can get data out of a black hole deterministically classically. But even if I am wrong, there are arguments that you can reconstruct the interior spacetime just from exterior measurements. But this reconstruction is real--- you can follow an object through mathematically, and this should reflect its internal state-transitions. – Ron Maimon Mar 24 '12 at 8:30
Even if Earth falls into it how you know that you are under the horizon rather than in a "string-decomposed" state? String-decomposed observer should see all the same, you pointed it yourself. Imagine the following: you being the Earth resident go with earth under the horizon, but then manage to go out (you claimed that it is possible). Now you talk to an Alpha Centauri resident. What measurements should you show him so to convince that you were under the horizon? Note, that he will reply that he all the way saw you with Earth outside(but very decomposed) and can reconstruct what you saw – Anixx Mar 24 '12 at 8:36

Because otherwise general relativity would contradict itself. The event horizon of a black hole is where not even light can escape. Below the horizon all photons must fall. In relativity theory all observers measure the speed of light the same, c; that's a postulate of the theory. Then all physical things (including observers) at and below the horizon must fall and keep falling, lest they measure the speed of light emitted upward to be something other than c. If you could stand on a very dense lump of matter of finite size at the center of a black hole, and pointed a flashlight upward, the photons would somehow have to fall to the ground (without moving upward at all) and you wouldn't measure the speed of light to be c in the upward direction. The theory would be broken. The singularity is the "can't fall further" point and the theory becomes inapplicable there.

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