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I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and "weird things happen".

Now to the question at hand, please follow my slow reasoning... As far as I've learned, to reach a black hole singularity, you must first cross an event horizon. The event horizon has this particular property of putting the external universe on an infinite speed to the falling observer. Now due to the Hawking radiation, and knowing that the cosmic background radiation is slowly dimming, sooner or later every black hole in this particular instance of inflation we are living in will evaporate, according to an external observer of said black holes.

This means that every black hole has a finite timespan, as long as this universe survives that long. Now if we go back to the falling observer, we had already established that such an observer would see the outside universe "speed up" infinitely. This means that when the falling observer "hits" the event horizon, he will (or it if we speak about particles, which is clearer in this case), be immediately transported in time towards the final evaporation moment of the black hole. Either this or the particle gets some weird treatment. My point is, such a particle never gets to the singularity, because it has no time to get to it. The moment it crosses the event horizon, the black hole itself evaporates.

Where am I wrong here?

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This question also bugged my mind and I've also asked the more or less same question, you may want to check the answer to it. –  Cem Jan 24 '11 at 0:57
You're not wrong. Physicists have no idea what actually happens to an observer falling into an evaporating black hole. There are arguments that reach contradictory conclusions, and this is one of them. –  Peter Shor Jan 24 '11 at 1:33
Could I suggest limiting this to a simple question, like the title! That is an excellent question that is already somewhat involved (it requires explaining what a Penrose diagram is). The rest of the logic contains some misconceptions that make answering the title question difficult. –  Columbia Jan 24 '11 at 2:36
I agree with Columbia. There is the essence of an interesting question here, but as is it's not very clear. –  David Z Jan 24 '11 at 4:53
The OP means "Cauchy horizon" not event horizon. You do not see things infinitely speed up at the event horizon, but at the inner Cauchy horizon. –  Ron Maimon Aug 19 '11 at 4:57

7 Answers 7

Indeed you made one mistake: the infalling observer does not see the outside universe "speed up". Look at what happens in a space-time diagram. At the spacetime point where your astronaut passes the horizon, he can only see what's in his past light cone, and that's the universe at early times only. It's the signals that he sends back (or tries to) that reach the outsine world only at infinite times.

Thus, the observing astronaut sees his black hole as it is long before any evaporation sets in, so his black hole is still there. Now leaving aside some other quantum issues, where opinions aren't completely settled and perhaps even our presently used language could be inapropriate, the observer just continues on, and in a finite amount of time, very quickly unless the black hole were more than millions of times heavier than the sun, he is killed by the central singularity.

In a black hole with high angular momentum (Kerr black hole), the singularity takes the form of a ring along the equator, and the astronaut might try to sail past it safely, and he would be able to enter into a strange new universe where he may or may not leave a negative mass black hole behind him, were it not that debris from other objects that fall in later will kill him before that happens, and while trying to pass a second horizon, he will be killed because that second horizon is unstable.

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The black hole the astronaut leaves in the new universe is not negative mass, it's the same as the original black hole, except sometimes reflected in angular momentum sense (depending on how you coordinatize the extension). Also, there is an "infinite speedup" for the external universe at the Cauchy horizon, and this does make it tough to cross the Cauchy horizon. It is not plausible anymore that the observer will go to another universe, since there is no information loss. I believe it just comes out in the same universe. –  Ron Maimon Aug 21 '12 at 3:29
An external observer sees the black hole evaporate in a finte time, but the same observer measures an infinite time before the infalling particle reaches the event horizon. That means the observer will see the black hole evaporate before the particle crosses the event horizon. Is this true, or are my assumptions wrong? –  John Rennie Aug 21 '12 at 10:25
@G.'tHooft: I am not wrong, I got it right. I did it myself, and I have no doubt. I disagree that the Cauchy horizon is unstable, this is Penrose's propaganda, and it is not supported by detailed calculation. The metric has a sheet discontinuity at the worst, there's no reason you can't pass this. The second horizon is crossed either in t going positive or negative (t is the spacelike coordinate, not r) and this is why I said the black hole is opposite spin (but this depends on how you coordinatize the exterior regions). There is never a negative mass black hole anywhere. –  Ron Maimon Aug 23 '12 at 0:55
@G.'tHooft: Since you carry authority, it helps if you admit when you are wrong, not leaving it unresolved. The mass is defined from the metric at infinity, and there is no exterior region in which gravity is repulsive, the exterior regions are all the same. Please don't repeat wrong things--- you can fix them by editing. I repeat: There is no negative mass black hole anywhere in the extended solution, this is wrong, wrong, wrong, for the benefit of people easily swayed by authority. –  Ron Maimon Aug 23 '12 at 4:18
You were talking of rotating black holes. If they don't rotate the negative $r$ region is closed off by a singularity. But if the holes rotates, you can get at $r<0$ because the singularity is only at the equator - you need to take a northern or southern route, also to avoid the region with closed timelike curves (at small negative $r$ and large $\sin^2\theta$. I'm talking of the Kerr and Kerr-Newmann metric (necessary for rotation). –  G. 't Hooft Sep 6 '12 at 16:17

Nothing is unusual to an observer falling into a black hole at the event horizon. He does not "hit" it. It is crossed with no fuss or bother. As he falls farther and farther into the black hole, though, tidal gravitational forces "spaghettifi" him. I do not know what you mean by saying that the external universe is "speeded up infinitely." The moment the particle crosses the event horizon, the black hole does not evaporate. The event horizon is not some sort of solid physical barrier. The particle will approach the singularity, but GR breaks down at/near singularities.

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Nice, Gordon! +1. –  Luboš Motl Jan 24 '11 at 7:30
This answer does not count the black hole evaporation and other concerns rised in the question. –  Anixx Feb 2 '11 at 22:20
@Anixx: The answer by Gordon is correct; this is the paradoxical thing about Hawking radiation. It is not observed by the observer falling in. –  G. 't Hooft Aug 22 '12 at 22:17
@G. 't Hooft the failing observer can always ask the distant observer. Also you're wrong: Hawking radiation is not detected by a close observer only if the BH diameter is much greater than the observer's dimensions, because its wavelength is about the diameter of the BH. –  Anixx Dec 7 '12 at 0:47

For some videos simulating what you would see falling into a black hole look at:


The outside universe does not speed up for an inertial observer falling into the black hole. If an observer hovers over the event horizon then the exterior world does appear to speed up. The observer who falls into a black hole will within a finite time period reach the singularity. However, tidal forces grow enormously and the observer is pulled apart before actually reaching it. In fact atoms and nuclei will be pulled apart before hand.

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What happens inside the event horizon of a black hole is at best a speculative question. Einstein's general relativity is the accepted gravitation theory. According to it, we cannot obtain information from an object dropped through the event horizon. So any answers you get will be sorely limited by the fact that no such experiment has ever been performed nor are we ever likely to perform one.

Current astronomical observations of the center of the galaxy suggest that Einstein's GR is working well fairly close to the event horizon. But GR blows up at the singularity so its predictions there are suspect at best.

It appears that the only way of obtaining information about that singularity is to jump into a black hole. You will not be able to get information back to your friends, but you might find out yourself. On the other hand, tidal effects might kill you before you get close enough.

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The interior state becomes important for quantum black holes. The horizon becomes uncertain, which puts the exterior states (holographic fields on the horizon etc) in some superposition with the interior states. So if one measures a qubit on a quantum black hole there is a probability that it may be an interior state. For an astronomical black hole it is clear nothing can be ascertained about the interior and quantum mechanics is FAPP inoperative. –  Lawrence B. Crowell Jan 24 '11 at 18:51

This puzzle is an aspect of the black hole information loss paradox and a proposed solution is the holographic principle and black hole complementarity.

The classical view of black holes is that any object which falls in ends its worldline on the singularity but an outside observer never sees this because the object appears to be frozen on the horizon. The paradox arises when this is considered in the light of quantum mechanics which tells us that the black hole can evapourate due to Hawking radiation. This means that the information about the object must be returned as part of the radiation.

The problem can only be resolved with a theory of quantum gravity and although our theories of quantum gravity are incomplete some theorists have worked out some principles that govern how the solution might work. One part of the solution is the holographic principle that requires that the information is stored on the event horizon. The second part is black hole complementarity which says that the fate of the object is observer dependent. To an observer outside, the object stops at the horizon and gradually returns its energy and information to the surroundings as Hawking radiation. To an observer who falls into the black hole with the object its fate is very different. the object continues to pass the horizon and is destroyed when it hits the singularity.

Since two such observers can never meet up and compared notes there is no physical contradiction between these complementary views.

Of course this is a speculative solution since it is far beyond anything we can currently test experimentally

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The idea that the observers will see different events is contrary to the very basic principles of any involved theory and the idea of singularities is contrary to cosmic censorship. Look yourself: by the time the external observer sees the falling observer destroyed (and information returned), the falling will find himself destroyed too. It is not evident why the falling observer cannot observe the information return the same way as the distant observer, other than he will be destroyed in the process. –  Anixx Feb 2 '11 at 23:04

It would seem that the area beyond the event horizon is disconnected from the rest of the universe. If it is then we must consider larger questions such as: is it even governed by the same laws? Besides this though, the singularity is not necessarily a real object, it is simply the expression that GR can give no information about what happens to space-time in the center of a black hole. We should further ask the question as to whether we can ever have a real theory of what goes on inside the black hole. We could model some equations, but what observer would be able to test them? Unless the area is not disconnected, in which case we need to reexamine GR. Changes to GR could change the very definition of what a black hole even is!!!! So to rephrase what seems to be the consensus: no one knows + it seems we don't have even the foundations to solve this problem yet

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I believe the issue is that external information from infinity will have difficulty catching up to the infalling object, ie, the light cones will not intersect. I have seen this answered in detail based on an infalling object having passed the event horizon, but since that can't happen, the person answering the question has oversimplified the problem. One should be able to approach the event horizon, and then try to accelerate away from it, but they risk being disintegrated by the aggregate of all infalling cosmic rays from infinity.

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