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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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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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    $\begingroup$ 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. $\endgroup$ – Ron Maimon Aug 21 '12 at 3:29
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    $\begingroup$ 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? $\endgroup$ – John Rennie Aug 21 '12 at 10:25
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    $\begingroup$ Your theory about leaving the black hole alive is not going to hold up at all, it's wild guesses that would create totally unnecessary conflicts with everything we do know about black holes. Simple thermodynamical arguments tell us that, in the real world, the only thing that can get out of a black hole is Hawking radiation. Astronauts would be suppressed by gigantic Boltzmann exponentials. The classical story would have you come out in some other universe; that's against any thermodynamical law so with hbar, it won't happen, even there. $\endgroup$ – G. 't Hooft Sep 6 '12 at 16:28
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    $\begingroup$ @Nathaniel, As for the formal nature of the question, you have to realise that the outside world sees the ingoing spaceship (or whatever) approach the horizon exponentially, that is, every microsecond or so, his distance from the horizon is reduced by a factor 1/2. Imagine what that is after millions of years when the black hole decays. But apart from that, you are right and the question is important. Today's understanding of physics at the ultrashort distance scale is quite incomplete, no matter what string theorists and AdS/CFT considerations try to tell you. I strongly suspect that ... $\endgroup$ – G. 't Hooft Aug 12 '15 at 7:44
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    $\begingroup$ ... what is needed is a fresh look at scale transformation symmetry and conformal symmetry. A way out could be that such symmetry transformations transform the singular situation with an eternally shrinking observer into a regular description of physical events. That would remove the contradiction, but the technical details are not understood - at least not by me. Questions we can answer are ones about the mathematical nature of any given space-time, like some solution to Einstein's equations, but as soon as new physical principles are asked for, we have to be very careful about what to say $\endgroup$ – G. 't Hooft Aug 12 '15 at 7:45
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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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  • $\begingroup$ Nice, Gordon! +1. $\endgroup$ – Luboš Motl Jan 24 '11 at 7:30
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    $\begingroup$ This answer does not count the black hole evaporation and other concerns rised in the question. $\endgroup$ – Anixx Feb 2 '11 at 22:20
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    $\begingroup$ @Anixx: The answer by Gordon is correct; this is the paradoxical thing about Hawking radiation. It is not observed by the observer falling in. $\endgroup$ – G. 't Hooft Aug 22 '12 at 22:17
  • $\begingroup$ @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. $\endgroup$ – Anixx Dec 7 '12 at 0:47
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For some videos simulating what you would see falling into a black hole look at:

http://jila.colorado.edu/~ajsh/insidebh/

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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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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  • $\begingroup$ 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. $\endgroup$ – Anixx Feb 2 '11 at 23:04
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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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    $\begingroup$ 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. $\endgroup$ – Lawrence B. Crowell Jan 24 '11 at 18:51
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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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In general relativity, (unlike in special relativity where time-space can be made universal) there is no concept of universal time-space, thus general observers have observations those are highly dependent at the space-time locations of the observers. Two observers who are standing apart in space-time may observe the same phenomena with astonishingly different results. The observer who is falling towards the event horizon will observe that he is speeding up towards the event horizon then reaches it and falls in the black hole. The observer who is at a safe distance from black hole will see that a person who was falling in towards the event horizon eventually slows down as he reaches the event horizon and stops there, never reaching the event horizon for billion and trillion of years(according to his watch).

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Perhaps the following description of the complete journey of the particle will bring clarity.

As a particle falls radially towards the event horizon of a black hole, time progresses differently depending upon the reference frame from which it is measured, because of the effect that gravity and motion has on the passage of time.

A first integral equation, derived from the Schwarzschild metric, allows calculation of the passage of coordinate time, as experienced by the distant observer. A second integral equation, derived from the Schwarzschild metric allows calculation of the passage of local time, as experienced by the particle. As long as the particle is outside the event horizon of the black hole, both the integrands are defined and the intregrals are perfectly well-behaved. So the journey of the particle can be tracked with certainty all the way until the event horizon.

As the falling particle gets closer to the event horizon, time goes quicker when measured in local time than the journey measured in coordinate time, because of the relativistic effects of motion and gravity.

Since velocity equals distance divided by time, there is a different perception of speed depending upon whether the distant observer or the particle is making measurements. For each location reached, the particle thinks it got there quickly, so it thinks it is going fast. The distant observer thinks it got there slower, so it thinks the particle is slowing down. The particle arrives at the same location, there is just a different perception of the amount of time it took to get to the location.

This perception of the speed slowing down from the perspective of the distant observer continues until the forward progress of the particle seems to be very slow. So slow, in fact, that at 10^60 years or so, when the black hole evaporates, the particle’s journey is at an end location that is outside the event horizon of the black hole.

Now, from the particle’s perception, it was going pretty fast when it reached the end location. It was trucking along at a fine speed, when all of the sudden the black hole instantly evaporated.

This is the scenario from the calculations of the integrals derived from the Schwarzschild metric. The integrals are perfectly well behaved, so there is no need to use any special coordinates. However, according to the theory of general relativity, the journey calculated from any reference frame should give the same result.

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  • $\begingroup$ If you want a more rigorous explanation, see my article: Weller D. "Five fallacies used to link black holes to Einstein’s relativistic space-time." Progress in Physics, 2011, v. 1, 93. $\endgroup$ – Doug Weller May 25 '17 at 3:41
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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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