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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.

G. 't Hooft
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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 03:29
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    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
  • @Ron and Killercam: sorry, you both got it wrong: in the Penrose diagram there are several spots where you might hope to cross the second horizon, at $r\rightarrow+$ or $-\infty$. The negative $r$ choice will land you in a negative mass black hole. But we agree that that horizon is unstable and will kill the astronaut. – G. 't Hooft Aug 22 '12 at 22:04
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    @John, you are right, but only in a formal sense: the infalling particle will close in to the horizon so rapidly, and the evaporation is so slow, that the external observer will not be able to distinguish the signals he receives from Hawking radiation. The signals from the particle will red-shift way too fast. – G. 't Hooft Aug 22 '12 at 22:09
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    @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 00:55
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    @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 04:18
  • I said that the SECOND horizon is unstable. It really means that objects falling in long after the observer himself, will get in his way, with infinite blue shift. There is also an issue about infalling objects in the new universe where your observer left the bh. But I thought we agreed there. What you did wrong was that you did not notice that, in the Kerr case, ignoring the problems with the second horizon, the observer can get out in two directions: $r\rightarrow\pm\infty$. The negative r direction corresponds with negative mass BH. Just look at the equs for metric, and Penrose diagram. – G. 't Hooft Aug 24 '12 at 09:47
  • The metric contains terms $(1-{2M\over r})$, so that if $r<0$ one must replace $r=-r',\ M=-M'$ to describe the extended region there. In Kerr, positive and negative r are smoothly connected. The singularity is only along the equator. We do have another pathology in that region: there is a region with closed timelike curves. Just look at the complete metric expression carefully. Now where is this wrong wrong wrong? – G. 't Hooft Aug 24 '12 at 09:53
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    @Ron I'm afraid you were saying some more incorrect things: you "believe" that the observer comes out in the same universe. But what will he (it) look like? Quite likely is that the observer will not be noticed by the other inhabitants in his (former) universe since he turned into Hawking radiation. All modern theories say that the best he could do is give some subtle twists in the Hawking particles of the same BH that he entered. He (it) turned into a ghost. – G. 't Hooft Sep 05 '12 at 15:36
  • And then, sorry, but the location of a horizon is a function of the $r$ variable, not $t$, so after passing that second horizon you will enter into the negative $r$ regime. The coordinates $r$ and $t$ do interchange, in the sense that, between the two horizons, $t$ becomes spacelike and $r$ timelike. But does the BH get opposite spin? Please think: how did you define the spin direction, in which coordinates? The statement is empty. – G. 't Hooft Sep 05 '12 at 15:37
  • @G.'tHooft: I did nothing wrong, the metric is the same in all sheets, and there is no "negative mass" region, as mass is defined asymptotically. I know what "current theories" say, I think they are wrong (AdS/CFT has it right, but nobody noticed because of the difficulty of taking the classical limit). For extremal or near extremal charged/rotating black holes, you come out cold, not thermalized. Here's a description of the idea: http://physics.stackexchange.com/questions/35506/reasons-to-suspect-that-matter-is-emitted-from-black-holes-nonthermally/35518#35518 . – Ron Maimon Sep 05 '12 at 17:01
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    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 06 '12 at 16:17
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    Mass is defined asymptotically, yes, but the mass term always comes in the combination $Mr$, so when $r\rightarrow-\infty$, where the angular part of the metric goes as $r^2$, you have to replace $r$ by $|r|$ to see what happens. So $M$ goes to $-M$. Look up the Penrose diagrams in Hawking and Ellis for example. And look up the Kerr metric (too long for these "comments"). If you do AdS/CFT you are doing quantum, and you won't avoid thermalization. Only the classical theory would seem to allow you to get out, God knows in which universe. – G. 't Hooft Sep 06 '12 at 16:18
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    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. – G. 't Hooft Sep 06 '12 at 16:28
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    @G.'tHooft I know this is an old post and you're not active here any more, but... when you said to John Rennie above that "you are right, but only in a formal sense...", it doesn't seem to quite answer the point. Even if the infalling observer becomes practically unobservable to the outside one, they are still technically outside the horizon, and remain so until after the hole has evaporated. (Right?) So, formally, there is no spacetime point at which the infalling observer passes the horizon, and the argument in your post seems on the face of it not to hold up. – N. Virgo Aug 11 '15 at 03:13
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    @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 ... – G. 't Hooft Aug 12 '15 at 07:44
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    ... 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

     – G. 't Hooft Aug 12 '15 at 07:45
  • There are 3 types of questions here. First, on what general relativity says about the geometry of space and time. The theory is extremely clear and unambiguous. We can answer all questions concerning the perception of space, time, and singularities by all sorts of observers. You get the answers by looking at the equations and their solutions. The only reason why I don't answer such questions is that I have no time for that, and anyway, you can figure it out yourself if your math understanding is above a certain minimum. Second, the question whether the theory is right at various points. – G. 't Hooft Apr 22 '16 at 10:10
  • There have been numerous observations that indicate very strongly that yes, the theory works up to minute detail, and furthermore, if there existed even a slight departure from the theory, the consequences would be huge and highly incredible (observers would be able to distinguish their coordinate fram from that of others, seeing lots of things that have never been possible.) I am conservative, I don't believe in any departure from general relativity, period. But third, there is something else, and that's quantum mechanics (QM). QM forces us to think differently about space and time, ... – G. 't Hooft Apr 22 '16 at 10:11
  • but this becomes physically important only at extremely tiny distance scales (tiny, even, compared to any of the particles studies at the LHC). In QM, the state we call "vacuum" has to be defined carefully, and according to QM, therefore, different observers call different states to be their "vacuum state". That's pretty serious: matter causes gravitational fields. Is there matter or not, at a given point x ? If you can read technical stuff, I advise you to read arXiv:1601.03447[gr-qc]. It tells you that, if you want to know the quantum states of a black hole, ... – G. 't Hooft Apr 22 '16 at 10:12
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    you first have to cut out the interior that isn't visible to the outsider, and then glue the rims together, like in a skin operation. On top of that, particles exchange the data about their positions with the data on their momenta. It's all pretty weird, but emerges from my calculations. G. 't H – G. 't Hooft Apr 22 '16 at 10:13
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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.

Gordon
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  • Nice, Gordon! +1. – Luboš Motl Jan 24 '11 at 07:30
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    This answer does not count the black hole evaporation and other concerns rised in the question. – Anixx Feb 02 '11 at 22:20
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    @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 07 '12 at 00:47
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    Two theoretical physicists endorsing a completely wrong answer is quite remarkable. – safesphere May 25 '20 at 05:52
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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.

Lawrence B. Crowell
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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

Philip Gibbs - inactive
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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 02 '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.

Carl Brannen
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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
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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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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.

Doug Weller
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  • 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. – Doug Weller May 25 '17 at 03:41
  • +1 to cancel the unjust downvote. – safesphere May 25 '20 at 04:31
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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).

M. Safeer Sadiq
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I do not believe there is even a true paradox here. The confusing factor is our fixation with time as measure. Proper time is the measure and predictor for local events, but it is the events themselves that must take priority. Consider clock pulses coming from outside the hole: it doesn't matter when the falling observer* receives them, only where it is at the time it receives them: and that is outside the event horizon. Proper time does not "progress" to the point that the observer penetrates the black hole. End of argument, I think.
Models in proper time embed the fallacy that proper time progresses beyond horizon penetration.

There are loads of other fallacies in the arguments above. Selecting one of the more egregious: the fact that the received light cone is narrow does not in any way restrict the places from which light that can reach you. Look at the ray diagram for a fish's eye-view of the world outside the pond as a simple (and perhaps surprisingly relevant) example

*For want of a better word

george storm
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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.

Ping Pong
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