Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question has been bothering me for a while. I have a crude hypothesis...

As I understand it, an observer falling into a black hole will cross the event horizon at some specific future (proper) time in, and that it will not be a traumatic event if the black hole is big enough (e.g. tidal forces will be quite mild).

Also, an outside observer will never observe the infaller quite reach the event horizon but it does not work the other way. The infaller will not observe times arbitrarily far into the future before crossing the event horizon..

Also, black holes evaporate, which may lead to some caveats about the previous two statements (which do not take evaporation into account).

So suppose we have a large black hole, destined to evaporate and vanish in the the year 1050 AD. And suppose I jump into it, equipped with a telescope that lets me observe the Earth. Before I reach the event horizon I will see 1050 AD arrive on earth. At that point I will see astronomers on earth waving flags to indicate that they have seen the black hole vanish. So if I look "down" I will see empty space with no black hole looming. So where am I? If I'm just adrift in space, am I in a cloud of all the other objects that ever fell into the hole?

Now for my crude hypothesis: as I fall, and the hole gets smaller, and the curvature near the horizon gets more acute, I'll be racked by tidal forces and blasted by Hawking radiation. Any extended body I happen to have will be disintegrated, so "I" will survive only if I'm an indestructible point, and the cloud of such particles is what astronomers see as the final flash of Hawking radiation. Is this even close to plausible?

share|cite|improve this question
Are there non-evaporating black holes? – Georg Feb 7 '11 at 11:40
@honeste_vivere was right; that edit makes no sense as an edit. – Beta Jun 22 at 0:48
up vote 17 down vote accepted

A few things:

1)Just because an observer crossing the event horizon doesn't necessarily feel ill effects AT THE TIME OF CROSSING the horizon, it doesn't mean that they won't inevitably end up at the singularity, where there will be plenty of ill effects--all timelike curves that cross the horizon end up at the singularity in a finite amount of proper time. For a particle falling into a non-spinning black hole, it's actually the same amount of proper time that it would take to fall into a Newtonian point mass.

2) You have to be very careful about what you mean by 'horizon' in the case of a black hole that eventually evaporates. There are several definitions of 'horizon', and depending on how you resolve the singularity, and upon how the hole evaporates these different definitions can differ in meaning--the most common difference is the apparent horizon-a 'point at which, for this given time, you can't go back', and the event horizon--'the point at which, you MUST end up at the singularity'. It might be possible that your evaporating black hole spacetime may have an apparent horizon but no event horizon, for instance. In that case, the whole paradox goes away.

3) A careful answer of this requires the careful drawing of a Penrose-Carter diagram of the relevant spacetime. If you managed to tweak it somehow so that you fell in, blasted your rockets for long enough to outlive the recontraction of the horizon, the short answer is that you wouldn't receive all of the information about all of the future, just that determined by the "null past" of the horizon--you would find out about all of the lightlike and timelike rays that fell into the horizon, but not those that would head toward the spot where the horizon used to be at times later than when the horizon was there.

share|cite|improve this answer
Sorry if this answer is too babbly. I can clarify in comments if need be. – Jerry Schirmer Nov 17 '10 at 2:04
+1 for your third point about the Penrose diagram, in particular – David Z Nov 17 '10 at 2:29
Not sure this answers the question. 1) all right, which happens first, I reach the singularity or I see the flags? 2) if we can agree that I don't cross the event horizon (because it doesn't last long enough or doesn't exist at all) then where do I end up? 3) I'll work on a diagram, but the sequence of events is still unclear to me. – Beta Nov 17 '10 at 2:46
Well, the point of drawing the diagram would be to see that you actually wouldn't see all of the flags--you would see a subset of them, corresponding to the 'null past' of the point at which you cross the apparent horizon (another weird consequence of evaporating black holes is that the stack of their apparent horizons actually forms a traversable timelike surface in the larger spacetime). Since at no point will you be able to outrun local light, you won't get any paradoxes any more severe than what you would get from the path curvature in the twin paradox. – Jerry Schirmer Nov 17 '10 at 2:50
Sorry, the above wasn't quite as clear as I'd like. There would be a spacetime point where the apparent horizon vanishes, or you cross the apparent horizon. Lightlike rays signaling that event would go to the future, and intersect with earth. Then, lightlike rays with information of hte fligs would go back. But, you'd already be out of the hole by then. So no contradiction. – Jerry Schirmer Nov 17 '10 at 4:56

There is an ongoing research regarding your question and some solutions have been proposed. I recommend

In short, it is proposed that the internal energy of infalling observer is fully transformed into kinetic energy and then into radiation. This radiation is called "pre-Hawking" radiation. Although there are some counter-arguments.

share|cite|improve this answer

My understanding is that a freely falling observer who falls into a black hole won't see any future date arrive at a distant point before crossing the event horizon. I think that's true only for an accelerating observer who hovers ever closer to a horizon.

share|cite|improve this answer

We never observe the event horizon begin to form. The entire event horizon is a region of space time outside our past light cone. I don't see how a firewall could form. For a firewall to form, radiation at the event horizon would have to cause the firewall at another part of the event horizon in its own past light cone. Even if we do observe the black hole evaporate away before we observe the infaller pass the event horizon, that doesn't necessarily mean they haven't passed the event horizon before we observe the black hole evaporate away. The event horizon is the region of space beyond which any photon is destined to reach the singularity. If evaporation occurs, it must occur by a process that's happening in all of space time. Any photon travelling directly away from the singularity of a black hole with no angular momentum that's where the event horizon would be if it weren't for evaporation will actually get closer to the singularity as a result of evaporation and therefore reach it. That means the event horizon is actually defined to be further because it's going to evaporate. We never observe the infaller reach the event horizon. Despite the fact that the event horizon is defined in terms of the black hole's future, that doesn't contradict the arrow of time going only in the forwards direction because the more time that goes by, the more precisely we can define the location of the event horizon at that time based on where we see the infaller now. Although the Hawking radiation is independent of what fell into the black hole, information is not lost. The information just goes outside the fabric of space entirely as a disorderly vacuum fluctuation once it reaches the singularity. At the macroscopic level, matter gets destroyed at the singularity and there's no mass of infinite density there. Despite the time reversibility of quantum determinism, there's a macroscopic arrow of time preventing the time reversibility of a singularity and therefore a white hole. After enough of the black hole evaporates, we actually should be able to see a small area of space just outside the event horizon and an even larger area of it closer to the event horizon where we see the infaller. That doesn't mean it is larger closer to the event horizon. It just means the region closer to the event horizon is more red shifted to us so we observe the part very close to the event horizon from longer ago when the black hole hadn't evaporated as much. It can't possibly be the case that virtual particles only become real particles when they're on either side of the event horizon because the location of the event horizon is defined by the black hole's future. Maybe for some complex reason, quantum mechanics just happens to be consistent in the absence of gravity despite the fact that it includes annihilation of virtual particles which is information travelling faster than light. That would explain why quantum mechanics predicts cause and effect going in the backwards direction and creating a firewall. Since the firewall would be cause and effect going in the backwards direction, maybe a full blown contradiction can be derived from quantum mechanics in the presence of a black hole let alone a black hole firewall. Just because the theory is consistent with observations and noncontradictory in the absence of gravity doesn't mean it's true. Maybe another quantum mechanical theory can properly explain an evaporating black hole without a firewall. On the other hand, maybe the physical laws can be explained by being simulated by quantum mechanics and in the case of a black hole, the simulation produces a result that simulates a firewall which contradicts general relativity. More information can be found at

share|cite|improve this answer
No offense, but this looks like gibberish to me, and the sentence beginning "Any photon..." (one of the few sentences that makes any sense to me) is clearly false. – Beta Jun 9 at 0:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.