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, the observer will see the universe above "speed up", and can see any future date arrive at a distant point 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?

  • $\begingroup$ Are there non-evaporating black holes? $\endgroup$ – Georg Feb 7 '11 at 11:40
  • 1
    $\begingroup$ @honeste_vivere was right; that edit makes no sense as an edit. $\endgroup$ – Beta Jun 22 '16 at 0:48

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.

  • $\begingroup$ Sorry if this answer is too babbly. I can clarify in comments if need be. $\endgroup$ – Jerry Schirmer Nov 17 '10 at 2:04
  • 4
    $\begingroup$ +1 for your third point about the Penrose diagram, in particular $\endgroup$ – David Z Nov 17 '10 at 2:29
  • $\begingroup$ 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. $\endgroup$ – Beta Nov 17 '10 at 2:46
  • $\begingroup$ 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. $\endgroup$ – Jerry Schirmer Nov 17 '10 at 2:50
  • $\begingroup$ 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. $\endgroup$ – 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.


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.

  • $\begingroup$ I finally see that you actually know stuff about an evaporating black hole so I upvoted your answer. I think there are a lot of idiots out there who made a mathematical mistake in thinking that the infaller will not even reach the event horizon before the black hole gets significantly smaller by Hawking radiation which is probably why this question had no more upvotes than downvotes before. That doesn't change the fact that a certain assumption of quantum mechanics still predicts that there is a firewall at the event horizon. $\endgroup$ – Timothy Jul 10 '18 at 4:19

Evaporation of a black hole is an event takes place before a particle can reach the event horizon in coordinate time. When the Schwarzchild metric is used to track events outside the event horizon, there is only one reality, regardless of the coordinates used to make measurements. That is, as calculated using the Schwarzchild metric, whether measured in coordinate time or local (proper) time, a black hole will evaporate before a particle can reach the event horizon. This suggests it is physically impossible for anything to cross an event horizon. If you want a more complete explanation, I wrote a short article on the topic: Weller D. "Five fallacies used to link black holes to Einstein’s relativistic space-time." Progress in Physics, 2011, v. 1, 93.

  • $\begingroup$ I think this is wrong. Just because an outside observer never observes you cross the event horizon doesn't mean you didn't cross the event horizon in a short time. The black hole actually finishes evaporating way later than you cross the event horizon. $\endgroup$ – Timothy Jul 10 '18 at 4:09
  • $\begingroup$ There is only one reality and one logical sequence of events. The vantage point from which observations are made does not change the logical sequence. If from outside the black hole it is observed the black hole evaporates before the event horizon can be crossed, then the same reality occurs wherever observations are made. Because of time dilation resulting from gravity, an object falling towards a black hole will perceive time passing (as measured by the frequency of light) at a different pace than will a distant observer. But both observers will see the same logical sequence of events. $\endgroup$ – Doug Weller Jul 11 '18 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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