If I fall into an evaporating black hole, where do I end up? 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?
 A: 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.
A: There is an ongoing research regarding your question and some solutions have been proposed. I recommend 


*

*http://arxiv.org/abs/gr-qc/0609024

*http://arxiv.org/abs/gr-qc/0701096

*http://arxiv.org/abs/1011.2219
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.
A: 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.
A: 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.
A: If the black hole evaporates, a distant observer will receive light signals from an object entering the event horizon before the evaporation signal, and there is no contradiction. The paradox arose because in the steady state case object light signals are received indefinitely into the future of the observer, and it was assumed this would be the case also for evaporation.
A: I believe my first paragraph below is incontrovertible.  The second paragraph is what I understand to be the situation - and may well be mistaken (like 99.9% of what is written about black holes)
What matters is only where the infalling "observer" is at the instant the hole evaporates.  You can (probably should) use proper time as a measure, but you need to use position (relative to event horizon) as the measure for this purpose. So we need somehow to answer the question "how far has proper time progressed at the instant the hole evaporates".
Qmechanic suggest we use the infalling observer's detection of events outside the black hole, and the correctness of this appears to be self-evident.
So the only question is whether light from every time prior to hole evaporation is capable of reaching the location of the infalling observer before it penetrates the horizon.  Ignoring Quantum Mechanical effects for now, it seems that GR predicts that the falling observer remains outside the horizon.
QM might actually 'transfer' the observer inside the horizon, but again the question is not what happens in proper time, but what where the observer is at the instant progressive externally-generated clock pulses reach it.  My understanding is that on this measure the 'falling observe' is actually getting closer to the event horizon - not falling to a central singularity as suggested by analyses that use proper time to the exclusion of all else.
