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Assume you are immortal, then say if you fell towards a blackhole, to an observer far away you will appear to slow down as you approach event horizon and gradually come to stand still.

However for you, your time flow will appear normal and never stopped.

Now as we know that there cannot be two different times in the fabric of spacetime , its same time flowing slower near heavier objects than away from them.

In other words you'll have some very real time flow difference like your 1 second = years for far away observer.

Now as you fall towards blackhole you should see the rest of the universe speeding up like a fast forwarded video tape (ignore lack of actual vision due to light bending and other things). By this logic when you hit the event horizon the universe's time would be going infinitely faster and hence the blackhole would have evaporated by that time and universe would end before you cross the event horizon.

So how can you ever fall inside one?

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marked as duplicate by Diracology, Qmechanic Jul 28 '16 at 5:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ See physics.stackexchange.com/q/21319/50583 $\endgroup$ – ACuriousMind Jul 27 '16 at 22:23
  • $\begingroup$ Time is that which the clock shows. Can there be more than one clocks? Yes. $\endgroup$ – CuriousOne Jul 27 '16 at 22:36
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    $\begingroup$ Possible duplicate of Can black holes form in a finite amount of time? $\endgroup$ – Diracology Jul 27 '16 at 23:31
  • $\begingroup$ "Now as you fall towards blackhole you should see the rest of the universe speeding up like a fast forwarded video tape " - If one hovers above the horizon, this is true. If one free-falls towards the horizon... $\endgroup$ – Alfred Centauri Jul 28 '16 at 0:20
  • $\begingroup$ The time on those two clocks would be inter related. When someone sees you stopped near horizon he's seeing just a second or two of your clock and its still the same time flowing slowly for you . The reality never splits, in all reference frames you'd be outside the horizon and once you fall through you'll fall through in all frames $\endgroup$ – user1062760 Jul 28 '16 at 8:00
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What you see depends on what light does before it gets to you. You can not discuss the reality inside a falling reference frame and "ignore lack of actual vision". Information can only travel with the speed of light. In General Relativity there is no instantaneous measurement.

You also have to keep in mind that you are describing a dynamical black hole; its mass changes due to to evaporation. So the metric/physics around the black hole change with time, wich makes the discussion much more involved.

I actually can not give a decisive answer because I lack the proper understanding of all the aspect in such a complex situation. First one would need to describe the geodesic of the falling observer and then geodesic of light signals coming to it. I find it very hard to give or even understand answers concering such motions; general relativity is very very different from our daily intuition.

EDIT: The authors of this paper Radial fall of a test particle onto an evaporating black hole state:

From the point of view of the test particle, the black hole seems to evaporate as soon as the particle approaches the horizon.

They use however a rather simple model for the evaporating blackhole and the "real" situation is still unclear.

It should be pointed out that the model used in this paper is mainly of academic interest, since the description of the physics near a black hole horizon still presents a difficult problem which is not yet fully understood...

The major problem I see with the whole paper is, that they use the Schwarzschild metric; which is derived assuming a static situation. So I am not convinced that the arguments made there for the point of view of the test particle have something to do with "reality". But since our whole understanding of black holes is still lacking a lot, I do not know if there is "the right answer" to such a problem.

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