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Suppose an object falls into a black hole that's so massive that it wouldn't get torn apart at the event horizon.

What happens to it after the black hole evaporates away?

According to the theory of general relativity without Hawking radiation, from observers outside a black hole, it takes an infinite amount of time for it to reach the event horizon but from the point of view of the object falling into the black hole, it passes the event horizon in a finite amount of time.

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    $\begingroup$ The same thing would happen that would happen if the object would fall into the sun: it would dissolve into a very hot thermal state of matter. In the sun it's plasma, we don't know what it is inside the black hole (the string theorists are predicting that it's a very hot string state, in a sense similar to a plasma), but the inescapable conclusion, so far, is that it's extremely hot in there. The enormous gravity of the black hole makes this process impossible to observe from the outside, but it's probably pretty trivial, in the end. "Things" lose their form and then come out as radiation. $\endgroup$
    – CuriousOne
    May 10, 2016 at 3:55
  • $\begingroup$ I think I solved it. The fabric of space is uniformly accelerating in 5-dimensional space time. There exists a hyperplane that in the refrence frame of any velocity in 5D space time, is travelling at c towards the fabric of space and the fabric of space is approaching it but never reaches it except where a blackhole forms. The event horizon is the intersection of the fabric of space with that hyperplane formed by a massive object. Since information can't travel faster than light in 5D space time, the event horizon can never shrink. Therefore, a blackhole never evaporates. $\endgroup$
    – Timothy
    May 12, 2016 at 21:27
  • $\begingroup$ Related: physics.stackexchange.com/q/970/2451 $\endgroup$
    – Qmechanic ♦
    Jun 7, 2016 at 20:41
  • $\begingroup$ Well, it is nice that this question has been bumped to homepage, but it contains invalid assumptions (mass of object somehow determines whether it gets ripped apart at horizon?) and from the comments it seems the author posted it only to promote their non-mainstream speculations. So I am not sure I want to put in the work to answer this properly because a valid mainstream answer will obviously not be accepted. $\endgroup$
    – Void
    Nov 19, 2019 at 12:28
  • $\begingroup$ @CuriousOne - I was under the impression that only very small black holes would be hot inside? $\endgroup$
    – BlackHoleSlice
    Feb 12, 2020 at 23:32

2 Answers 2

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The hypothesis doesn't make much sense. The resistance of an object to being ripped apart is given by its elasticity, not by how much is massive.

Anyway, to the central question there is no answer yet. This is the so called information loss paradox, one of the greatest unsolved problem in theoretical physics.

In general relativity the object falls in a finite amount of proper time and hit a singularity (assuming the most simple black hole, Schwarzschild). Due to Hawking radiation the hole evaporates completely and information is lost because a pure state has evolved into a mixed state.

One of the most promising resolutions to the paradox is given by string theory: the fuzzball proposal. Basically the hole is a sort of very degenerate stringy star (there is an over-over simplification here) free of horizon and singularites that radiates more like an usual piece of burning coal, so the information is not lost. The crucial point is that (in the usual simplified picture of pair particles produced near the horizon) the place in which the particle are produced is NOT vacuum, so the there is not information paradox in the first place.

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  • $\begingroup$ Space is a uniformly accelerating fabric in an uncurved 5-dimensional space time with only one time like dimension. Clearly in the 5-dimensional space time, the reaching of the event horizon does not occur in the future light cone of the disappearence of the blackhole so it must eventually reach the event horizon from its own perspective. Since the fabric of space at the event horizon can't move faster than c, the only way for the blackhole to evaporate away is for space to bud off above the event horizon. $\endgroup$
    – Timothy
    May 12, 2016 at 4:02
  • $\begingroup$ 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. $\endgroup$
    – Timothy
    Jun 7, 2016 at 19:47
  • $\begingroup$ @Rexcircus: Not sure why you got downvoted. While I can't vouch for the last paragraph about string theory, the middle two paragraphs answer OP's question. As for the the first paragraph: Maybe you misread OP's question or I'm not understanding your argument but I think the idea behind OP's assumption of a very massive black hole was that the curvature at the event horizon would be so small, i.e. the region would be so close to Minkowski space in local inertial coordinates, that other forces (like EM forces, self-gravity etc.) would be prevail and the object wouldn't get ripped apart. … $\endgroup$
    – balu
    Nov 15, 2019 at 10:33
  • $\begingroup$ …Sure, how strong these internal forces are, is related to the object's elasticity. But if the object doesn't easily fall apart in Minkowski space (when e.g. interacting with other objects), then it shouldn't in any very-low curvature vacuum region of spacetime, either. $\endgroup$
    – balu
    Nov 15, 2019 at 10:37
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from observers outside a black hole, it takes an infinite amount of time for it to reach the event horizon

Turn on your logic. If the black hole evaporates in finite time, it will evaporate before the faraway observer will see the infalling observer reaching the horizon. So, after the final evaporation event the faraway observer can meet the infalling observer and ask how he feels.

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