Suppose I have a classical schwarzchild Black hole $B$ of mass $M$. And consider a spherical-subset (or sub-black-hole if you will) $B'$ (sharing the same singularity and having mass $M'$ (We will bound that $M' >10^{17} \text{grams}$).

In reading:


One has that the time to evaporation of the black hole is given by:

$${\displaystyle \tau =8.66\times 10^{-27}\;\left[{\frac {M}{\mathrm {g} }}\right]^{3}\;\mathrm {s} \,.}$$

Now suppose I have an observer $O$ that enters the event horizon of $B$ but has not yet crossed the event horizon of $B'$ (where this horizon is defined as the horizon-that would have existed had all the outer layers of $B$ been removed) .

Now the radial distance from $O$'s current locatino to the target event horizon can be given by some distance $d$. And the direction vector of $O$ is such that it must have a non-zero component pointing towards the singularity (call the direction S)..

But suppose $O$ begins to accelerate extremely hard away from this direction towards the singularity. It's not possible of course to actually move away from the singularity but one could reduce the velocity $V_s$ by expending enough energy such that

$$ \frac{d}{V_S} \ge 8.66\times 10^{-27} \left[{\frac {M'}{\mathrm {g} }}\right]^{3}$$

Then $B'$ would have completely evaporated before $O$ arrived at it. Now $B$ does have a finite amount of time before it will in its entirety evaporate. Yet $O$ appears now to be able to stay INDEFINITELY inside $B$ without reaching the singularity.

So here is my confusion: there are 2 interpretations of what is happening now. Observers on the outside say that $O$ went inside, wasn't seen again, and like all things was ejected as hawking radiation. But $O$ could claim they went inside, somehow managed to avoid the singularity by falling too slowly and now can't escape, but aren't destroyed.

So either:

  1. In order to avoid this contradiction there has to be a minimum speed that everything must FALL to the center regardless of what they do

  2. Is it possible for 2 contradictory events to take place and both be valid?

What is the correct deduction here?

  • $\begingroup$ The mass of a black-hole is concentrated at the point-like center. You can't have a 'sub' black-hole of lesser mass. $\endgroup$ Commented Feb 25, 2017 at 2:55
  • $\begingroup$ Ok @DilithiumMatrix this doesn't break my thought experiment but motivates some questions that can help me break it: consider then just sub-spheres of the event horizon, there should be a hawking radiation like behavior at the surface of each sub-sphere (except the radiation generated by the sub-spheres that are contained in the outer event horizon will just fall right back towards the singularity). Then is it correct to say it is impossible for an observer to avoid hitting the singularity BEFORE the black hole evaporates? $\endgroup$ Commented Feb 25, 2017 at 3:08
  • $\begingroup$ Possible duplicate of If I fall into an evaporating black hole, where do I end up? $\endgroup$ Commented Feb 25, 2017 at 4:00
  • $\begingroup$ @sammygerbil its close but doesn't quite address since teh answer essentially asks to the OP to just be more clear and doesn't address their perceived paradox. Here I've made it explicit i refer to event horizons, DilithiumMatrix has brought up a point which might reveal a flaw in my assumptions but till thats sorted out this question will necessitate a different answer than the one you have linked $\endgroup$ Commented Feb 25, 2017 at 4:05
  • $\begingroup$ So a schwarzchild black hole B can be identified with a sphere of some radius $R$ (dictated by the radius of the event horizon). In the volume of space of B exists a sphere of space of radius $R' < R$ which shares the center of B. Thats what I mean by subsphere. The surface of this sphere (a region of space that doesn't have any otherwise interesting properties) would also be a horizon (though not interesting because it is contained in an event horizon). But basically I'm considering if an observer reaches this new horizon only after the original black hole has evaporated, what happens? $\endgroup$ Commented Feb 25, 2017 at 5:50

1 Answer 1


No Schwarzschild black holes exist in our universe so the situation you describe cannot arise. A Schwarzschild black hole is time independent, meaning it has to have existed for an infinite time and continue to exist for another infinite time.

This isn't just a bit of quibbling terminology because no true horizon can form in a finite coordinate time, where coordinate time means the time recorded by an observer far (effectively infinitely far) from the mass. However the black hole can evaporate in a finite coordinate time, in effect meaning that a true horizon cannot ever form. This is discussed in Why does Stephen Hawking say black holes don't exist?

A related question has been raised in Does any particle ever reach any singularity inside the black hole? but that has no definitive answer despite one of the answers being from a Nobel prize winner. The impression I get is that we simply don't understand black hole evaporation well enough to give a definitive account of what would happen. Hawking's calculation is semiclassical so it just takes the (nonexistant) Schwarzschild geometry as an existing background. A full treatment would presumably be based on something like the Oppenheimer-Snyder metric and calculate how the Hawking radiation modified its evolution, but I know of no such calculations.


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