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Let's say you have a very powerful spaceship and decide to travel to the event horizon of a supermassive black hole, but just before crossing you reverse your thrusters and escape. According to my understanding, you could travel to an arbitrarily distant date in the future using this method. So lets say you travel to the heat death of the Universe, say $10^{100}$ years in the future. In this case all the black holes will have evaporated, but that would mean your black hole should have evaporated while you used it to travel, not allowing you to travel that far in the future. Is this a paradox?

This question has some good related answers and discussion. How can anything ever fall into a black hole as seen from an outside observer?

user289980
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You can't travel past the point that the black hole evaporates using your strategy. The commonly-made statement that you can travel arbitrarily far into the future is only true in the approximation that the black hole obeys classical general relativity—that is, with no evaporation (evaporation being a quantum effect).

(Note that, strictly speaking, an evaporating black hole also has no event horizon, because all the energy that went in will ultimately come back out.)

Buzz
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  • Wouldn't this mean it is impossible to cross the event horizon as well? – user289980 Mar 21 '22 at 05:12
  • @user289980 no, it is theoretically possible to cross the horizon of a non evaporating black hole – Dale Mar 21 '22 at 11:37
  • @user289980 No, because time doesn't stand still for the person falling. Time for the person falling stands still from the perspective of someone outside watching. – DKNguyen Mar 21 '22 at 17:30
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    @DKNguyen So if the outside observer was immortal, they would see the ship take an infinite amount of time to reach the event horizon. Concurrently, the black hole is also evaporating, and, ignoring surrounding radiation, in a finite amount of time the black hole will have evaporated, meaning that to the outside observer the ship never crossed. However, the ship thinks they crossed, leading to a disagreement. – user289980 Mar 21 '22 at 18:05
  • @Dale What about crossing an evaporating on? It seems like there would be a disagreement if my response to DKNguyen is accurate. – user289980 Mar 23 '22 at 02:08
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    @user289980 there is some indication that an evaporating black hole doesn’t actually have a horizon: https://arxiv.org/abs/1102.2609 Nevertheless, there are worldlines for an evaporating black hole spacetime which go to the singularity. Those end and do not appear after evaporation – Dale Mar 23 '22 at 02:30
  • @safesphere So you think it is impossible to cross the event horizon? Just trying to be clear. – user289980 Mar 24 '22 at 00:34
  • @user289980 See my edit. – Buzz Mar 24 '22 at 01:35
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    @user289980 Yes, in semi-classical gravity, nothing can cross the horizon of an evaporating black hole, because it evaporates and disappears before anything can approach the horizon. You can find popular Penrose diagrams that show otherwise, but these diagrams are incorrect. – safesphere Mar 24 '22 at 07:03
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    “an evaporating black hole also has no event horizon” - The $r=2M$ horizon is still there. It just is not of the “event” type. There are at least a dozen different types of horizons in GR. – safesphere Mar 24 '22 at 17:53
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When the black holes evaporates it looses mass which means that its radius shrinks (remember that radius and mass of a back hole are linearly related, consider the Schwarzschild Radius). From this you can calculate the lifetime of a black hole, simply by calculating the time when the mass of the black hole vanishes. You could calculate this time in the frame of an observer very far away from the black hole.

So, first of all, this means that you have to constantly adapt your radius to the black hole with your spaceship. Now, from the Schwarzschild metric you can calculate the difference between your proper time and the proper time of an observer far away. This is given by

$$\frac{ t_{\rm spaceship}}{t_{\rm infinity}} = 1-\frac{r_s(t)}{r_{\rm spaceship}}$$, where $r_s(t)$ is the (time-dependent) Schwarzschild Radius. To calculate the time that passes you would have to integrate over the time you spend at the horizon of the black hole. All of this is an approximation, since the metric of an evaporating black hole is not known. I approximate it here as Schwarzschild metric that changes its mass in time. So, you can make time slow down with respect to the observer at infinity, but of course only as long as that black hole exists. It is a dynamical process.

Of course this is nothing new, when you are closer to the sun you can slow down your time with respect to someone far away, but of course only as long the sun is there!

konstle
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You are overlooking two points which make your supposed paradox disappear. The first is that while time slows for you when you are close to the black hole, it also slows for the black-hole itself. The black hole cannot age enormously while a short time passes for you, since it too is time dilated. Secondly, if the black hole were to evaporate in some finite time, then so too would the cause of your time dilation, so you could not remain time dilated for longer than the life of the black hole itself.

Marco Ocram
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  • Wouldn't' this mean that a blackhole would have an infinite lifespan to an outside observer, even ignoring outside radiation, since time dilation goes to infinity at the horizon? That goes against what I have read on the subject and also other answers here that say the blackhole will evaporate while you use it. – user289980 Mar 24 '22 at 00:30