3
$\begingroup$

Everybody knows that you can only fall into a black hole and you can never get out. But on second thought, this flies in the face of time reversal invariance, which says that every process that goes forward and also go backwards. Moreover, the standard Schwarzschild black hole is invariant under time reversal because it's a static spacetime.

In Newtonian physics, if a ball falls into a deep well, it can bounce around inside for a long time. But if it ever hits anything that reverses its velocity, it'll just retrace everything it did in reverse, then come right back out of the well. Does this argument also apply for something that's fallen into a black hole?

$\endgroup$
9
  • 1
    $\begingroup$ Didn't you answer your own question by describing the Schwarzschild spacetime? In the frame of a remote observer, objects never cross the event horizon. Thus, if an object's speed is reversed before the horizon, the object would return at any time up to infinity. So the time reversal works up to the point where time stops at the horizon. $\endgroup$
    – safesphere
    Commented Jan 12, 2018 at 17:45
  • 2
    $\begingroup$ "Moreover, the standard Schwarzschild black hole is invariant under time reversal because it's a static spacetime." - the exterior geometry is static but within the horizon, the geometry is dynamic. $\endgroup$ Commented Jan 12, 2018 at 19:19
  • $\begingroup$ @AlfredCentauri Can you explain a bit more what that means? $\endgroup$
    – knzhou
    Commented Jan 12, 2018 at 19:21
  • $\begingroup$ Also, I think it is the case that the geodesics that cross the black hole horizon in the future crossed the white hole horizon in the past. $\endgroup$ Commented Jan 12, 2018 at 19:21
  • 1
    $\begingroup$ @Rococo No, my question is entirely classical, so I think it should be simpler to answer. There are none of the complications of Hawking radiation! $\endgroup$
    – knzhou
    Commented Jan 13, 2018 at 7:38

5 Answers 5

4
$\begingroup$

An existing answer and a comment have both pointed out that it takes an infinite coordinate time for anything to reach the event horizon, so let us instead consider the perspective of an infalling spaceman. The question is then whether our spaceman falling through the even horizon can experience passage pass back out through it.

And the answer is that provided they do not hit the singularity then they can indeed pass back out. In both the Reissner-Nordstrom and Kerr black holes it is possible to fall in, miss the singularity and escape back out. This happens in a finite proper time for the spaceman. However since it takes an infinite coordinate time to reach the horizon our spaceman emerges into a region of spacetime that is causally disconnected from the original departure point.

In the special case of a Schwarzschild black hole it is not possible to miss the singularity, and this would of course not be good for our spaceman's health. But even leaving this aside we cannot integrate the equations of motion through a singularity so GR cannot tell us what happens at or beyond this point.

$\endgroup$
6
  • $\begingroup$ Are the Kerr/ Reissner-Nordstrom solutions actually physically stable though? I tend to find various sources that suggest they are not (e.g. jila.colorado.edu/~ajsh/insidebh/waterfall.html) but I'm quite out of my depth to assess them. $\endgroup$
    – Rococo
    Commented Jan 13, 2018 at 21:19
  • $\begingroup$ No, the answer is not right. See my answer below. A rotating BH, with a Kerr solution, has an event horizon and when inside it they can not come back out. More below $\endgroup$
    – Bob Bee
    Commented Jan 14, 2018 at 4:01
  • $\begingroup$ And yes they are stable, we've seen the merging of BHs into a strongly rotating BH. If you have a specific question that asks about stability of the Kerr BHs, with some description of what those papers say, please post it as a question. $\endgroup$
    – Bob Bee
    Commented Jan 14, 2018 at 4:27
  • $\begingroup$ @Rococo an isolated Reissner-Nordstrom black hole is certainly stable but there is some argument about whether the energy flux from matter falling into then back out would destabilise it. In any case the point is moot since no black holes of any form exist. The Schwarzschild, RN and Kerr solutions are time independent so can only exist in an infinitely old universe that will exist for an infinite time into the future. These solutions are only approximations to real black holes, and real black holes probably don't allow trajectories in and out again. $\endgroup$ Commented Jan 14, 2018 at 5:38
  • $\begingroup$ @BobBee- I was referring to the claim from the link provided above that "from its inner horizon inward, the Kerr geometry is not a realistic model of an astronomical black hole, because the Kerr geometry is subject to the inflationary instability at its inner horizon." Of course, this would not be ruled out by our observations. $\endgroup$
    – Rococo
    Commented Jan 14, 2018 at 17:27
3
$\begingroup$

A rotating BH is a Kerr BH, with the Kerr metric the solution. The Kerr solution has a ring singularity, but for the puRposes of the question the important fact is that it also has an event horizon. The confusion arises because it also has an external mathematical surface, called the ergosphere, where in fact it is possible to come in to and still come out.

This ouTer surface in not a true horizon, but it is a surface inside of which a particle (or ideal observer) has to rotate with the BH. Penrose determined that it is indeed possible to come into the ergosphere with a certain energy and come out faster, with a higher energy. This is conceptually a way to extract energy from the BH.

But the more important surface for the question is the event or inner horizon. It is a true horizon and if you go inside it you can never. Come out. If that was not true it would not be a BH.

See the simplest description of those in the Wikipedia article at https://en.m.wikipedia.org/wiki/Kerr_metric.

The astrophysically discovered merging BHs almost surely were Kerr BHs, and once they get inside the event or true horizon of the other they just merge, very quickly, into a single rotating Kerr BH.

I've not checked about a charged or charged rotating BH but am pretty sure the answer is the same. It's more or less irrelevant because it's unlikely to find an astrophysical object with macroscopic charge, but possible and answer should be the same.

The consideration of how long it takes for an observer at infinity to see the particle (or whatever fall into the horizon is also irrelevant for practical purposes. That whole topic of the infinite time it would take has been answered plenty times on the PSE, but a short version is that whether it falls into the event horizon or just comes infinitely close to it has no observable difference at infinity, and for all purposes you can consider it as just falling in. We do keep detecting merging BHs with LIGO, so for these purposes they exist.

So the OPS question, why does time symmetry not allow particles coming back out after they go into a true event horizon? It's not the same, there can be white holes and BHs, we know how BHs may tend to form in gravitational collapse, we still have not seen a white hole (the Big Bang is similar to one but not the same).

$\endgroup$
2
$\begingroup$

The problem here is that one is not being complete enough in taking the time reversal, that is, not accounting for that in a time-reversed picture everything in the universe is reversed and so inadvertently allowing some objects to still be forward-oriented in time. A trajectory that "dips" into the horizon and comes back out is as impossible in forward time as it is in reverse time (the "dip" just goes the other way), which is at first sight what one might be thinking of. But the true time reverse of something falling into and being destroyed at the hole singularity is something quite different from this. You have to remember the reversal applies to literally the entire motion, not just part of it. That is, the initial point of the motion becomes the final point in the time reversal and the final point of the motion becomes the initial point. Furthermore it also applies to the black hole itself, changing the sense of the gravitational field to be anti-gravitational.

Thus, taking all of the motion into consideration, one sees that this only "valid" "reverse fall" is one involving an object who is born at the singularity and then rises up to escape, of a reverse black hole which is a "white hole". Rising up from a time-forward black hole is equivalent in time reversal to sinking into a time-reverse white hole, which is as impossible as you'd think the time-forward scenario was. Failing to take into account the black hole itself is also time-reversed is important.

It is also helpful to look at it from the point of view of an outside observer. An outside observer watching something fall into a black hole will see that something appear to slow down as it nears the horizon, so as to never quite reach it until infinite future time. In the time reverse scenario, the final time becomes the initial time and the initial time becomes the final time, so in that case the observer sees an object that has been rising up from the white hole since infinite past time, and then only now when sie happens to be present, does it finally make it out far enough to acquire appreciable speed and fly past hir spacecraft.

(Now you might be slick and ask well how can we have an observer at all, wouldn't sie be perceiving things backwards too since we time-reversed everything in the universe and thus also the observer and then sie would think sie is perceiving a black hole situation as hir consciousness is evolving backwards in time as well? Well that's the beauty of time reversal symmetry. We can still have the time-forward observer in with the time-reversed black hole since that is a possible dynamical path because the laws of physics work the same in both directions so they will still evolve the time-forward observer forwards. It's quite "unlikely" in the time reversed universe to occur naturally, but it's still possible! Just as we could in theory have a white hole in our own universe - very unlikely, but still possible, and definitely will work if we "cut and pasted" the observer from a forward universe as one might be imagining in one's head. (FWIW in our own universe, that is, the time reversal of the reversal, what it corresponds to is the formation of an organism that lives its life going from its death to its birth. That is possible, just extraordinarily unlikely, and is the same thing as a shattered coffee mug spontaneously getting up and reassembling itself to an intact one on the table.))

$\endgroup$
0
$\begingroup$

From the point of view of an outside observer, the object never reaches the event horizon, so it can come back. Also, it's possible for the black hole to emit Hawking radiation that coalesces into a spaceship emerging from the black hole. The probability of this happening is astronomically small. Thus, this provides a "reversibility" of a spaceship flying into a black hole that is analogous to the sense in which an ice sculpture melting is "reversible"; the reverse process is physically possible, but thermodynamically impossible.

$\endgroup$
0
$\begingroup$

In the case of black holes that are rotating and charged, a repulsive potential is generated inside, which throws the bodies that have fallen below the event horizon backwards, out of the horizon. On certain orbits and with certain energies. enter image description here

https://www.oocities.org/zcphysicsms/

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.