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One can fall into a black hole but not fall out of it. Does this mean black holes violate T-symmetry?

The closest thing I found to this is this section on Wikipedia, but it doesn't cite any sources and sounds dubious.

  • Our laws of physics might break down at the singularity, but not at the event horizon, and I remember reading that in a Schwarzschild black hole, all geodesics lead to the singularity - i.e. there's still a forward concept of time.
  • The section also mentions white holes, but that doesn't seem to solve the problem: even if white holes exist, they would just violate T-symmetry in another way - one can fall out of it but not into it.
  • It says that the modern view of black hole irreversibility is related to the second law of thermodynamics, but as far as I understand it, the second law of thermodynamics doesn't violate T-symmetry because on a microscopic level all the motion is T-reversible.

Other results I found focus on the black hole information paradox, which is not what I'm asking about.

If the answer is "yes", how is this possible given that only the weak force of the four known forces violates T-symmetry?

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    $\begingroup$ (1) Even in the context of plain old Newtonian mechanics, there is a distinction between symmetry of the laws and symmetry of solutions to the equations of motion. (2) General relativity does not have discrete symmetries such as time-reversal symmetry. In a general spacetime, we cannot even define a time-reversal operation. $\endgroup$ – Ben Crowell Jan 16 '18 at 22:01
  • $\begingroup$ @BenCrowell - I don't understand both points. Can you elaborate? $\endgroup$ – Allure Jan 16 '18 at 23:03
  • $\begingroup$ Related (not a duplicate, but you may or may not find my answer there useful) physics.stackexchange.com/questions/27922/… In short, I claim in that answer that black holes do violate $T$ symmetry, in the same kind of way that pouring milk into coffee does - they're a macroscopic approximation of the true physical system, and are subject to the second law. $\endgroup$ – Nathaniel Jan 17 '18 at 4:55
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EDIT: the short answer to this question is that a time-reversed black hole is a white hole, full stop, so if you apply time-reversal to a particle falling into a black hole, you get a particle falling out of a white hole, but we don't physically expect to observe white holes.

Original text:

A blackhole space-time does not violate T-symmetry because, the extended Kruskal solution also contains a white hole:

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

so, if you time-reverse the portion of a curve falling into the black hole, it will become a portion of a curve falling out of the white hole.

Now, we expect that the universe was created with initial conditions that don't allow white holes to exist, but this would mean that the T-symmetry in GR is spontaneously broken by some quantum theory that is not GR. It is absolutely present in the schwarzschild and Kerr spacetimes, though, thanks to the extend kruskal coordinates trick.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jan 18 '18 at 20:15
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Black holes do not violate T-symmetry, but as macroscopic systems interacting with an environment and subject to the laws of thermodynamics they do have a thermodynamic arrow of time for processes around them.

White hole is simply statistically improbable black hole. Hawking radiation has a thermal spectrum and that means that even high energy and complex states could be radiated away from it. Such energy states would include (for sufficiently large black hole), for example, an astronaut in a spaceship flying out of such a black hole. Of course, the probability of such an event would be immeasurably tiny to occur in our universe, with a realistic black hole much more likely to evaporate by emission of long wavelength photons over the course of $10^{68}$ to $10^{99}$ years (plus the high-energy explosion at the end of it) rather than ever emitting something interesting.

In the (conjectured) distant future of our universe after the baryon matter has decyed there would be a Black Hole Era. At this point it would not even be right to call them black since they would be only things providing illumination (by means of Hawking radiation) to the universe. At this point the thermodynamic arrow of time makes them white holes.

So the really T-symmetric setup would be the black hole at equilibrium with the surrounding space (and so with the same ambient temperature). For such a setup for every photon falling into the black hole there would be (on average) a photon Hawking-radiated away. And for every astronaut Hawking-radiated out of the black hole there would be one manifesting itself as a Boltzmann brain outside of it and then falling in.

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  • $\begingroup$ This is just wrong. White holes are time-reversed black holes, and only the extended kruskal solution is time -symmeteic. $\endgroup$ – Jerry Schirmer Jan 17 '18 at 14:46
  • $\begingroup$ @JerrySchirmer: I believe that OP is talking about black holes as realistic physical systems and not about Ricci-flat metric. In particular, they would evaporate if left alone. My setup offered at the end (BH in thermal bath) is time reversal invariant in the same sense as any thermodynamic system in thermal equilibrium would be from the point of view of external observer (outside horizon for BH) studying the system for finite amount of time. But I would clarify the formulations (soon). $\endgroup$ – A.V.S. Jan 17 '18 at 15:43
  • $\begingroup$ Test orbits around a black hole in a thermal bath will not be time-reversal invariant, though. In particular, infalling orbits would not have a good time reverse analogue. You apply time reversal to a black hole, and you get a white hole. There is no way around this -- it sends you to a different patch of the extend Kruskal space. $\endgroup$ – Jerry Schirmer Jan 17 '18 at 17:15
  • $\begingroup$ @JerrySchirmer: Like I said, outside observer, what happens under horizon is not observable from outside. I would try to clarify this. $\endgroup$ – A.V.S. Jan 17 '18 at 17:51
  • $\begingroup$ It doesn't matter, though, you get different behaviour outside the horizon in white and black hole solutions -- the $r=6M$ instability being the most obvious example of this. $\endgroup$ – Jerry Schirmer Jan 17 '18 at 18:01
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No, black holes do not violate time symmetry. Allow me to explain.

If we throw a massless particle into a black hole, it'll never pass the event horizon. It'll only get closer and close and slow down to a halt at the event horizon.The only way for it to go through the event horizon is to have the event horizon expand around it. This can be done by lowering the gravitational potential around the test particle to below -c^2/2. This could be done by throwing a massive object towards the test particle.

If we throw a 3D massive object into a black hole, the front of it will have the event horizon expand around it due to the gravitational potential of the back of the object. The very back of the object will not enter the black hole, only approach the event horizon at slower and slower speeds. If you were to watch this in reverse(reverse the velocities), the back of the object would move away from the event horizon, the event horizon would shrink past some more of the object's mass, and some mass would escape the black hole. This would continue until the whole object escapes the event horizon and exits the black hole.

So no, black holes do not violate time symmetry.

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    $\begingroup$ The description presented in this answer doesn't actually seem to have anything to do with time-reversal symmetry. $\endgroup$ – Ben Crowell Jan 16 '18 at 22:48
  • $\begingroup$ Oh, I realize this now. I guess I misinterpreted the question... $\endgroup$ – Laff70 Jan 16 '18 at 23:11

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