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Assume you have two high energy particles approaching each other and forming a black hole even before colliding (but before a singularity is formed, which I am not sure that is possible). If the laws of physics are time reversible, then I could start my problem with these two same particles with their momentums reversed, and the solution should be a black hole that splits into the two particles. Is this picture correct? I suspect it is not for some reason I am missing. Or is this the way a black hole evaporates?

Note: actually, we can restrict ourselves to analyze two classical (but relativistic) particles that do not interact, let us forget about quantum mechanics here, so there should be no black hole evaporation or entropy, I believe.

Qmechanic
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  • You could view a time reversed black hole as a white hole, such an object pushes matter outward. But time doesn't run backwards. – Triatticus Sep 03 '22 at 00:18
  • But that is only for eternal black holes, right? –  Sep 03 '22 at 00:22
  • @safesphere I will look at it, thanks! –  Sep 03 '22 at 17:10
  • @safesphere I understand your analogy, but I do not understand why that can happen if you have reversible laws and just two particles, no obvious thermodynamical irreversibility here. So why this apparent contradiction? What makes the behavior of the system not time symmetric? –  Sep 03 '22 at 17:40
  • I think @safesphere should convert the first comment into an answer – RC_23 Sep 05 '22 at 23:53

2 Answers2

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The situation is not realistic.

Two elementary particles cannot form a black hole.

Two composite particles can, if the black hole mass is larger than the Planck mass. (Black holes cannot be less massive than the Planck mass.)

If the particles do form a black hole, then a temperature arises. Black holes are thermodynamic systems with a bath. They do not show reversibility. So the black hole will not split.

KlausK
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    "Two elementary particles cannot form a black hole" why is that? It seems at odds with everything I read before. My question is why time reversible equations cannot have a time reversible solution, and there are two particles here, no thermodynamics involved. –  Sep 05 '22 at 20:31
  • A black hole is a thermodynamic system. There is no way to have a black hole and no thermodynamics. That is also the reason that there are no white holes. – KlausK Sep 06 '22 at 12:31
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If the laws of physics are time reversible, then I could start my problem with these two same particles with their momentums reversed, and the solution should be a black hole that splits into the two particles.

There is a really subtle, and in my opinion beautiful, detail in this statement: the definition of black hole is not time-reversible.

From the moment you say "black hole", you gave up on reversibility. By definition, a black hole is the region of spacetime which no observer that goes to infinity in infinite time can see (see this wonderful PBS Spacetime video for more details). This definition assumes that the black hole region can't be viewed from the future, and it is not symmetric with respect to time reversal. An analogue definition is that of a white hole.

This asymmetry in the very definition of black hole is what allows, for example, the result that a black hole's area can only increase over time, which explicitly distinguishes past from future. The answer to your question is then essentially the same: since the very definition of a black hole already distinguishes past and future, no, you can't find a black hole splitting into two particles by attempting to invoke time reversal symmetry.

Notice that this result that the area of a black hole always increases is fairly similar to the Second Law of Thermodynamics. In modern days, it in fact is interpreted as the application of the Second Law of Thermodynamics to systems involving black holes, as I discussed in this a bit more technical post.

Níckolas Alves
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  • However, the second law is an emergent law. The fundamental laws are still considered reversible. The question, then, is whether or not a black hole represents fundamental irreversibility, which is very much different. – The_Sympathizer Sep 05 '22 at 23:55
  • @The_Sympathizer I believe that would enter the realm of quantum gravity, and hence there is no possible answer. In this answer I'm assuming no quantum effects, and hence there is some limitation to what can be done. In classical General Relativity, the very definition of black hole is already irreversible – Níckolas Alves Sep 05 '22 at 23:58
  • I think I understand your point. So it is fair to say that a back hole does not form until a singularity is formed? otherwise the solution before the singularity is formed should be reversible? But this would imply that a given mass within its associated Schwarzschild radius is not doomed to become a black hole if the momentums have the right direction? Or what makes the problem irreversible? If GR is time reversible, can it have non-time-reversible solutions? –  Sep 06 '22 at 00:12
  • @CarlosGauss If a mass is within its own Schwarzschild radius, it is a black hole. However, I'd say "until a singularity is formed" is an ill-posed statement. There is not a single observer who can state the singularity has been formed. As safesphere pointed out elsewhere in this post, from the point of view of an external observer, the black hole takes infinitely long to form, and the same can goes for the singularity and any observer outside or inside the hole. – Níckolas Alves Sep 06 '22 at 00:17
  • I see, thanks, I still have a confusion, but I will post it as a different, more precise question. –  Sep 06 '22 at 00:19
  • GR definitely has non-time-reversible solutions. Standard cosmology assumes we live in one of them: the Big Bang cosmology with an ever expanding Universe is quite explicitly non-time-reversible. – Níckolas Alves Sep 06 '22 at 00:19
  • @safesphere My point is precisely that the time reversed version of a black hole is not a black hole (I do agree that it is a white hole). Furthermore, OP's question doesn't necessarily concern only the outside: the question is whether a black hole can divide into two particles, which it can't. I also disagree that i did not answer the question: I argued that the premise is wrong, which explains why the conclusion is weird – Níckolas Alves Sep 06 '22 at 19:19
  • After re-reading the question, I deleted my comment, because this scenario is impossible. Relativistic bullets with mostly kinetic energy don’t bend spacetime, so two particles won’t create any space curvature, much less “a black hole”, before they collide. The question is based on a typical misconception that the Schwarzschild radius is spacelike. It is not. The spatial part of the Schwarzschild radius is zero. Two particles can come arbitrarily close to each other, but nothing will happen until they collide and their kinetic energy converts to something else that actually curves spacetime. – safesphere Sep 07 '22 at 07:03
  • “If a mass is within its own Schwarzschild radius, it is a black hole” - This is a very troublesome statement that leads to a wide-spread misconception. The radial distance from the horizon to the origin is zero meters. The Schwarzschild radius is timelike, it is measured in seconds, not in meters. It is zero meters for any black hole. You can put the particles arbitrarily close to each other, but they still won’t be inside their Schwarzschild radius. – safesphere Sep 07 '22 at 17:13
  • @safesphere That is a very interesting point. While I agree the Schwarzschild radius is timelike, I believe I disagree with your conclusion. I've created a chat room to continue this discussion, if you're interested in doing so. – Níckolas Alves Sep 07 '22 at 18:38
  • @NíckolasAlves Please see my reply in chat. – safesphere Sep 08 '22 at 06:20