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Could white holes undergo reverse hawking radiation? This is because of quantum mechanical uncertainty. So for example a pair of particles one a positive and one a negative energy particle. The negative energy escapes the white hole. So could white holes also reverse hawking radiate and get mass through this process. Am I right or is there some unknown physics that would prohibit it?

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The answer depends on what you mean by "radiation." A white hole is the time-reverse of a black hole. If you take Hawking's original calculation and time-reverse the whole thing, you get a white hole that absorbs radiation from the vacuum. That's just as interesting as a black hole emitting radiation, because a white hole isn't supposed to be able to absorb anything in classical general relativity, just like a black hole isn't supposed to be able to emit anything.

If you were asking whether a white hole would emit Hawking radiation, then the answer would be less interesting: even in classical general relativity, white holes emit everything. It's the time-reverse of a black hole: you can't escape from a black hole (classically), and you can't stay inside a white hole (classically).

could white holes also reverse hawking radiate and get mass through this process?

The answer is presumably yes, but there's a catch. Hawking's calculation considers quantum fields in a background spacetime, which means the spacetime doesn't react to the quantum fields. That's just an approximation, of course, and in reality we expect that the spacetime will react, resulting in the loss of mass from a black hole. To implement that in a self-consistent theory requires some kind of quantum gravity theory, and I expect that the existence of white holes in quantum gravity would be exceedingly unlikely, for statistical reasons.

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    $\begingroup$ Naive question. Wouldn't time reversing the Hawking process be like time reversing the breaking of an egg? It's a thermodynamic process so I'm not sure if time reversing makes sense? $\endgroup$ – Dvij D.C. May 19 at 23:42
  • $\begingroup$ Ha, interesting. I didn't think of it this way but it certainly seems related for if we wouldn't neglect the back-reaction, it'd be an exact calculation and thus, presumably (hopefully? xD) fully reversible. Edit: Or maybe not, I'm confused now. Because the thermodynamic nature comes from taking the partial trace over all the degrees of freedom falling into the black hole (correct me if I'm wrong). So even if we took the gravitational back reaction on spacetime into account, to an outside observer, the process would still look thermodynamic. $\endgroup$ – Dvij D.C. May 19 at 23:54
  • $\begingroup$ @DvijD.C. The time-reverse doesn't require switching "inside" and "outside". We would still take the partial trace over the "inside", where the singularity is. In QFT, observables are tied to regions of spacetime, not to particles (or propagating modes), and the partial trace is just an instruction to ignore the observables in a given region. In this case, that region is the "inside", where the singularity is. Time-reversal doesn't change that. Does that address your comment, or did I misunderstand what you meant? $\endgroup$ – Chiral Anomaly May 28 at 0:35
  • $\begingroup$ @DvijD.C. If that's not what you meant, then maybe this: the time-reverse of a conventional blackbody still looks thermal but with ingoing radiation instead of outgoing. Is that what you meant by still looking thermodynamic? $\endgroup$ – Chiral Anomaly May 28 at 0:39

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