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We usually think of white holes as 'thermodynamically reversed black-holes', and this kind of membranes have not been observed in our universe. However, there is some other kind of 'topologically reversed black hole' which we know exists: our cosmological event horizon (CEH). It is reverse in the sense of the membrane direction where light cannot come out, the CEH allows outsiders to look in, but doesn't allow insiders to look out.

Question: How do GR describe in general reversed-orientation black holes like the example of our CEH? Please discuss the possibility that exact GR solutions where 'white holes' exist, we might be wrongly interpreting the solution, and what we rather should expect is a membrane-inverted black hole

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A cosmological horizon isn't the same thing as a black hole horizon--the black hole horizon is an essential feature of the spacetime that is located where it is due to special geometry. A cosmological horizon is an observer-dependent phenomenon that describes when two observers are out of causal contact with each other. The only sense in which white hole solutions are the same as cosmological horizons is that both are past trapping horizons, which you have already described qualitatively.

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  • $\begingroup$ i really don't see a physical difference between one kind of horizon (black holes) or the other (CEH), besides the sense in which useful information flows $\endgroup$
    – lurscher
    Commented Jun 10, 2013 at 16:32
  • $\begingroup$ the CEH horizon is observer-dependent--a horizon only exists relative to a point. In a black hole spacetime, all observers agree on where the horizon is. $\endgroup$
    – Zo the Relativist
    Commented Jun 10, 2013 at 17:30
  • $\begingroup$ well, that is debatable. Classically yes, but as this question (physics.stackexchange.com/q/22498/955) highlights very clearly, the physical event horizon of black holes that have formed in a finite time are diffuse, they are not infinitely resolved as the eternal black hole solutions we are used to studying on classical gravity. Real black holes evaporate, so the physical horizon is not precisely located. Hence, I don't see this difference as clear-cut. Maybe we should think about this some more? $\endgroup$
    – lurscher
    Commented Jul 10, 2013 at 18:09
  • $\begingroup$ @lurscher: within a particular time slicing every point in a FRLW spacetime has a distinct cosmological horizon. In a Schwarzshild spacetime, and for a given time slicing, there is only one apparent horizon. The fact that we're fixing the time slicing and using apparent horizons and not event horizons gets rid of the issue of shrinking black holes, since the horizon may have shrunk at a future time, but it is still there. They are distinct events. $\endgroup$
    – Zo the Relativist
    Commented Jul 10, 2013 at 19:14
  • $\begingroup$ I would also argue that Ben Crowell's second penrose diagram is probably not the right one to be using in this case. You'd be better off doing something like writing down a Vaidya solution for a shrinking mass profile, and then going through the whole conformal transform procedure again. A spacetime that has a black hole that has evaporated is not a perturbation of a schwarzshild spacetime, it's a completely new one. $\endgroup$
    – Zo the Relativist
    Commented Jul 10, 2013 at 19:21

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