Timeline for Flat space limit of the Schwarzschild metric and Hawking temperature
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2011 at 23:36 | comment | added | Zo the Relativist | You are right to say that it is most assuredly not the $r=3M$ surface of photon orbit in the Schwarzschild solution. | |
| Jan 25, 2011 at 23:35 | comment | added | Zo the Relativist | @dbrane: sorry, I was getting a little cavalier about what I meant by orbit. I meant the photon path that tries to escape the black hole by pointing radially outward, but neither falls into the black hole nor escapes. Formally, an apparent horizon is a closed two-surface with two perpendicular null vectors $k$ and $\ell$ that satisfy $k^{a}\ell_{a}=-1$ and $\left(g^{ab}+k^{a}\ell^{b}+\ell^{a}k^{b}\right)\nabla_{a}\ell_{b}=0$. What this means is that $\ell_{a}$ has no component of its gadient lying on the AH, and therefore, at some infinitesimally later time, will still be at the 'same place' | |
| Jan 25, 2011 at 23:31 | comment | added | dbrane | Hang on, the innermost-stable-photon-orbit (ISPO) definition can't possibly be the same as the way you defined it in the link, which I'm going to paraphrase as "the surface that traps you if the BH were to stop evaporating once you crossed this surface". Because assuming no evaporation, you could have a photon that crosses the ISPO and still escapes to infinity whereas you couldn't have a photon that can cross the surface as defined in your link and then escapes to infinity (assuming no evaporation). Right? | |
| Jan 25, 2011 at 23:19 | comment | added | Zo the Relativist | @dbrane: that's not really a bad definition of an apparent horizon, the formal one basically just formalizes this idea--the definition of the event horizon,however, is the one that's a bit trickier--it depends on the end state of the spacetime, and you take all of the light rays that hit infinity, and then you take their topological boundary. | |
| Jan 25, 2011 at 23:08 | comment | added | dbrane | Wow, I should have realized that. I hadn't heard of the distinction between an apparent horizon and an event horizon before (one of the relativity courses I took defined the apparent horizon as the radius of the smallest stable photon orbit - bah, misleading). I read your answer at physics.stackexchange.com/questions/970/… and that really helped a lot. Thanks! | |
| Jan 25, 2011 at 22:53 | comment | added | Zo the Relativist | @dbrane: The evaporation happens due to an apparent horizon. If the black hole goes away at late times, then the final state is a horizonless, singularity-free spacetime--any geodesic inside the horizon will eventually be able to exit it (at least in the case of a timelike singularity, which is the case for a hole with even an infinitesimal charge or angular momentum). Another way of phrasing this is that a black hole with a decreasing radius forms a timelike surface in the whole spacetime, and there is a formal result that shows that timelike surfaces cannot be trapping horizons. | |
| Jan 25, 2011 at 22:14 | comment | added | dbrane | "a spacetime with an evaporating black hole wont' have an event horizon" Why not? How can you even have evaporation with out an event horizon? | |
| Jan 25, 2011 at 21:47 | comment | added | Zo the Relativist | @dbrane: depends on how you feel about cosmic censorship, for one thing. (though I actually have no idea how the Hawking calculation works out for something like a $a>M$ spacetime). And a spacetime with an evaporating black hole wont' have an event horizon anyway. | |
| Jan 25, 2011 at 17:35 | comment | added | dbrane | When I said "in the complete quantum gravity theory", I meant the complete QG theory that exists out there Platonically but which we don't yet know :) It's clear to me that a topology change would make a huge difference but not so much for something that simply mimics it. In a complete QG theory, what should still survive is the fact there is a horizon in the black hole case and none in the M→0 limit. | |
| Jan 25, 2011 at 2:32 | comment | added | Zo the Relativist | @dbrane: But, there is no complete theory, so we're stuck with semiclassical results. And in a complete theory, you would still expect a qualitative difference between a state with a classical black hole limit and one that is without one, and that this would mimic the topological difference you see classically. | |
| Jan 24, 2011 at 3:25 | comment | added | dbrane | Is the topology change relevant? Because presumably, in the complete quantum gravity theory of a black hole, there will be no distinction between the black hole scenario and the post-evaporation scenario in terms of singularities, since singularities don't exist in the complete theory. The second explanation makes sense to me though. +1 for that :) | |
| Jan 23, 2011 at 20:28 | history | answered | Zo the Relativist | CC BY-SA 2.5 |