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The famous Hawking calculation is done with an assumption that the background is static, i.e. the evaporation doesn't change the mass parameter in the metric. Thus, we simply describe the geometry using the static Schwarzschild (or, generically, Kerr-Newman) metric. But clearly, the evaporation actually makes the geometry non-static and thus, the geometry should actually be described using a non-static metric. I am finding a hard time finding out what metric this is.

I think that even if the Hawking calculation is done within the assumption that the background metric is static, one can safely assume that a spherically symmetric radiation will still be being emitted from an evaporating black hole even during the stages where the static assumption is inappropriate. Thus, the natural guess for a non-static metric that describes the geometry of an evaporating black hole would be the Vaidya metric.

But, as discussed in this answer, an outgoing Vaidya metric describes a metric for which the mass parameter is continually decreasing--but this doesn't describe a black hole geometry, instead, it describes a white hole geometry. Further, as discussed in the same answer, an ingoing Vaidya metric describes a black hole geometry--but with a monotonically increasing mass parameter. Thus, none of the two Vaidya metrics qualify to describe an evaporating black hole.

So, my question is, is there any known metric that can describe a spherically symmetric geometry whose mass parameter decreases with time and the horizon is of the nature that resembles a black hole horizon? If so, then it can be considered as a metric that describes an evaporating black hole.

Edit

I recently read a comment by @JerrySchirmer that the Hawking radiation violates the energy conditions. If so is the case then the argument that an ingoing Vaidya metric has a monotonically increasing mass parameter doesn't work (as this argument relies on the null energy condition). If someone can provide some canonical references in this regard then it would be truly helpful.

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  • This is related to a question of mine that never got an answer: http://physics.stackexchange.com/q/240627/109928 – Stéphane Rollandin Nov 28 '17 at 20:35
  • @StéphaneRollandin Yes, they are quite related. I don't think there is any widely accepted answer to your question as far as what I gathered from a conversation with the user John Rennie a few days back. –  Nov 28 '17 at 20:46
  • BTW, good luck with your application for the PhD programme! From what I've seen here on PSE, you are a great physicist :-) – AccidentalFourierTransform Nov 29 '17 at 15:21
  • @AccidentalFourierTransform Thanks a lot! That really boosts the morale. :-) I hope you read this before a moderator (though rightly) deletes these comments! –  Nov 29 '17 at 15:55
  • Just adding that the Vaidya metric is definitely not a white hole. – Zo the Relativist Aug 29 '19 at 15:16
  • @JerrySchirmer In my understanding, an outgoing Vaidya metric might describe some non-blackhole non-whitehole astrophysical body outside a certain radius larger than the implied horizon radius. But, if we put this radius cut-off then doesn't an outgoing Vaidya metric mean a whilehole? Thanks for your time. –  Aug 29 '19 at 15:30
  • @DvijMankad : generically, the distinction between a black hole and a white hole is whether the singularity is in the future or the past, not the traversibility of the horizon. You could probably transform the standard Vaidya metric into a white hole solution the same way you would a Schwarzshchild solution. (by time-reversing), and this might amount to just a choice of the matter function, but it is definitely a black hole and not a white hole in the standard way it's written (though that is, admittedly, usually involving a monotonically increasing mass function). – Zo the Relativist Aug 29 '19 at 18:21
  • @JerrySchirmer Ah, I see. But now I am confused as to whether an outgoing Vaidya metric with a singularity makes any sense at all. If the singularity is a future singularity just like in the case of a black hole singularity then the horizon must not be traversable in the ingoing direction, right? Or is the outgoing Vaidya metric a naked singularity with a weird traversability property of the horizon? I just can't exactly stomach how can a singularity lie in future but the horizon is not traversable towards the singularity? I feel I am just relying on intuition without much thinking. –  Aug 29 '19 at 20:07
  • @DvijMankad you'd have to solve the geodesic equation to know whether the horizon is traversible, but you already know that since such an object violates the area theorem, it cannot satisfy the normal energy conditions. – Zo the Relativist Aug 29 '19 at 22:23

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I don't have time to look it up, but there is a paper from 1982 where the authors perform Hawking's calculation against a Vaidya background (the Vaidya metric has a free mass function in the version I've seen), and then solve for the mass function so that it matches the outgoing Hawking radiation in the radiative limit. I believe they get a different power law for decay than the standard one that you calculate from just using Boltzmann's law on the Hawking temperature.

But yeah, the horizon in this spacetime will be an apparent horizon but not a proper event horizon, and it will be two-way tranversible (and it has to be for any shrinking black hole. Imagine the frozen null generator of the hole. It just sits at the horizon. When the horizon shrinks, it will be outside the horizon, suddenly, and will be able to reach null infinity).

Zo the Relativist
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Just found this question through a related link, but Hayward 2006 or (if you prefer to have the singularity) Hiscock 1981 are typical references on this topic.

Joe Schindler
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  • Although one thing both of these neglect is that the formation dynamics depend on $m(u)$, but the evaporation dynamics depend on $m(v)$, so to really get the details right you actually need a more complicated $m(u,v)$. – Joe Schindler Nov 14 '19 at 21:22