I read about Hawking radiation in Wikipedia and understood black holes may disappear over a long period of time. But my question is why does it have to disappear, because after some time of radiation the mass of black hole won't be enough to maintain its "black holeness" (Hawking radiation stops) and it could become another planet or star or some other object. So Technically speaking we should be saying black holes get converted to some other object because of Hawking radiation. What is wrong in this assumption?
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$\begingroup$ A black hole is not something with very high mass, but rather something with very high density. In principle it is possible for very light black holes to exist. As a black hole loses mass through Hawking radiation, its density remains the same (infinite), and so it remains a black hole. $\endgroup$– gj255Commented Oct 21, 2016 at 11:59
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$\begingroup$ See this in the Wikipedia article: "One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy". The trouble is that we don't know of any negative-energy particles. $\endgroup$– John DuffieldCommented Oct 23, 2016 at 12:02
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2 Answers
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We need to be careful to distinguish the idealised black hole, e.g. the Schwarzschild metric, from the black holes that we might see in the real universe. The reason for the difference is that the idealised black holes take an infinite time to form so we'll never see one. For more on this see the answers to Why does Stephen Hawking say black holes don't exist?
Let's start with the idealised black hole since this is simpler. In this case all the mass is concentrated at the singularity in the centre of the black hole and the density is infinite. So as energy is lost to Hawking radiation the black hole remains a black hole since it still has an infinite density mass at the centre. All that happens is that the event horizon radius decreases.
Now a real black hole is a rather more complicated object since what we would see were we next to one is an object that isn't a black hole yet, but that is destined to become one - or at least would be destined to become one if it weren't for Hawking radiation.
The object we see has a shell of infalling matter but no event horizon. The Hawking radiation decreases the mass of the infalling matter, but the remaining matter is still on a trajectory that would form an event horizon given infinite time. So if you were able to somehow magically turn off the Hawking radiation at any point the infalling matter left would still go on to form an event horizon. Since we can't turn off the Hawking radiation what happens is that all the infalling matter radiates away before the event horizon forms. So in this case the object never forms an event horizon at all.
answered Oct 21, 2016 at 9:42
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$\begingroup$ I thought Hawking radiation assumes the existence of an event horizon. $\endgroup$ Commented Oct 21, 2016 at 9:48
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$\begingroup$ @flippiefanus: The original calculation did because it's too hard to do the calculation for a time dependent object like a real black hole. However we now know that Hawking radiation does not require a (true) event horizon. See the question I linked in my answer and the links therein. $\endgroup$ Commented Oct 21, 2016 at 9:50
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$\begingroup$ How does this work in Kruskal coordinates? As far as Rindler describes it in his textbook, the collapsing object ought to form a singularity in finite proper time, and I suppose it is only much later (according to the Kruskal $T$ coordinate too?) that the black hole manages to evaporate away. $\endgroup$ Commented Oct 21, 2016 at 9:58
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$\begingroup$ @HenningMakholm The KS coordinates are a way of writing the Schwarzschild metric. A real black hole is not described by the Schwarzschild metric so the KS coordinates are not a useful way of describing it. The question of whether the black hole evaporates before the observer reaches the singularity is a hard one. My understanding is that the black hole evaporates in a finite, and shorter, time as measured by the infalling observer. So the singularity is never reached. $\endgroup$ Commented Oct 21, 2016 at 10:03
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$\begingroup$ @JohnRennie: Hmm, yes. I suppose what I was really asking is whether there's a known metric that describes the evaporating black hold inclusive of Hawking radiation, which could be written in a Kruskal-like enough (say, conformal) way to suggest an understanding of what it even means for the horizon to shrink. $\endgroup$ Commented Oct 21, 2016 at 10:12
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The error in your assumption is that a black hole has any memory of what it started out as (b.t.w. the jury is still out on that one).
From a general relativity point of view once a black hole has formed, there is no information about what it was before, as it now only carries three pieces of information, its mass $M$, its angular momentum $L$ and its charge $Q$.
The process of hawking radiation is to loose (at least) mass by radiating photons at a temperature proportional to $1/M$. This process will continue until $M$ is zero, and at no time will the black hole stop being a black hole.
