Theoretically, does an evaporating black hole remain a black hole until it evaporates to nothing? Or does it change into something else once it evaporates to below a certain mass?


3 Answers 3


Utterly unknown.

So Hawking proposed radiation, that reduces the mass of the black hole (BH) by black body radiation. The effective temperature is antiproportional to the mass of the BH. In the linked paper, Hawking concludes that BHs end in "fairly small explosions by astronomical standards."

There are different solutions to the equations of general relativity that produce black holes, not only the common Schwarzschild metric. Whether the curvature singularity or the event horizon vanish before the BH completely evaporates, depends on the internal structure of the BH. As any mass, even very small ones, can become BHs if you compress them enough, there might be a singularity until the very end (not taking into account quantum mechanics). On the other hand a "vanishing event horizon" is not something we should be strictly speaking about, because then it wasn't an event horizon in the first place. You may see the looming problems with black hole remnants.

Also: according to the no-hair theorem, BHs are completely characterized by mass, electric charge and angular momentum. If you can see a remnant, e.g. a clump of matter, could you derive its complete properties from the previous three only?

In research, a remnant seems to be generally undesired but possible. This fresh and unreviewed paper jumps on the bandwagon and proposes white holes as remnants. So the debate about the end of BHs and their information theoretical implications is still a hot topic.

Disclaimer: I'm not a BH specialist. (And I'm waiting for the time where "black hole specialist" is a craft job designation.)

  • $\begingroup$ The main web page for the linked paper is arxiv.org/abs/1412.8366 , and it appears the paper was published in 2015. $\endgroup$
    – Bit Chaser
    Apr 5, 2018 at 17:35
  • $\begingroup$ There are two papers linked in adjacent sentences. (I've tried to rephrase it to avoid that, but then it didn't sound quite right). I agree that linking to the abstract page might be better and changed the links. $\endgroup$
    – Hannes
    Apr 5, 2018 at 18:09
  • $\begingroup$ "As any mass, even very small ones, can become BHs if you compress them enough" - what? Isn't the definition of a BH that it is so heavy that even light is sucked into it. If the Moon become as dense as a BH, it would still only weigh as much as it does today and today it doesn't attract light in any significant way. $\endgroup$
    – d-b
    Apr 13, 2021 at 19:46
  • $\begingroup$ @d-b It's due to the Schwarzschild radius. The mass stays the same, but the space in which it is packed is tiny, which is what causes the high gravity. The Schwarzschild radius is the radius in which the object's mass would need to be for it to collapse into a black hole. $\endgroup$ Nov 16 at 14:30

Any object whose radius is smaller than its Schwarzschild radius is called a black hole. The Schwarzschild radius is given as $$ r_s = \frac{2GM}{c^2}$$ where G is the gravitational constant, M is the object mass and c is the speed of light.

What's important to note here is that this is a linear relationship; in other words, the radius decreases in exact proportion to the decrease in mass. We would therefore expect the black hole to continue to behave as a black hole pretty much right up to the point that it completely evaporates.

I say "pretty much", because what happens at the point of final evaporation is that the density and temperature of the black hole approach infinity while its mass and Schwarzschild radius approach zero. The physics at that point is well beyond my modest abilities/knowledge; I suspect our current models break down at a radius below the Planck length.


According to theory of general relativity mass warps the spacetime. In black hole, its mass causes high curvature of spacetime (infinite curvature at its centre called singularity). When black hole evaporates,its mass decreases therefore its ability to wrap spacetime decreases and when it evaporates fully there will be no distortion of spacetime. So it will become normal space without zero/very less curvature of space time.

  • $\begingroup$ I think you mean "warps", not "wraps". $\endgroup$
    – PM 2Ring
    Apr 6, 2018 at 14:46

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