We know that the temperature $T$ of an uncharged non-rotating black hole is inversely proportional to the radius $r$ of the event horizon of the black hole, i.e., $$T \propto r^{-1}$$ Therefore, $$\dfrac{dr}{dt}\propto \dfrac{dM}{dt}\propto (r^2)(r^{-4})\propto r^{-2}$$ where $M$ is the mass of the black hole, and $t$ is the time in the clock of an asymptotic static observer. The $r^{-4}$ comes from the $T^4$ dependence of the Stephen-Boltzmann law and $r^2$ arises as the area of the surface from which the black hole emits radiation is $4\pi r^2$.

Now, the following is a leap of faith taken without an explicit calculation but I think it can be expected that although the $t$ used in the previous calculation is the asymptotic static observer's time, the rate of evaporation of the black hole depends on its radius in a qualitatively similar way for other observers as well, i.e., the smaller the black hole, the faster the event horizon shrinks.

If this is true then for sufficiently small black holes, in the free-falling frames near the horizon, the rate of shrinking of the event horizon should become superluminal. This would mean that no particle could ever cross the event horizon as the particles are not allowed to be superluminal. This phenomenon of nothing crossing the horizon would not be due to any time-dilational effects as in the case of an asymptotic static observer. This would be a genuine inability of an observer to cross the horizon. So, my question is, is it true that if the black hole is small enough, nothing can enter the black hole?

If I stretch the argument a bit further, maybe I can argue that black holes would actually stop evaporating at this stage using the popular (and yes, I know not completely accurate) picture of the Hawking radiation where a particle-antiparticle pair is created near the horizon and one falls into the black hole. In this picture, one particle getting lost into the black hole is crucial otherwise it would just be a mundane quantum fluctuation that happens everywhere. So, if the black hole is evaporating fast enough, the particle at the horizon couldn't fall in. This would mean no Hawking radiation and thus, no further evaporation. Now, one might think that at this point where the black hole stops evaporating, the evaporation rate is zero and thus, the particles can again do their "one particle falls in and the other comes out" thing but I think it is not possible. If it were to happen then the temperature (being restricted to be high by the black holes small radius) would have to be so high that it would again have to stop evaporating instantaneously. I would like the responses to give some comments on whether this speculation has any sensible ground. I believe this argument could also mean that the Hawking temperature formula requires a serious revision for a small black hole. Is this true?


The Hawking calculation was an approximation using semiclassical gravity i.e. it treats the curved spacetime as a fixed background. So it is only useful when the power emitted as Hawking radiation is a small fraction of the black hole mass. You are quite correct that the Hawking formula would not work in the final stages of the evaporation where the rate of change of the black hole mass became significant.

However I don't think anyone has really improved on this and the final stages of black hole evaporation are still poorly understood. I have seen attempts to describe it using string theory, but these are extremely speculative to say the least. At the moment we simply don't know what happens.

But I don't think your comments about the evaporation stopping are well founded because the evaporation is not due to virtual particles falling into the black hole. That is a popular analogy and is little better than a caricature of the real process. See An explanation of Hawking Radiation for an attempt to explain Hawking radiation in a non-technical way (though I fear this was only partially successful).

The arguments about objects falling into the black hole are also ill founded because you need to distinguish between the proper time of the infalling object and the coordinate time of the observer watching from infinity. Even for large black holes, where evaporation can be ignored, the observer far from the black holes never sees anything cross the event horizon, while for the infalling observer the infall speed exceeds $c$ once the horizon has been crossed. The simple argument you present isn't a useful way to approach the question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.