# How does Hawking radiation grow as a black hole evaporates?

The temperature of Hawking radiation is inversely proportional to the mass of a black hole, $T_{\rm H}\propto M_{\rm BH}^{-1}$, and so as the black hole shrinks the temperature of the radiation should grow. What happens as the mass shrinks to zero?

Naive application of the above formula leads to arbitrarily energetic Hawking radiation. Do assumptions in the Hawking Radiation derivation just break down in this limit? If so, is it known how to handle this limit?

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Naive application of the above formula leads to arbitrarily energetic Hawking radiation. What do you mean by this? Any blackbody spectrum contains photons with arbitrarily high energies. –  Ben Crowell Oct 10 '13 at 1:04

The calculation of the Hawking temperature is semi-classical i.e. the spacetime curvature is treated classically while the radiation is modelled using QFT. As the black hole mass approaches zero we expect the semi-classical approximation to break down and a full quantum gravity calculation will be needed. At the moment we have no theory for quantum gravity so we cannot model the last few moments of the evaporation.

I have seen a presentation by (I think) Arkani-Hamed where he talked about such a calculation using string theory. As I recall the event horizon disappeared to leave a number of highly excited strings. I don't know if this work was published anywhere, but a quick Google found several publications e.g. this pdf.

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