How fast would a hypothetical microscopic quantum black hole evaporate with an effective mass of two protons?

Question

According to the analysis shown in this research here, see link the evaporation time can be calculated in seconds using this equation:

$$ \begin{array}{l} t_{\text {evap }}=\left(\frac{5120 \pi G^2}{\hbar C^4}\right) M^3 \\ \\ t_{\text {evap }} \approx 10^{-16} M^3 \end{array} $$

The smaller the mass of the BH the faster the BH evaporates into nothingness.

This would result in the calculation of a hypothetical quantum microscopic BH with an effective mass equal to two proton masses of an extremely tiny evaporation time of:

$$\\t_{\text {evap }} \approx 2 \cdot 10^{-16} m_{p}^3 \approx 3.75 \cdot 10^{-96} s $$ Which is less than the Planck time $t_{p}$ and therefore an impossibility.

Is the above analysis and result presented in the above linked research valid and is this conclusively proving that the formation of stable microscopic quantum BHs is impossible in our c bound spacetime universe?

Answer 1

Hawking's analysis is semi-classical. It assumes the geometry of spacetime is fixed and calculates how the quantum fields behave in this curved but fixed spacetime.

When the black hole is small its mass will be changing fast and therefore the spacetime geometry will also be changing fast, and under those circumstances the assumptions Hawking made are no longer valid. So you cannot apply Hawking's results to a black hole with the mass of two protons and expect it to work. This means your calculation probably is not physically meaningful.

To describe what happens in the final stages of black hole evaporation we need a full quantum theory of gravity but as of writing this no such theory exists. I have seen ideas for what happens based on string theory, but I would class these as pure speculation untainted by any form of experimental evidence. The harsh reality is we have no idea what happens in these extreme situations.