# Hawking Radiation as $M(t) \rightarrow 0$

The luminosity $P$ of a Kerr-Newman black hole with charge $Q$ and angular momentum $J$ is given by $$P = \frac{1}{240} \frac{\hbar c^6 \left( 1 -\frac{Q^2 }{ 4 \pi \epsilon_0 G M^2} -\left( \frac{J c}{M^2 G} \right)^2\right)^2 }{ \pi G^2 M^2 \left( 2 +2 \sqrt{ 1 - \frac{Q^2 }{ 4 \pi \epsilon_0 G M^2} - \left( \frac{J c}{M^2 G} \right)^2 }-\frac{Q^2 }{ 4 \pi \epsilon_0 G M} \right)^3}.$$ For a Schwarzschild black hole, we set $Q = 0$ and $J = 0$: $$P = \frac{\hbar c^6}{15360 \pi G^2 M^2}.$$ Moreover, through mass-energy equivalence, $P$ gives the rate $\dot{E}$ at which energy is radiated away in particles of mass $m$ (here we assume that all the particles created have mass $m$): \begin{align} P & = \frac{dE}{dt} \\ & = \frac{d}{dt} \frac{Nmc^{2}}{\sqrt{1-\beta^2}}, \end{align} where $N$ is the number of particles created. Given that $N \propto A$ (and in turn $N \propto M^2$), we write $$N = kA = 16\pi k \frac{G^{2}M^{2}}{c^{4}},$$ where $k$ is an arbitrary proportionality constant. Hence, \begin{align} P & = 16\pi k \frac{G^{2}M^{2}}{c^{4}} mc^{2} \frac{v}{\left ( 1 - \beta^2 \right )^{3/2}} \frac{dv}{dt} \\ \frac{\hbar c^8}{245 760 \pi^2 k m G^4 M^4} & = \frac{v}{\left ( 1 - \beta^2 \right )^{3/2}} \frac{dv}{dt}. \end{align} It is clear that as $M \rightarrow 0$, $\beta^2 \rightarrow 1$ and therefore $v \rightarrow c$. What gives?

Note that the evaporation time $t_{ev}$ $$t_{ev} = \frac{5120\pi G^2M_{0}^3}{\hbar c^4}$$ is finite. Given enough time, $M_{0}$ is radiated away in its entirety. But this would mean that the black hole would produce massive radiation with $v = c$, which is prohibited. If photons were emitted, then $\lambda \rightarrow 0$, which is nonsensical. Does this issue have to do with the fact that we do not know how Hawking radiation would behave on the Planck scale. Would we expect there to be a lower bound on the mass of black holes in a quantum theory of gravity?

The first one that comes to mind is that you keep the mass of the particles being emitted fixed. Clearly this can not be the case: at some point the mass of the black hole will be less than any fixed $m$.
Your assumption that $N\propto A$ seems totally unfounded. In fact, the rate of particles being created very obviously increases with decreasing $A$, so that assumption is clearly incorrect.
• Makes sense. However, if you consider only photons with energy $E = hf$, you get $P = h \dot{f}$, which thereby gives the equation $\frac{\hbar c^6}{15360 \pi G^2 M^2} = 16\pi k \frac{G^{2}M^{2}}{c^{4}} hf$ which suggests that as $M \rightarrow 0$, $\lambda \rightarrow 0$ – Colliding Particles Jun 5 '18 at 0:26
• You're making strange assumptions about the continuity of $N$ as well. Clearly at some point $Mc^2 <hf$ for the frequency your formula predicts. I also don't get your justification for why you think $N\propto A$. That seems totally unfounded to me. – Chris Jun 5 '18 at 0:32