# Black hole evaporation: remnants of the big bang

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.2.

We’ve seen that through radiation, a black hole can eventually evaporate in a time ${t_{0}}$ :

$\displaystyle \boxed{t_{0}=2.0903\times10^{67}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ years}} \ \ \ \ \ (1)$

where ${M_{s}}$ is the solar mass. Clearly if we’re going to observe black hole evaporation, its mass ${M}$ must be considerably less than the sun’s. Suppose black holes were formed during the big bang at ${13.7\times10^{9}}$ years ago. If the black hole is just evaporating now, its mass would have been:

 $\displaystyle M$ $\displaystyle =$ $\displaystyle M_{s}\left(\frac{13.7\times10^{9}}{2.0903\times10^{67}}\right)^{1/3}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8.69\times10^{-20}M_{s}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(8.69\times10^{-20}\right)\left(1.989\times10^{30}\right)\mbox{ kg}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.728\times10^{11}\mbox{ kg} \ \ \ \ \ (5)$

To put this in perspective, this is equivalent to an asteroid, with a typical rocky density of ${\rho=5000\mbox{ kg m}^{-3}}$, with a radius of

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \left(\frac{3M}{4\pi\rho}\right)^{1/3}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 202\mbox{ m} \ \ \ \ \ (7)$

To see how much energy is released in the final second of the black hole’s life, we can start with the equation we had earlier from the Stefan-Boltzmann relation:

$\displaystyle M^{2}dM=-\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}dt \ \ \ \ \ (8)$

If we integrate this from ${t=0}$ to a time ${t_{1}}$ one second before the present, at which time the black hole’s remaining mass is ${M_{1}}$, then

 $\displaystyle \int_{M_{0}}^{M_{1}}M^{2}dM$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\int_{0}^{t_{1}}dt\ \ \ \ \ (9)$ $\displaystyle \frac{1}{3}\left(M_{1}^{3}-M_{0}^{3}\right)$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{1}\ \ \ \ \ (10)$ $\displaystyle M_{1}$ $\displaystyle =$ $\displaystyle \left[M_{0}^{3}-\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{1}\right]^{1/3} \ \ \ \ \ (11)$

However, ${M_{0}^{3}}$ is just ${\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}t_{p}}$, where ${t_{p}}$ is the present time, so that ${t_{p}-t_{1}=1\mbox{ s}}$. Therefore the mass remaining 1 second before complete evaporation is:

 $\displaystyle M_{1}$ $\displaystyle =$ $\displaystyle \left[\frac{3\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\left(t_{p}-t_{1}\right)\right]^{1/3}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 341.38\left(t_{p}-t_{1}\right)^{1/3} \ \ \ \ \ (13)$

In GR units, a time of 1 second is ${3\times10^{8}\mbox{ m}}$ so we get for the mass

$\displaystyle M_{1}=2.28\times10^{5}\mbox{ kg} \ \ \ \ \ (14)$

This is equivalent to

$\displaystyle E=M_{1}c^{2}=2.06\times10^{22}\mbox{ J} \ \ \ \ \ (15)$

which is about 500,000 times the energy released in an atomic bomb blast.