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.

3 thoughts on “Black hole evaporation: remnants of the big bang

  1. Pingback: Black hole radiation: mass as a function of time | Physics pages

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