# Black hole evaporation: how long will a black hole live?

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Boxes 16.3, 16.4.

Since a black hole can radiate its energy away and the energy of a black hole is equivalent to its mass, it’s possible that a black hole will simply evaporate after a long enough time. We can get an estimate of this time as follows. First, we recall that, for particle-antiparticle pairs that are produced very close to the event horizon, the energy of the particle at infinity tends to a constant:

$\displaystyle E_{\infty}=\frac{\hbar}{4GM} \ \ \ \ \ (1)$

This is only a crude estimate based on oversimplifying the situation and a more accurate calculation, due to Stephen Hawking, results in

$\displaystyle E_{\infty}=\frac{\hbar}{8\pi GM} \ \ \ \ \ (2)$

In thermodynamics, a blackbody is a body that absorbs all frequencies of electromagnetic radiation (hence ‘black’) and radiates with a spectrum that depends only its temperature ${T}$. The wavelength of radiation peaks at a shorter value the higher the temperature (which is why an iron bar, for example, glows red when it gets hotter; at cooler temperatures it radiates in the infra-red). Hawking showed that the energy spectrum of a black hole is actually the same as that of a blackbody with temperature ${T}$ if we set ${E_{\infty}=k_{B}T}$, where ${k_{B}}$ is Boltzmann’s constant. We can therefore define the temperature of a black hole as

$\displaystyle \boxed{T=\frac{\hbar}{8\pi k_{B}GM}} \ \ \ \ \ (3)$

To use this in calculations, it’s best to convert the constants to relativistic form, where ${c=1}$. We start with their values in SI units. For Planck’s constant:

 $\displaystyle \hbar$ $\displaystyle =$ $\displaystyle 1.0546\times10^{-34}\mbox{ J s}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.0546\times10^{-34}\mbox{ kg m}^{2}\mbox{ s}^{-1}\times\left(3\times10^{8}\right)^{-1}\mbox{ s m}^{-1}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.5153\times10^{-43}\mbox{ kg m} \ \ \ \ \ (6)$

For Boltzmann’s constant:

 $\displaystyle k_{B}$ $\displaystyle =$ $\displaystyle 1.3807\times10^{-23}\mbox{ J K}^{-1}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.3807\times10^{-23}\mbox{ kg m}^{2}\mbox{ s}^{-2}\mbox{ K}^{-1}\times\left(3\times10^{8}\right)^{-2}\mbox{ s}^{2}\mbox{ m}^{-2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.534\times10^{-40}\mbox{ kg K}^{-1} \ \ \ \ \ (9)$

Thus the temperature is

$\displaystyle T=\frac{9.118\times10^{-5}}{GM}\mbox{ K} \ \ \ \ \ (10)$

In terms of the solar mass ${M_{s}}$, we can use ${GM_{s}=1477\mbox{ m}}$ so

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{9.118\times10^{-5}}{GM_{s}}\frac{M_{s}}{M}\mbox{ K}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.173\times10^{-8}\frac{M_{s}}{M}\mbox{ K} \ \ \ \ \ (12)$

A one solar mass black hole is thus almost at absolute zero.

The rate at which an object radiates energy is given by the Stefan-Boltzmann law, which states

$\displaystyle \frac{dE}{dt}=A\sigma T^{4} \ \ \ \ \ (13)$

where ${A}$ is the surface area of the object and ${\sigma=2.105\times10^{-33}\mbox{ kg m}^{-3}\mbox{ K}^{-4}}$ is the Stefan-Boltzmann constant, in relativistic units. In the case of a black hole, the energy is just the mass, so

$\displaystyle \frac{dE}{dt}=-\frac{dM}{dt} \ \ \ \ \ (14)$

and the area is that of a sphere with a radius ${r=2GM}$, so

$\displaystyle A=4\pi\left(2GM\right)^{2}=16\pi\left(GM\right)^{2} \ \ \ \ \ (15)$

Using 3, we then have

 $\displaystyle \frac{dM}{dt}$ $\displaystyle =$ $\displaystyle -16\pi\left(GM\right)^{2}\sigma\left(\frac{\hbar}{8\pi k_{B}GM}\right)^{4}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\frac{1}{M^{2}}\ \ \ \ \ (17)$ $\displaystyle M^{2}dM$ $\displaystyle =$ $\displaystyle -\frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}dt\ \ \ \ \ (18)$ $\displaystyle \frac{1}{3}M^{3}$ $\displaystyle =$ $\displaystyle \frac{\sigma\hbar^{4}}{256\pi^{3}k_{B}^{4}G^{2}}\Delta t\ \ \ \ \ (19)$ $\displaystyle \Delta t$ $\displaystyle =$ $\displaystyle \frac{256\pi^{3}k_{B}^{4}}{3\sigma\hbar^{4}G}\left(GM\right)^{3}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{256\pi^{3}k_{B}^{4}}{3\sigma\hbar^{4}G}\left(\frac{GM}{GM_{s}}\right)^{3}\left(GM_{s}\right)^{3} \ \ \ \ \ (21)$

where ${\Delta t}$ is the time required for the black hole to evaporate completely. We can plug in the numbers, using ${G=7.426\times10^{-28}\mbox{ m kg}^{-1}}$ and ${1\mbox{ year}=9.461\times10^{15}\mbox{ m}}$ to get

 $\displaystyle \Delta t$ $\displaystyle =$ $\displaystyle 1.9777\times10^{83}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ m}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.0903\times10^{67}\left(\frac{M}{M_{s}}\right)^{3}\mbox{ years} \ \ \ \ \ (23)$

In other words, for a one solar mass black hole, we’re not going to see it evaporate any time soon.

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