# Calculating the black hole evaporation time

Black holes evaporate very, very slowly by emitting Hawking radiation, and eventually they disappear, maybe even during the lifetime of the universe ($$\sim 1.5 \times 10^{18}{\rm s}$$). That's what I recall from the lectures I attended.

I now stumbled upon a Black Hole Evaporation Time Calculator which uses this formula:

$$\tau_{\rm evaporation} \sim \left( \frac{M_{\rm black~hole}}{M_\odot}\right)^3 \times 10^{66} {\rm ~years}$$

As usual, $$M_\odot$$ is the solar mass. My questions: How is that formula deduced? In which limiting case(s) does it hold?

Edit Just as a reminder: Right now, $$4.3 \times 10^{17} {\rm s}$$ passed since the big bang.

• There's a more extensive Hawking radiation calculator here: vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator with brief explanations of the formulae. Jan 23, 2021 at 0:02
• As the note at the bottom of that page mentions, the universe is still far too hot for stellar mass or larger BHs to evaporate. OTOH, that's not all that relevant to the huge evaporation times of such BHs. On that time-scale, the current age of the universe is barely an eyeblink since the Big Bang. Jan 23, 2021 at 0:06
• I don't know where that "lifetime of the universe" number comes from. The universe is expected to last a lot more than 47.5 billion years. I guess it could die earlier if the Big Rip happens. The other number looks fine. Jan 23, 2021 at 15:13
• BTW, Google is pretty good at doing unit conversions and calculations involving units. Just paste the conversion request or calculation into the search bar, eg (4.3*10^17 s) in years, and that'll invoke the Google Calculator. You sometimes need to get creative so that Google realises that it's a calculation and not a search request. Jan 23, 2021 at 15:14
• The vttoth calculator says that a 1 $M_\odot$ BH has a lifetime of 6.61817E74 seconds, or 2.09717E67 years. We don't expect astrophysical BHs, formed as supernova remnants, to be that small: the smallest is somewhere around 2.5 to 3 $M_\odot$. See en.wikipedia.org/wiki/… Jan 23, 2021 at 15:22

The derivation of life time formula for black holes:

It is not difficult to approximately estimate the life time of a black hole. Since Hawking radiation is indeed a black-body radiation, the rate at which the relativistic energy/mass of black hole ($$M$$) radiated away by its Hawking radiation can be computed by use of the Stefan-Boltzmann law as

$$- \frac{{dM}}{{dt}} = \frac{{dE}}{{dt}} = A\sigma {T^4},$$

where $$\sigma = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$$ is the Stefan–Boltzmann constant, $$A$$ is the object's surface (here black hole's horizon area, $$A=4 \pi r_s^2$$, where $$r_s=2GM$$ is the horizon's radius), and $$T$$ is the (Hawking) temperature. Note that this is the rate at which any blackbody radiates energy, no matter what is that object. Also, in this formula, $$- \frac{{dM}}{{dt}}$$ means that when black hole radiates, its mass decreases. Having this together with the Hawking temperature of the (Schwarzschild) black hole, i.e.,

$${T_{Hawking}} = \frac{{\hbar}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}^\circ },$$

we can evaluate the life time of the black hole ($$M_\odot = 2×10^{30} \rm{kg}$$ is the solar mass). For this purpose, we shall assume that the black hole initial mass ($$M$$) will eventually evaporate to nothing. So, integrating the Stefan-Boltzmann law, i.e. $${\tau _{{\rm{life}}}} = - \int_M^0 {{{(A\sigma {T^4})}^{ - 1}}dM}$$, yields

$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$

This is much greater than the age of the universe!

In addition, I should emphasize that these computations have been performed in units in which the speed of light is set to unity ($$c=1$$). For example, in SI units, the Stefan-Boltzmann constant is given by $$σ=5.67 \times 10^{-8}\,\mathrm {W\,m^{-2}\,K^{-4}}$$), but in this answer we had $$\sigma = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$$.

Edit: In comments, it has been discussed that there are other approximations. Of course, for example, the Hawking temperature for astrophysical black holes would be extremely small, even much smaller that the CMB radiation (see more details in my answer here, please), so black hole can absorb more matter and then they grow more and more! In order to derive that formula, you should omit this challenging part. Another approximation comes from the number of particles (scalars, photons, vectors, fermions). However, the Hawking temperature (e.g., according to surface gravity definition) is same for every type of particle, and the number of species only adds a numerical factor to the Stefan-Boltzmann law, but still the order of magnitude of the final answer is approximately valid. In addition, the particles scaping from the black hole need to pass trough the potential and this change the resulting spectrum by a greybody factor ($$\Gamma(\Omega)<1$$). In this answer, which is very common in GR books, I utilized the Hawking radiation of Schwarzschild black hole with $$\Gamma(\Omega)=1$$. For more (another) details, also see the following related SE links/papers/books: