<p><strong>The derivation of life time formula for black holes:</strong></p>
<p>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 (<span class="math-container">$M$</span>) radiated away by its Hawking radiation can be computed by use of <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" rel="nofollow noreferrer">the Stefan-Boltzmann law</a> as</p>
<p><span class="math-container">$$- \frac{{dM}}{{dt}} = \frac{{dE}}{{dt}} = A\sigma {T^4},$$</span></p>
<p>where <span class="math-container">$\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$</span> is the Stefan–Boltzmann constant, <span class="math-container">$A$</span> is the object's surface (here black hole's horizon area, <span class="math-container">$A=4 \pi r_s^2$</span>, where <span class="math-container">$r_s=2GM$</span> is the horizon's radius), and <span class="math-container">$T$</span> 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, <span class="math-container">$- \frac{{dM}}{{dt}}$</span> means that when black hole radiates, its mass decreases. Having this together with the Hawking temperature of the (Schwarzschild) black hole, i.e.,</p>
<p><span class="math-container">$${T_{Hawking}} = \frac{{\hbar}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}^\circ },$$</span></p>
<p>we can evaluate the life time of the black hole (<span class="math-container">$M_\odot = 2×10^{30} \rm{kg}$</span> is the solar mass). For this purpose, we shall assume that the black hole initial mass (<span class="math-container">$M$</span>) will eventually evaporate to nothing. So, integrating the Stefan-Boltzmann law, i.e. <span class="math-container">${\tau _{{\rm{life}}}} =  - \int_M^0 {{{(A\sigma {T^4})}^{ - 1}}dM}$</span>, yields</p>
<p><span class="math-container">$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$</span></p>
<p>This is much greater than the age of the universe!</p>
<p>In addition, I should emphasize that these computations have been performed in units in which the speed of light is set to unity (<span class="math-container">$c=1$</span>). For example, in SI units, the Stefan-Boltzmann constant is given by <span class="math-container">$σ=5.67 \times 10^{-8}\,\mathrm {W\,m^{-2}\,K^{-4}}$</span>), but in this answer we had <span class="math-container">$\sigma  = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$</span>.</p>
<p><strong>And your final question:</strong></p>
<blockquote>
<p>In which limiting case(s) does it hold?</p>
</blockquote>
<p>This estimation has been derived for nonrotating black holes. For rotating (Kerr) black holes, it is expected the order of magnitude of this estimation is approximately valid, at least in the case of slowly rotating black holes. I guess, the computation in such cases would be more difficult.</p>
<p><strong>Edit:</strong> 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 <a href="https://astronomy.stackexchange.com/questions/43376/can-hawking-radiation-ever-be-detected-and-does-it-exist/43380#43380">my answer here</a>, 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 (<span class="math-container">$\Gamma(\Omega)&lt;1$</span>). In this answer, which is very common in GR books, I utilized the Hawking radiation of Schwarzschild black hole with <span class="math-container">$\Gamma(\Omega)=1$</span>. For more (another) details, also see the following related SE links/papers/books:</p>
<ol>
<li><p><a href="https://astronomy.stackexchange.com/questions/35260/is-hawking-radiation-detectable">Is Hawking radiation detectable?</a></p>
</li>
<li><p><a href="https://rads.stackoverflow.com/amzn/click/com/0521868343" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Introduction-Quantum-Effects-Viatcheslav-Mukhanov/dp/0521868343</a></p>
</li>
<li><p><a href="https://arxiv.org/abs/2011.03486" rel="nofollow noreferrer">https://arxiv.org/abs/2011.03486</a></p>
</li>
<li><p><a href="https://arxiv.org/abs/1711.01865" rel="nofollow noreferrer">https://arxiv.org/abs/1711.01865</a></p>
</li>
</ol>

CC BY-SA 4.0
2021-04-16T07:21:03.950Z
There is a few more approximations made in your calculation (which would be good to mention): 1) Assumes only a single species of massless particle with a single polarization. In reality it should be at least 4. 2) There is also Hawking radiation into fields with mass. This highly suppressed when the Hawking temperature is small compared to the mass of the fields, but will become relevant in the final stages of evaporation.  3) It ignores so called "gray body" factors due to part of the radiation scattering back into the black hole. (...)
TimRias
CC BY-SA 4.0
2021-04-16T07:23:12.637Z
4) It ignores that any black hole is emerged in the temperature bath provided by the CMB. As long as the temperature of the CMB is higher than that of the black hole, it will experience net absorption of photons and grow rather than evaporate.
TimRias
CC BY-SA 4.0
2021-04-15T20:58:09.650Z
2021-04-16T11:13:39.050Z
2021-04-16T11:13:39.050Z
Ad Astra
<p>Black holes evaporate very, very slowly by emitting <a href="https://en.wikipedia.org/wiki/Hawking_radiation" rel="nofollow noreferrer">Hawking radiation</a>, and eventually they disappear, maybe even during the lifetime of the universe (<span class="math-container">$\sim 1.5 \times 10^{18}{\rm s}$</span>). That's what I recall from the lectures I attended.</p>
<p>I now stumbled upon a <a href="http://eguruchela.com/physics/calculator/Black-Hole-Evaporation-Time-Calculator" rel="nofollow noreferrer">Black Hole Evaporation Time Calculator</a> which uses this formula:</p>
<p><span class="math-container">$$\tau_{\rm evaporation} \sim \left( \frac{M_{\rm black~hole}}{M_\odot}\right)^3 \times 10^{66} {\rm ~years}$$</span></p>
<p>As usual,  <span class="math-container">$M_\odot$</span> is the solar mass. My questions: <strong>How is that formula deduced? In which limiting case(s) does it hold?</strong></p>
<p><strong>Edit</strong> Just as a reminder: Right now, <span class="math-container">$4.3 \times 10^{17} {\rm s}$</span> passed since the big bang.</p>

CC BY-SA 4.0
2021-01-23T00:02:24.460Z
There's a more extensive Hawking radiation calculator here: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator with brief explanations of the formulae.
PM 2Ring
CC BY-SA 4.0
2021-01-23T00:06:01.650Z
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.
PM 2Ring
CC BY-SA 4.0
2021-01-23T15:13:49.140Z
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](https://en.wikipedia.org/wiki/Big_Rip) happens. The other number looks fine.
PM 2Ring
CC BY-SA 4.0
2021-01-23T15:14:10.963Z
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.
PM 2Ring
CC BY-SA 4.0
2021-01-23T15:22:18.093Z
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 https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit
PM 2Ring
<black-hole><hawking-radiation>
Calculating the black hole evaporation time
B--rian
# Calculating the black hole evaporation time

(Asked by: username_0 on 2021-01-22)

Black holes evaporate very, very slowly by emitting [Hawking radiation](https://en.wikipedia.org/wiki/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](http://eguruchela.com/physics/calculator/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.

**Tags:** black-hole, hawking-radiation
> **Comments:**
> **username_1**: There's a more extensive Hawking radiation calculator here: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator with brief explanations of the formulae.  
> **username_1**: 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.  
> **username_1**: 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](https://en.wikipedia.org/wiki/Big_Rip) happens. The other number looks fine.  
> **username_1**: 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.  
> **username_1**: 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 https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit  

---

## 1 answers

### Answer by username_2 on 2021-04-15. Score: 4 | Accepted answer

**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](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_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 $\sigma = 2.105 \times {10^{ - 33}}\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{{\rm{ - 3}}}}{{\rm{K}}^{{\rm{ - 4}}}}$.

**And your final question:**

> In which limiting case(s) does it hold?

This estimation has been derived for nonrotating black holes. For rotating (Kerr) black holes, it is expected the order of magnitude of this estimation is approximately valid, at least in the case of slowly rotating black holes. I guess, the computation in such cases would be more difficult.

**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](https://astronomy.stackexchange.com/questions/43376/can-hawking-radiation-ever-be-detected-and-does-it-exist/43380#43380), 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:

1. [Is Hawking radiation detectable?](https://astronomy.stackexchange.com/questions/35260/is-hawking-radiation-detectable)
2. [https://www.amazon.com/Introduction-Quantum-Effects-Viatcheslav-Mukhanov/dp/0521868343](https://rads.stackoverflow.com/amzn/click/com/0521868343)
3. <https://arxiv.org/abs/2011.03486>
4. <https://arxiv.org/abs/1711.01865>

> **Comments:**
> **username_3**: There is a few more approximations made in your calculation (which would be good to mention): 1) Assumes only a single species of massless particle with a single polarization. In reality it should be at least 4. 2) There is also Hawking radiation into fields with mass. This highly suppressed when the Hawking temperature is small compared to the mass of the fields, but will become relevant in the final stages of evaporation.  3) It ignores so called "gray body" factors due to part of the radiation scattering back into the black hole. (...)  
> **username_3**: 4) It ignores that any black hole is emerged in the temperature bath provided by the CMB. As long as the temperature of the CMB is higher than that of the black hole, it will experience net absorption of photons and grow rather than evaporate.  

---

