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Reading about Hawking's Radiation I understood that black holes lose energy over time - which is logical in some way (otherwise they would be there forever and heat death would never technically happen)

But - how exactly does it "evaporate"? What happens when it no longer contains enough mass within it's Schwartzshild radius ? Does it explode somehow? Transforms into "regular matter"? Simply vanishes over time? Or?

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    $\begingroup$ The radius shrinks as the mass shrinks. $\endgroup$ – G. Smith May 24 at 20:10
  • $\begingroup$ Oh, thanks, that's that "small" detail that I missed. If you would like to post it as an answer - I'd gladly accept $\endgroup$ – Alma Do May 24 at 20:10
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    $\begingroup$ There’s more to it than that. At the end the radiation flux increases so much that there is an “explosion”, which is why one of Hawking’s papers was titled “Black hole explosions?” And no one understands whether some kind of Planck-scale remnant is left behind. $\endgroup$ – G. Smith May 24 at 20:12
  • $\begingroup$ I’ve posted an answer. $\endgroup$ – G. Smith May 25 at 3:32
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A black hole evaporates by radiating away energy in the form of photons, gravitons, neutrinos, and other elementary particles in a process that involves quantum field theory in curved spacetime. This causes it to lose mass, and so its radius shrinks. It remains a black hole as it shrinks. The increased spacetime curvature at the horizon makes it radiate more and more powerfully; its temperature gets hotter and hotter. The more mass it loses, the faster it loses what it has left!

I agree with Michael Walsby that small black holes are speculative and have not been detected. I am not so sure that they never will be, and it is important to understand how they behave.

As the Wikipedia article explains, for a non-rotating black hole of mass $M$, the radius of the event horizon is

$$R=\frac{2G M}{c^2}$$

and the Hawking temperature is

$$T=\frac{\hbar c^3}{8\pi k_B G M}.$$

If you make the approximation that the black hole is a perfect blackbody, then the radiated power is

$$P=\frac{\hbar c^6}{15360\pi G^2 M^2}$$

and the lifetime of the hole is

$$t=\frac{5120\pi G^2 M^3}{\hbar c^4}.$$

Notice the simple power dependence of all these quantities on $M$. Everything else is just constants. It is easy to substitute numerical values and compute the following table for black holes whose masses range from that of an asteroid down to that of a bowling ball:

$$\begin{array}{ccccc} M\text{ (kg)} & R\text{ (m)} & T\text{ (K)} & P\text{ (W)} & t \text{ (s)}\\ 10^{20} & 1.49\times10^{-7} & 1.23\times10^{3} & 3.56\times10^{-8} & 8.41\times10^{43}\\ 10^{19} & 1.49\times10^{-8} & 1.23\times10^{4} & 3.56\times10^{-6} & 8.41\times10^{40}\\ 10^{18} & 1.49\times10^{-9} & 1.23\times10^{5} & 3.56\times10^{-4} & 8.41\times10^{37}\\ 10^{17} & 1.49\times10^{-10} & 1.23\times10^{6} & 3.56\times10^{-2} & 8.41\times10^{34}\\ 10^{16} & 1.49\times10^{-11} & 1.23\times10^{7} & 3.56\times10^{0} & 8.41\times10^{31}\\ 10^{15} & 1.49\times10^{-12} & 1.23\times10^{8} & 3.56\times10^{2} & 8.41\times10^{28}\\ 10^{14} & 1.49\times10^{-13} & 1.23\times10^{9} & 3.56\times10^{4} & 8.41\times10^{25}\\ 10^{13} & 1.49\times10^{-14} & 1.23\times10^{10} & 3.56\times10^{6} & 8.41\times10^{22}\\ 10^{12} & 1.49\times10^{-15} & 1.23\times10^{11} & 3.56\times10^{8} & 8.41\times10^{19}\\ 10^{11} & 1.49\times10^{-16} & 1.23\times10^{12} & 3.56\times10^{10} & 8.41\times10^{16}\\ 10^{10} & 1.49\times10^{-17} & 1.23\times10^{13} & 3.56\times10^{12} & 8.41\times10^{13}\\ 10^{9} & 1.49\times10^{-18} & 1.23\times10^{14} & 3.56\times10^{14} & 8.41\times10^{10}\\ 10^{8} & 1.49\times10^{-19} & 1.23\times10^{15} & 3.56\times10^{16} & 8.41\times10^{7}\\ 10^{7} & 1.49\times10^{-20} & 1.23\times10^{16} & 3.56\times10^{18} & 8.41\times10^{4}\\ 10^{6} & 1.49\times10^{-21} & 1.23\times10^{17} & 3.56\times10^{20} & 8.41\times10^{1}\\ 10^{5} & 1.49\times10^{-22} & 1.23\times10^{18} & 3.56\times10^{22} & 8.41\times10^{-2}\\ 10^{4} & 1.49\times10^{-23} & 1.23\times10^{19} & 3.56\times10^{24} & 8.41\times10^{-5}\\ 10^{3} & 1.49\times10^{-24} & 1.23\times10^{20} & 3.56\times10^{26} & 8.41\times10^{-8}\\ 10^{2} & 1.49\times10^{-25} & 1.23\times10^{21} & 3.56\times10^{28} & 8.41\times10^{-11}\\ 10^{1} & 1.49\times10^{-26} & 1.23\times10^{22} & 3.56\times10^{30} & 8.41\times10^{-14}\\ 10^{0} & 1.49\times10^{-27} & 1.23\times10^{23} & 3.56\times10^{32} & 8.41\times10^{-17}\\ \end{array}$$

As you can see, as the hole shrinks, it gets tremendously hot and radiates enormous amounts of power. This is why Hawking titled one of his papers "Black hole explosions?"

As far as I know, no one is sure whether a hole evaporates completely or leaves behind a Planck-scale remnant.

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  • $\begingroup$ We do not know what the rest masses of neutrinos are, but once the temperature approaches these values neutrinos would be radiating most of the power. When the black hole temperature approaches electron, muon, pion etc. rest mass this channel also becomes relevant and adds to overall power. So possibly all rows of your table are incorrect. $\endgroup$ – A.V.S. May 25 at 5:50
  • $\begingroup$ I agree. I used the simple formulas from Wikipedia’s “Hawking radiation” article, which calls them a crude analytic estimate. They give the general idea that the hole shrinks faster and faster and radiates more and more power. Accurate calculations require detailed calculations such as those initially done by Don Page in 1976, taking the transmission coefficients for fields of various spins and masses into account. $\endgroup$ – G. Smith May 25 at 6:04
  • $\begingroup$ How is Hawking calculation of power, evaporating time and temperature affected by the reaching of the corresponding particle masses?References? $\endgroup$ – riemannium Jun 30 at 17:04
  • $\begingroup$ journals.aps.org/prd/abstract/10.1103/PhysRevD.16.2402 $\endgroup$ – G. Smith Jun 30 at 17:09
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No black hole has ever yet evaporated;the energy they absorb from their surroundings far exceeds what they lose by Hawking radiation. It may well be the case that the universe will collapse & be recycled in a Big Crunch before the first black hole has had time to evaporate. To those who say the universe is expanding too fast to collapse,I say there is not unanimous agreement among cosmologists for that. At present,a Big Crunch can't be ruled out.

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  • $\begingroup$ Doesn't this statement depend on the size of the black hole? $\endgroup$ – BowlOfRed May 24 at 21:43
  • $\begingroup$ Yes,but stories of marble-sized black holes containing about the same amount of mass as the Earth are pure speculation. No one has ever detected such a black hole & almost certainly never will. In the highly unlikely event they exist,they will still be mopping up matter from their surroundings & becoming larger rather than smaller $\endgroup$ – Michael Walsby May 24 at 21:53
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    $\begingroup$ You introduce fair points, but "No black hole has ever yet evaporated" is a bold statement, unless you refer to macroscopic black holes only. Primordial microscopic black holes may have evaporated already. $\endgroup$ – Rexcirus May 26 at 10:13
  • $\begingroup$ Yes,that could be,but I have never yet seen any evidence that microscopic black holes ever existed;it's all mere speculation. $\endgroup$ – Michael Walsby May 26 at 10:19
  • $\begingroup$ I suspect some of the sources of gamma rays, cosmic rays or even FRB could be Hawking radiation emissions...No proof though... $\endgroup$ – riemannium Jun 30 at 17:06

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