Schwarzschild solution is permanent and unchanging in time. What would happen if we modify it so to account for the change of BH mass over time due to evaporation?

  • $\begingroup$ Real black holes are very cold. Stellar mass BHs have Hawking temperatures in the nanokelvin range, SMBHs are even colder. The CMB will be much hotter than that into the far distant future. So no black holes will be losing mass for a very long time. (Also, real black holes have significant angular momentum, which means you need the Kerr metric, not the Schwarzschild). $\endgroup$ – PM 2Ring Jan 6 at 12:13
  • 1
    $\begingroup$ @PM2Ring so what? Any black hole should completely evaporate by the time any external object reaches its event horizon. So when we are discussing objects near the event horizon we have to account for evaporation. $\endgroup$ – Anixx Jan 6 at 12:17
  • 2
    $\begingroup$ To a distant observer, it appears that stuff slows down & "freezes" at the event horizon (where it rapidly redshifts to invisibility), but it does not look that way to a body freefalling into the BH. However, you need to use appropriate coordinates to prove that. Schwarzschild coords aren't any good for this, since they have a coordinate (i.e., removable) singularity at the EH. You need to use something like Lemaître coords, or Gullstrand-Painlevé. $\endgroup$ – PM 2Ring Jan 6 at 12:34
  • $\begingroup$ That would be the Vaidya metric : link.springer.com/article/10.1023%2FA%3A1018871522880 $\endgroup$ – Slereah Jan 6 at 14:00

You use the word "solution" in your title, which implies a solution to the Einstein field equations. Given that the initial condition on a Cauchy surface is a Schwarzschild black hole, the solution is unique, and there is no radiation. They Einstein field equations are classical, and they are not compatible with Hawking radiation.

If you want radiation coming out of a classical black hole, there are probably solutions that are very close to the maximal extension of the Schwarzschild spacetime, so that they include both a white hole and a black hole. But these are not compatible with formation by astrophysical collapse, and any radiation from the white hole is arbitrary and unpredictable, and is not the same as Hawking radiation.

| cite | improve this answer | |

You can model the time-dependent spherically symmetric space-time (with a nonzero stress-energy tensor) of an evaporating black hole by the metric $$ds^2 = - e^{2\psi(r,v)}(1 - 2m(v,r)/r)dv^2 + e^{2\psi(r,v)} dv dr + r^2 d\Omega^2$$ From Einstein equations you then get differential equations for $m,\psi$ based on the stress-energy matter content of the space-time. You can take this content to be the expectation value of the (renormalized) stress-energy tensor of your quantum fields, and solve the corresponding set of equations in a semi-classical regime. This is impossible exactly, and it is thus necessary to pass to some kind of adiabatic or iterative approximation. I found a nice summary of this in the paper "The Semi-Classical Back Reaction to Black Hole Evaporation" by S. Massar (1994).

Beware, however, that studying back-reaction due to Hawking radiation beyond simple models or first perturbative corrections is not physical. This is because the relevant dimensionless parameter that controls the magnitude of back-reaction is $\dot{M} \sim \hbar/M^2$ (in $G=c=1$ units). But notice that the same parameter controls quantum-gravity contributions at the horizon! So if you are studying back-reaction from Hawking evaporation on the space-time, you should be including quantum-gravity contributions to the space-time geometry as well. In other words, there is no self-consistent treatment of the late time of black-hole evaporation, and taking back-reaction into account is not going to change that.

In the light of the discussion under the post, I will show the Penrose diagram that you get from various toy models of black hole evaporation. In other words, this is the expected scenario that has not been proven. It corresponds to a black hole that has been formed by collapse, evaporates, and then disappears in an "evaporation event", leaving empty space behind.

enter image description here

(taken from Okon & Sudarsky, 2018)

On this diagram you have the collapse of matter and formation of the black hole sketched in the lower part of the diagram (matter of the spherically symmetric collapsing star in shaded gray). $\Sigma_1$ corresponds to a hypersurface of some reasonable coordinate time that penetrates the formed event horizon marked by a black diagonal line. (Note that Schwarzschild-like time $t$ has constant slices that penetrate the horizon only in a funny singular way so this is not a $t$-slice!) There are regions in the lower part of the diagram that are very well described by the Schwarzschild metric, namely the exterior of the star (even after horizon formation) up to some time before the evaporation event.

Once you understand the actual causal structure of Schwarzschild space-time and the language of Penrose diagrams, this picture will make a lot of sense. But once again, what happens around the "evaporation event" is unresolved, and the upper right part of the diagram might actually look very differently in nature.

| cite | improve this answer | |
  • $\begingroup$ I thinjk back-reaction contributes nothing. $\endgroup$ – Anixx Jan 6 at 14:34
  • $\begingroup$ @Anixx What happens in Schwarzschild coordinates is not a good representation of what actually happens in the space-time, especially around the horizon. I recommend to follow the Kruskalization and the construction of the Penrose diagram of Schwarzschild space-time as presented in the canonical book "Gravitation" of Misner, Thorne & Wheeler. $\endgroup$ – Void Jan 6 at 14:49
  • $\begingroup$ Schwartzschild space-time is useless for discussion on what happens around the horizon because it is not applicable to the time scales comparable with the black hole evaporation (it is static). Thus I am asking for a more suitable solution. $\endgroup$ – Anixx Jan 6 at 15:28
  • $\begingroup$ I added a discussion of the global structure of the evaporation scenario and where the Schwarzschild space-time should be a good approximation. Of course, taking a patch of Schwarzschild space-time in the usual Schwarzschild coordinates is a bad idea, you have to take a patch in terms of different space-time coordinates. A black hole is a space-time event, it is not an object, and you can see that if you examine the solution more closely in various coordinates. $\endgroup$ – Void Jan 7 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.