Suppose we have two black holes of radius $R_b$ orbiting at a distance $R_r$. I believe semi-classical approximations describe correctly the case where $R_r$ is much larger than the average black body radiation wavelength due to Hawking radiation.

Do we have approximations for Hawking radiation temperature where the distance $R_r$ is of the same order, or in the case where it is much shorter than the radiation average wavelength?

In the absence of a concrete analysis for either one, Do we have any physical insight to affirm if Hawking radiation will be either inhibited or increased in the above situations?

  • $\begingroup$ I guess the radiation would be inhibited, since the black holes absorb radiation from each other, thus lose mass and hawking-radiate more slowly. $\endgroup$ – resgh Dec 8 '12 at 14:27
  • $\begingroup$ remember the distance between the BHs is the same order or smaller than the wavelength of the radiation. There might be nontrivial boundary effects that qualitative change the Bogoliubov transformation $\endgroup$ – lurscher Dec 8 '12 at 14:40
  • $\begingroup$ Just a guess. I'm not bothering to do any serious mathematics so I can't prove anything. $\endgroup$ – resgh Dec 8 '12 at 14:42
  • $\begingroup$ What exactly do you mean by at a distance $R_r$? $\endgroup$ – MBN Dec 8 '12 at 14:55
  • $\begingroup$ @MBN Read the first sentence of the question. $\endgroup$ – resgh Dec 8 '12 at 14:58

This problem would be very difficult. Obviously for it to be physically relevant this would have to pertain to two small black holes in a mutual orbit. Large astrophysical black holes are are at temperature $6.1\times 10^{-8}K(M/M_{sol})$ and Hawking radiation is insignificant,

General relativity can only solve the orbit of a test mass. The general two body problem is not integrable. Hence the motion of two black holes in a mutual orbit is not solved by exact means, but must be numerically evaluated. Then if you throw Hawking radiation into the picture it becomes complicated.

Geroch looked at the issue of the shape of a black hole horizon due to external distributions of matter. This was way back in the 1970s. Two mutually interacting black holes will also distort their horizons. In a sense the more curvature there is to the horizon per unit area the more radiation there will be. Think analogously with a flat radiating surface and one with a rough surface, or why fins and the rest are used in units to dissipate heat. The gravity field near a horizon area that is curved will have a larger gradient to the gravity field. The surface gravity $g^2~=~\nabla^a\xi_b\nabla_a\xi^b$ will be larger if the Killing fields have a greater divergence.

You would then have two competing processes. The first is that by distorting the horizons of the black holes the Hawking radiation might actually increase in rate. However, the two black holes will radiate some of that radiation at each other. How that detailed balance works out is hard to conjecture on.

| cite | improve this answer | |

The interesting form of radiation here is not Hawking radiation, but gravitational wave radiation. For astrophysically-sized black holes, the Hawking radiation is completely negligible relative to other processes. For example, for a solar mass sized Schwarzschild black hole, the black hole radiates like a black body at 60 nano Kelvins (far below the background CMB temperature). Certainly the calculation of the Hawking effect in this more complicated setting will be more difficult, but of course the Hawking radiation won't suddenly become important once you have two black holes that are orbiting each other closely.

| cite | improve this answer | |
  • $\begingroup$ Hawking radiation is (in part) electromagnetic waves; gravitational radiation is gravity waves. They're different. Why is one more interesting than the other? $\endgroup$ – Peter Shor Aug 16 '16 at 21:12
  • $\begingroup$ By interesting I mean physically dominant, in the sense that the physical effects of the Hawking radiation will be swamped by the effects of the gravitational radiation. $\endgroup$ – Surgical Commander Aug 17 '16 at 1:52

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.