5
$\begingroup$

Black holes are expected to radiate like a perfect black radiator at the Hawking temperature, which means that they'll emit all particles according to the relevant formulas one can derive using statistical mechanics. So, black holes should emit not only photons, but also gravitons, axions (if they exist), and also more massive particles like electrons, protons, and even smaller black holes (with extremely small probability).

If we focus on the emission of gravitons then this seems paradoxical, because the power emitted in the form of gravitational waves by an object is related to the second time derivative of the quadrupole moment. But where does the fluctuation of the quadrupole moment of a black hole come from?

Improve this question
$\endgroup$
1

3 Answers 3

Reset to default
2
$\begingroup$

You should think of it the following way: for each field (scalar field, Maxwell field, gravitational, etc), there are linearized perturbations off the black hole which vary in time and in space (both the radial direction and along the horizon). Since black holes radiate, each of these linearized perturbations are "turned on" or present, as you say.

In gravity, there is no monopole or dipole radiation, and so the fluctuations for the gravitational field begin with the $\ell = 2$ mode where $\ell$ is the usual spherical harmonic angular momentum quantum number. So for all modes of the gravitational radiation, the horizon is deformed through the addition of a quadruple or higher moment. For other types of radiation such as a scalar field, all the higher moments are present also, but so are the $\ell =0, 1$ ones.t

Improve this answer
$\endgroup$
2
$\begingroup$

you can say the same thing about the emitted electromagnetic radiation, especially for a neutral spherically symmetric black hole, where classically there can not be any radiation. The explanation to both phenomena is that the source of fluctuation is quantum mechanical, and the statement that the power of gravitational waves is proportional to the second time derivative of the quadrupole is an entirely classical formula.

Improve this answer
$\endgroup$
0
$\begingroup$

So, black holes should emit not only photons, but also gravitons, axions (if they exist), and also more massive particles like electrons, protons, and even smaller black holes (with extremely small probability).

There exists a misunderstanding here, I think.It is not the black hole that is emitting the radiation, but the environment at the event horizon, before the radius of inevitable trapping of any matter or radiation.

Black body radiation is connected with electromagnetic radiation, and has been derived and fitted to experiments for photons. It is not a classical phenomenon, it is one of the main reasons that quantum mechanics was established for the study of the microcosm of atoms molecules and particles, and declared to be the underlying framework of all nature.

Hawking showed that the radiation leaving the region above the event horizon of a black hole is mathematically the same as black body radiation. but it should be kept in mind that nothing escapes the event horizon of a black hole, so the black hole itself is not a black body radiator, but the region above the event horizon and the photon sphere ( trapped photons in orbit) is.

This radiation does not come from dipoles and quadrupoles radiating but from elementary particle interactions. These can in principle be computed using feynman diagrams of all orders and with vacuum fluctuations and particle antiparticle loops. Any particles radiated will be from a quantum mechanical effect.

Let us assume that gravity is quantized and gravitons exist.

The feynman diagrams describing the interactions of gravitons with the rest of the zoo of particles have vertices which depend on a coupling constant that is orders of magnitude smaller than the strong , and electromagnetic

gravitational coupling constant is the gravitational coupling constant

to be compared to 1/137 of the electromagnetic one.

As the couplings become squared even for first order interactions, considering the complexity of the feynman diagrams that will produce a real, leaving elementary particle, the probability of gravitons contributing appreciably to the radiation of a black hole is infinitesimally small.

Here is a heuristic drawing of Hawking radiation

hawkinrad

Showing vacuum fluctuation pairs one of them falling into the hole , the other escaping. It is not a simple situation.

If we focus on the emission of gravitons then this seems paradoxical, because the power emitted in the form of gravitational waves by an object is related to the second time derivative of the quadrupole moment. But where does the fluctuation of the quadrupole moment of a black hole come from?

Hawking radiation is a quantum mechanical phenomenon. It is not the result of a variation in a collective electric or magnetic field, as far as photon emission goes, and equally it does not depend on the variations of masses as far as graviton emissions go. It will be the result of particle interactions modeled with feynman diagrams with virtual exchanges to the available fields at the horizon of the black hole and the emission of a real particle with some probability.

.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ The final paragraph is very fuzzy. How can a particle interact with a Feynman diagram??? $\endgroup$
    – Arnold Neumaier
    Apr 26, 2016 at 16:42
  • $\begingroup$ @ArnoldNeumaier I meant modeled of course. I will correct it. $\endgroup$
    – anna v
    Apr 26, 2016 at 16:49
  • $\begingroup$ Gravitational radiation are harder to generate than electromagnetic radiation not only because gravitational coupling constant is small, but also the lowest order of gravitational radiation is quadripole radiation compared to dipole radiation of electromagnetic wave. Does it make black body gravitational radiation even weaker? $\endgroup$
    – Ballistics
    Apr 26, 2016 at 17:08
  • 1
    $\begingroup$ @Ballistics let me repeat: black body radiation is a quantum mechanical phenomenon, it cannot be explained classically even for electromagnetic waves. Therefore gravitons are the game ( if they exist) with their spin two and the corresponding diagrams. Quadrupole gravitational radiation is built up by a confluence of gravitons from asymmetric masses, similar to the way that dipole is built up by a confluence of photons from an antenna, but that is another story. not black body which is quantum. $\endgroup$
    – anna v
    Apr 26, 2016 at 17:16
  • $\begingroup$ You can define a gravitational coupling constant if you like, but the mass of the electron is not relevant in this context. This answer reads to me like a lot of words with no underlying logic. The result you claim is also simply wrong. See arxiv.org/abs/0812.0825 , text following equation 4. $\endgroup$
    – user4552
    Aug 26, 2017 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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