Do black holes emit equal amounts of gravitons in addition to photons via Hawking radiation?

Question

According to Hawking’s theory, black holes have temperatures inversely proportional to their masses and emit photons like an ideal black body. However, besides EM radiation there is also gravitational wave (GW) radiation made of hypothetical gravitons (which are not yet observed). Because gravitons are also massless, the ratio of EM and GW radiation should be irrelevant to the masses of black holes (unlike massive particles which have cutoff temperatures).

So what’s the exact proportion of these two terms? While it’s intuitive that the proportion of GW radiation should be orders of magnitude lower due to the extremely weak coupling of gravity, Hawking’s calculations revealed that the magnitude of the EM radiation is irrelevant to the coupling constant of electromagnetism (aka fine structure constant). This is not hard to understand. Ideal black body radiation is only determined by the energy partitioning between different degrees of freedom at thermal equilibrium. The extremely weak coupling of gravity only delays but not prevents the equilibrium. Considering that gravitons are massless and have two independent polarizations like photons, they should accommodate equal degrees of freedom. As a result, black holes should emit an equal proportion of EM and GW radiation. Although currently we don’t have a viable theory of quantum gravity yet, some candidate theories like string theory can deduce the same result of Hawking’s theory. Can we do similar calculations on GW radiation using these theories?

Answer 1

Your intuition is roughly correct. If perturbations of the gravitational field act like other quantum fields (i.e. gravitons exist), then Hawking radiation will also consist of gravitons. However, the amount of graviton Hawking radiation will not be the same as the amount of photon Hawking radiation. The amount of Hawking radiation emitted in a particular massless field, primarily depends on the spin of that field.

There are two main physical differences:

1. Low order multipoles. The graviton is a spin-2 field. Consequently, it cannot produce any monopole or dipole modes.

2. Greybody factors While Hawking radiation starts out as a perfect black body spectrum, it doesn't stay that way. As the Hawking radiation travels out of the potential well of the black hole, parts of the radiation get reflected back to the black hole. The fraction of the radiation in a particular mode that reaches infinity is called its greybody factor. This greybody factor depends on the spin of the field.

Consequently, the amount of Hawking radiation emitted into a particular massless field is different for gravitons (spin 2), photons (spin 1), or a hypothetical massless scalar field (spin 0).

By numerically calculating the greybody factors Don Page calculated the strengths of the Hawking radiation in different types of fields in this paper.

According to his calculations for a non-rotating black hole the energy emitted into Hawking photons is roughly six times as large as the energy emitted into Hawking gravitons. This ratio changes as the spin of the black hole is increased, since due to super-radiance it is easier for higher spin fields to reach infinity (and can even lead to amplification of the Hawking radiation). When the spin parameter $a\sim 0.6M$ the photon and graviton flux are equal. For an extremal Kerr black hole the graviton flux is almost 28 times as large as the photon flux.