# If photons have a finite mass, will black holes emit a factor 3/2 more electromagnetic Hawking radiation?

An old but wrong argument that photons must be exactly massless is to observe that black bodies radiate according to the Stefan–Boltzmann law and then argue that if photons were massive they would have 3 polarizations and therefore would have to radiate 3/2 times the power predicted by the Stefan–Boltzmann law, no matter how small the photon mass is. This argument is flawed because in the limit of zero photon mass the transitions involving changing the number of longitudinal photons are suppressed by a factor equal to the squared ratio of the photon mass and the frequency. So, black bodies made out of ordinary matter will actually be (almost) transparent to longitudinal photons.

But it seems to me that since black holes are ideal black bodies, they will emit longitudinal photons and thus radiate a factor 3/2 more if photons have a finite mass.

• "black holes are ideal black bodies" - does that, in fact, follow from the derivation of Hawking radiation? For one, I think the original derivation was for a scalar field - do you have a reference where Hawking radiation is actually derived for photons, instead of assumed by analogy? I see no reason one wouldn't expect a similar suppressing factor. – ACuriousMind Feb 17 '17 at 12:51
• Interesting idea. Note that black holes don't have unit emissivity: arxiv.org/abs/0812.0825 . Also, if photons have mass, then for black holes of all but the smallest sizes we would expect Hawking radiation to be completely dominated by gravitons. – Ben Crowell Aug 26 '17 at 23:58