# Why Hawking radiation contains other particles besides photon?

1. Nov 20, 2009

### magnetar

Hawking radiation is standard thermal spectrum, thermal spectrum means it only can emit photon.

why Hawking radiation contains other particles?(electron neutrino neutron proton etc)

2. Nov 20, 2009

### clamtrox

'Thermal' does not mean only photons... What it means that you have a probability distribution for the energy of the particles, of the form of

$$\frac{1}{e^{E/T}-1}$$

Black holes radiate every kind of particle because all quantum fields behave in a similar way near the event horizon -- and why wouldn't they.

Last edited: Nov 20, 2009
3. Nov 20, 2009

### Naty1

Magnetar: You make an interesting point insofar as "thermal" and "black body" Hawking radiation terminology seems inconsistent insofar as more than photons are involved.

I thought typical "thermal" and "blackbody" terminology usually refers to electromagnetic (photon) absorption. But post #2 correctly reflects my understanding that particles of all types are emitted from black holes. So I infer that the terminology applies to the spectrum rather than the entities carring the energy..

4. Nov 20, 2009

### bcrowell

Staff Emeritus
Interesting question, magnetar. The Stefan-Boltzmann constant can be expressed purely in terms of kB, h and c, so that makes me think that there is nothing special about photons or electromagnetism here. If you look at the Feynman diagram for vacuum fluctuations to produce a pair of photons, there is no electron involved, and no vertex at which an electron's world-line enters, so I think it makes sense that the rate of radiation is independent of e.

Many of the other possibilities wouldn't seem to lead to any observable effects. For instance, electron-positron pairs could contribute, but there's no way you could observe them at a distance, because the positrons would annihilate with electrons before they could cross interstellar distances.

If I'm understanding what Clamtrox is saying, I think E would have to be the mass-energy of the particle here, not just its kinetic energy. So unless the black hole is extremely small and hot, exp(E/T) should be extremely large for a particle of any significant mass. So the probability of emitting a neutron, etc., would be negligible except maybe for a black hole that was in its final burst of radiation.

If I'm thinking straight this morning, I think the link between the basic thermodynamic expression 1/(exp(E/T)-1) for bosons and the standard blackbody curve requires counting the states of the photon field. For other types of bosons, e.g., bosons with mass or different spin, you'd have different statistics. In particular, you're not going to get emission that goes down to E=0 if the particle has nonzero rest mass, since E>=m. For fermions, you're going to get Fermi-Dirac statistics, with 1/(exp(E/T)+1).

Neutrinos seems like the most reasonable candidate for something interesting. But the temperature of a solar-mass black hole is so low that I would expect the rate of emission of neutrinos to be essentially zero, since they do have nonzero rest mass.

5. Nov 21, 2009