# Hawking radiation (black hole evaporation) [duplicate]

I understand that one of the simplified ways of looking at Hawking radiation is a pair of virtual particles close to the event horizon (but outside of it). The particle with negative energy falls into the black hole, thus decreasing its mass/energy, while the one with positive energy escapes into the Universe outside of the black hole. My question is: Why is it that the negative energy virtual particle always falls into the black hole (with respect to the observer)? That is, why is there not a random distribution of either a negative particle or a positive particle falling into the black hole?

## marked as duplicate by John Rennie, JamalS, Qmechanic♦Feb 2 '15 at 12:17

You have to keep in mind the picture is heuristic. Having said that, in curved spacetime in general there is no global definition of time. The notion of energy is tied to time evolution - it is the Noether charge under time translation symmetry. For black holes, the energy you are referring to is the energy defined at infinity where time translation is indeed a symmetry as spacetime becomes flat.

In flat space negative energy particles do not exist (as the vacuum is lowest energy).

In this heuristic picture statistically the number of particles with positive energy is the same as the number of anti-particles with positive energy (and similarly for negative energy). However, only the positive energy ones can make it out to asymptotic infinity for aforementioned reasons.