How particles escape black holes

Black holes that emit Hawking radiation

The escape velocity is c at the event horizon of a black hole. Above the event horizon, the escape speed is below c. Exiting Hawking radiation particles form above the event horizon. Therefore, they can escape if they point up at a sufficiently narrow angle from the vertical and if they are sufficiently energetic.

Escaping particles form as virtual particle-antiparticle pairs in the "infalling" coordinate system: One of the two particles forms outside the event horizon; The counterpart forms below the event horizon. Thus the originally virtual particles cannot annihilate and therefore become real particles; a particle can escape; the counterpart falls in the direction of the singularity.

The energy required to form the exiting particle and its remaining kinetic energy after it has escaped are subtracted from the mass of the black hole.

The mechanism described will likely work for both particles that form very close to the event horizon when the tidal forces are high enough to separate the virtual particle pair before it can be annihilated.

The formation of virtual particles is based on the Heisenberg uncertainty with regard to time and energy: very short time intervals require energy uncertainty, which leads to short-lived particle-antiparticle pairs.

user748

But if the Hawking radiation is inside and everything else arises inside, how can a particle, even if it is virtual, form outside? How exactly are they "virtual"?

Gerald

Virtual particles form spontaneously and usually only live for a very short time, less than 1 to 22 seconds. For details see en.wikipedia.org/wiki/Virtual_particle. Only half of the radiation is inside, the outer (anti) particle can become Hawking radiation. A particle that looks virtual from an infalling coordinate system may look real from the outside due to the extreme time dilation near the event horizon.