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Comment by beaner | Hacker News Reader

One thing I never understood is, how does a black hole evaporate, even via hawking radiation? Doesn't the matter it ejects just fall back in?

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Oversimplifying, but the majority of the particles created are massless particles (usually photons), and a photon outside the event horizon can, by definition, escape.

For a more detailed answer with references, see the last answer here at https://physics.stackexchange.com/questions/89983/how-many-o... .

My limited understanding is that the Hawking radiation is caused by virtual particle-antiparticle pairs spawning just outside the event horizon. The event horizon is by definition the surface on or inside of which not even light can escape. This seems to imply that outside the event horizon some non-zero mass objects can still escape, if they're on a favourable trajectory. E.g. aimed directly away from the event-horizon isosurface.

My layman's understanding mirrors this.

Because one part of a pair of virtual particles crosses the horizon, and the other one does not, they become separated by said horizon and cannot recombine and disappear. The particles become actual and not virtual, but one is lost to the hole. Hence, it looks to the external world as if a particle was emitted by the hole.

Note that to my (layman's) knowledge, this has not been actually observed, but does seem to be a modern consensus and involved a paid bet.

There's lots of different ways to approach explaining the phenomenon of hawking radiation, PBS space time does a good job here: https://www.youtube.com/watch?v=qPKj0YnKANw

Ultimately, hawking radiation does not come from within the event horizon, so it can escape into the wider universe.

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