| ▲ | metalliqaz 3 hours ago | ||||||||||||||||
Please correct my understanding as I am not a scientist, but why do you assert that Hawking radiation "escapes" or "comes out of" a black hole? Nothing crosses the event horizon. Isn't it more correct to say that Hawking radiation is produced by the extreme warping of the spacetime very near the event horizon, and the black hole shrinks to conserve the energy? Black holes can also shrink as gravitational waves are produced, but we don't say that the waves "escape" the black hole. | |||||||||||||||||
| ▲ | auntienomen an hour ago | parent | next [-] | ||||||||||||||||
Strangely, it is _not_ the instrinsic curvature of spacetime that produces Hawking radiation. There's a variation on Hawking radiation -- called Unruh radiation -- experienced by accelerated observers in flat spacetime. Any two observers will agree on the value of a field (e.g. the electric field) at a point, but an accelerated observer will experience empty spacetime as a _thermal bath of particles_, with a temperature proportional to the acceleration. Relatively accelerated observers don't agree on what the vacuum is. | |||||||||||||||||
| ▲ | rhdunn 2 hours ago | parent | prev [-] | ||||||||||||||||
My understanding is that Hawking radiation is where: 1. a particle/anti-particle pair is created at the event horizon; 2. the particle is on the outside edge of the event horizon, so "escapes" the black hole; 3. the anti-particle is on the inside edge of the horizon, so decreases the size of the black hole due to particle/anti-particle annihilation (with a corresponding particle on the inside of the black hole). | |||||||||||||||||
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