Understanding Black Hole Evaporation with Hawking Radiation

[cbd1]

With the theory of Hawking Radiation, people generally say to imagine that the black hole has a temperature, so it must radiate heat. But, this really is not how it is said to work.

The means by which the black hole evaporates is not that particles are actually coming out of it, or that it is radiating heat. But rather, the black hole is losing mass by taking in anti-particles from virtual particle pairs. So it is not that mass leaves the black hole, but that the black hole sucks up anti-mass. What comes away from the hole is not actually anything extracted from the hole, but a particle that was created outside of the hole. So what has happened is there has been an exchange, the black hole's taking in anti-mass allows for the mass of the new particles Is this basically correct, actually more accurate than saying that something is coming "out of" the black hole?

On a side note, why again is it that there is not an equal balance. You would think that there would be a 50% chance that the positive particle go in the hole and the anti-particle to fly away. That is, why would it not balance, half the time taking in the normal particle and half the time taking in the other?

 

[phinds]

There is a fundamental problem with your question and it is this: you are taking literally a heurestic. The "particle / anti-particle" description of Hawking Radiation is just a heuristic posited by Hawking himself, who described it as being just the best he could do to explain in English something that really can only be explained by math.

As I understand it, BH's DO have a temperature and they do radiate but until they get very small their outgoing radiation is totally swamped by incoming radiation, even if that is from nothing but the CMB. As to the exact methodology of the radiation, I do not know the math and can only refer you to Hawking's formal papers.