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by kovrik 2481 days ago
> Stephen Hawking proved that the area of a black hole’s event horizon — the surface that marks its boundary — cannot decrease.

I thought he proved the opposite: that black holes (slowly) evaporate through Hawking radiation, lose mass, hence their event horizon shrinks. Could anyone please elaborate?

Also, do scientists actually say that thermodynamics and black holes are the same thing? Never heard of that. I think they only mean that laws of thermodynamics are universal and should be applicable to black holes as well.

4 comments

He proved both. He and Penrose proved theorems that showed that, under certain assumptions, the area of a black hole's event horizon can never decrease. But he later showed that quantum fields near the horizon can violate some of the assumptions that are required for the area theorem, so when those quantum effects are taken into account, black holes can radiate and this can cause the area of their event horizons to decrease.
So he proved "if x then y", then later proved "!x"
Well, perhaps, but in physics we like to think of our field as containing many separate, logically independent models of reality. Yes, "x" probably isn't true in our universe, but it is true in an exceptionally good model for our universe, and that's what makes it an important result. It's like how the round Earth is wrong, but way less wrong than the flat Earth.
No. x -> y && !x does not give !y, it just gives nothing.

He proved y under some set of assumptions, and then later !y under another set of assumptions. So he proved a -> y && b -> !y.

But, given x->y, if you prove !y, you can conclude !x, so in a sense kibibu is right (possibly requiring "proved" to be read as "indirectly proved"), and we should give them the benefit of the doubt.
> given x->y, if you prove !y, you can conclude !x

But given x -> y, if you prove !x, you can't conclude anything. And kibibu said !x, not !y.

Yeah, I interpreted:

> quantum fields near the horizon can violate some of the assumptions that are required for the area theorem

as !x

I don't think it is correct to say he proved both. A more correct statement would be something like "Stephen Hawking proved that the area of a black hole’s event horizon cannot decrease within the realm of classical mechanics".
He proved this in the context of classical general relativity. Hawking radiation is a (semiclassical) quantum gravity effect which is remarkable for being able to break this classical result.

The identification of area with entropy doesn't mean that entropy decreases as a black hole evaporates, because the emitted particles carry the entropy.

But mass decreases which implies that event horizon shrinks, right?

Also, I thought never-increasing entropy was only applicable to closed systems. In that sense, when particles leave black hole through Hawking radiation, total entropy of the Universe remains the same. Am I missing something?

> mass decreases which implies that event horizon shrinks, right?

Yes. So the entropy of the hole by itself decreases. But the entropy of the whole system, including the hole and the Hawking radiation it emits, increases.

> when particles leave black hole through Hawking radiation, total entropy of the Universe remains the same

We have no way of evaluating the total entropy of the universe. The concept itself might not make any sense, since we have no way of making measurements or running experiments on the universe from the outside.

Use the word proof carefully. We've never observed Hawking radiation, we just think it's likely to exist.
The proof is theoretical. It doesn't mean that Hawking radiation must exist, but it means that if the theories that the proofs are based on are sufficiently correct, then Hawking radiation must exist.
> I think they only mean that laws of thermodynamics are universal and should be applicable to black holes as well.

That's correct; that's all they mean.