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> 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 opposi
by kovrik 7y 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.
- knzhou 7y agoHe 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.
- kovrik 7y agoBut 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?
- pdonis 7y ago> 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.
- pdonis 7y agoHe 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.
- deleted 7y ago[deleted]
- kibibu 7y agoSo he proved "if x then y", then later proved "!x"
- knzhou 7y agoWell, 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.
- nightcracker 7y agoNo. 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.
- pdonis 7y ago> 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.
- platz 7y agoUse the word proof carefully. We've never observed Hawking radiation, we just think it's likely to exist.
- antonvs 7y agoThe 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.