From what I understand 2nd law of thermodynamics is just saying that information cannot be destroyed, ok got it. When a massive star collapses under it's own weight, it turns into a black hole and all those physical information of the original star simply vanished into oblivion until quantum mechanics come into the rescue by suggesting black hole has a temperature! So the entropy of black hole is related to it's surface area and it should be decreasing over time, now I'm super confused why bother giving black hole entropy to have it blatantly violates the 2nd law of thermodynamics which also saying entropy must not decrease?

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    $\begingroup$ There's no 2nd Law violation implied by an object having a decreasing entropy. The object is not the whole system. Even if the object is alone in a vacuum, the vacuum itself is still a real thing. $\endgroup$
    – g s
    Dec 11, 2021 at 8:51

1 Answer 1


The second law of thermodynamics actually requires black holes to have entropy, otherwise you could reduce $S$ by throwing mass into one.

Hawking radiation may shrink a black hole, but the radiation has entropy of its own, so $S$ still isn't reduced. Indeed, if you throw something into a black hole, then wait for Hawking radiation emission to reduce the black hole's mass to its original value, the law predicts the outgoing radiation will have at least as much entropy as the object you destroyed.

So what the law really tells us is blackbody radiation has an especially high entropy for a given mass-energy.

  • $\begingroup$ To apply the second law of thermodynamics, one should start arguing why a BH can be considered a system at equilibrium. $\endgroup$ Dec 11, 2021 at 11:03
  • $\begingroup$ @GiorgioP Indeed, which is what the zeroth law of BH thermodynamics ends up claiming. The argument as to why is the BH's temperature is proportional to its surface gravity. $\endgroup$
    – J.G.
    Dec 11, 2021 at 11:34
  • $\begingroup$ Still, the fact that "the surface gravity is analogous to temperature" does not show that it coincides with the thermodynamic temperature. If it would coincide, there should be a normal thermodynamic system with some well-defined temperature that would be in thermal equilibrium with a BH. I have a hard time imagining a normal system in thermal equilibrium with any gravitational system. Missing evidence for that, I can only consider the "BH entropy" as a funny name for something analogous, but not coinciding, with the usual entropy. $\endgroup$ Dec 11, 2021 at 17:07
  • $\begingroup$ @GiorgioP That is a big topic beyond the scope of comments here, but it's essentially where the thermodynamics of the virial theorem (itself a topic that takes some reading, since the VT isn't usually considered from that perspective) meets relativistic quantum field theory in curved spacetime. $\endgroup$
    – J.G.
    Dec 11, 2021 at 18:39

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