Is black hole entropy, computed by means of quantum field theory on curved spacetime, the entropy of matter degrees of freedom i.e. non-gravitational dofs? What is one actually counting?
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No it isn't. This is a mysterious thing in quantum field theory on curved space, as first noted by 't Hooft. If you assume there is a certain amount of entropy in the quantum fields surrounding the black hole, due to their thermal nature, you might estimate that there is a local contribution to the entropy from each approximate mode at the correct local Hawking temperature of the black hole. This entropy is divergent in quantum fields in curved space, because the time dilation factor makes it that at a fixed energy, the number of modes diverges as you approach the horizon. This is one of the paradoxes that led t'Hooft to the holographic principle. Within AdS/CFT models, it is easy to give an answer-- the entropy of a black hole is the entropy of it's CFT description. This includes systems like stacked branes, in which case, the entropy of the black hole is the number of vacuum states. This is Strominger and Vafa's famous calculation of 1995-96. This entropy coincides with the extremal horizon area (although in this case, the black hole is extremal, so the temperature is zero). Within string theory, this mystery is essentially resolved. The entropy is the entropy of the microscopic constituents of the black hole. It is not resolvable in curved-space QFT because of the 't Hooft divergence, and it is not well resolved in an agreed upon manner in any other approach (this means loops). |
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There are a multiplicity of ways of deriving the Hawking formula for black hole entropy. Some techniques, like Bekenstein's argument, do equate the entropy of matter falling into the hole with the entropy of the hole. Some actually count gravitational microstates in various quantum gravity schema. The result $S\propto A$ seems to be quite generic in all of these approaches, however, at least to the first order of approximation in $\hbar$ (I believe the LQG result has $\mathbb{O}(\hbar^{2})$ corrections to the Hawking formula). |
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