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?

  • $\begingroup$ There is no agreement about the origin of BH entropy. You can get the Bekenstein-Hawking result —or at least the "area law"— by means of different ways. See here scholarpedia.org/article/… $\endgroup$ – Diego Mazón Aug 17 '12 at 22:23
  • $\begingroup$ why is one not just considering all kinds of field dofs? That would probably take a theory of quantum gravity I guess?? $\endgroup$ – Hamurabi Aug 17 '12 at 22:27

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).

  • $\begingroup$ String theory is not able to compute the entropy of 4-dimensional Schwarzschild or Kerr BH. Or is there any recent progress? Wald computed the BH entropy as the Noether charge under diff. in 1993. $\endgroup$ – Diego Mazón Aug 18 '12 at 6:04
  • $\begingroup$ By the way, just a minor thing: 't Hooft instead of t'Hooft. $\endgroup$ – Diego Mazón Aug 18 '12 at 6:06
  • $\begingroup$ So the picture that String theory gives, is less clear and complete than mostly stated? $\endgroup$ – Hamurabi Aug 18 '12 at 6:43
  • $\begingroup$ @drake: I agree that there is no analytical computation, but there is no absolutely no doubt that if you sat down and counted microstates you would get the right answer. This is not true of other approaches. I don't know the Wald thing, but it is not possible to do it from his methods in '93 by counting states, just by doing formal identifications. From guessing, it's probably just identifying the natural periodicity of imaginary time in the diff group of the exterior, and making some formal trick. It can't be real entropy, like in strings. $\endgroup$ – Ron Maimon Aug 18 '12 at 7:03
  • $\begingroup$ @Hamurabi: I don't get it--- I thought I said it's complete and clear. $\endgroup$ – Ron Maimon Aug 18 '12 at 7:03

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).

  • $\begingroup$ Bekenstein's argument is heuristic, it doesn't actually equate the entropy of the infalling matter to the black hole, it just uses the second law and the arbitrariness of the infalling matter to get a good understanding of what the Black hole entropy should be. $\endgroup$ – Ron Maimon Aug 18 '12 at 5:18
  • $\begingroup$ This is not really true--- you get $S = cA + b$, where b is sometimes a huge constant, and c depends on adjustible parameters in LQG, last I looked (I didn't look too hard). $\endgroup$ – Ron Maimon Aug 18 '12 at 7:06
  • $\begingroup$ @ronmaimon : I'm citing this from a seminar a long time ago. I know that Ashtekar has a paper deriving the Hawking formula, and the result is that you get the Hawking formula to first order, and then there are higher order corrections. And I don't see where the contradiction with what I said about the Bekenstein bound is--you talk abuout lowering mass into the BH in a way that minimizes entropy gain, you lose the mass in the horizon, and then note that $S\propto A$. $\endgroup$ – Jerry Schirmer Aug 18 '12 at 7:10
  • $\begingroup$ I agree, but the constant is not determined, and the entropy is not in the object once it dissolves into the black hole, and OP's question is, "where is it?" $\endgroup$ – Ron Maimon Aug 18 '12 at 7:13
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    $\begingroup$ as i have written in lqg currently above the immirzi parameter disappears in the first term. it gives altogether $S\sim A+logA+\hbar f(\gamma)$, where $f(\gamma)$ is a function of the immirzi parameter $\gamma$. apparently one is counting the dofs of the quantum gravitational field of the horizon. not the matter dofs. $\endgroup$ – Hamurabi Aug 18 '12 at 23:49

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