4
$\begingroup$

According to the wikipedia article http://en.wikipedia.org/wiki/Naked_singularity: "Some research has suggested that if loop quantum gravity is correct, then naked singularities could exist in nature, implying that the cosmic censorship hypothesis does not hold. Numerical calculations and some other arguments have also hinted at this possibility."

The entropy of a black hole is proportional to its horizon area. What is the entropy of the black hole if the cosmic censorship hypothesis fails to hold? Do we assume this hypothesis for the expression of the entropy?

$\endgroup$
1
  • $\begingroup$ "What is the entropy of the black hole if the cosmic censorship hypothesis fails to hold?" Just a note on terminology: by definition, a naked singularity isn't a black hole. $\endgroup$
    – user4552
    Commented Aug 4, 2013 at 15:32

2 Answers 2

5
$\begingroup$

The first paragraph of the question seems irrelevant to me. Plain old classical GR allows naked singularities. It just makes it difficult to produce one by gravitational collapse, from generic initial conditions, without exotic matter.

In terms of classical GR, Hawking proved an area theorem (Hawking 1973) that says that when two or more black holes merge, the area of the event horizon always increases, assuming the weak energy condition. Hawking, Carter, and Bardeen realized that they could form a set of laws that were exactly analogous to the laws of thermodynamics, and in this system of analogies, the area of an event horizon was the analog of entropy. In terms of tested physical theories, this is the only justification for associating an entropy with the area of the event horizon. The Hawking area theorem doesn't include contributions from the singularities, so that's the justification for not including them in the definition of entropy.

However, the area theorem requires the assumption of "a regular predictable space," which is defined as one that is "strongly future asymptotically predictable" from a partial Cauchy surface (and also satisfies some topological conditions). The whole reason that cosmic censorship is of such great theoretical interest is that if it's violated, then causality is violated, in the sense that we don't have uniqueness and existence of solutions to initial-value problems, i.e., Cauchy problems. So although I haven't dug carefully into the technical details, it sounds to me like a naked singularity violates the assumptions of the Hawking area theorem.

Therefore there appears to be no way, based on tested physical theories, to discuss the entropy of a naked singularity. I don't think it should be a big surprise that entropy fails to be a well-defined concept in a spacetime that isn't causally well-behaved. For example, GR allows spacetimes that have closed timelike curves or that aren't even time-orientable, and in such a spacetime we obviously can't have any sensible analog of the second law. Also, causality breaks down in the case of a naked singularity, as expressed by John Earman's famous observation that it would be consistent with the laws of physics if we imagine that "all sorts of nasty things -- TV sets showing Nixon's 'Checkers' speech, green slime, Japanese horror movie monsters, etc. -- emerge helter-skelter from the singularity." How do you count up the entropy of the green slime that a naked singularity is intending to output? Obviously you can't.

Hawking and Ellis, The large scale structure of space-time, 1973, Proposition 9.2.7, p. 318

$\endgroup$
7
  • $\begingroup$ This point is extremely interesting "The whole reason that cosmic censorship is of such great theoretical interest is that if it's violated, then causality is violated, in the sense that we don't have uniqueness and existence of solutions to initial-value problems, i.e., Cauchy problems.". Dont black holes(in classical GR) also have the same issue, in the sense multiple initial conditions will end up with the same final configuration. Also Why is causality violated? Perhaps you mean that evolution of the singularity itself is indeterminate. $\endgroup$
    – Prathyush
    Commented Aug 5, 2013 at 8:28
  • $\begingroup$ @Prathyush: I don't understand the black hole example. In classical mechanics you can throw a stone at different angles with different velocities and they all can fall on the same spot. That doesn't change the predictability of the theory. $\endgroup$
    – MBN
    Commented Aug 5, 2013 at 9:26
  • $\begingroup$ @MBN Even if the stone lands on the same spot everytime, The final velocity will not be the same. One will still be able to recover information about how the ball was thrown and so on... It seems to me that once a black hole is formed on can say that several initial conditions can give rise to that particular black hole(classically) and if that is correct, The usual assumption that one can determine the past uniquely from the present will not hold. $\endgroup$
    – Prathyush
    Commented Aug 5, 2013 at 9:36
  • $\begingroup$ Sorry, I am talking nonsense. $\endgroup$
    – MBN
    Commented Aug 5, 2013 at 13:21
  • $\begingroup$ @Prathyush: Interesting point; I hadn't thought of it that way before. But if you look at the Penrose diagram for an astrophysical black hole, there clearly are Cauchy surfaces. The singularity is spacelike, so Cauchy surfaces don't have to intersect it. $\endgroup$
    – user4552
    Commented Aug 5, 2013 at 16:00
1
$\begingroup$

Black hole entropy is not dependent on the singularity or its structure, it is a feature of horizon, and represents our inability to observe microstates from outside. Even 'black hole' isn't necessary for the entropy: cosmological horizons also have the entropy as indicated by the same Bekenstein-Hawking formula: $$ S = \frac{kA}{4\ell_{\mathrm{P}}^2}, $$ $k$ -- Boltzmann constant, $A$ -- horizon area, $\ell_\text{P}$ -- Planck length.

Therefore, no horizon -- entropy is zero. Cosmic censorship hypothesis is not needed for the derivation of Bekenstein-Hawking formula.

$\endgroup$
2
  • 2
    $\begingroup$ But lots of things have a non-zero entropy without having a horizon. Any object from your day to day life will serve as an example. So this argument by itself doesn't show that a singularity has zero entropy. $\endgroup$
    – N. Virgo
    Commented Aug 4, 2013 at 15:15
  • $\begingroup$ But here we are talking about the entropy of an exact solution in GR. So any entropy will be associated with matter and not with singularity. $\endgroup$
    – user23660
    Commented Aug 5, 2013 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.