Friday, September 26, 2008

Microscopic entropy of Kerr black holes

Monica Guica, Thomas Hartman, Wei Song, and Andrew Strominger (click!) show that the string theorists' ability to compute black hole entropy microscopically goes well beyond the canonical, supersymmetric, higher-dimensional, and "stringy" cases such as OSV and split attractors.

What about the black holes we see in the telescopes?

All of them are almost exactly electrically neutral - so you don't need the Reissner-Nordström solutions if you're an astrophysicist. But many of them have a huge angular momentum which is why the Kerr solution is important.

The Kerr solution prevents the angular momentum from exceeding a limit:
J < GM2
The rotating black holes that saturate the bound above are called extremal Kerr black holes. And they're the focus of their paper.



Video 1: GRS 1915+105, the heaviest known stellar black hole in the Milky Way, including fake pictures and its authentic sound translated from its X-ray spectrum to somewhat more melodic frequencies. (MSNBC about Wheeler.)

Are there any extremal black holes observed in the telescopes? Well, not exactly extremal ones. But can you get close? You bet. For example, a binary X-ray source GRS 1915+105 (wiki) has
J / GM2 > 0.98.
Wow, that's almost one. It's useful to know something about this object as well as extremal Kerr black holes in general. The object has mass about 14 solar masses.




Extremal Kerr black holes have entropy equal to A/4G, as all large black holes. But because the area scales like the squared radius and the radius scales like the mass, the area also scales like the squared mass i.e. like the angular momentum. The entropy is, in fact,
S = 2 pi J / hbar.
Note that in proper physics, the entropy of all simple enough black holes is naturally an integer (or rational) multiple of pi (and not of log(2) or log(3) or log(1917) as loop quantum gravity nuts would like you to believe).

Can they recover this entropy microscopically? Yes, they can. They first isolate the NHEK near-horizon limit of the geometry. This limiting geometry is written in coordinates "t, phi, y" and "theta". The "t, phi, y" portion is AdS3-like and it is fibered and squeezed over a compact "theta". The "y" coordinate is radial and "t, phi" are coordinates spanning a cylinder in which a CFT2 becomes chiral (left-moving).

Let's move to the boundary description psychologically

Consequently, the situation has a CFT2 boundary dual: it is the recently discussed chiral CFT of Strominger et al.: see the full video and brief transcript of his Strings 2008 talk. You may be surprised that a four-dimensional geometry has a two-dimensional dual. But you shouldn't worry too much because the extra coordinate "theta" is compact: you may imagine that it is analogous to the five-sphere in the AdS5 x S5 case.

The reparametrizations of "phi" generate a classical Virasoro algebra. If you construct the corresponding generators and compute their classical Dirac brackets, then you get the Virasoro algebra with a central extension. It may be another surprise that a classical calculation gives you the "quantum" central term but it does, roughly because the left-moving central charge is written as
cL = 12 J / hbar
in terms of the classical quantity "J". In other words, the "hbar c_L" coefficient in the central extension can be written simply as "12 J" which has no Planck's constant. That's why it can be seen in classical Dirac brackets. It works because the angular momentum is so large (in units of hbar).

When they compute the central term (and thus the entropy, because of Cardy's formula), they must be careful about the boundary conditions for different components of the metric at "y=0". They scale as different powers of "r" (or, essentially, "y"), depending on how "chiral" these components are.

At any rate, when the dust is settled, the (approximate, plus minus 2%) entropy of GRS 1915+105 calculated from the angular momentum (2 pi J) works, is close to 1 x 10^{79} (times Boltzmann's constant), and can be extracted from a chiral two-dimensional conformal field theory whose left-moving central charge is 2 x 10^{79}, including the numerical coefficient. ;-)

I've checked the numbers (using Planck units and assuming the central value of the mass) but yes, the error margin is about 50% because of the uncertain mass of the black hole.

5 comments:

  1. Please enlighten me: Why do you expect the entropy to be an integer given that at least micronanonically it is the log of then degeneracy which itself is definitely an integer.

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  2. Dear robert,

    this is not a matter of "expectation" or debates. It's about our knowledge of facts about physics. My statement is valid and yours is not because I am talking about black holes, not about some tiny systems with a very small number of microstates.

    The statement that the entropy is log of an integer becomes completely meaningless for large systems such as black holes because its validity depends on having the microcanonical ensemble. In more normal, canonical or mixed ensembles, there is no trace of the integrality of the number of microstates.

    Moreover, the microcanonical ensemble itself becomes unnatural for real, non-SUSY black holes because the microstates never have exactly the same masses. The spacing between the microstates is exponentially tiny and the statement that exp(S) is an integer would require an unrealistic, exponential accuracy and an unnatural, sharp decision which energy intervals are included. The microstates also have a width (imaginary part) which makes it a mess.

    I was - and I am - talking about actual, large black holes such as the extremal Kerr black hole or the original 5D Strominger-Vafa black hole or any similar, sufficiently simple - usually extremal - black hole.

    Their entropy is a "simple" multiple of "pi" simply because the factor of "pi" appears as a part of Cardy's formula while the remaining factors are "simple" and they have no pi-like factors if written in terms of (integral) charges and angular momenta.

    In the extremal 4D Kerr case, the entropy is 2.pi.J. In the Strominger-Vafa case which is more general, the entropy is 2.pi sqrt(Q_H Q_F^2 / 2) where the square root is generally not an integer but it may be.

    The entropy divided by pi is an integer or at least a square root of a half-integer which is still an exponentially more strict a constraint than yours. Moreover, unlike yours, it is valid for large black holes and independent of conventions about ensembles.

    For BPS black holes, it is true that the full degeneracy at a certain level is an integer that can be computed "combinatorially". But it is certainly not true that this integer is a power of 2, 3, or 1917 which would be needed to claim that the entropy of a large black hole is a multiple of log(2), log(3), or log(1917).

    Where have you been in the last 12 years if you're asking similar questions? These are perhaps the most important formulae in physics since the mid 1990s. You sound like an LQG crackpot.

    Best
    Lubos

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  3. Incidentally, when it comes to integer degeneracies, you might think that the fact that the entropy is "naturally" a sqrt(integer)-multiple of pi is incompatible with it.

    You might think that I am saying that the exponential of pi times square root of an integer should be close to an integer which can't be the case at the same moment.

    However, it often is.

    I wrote about it in numerical coincidences. For example, exp(pi.sqrt(163)) is equal, believe me or not, to 262537412640768743.9999999999993...

    That differs from an integer by 10^{-12} or so. The number "163" is related to a macroscopic, Bekenstein-Hawking-like calculation of the entropy of a 3D black hole while the number 262537412640768744 is the number of microstates at a certain level.

    It is not a coincidence that exp(pi.sqrt(163)) is crazily close to an integer: one can actually give an argument (related to j-functions) that doesn't rely on any brute force numerical calculation.

    The number pi.sqrt(163) for this entropy would come from the Bekenstein-Hawking formula while the corrections needed to get ln(262537412640768744) are quantum corrections in quantum gravity. The latter become very small for large black holes. The example with 262537412640768744 is not quite a "large" black hole but this number of microstates is large enough for the quantum corrections to be small and give you only 10^{-12} deviation from an integer.

    Best
    Lubos

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  4. Is there some relationship between Heegner numbers and maximal supergravity? N=8 supergravity has 163 fields.

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  5. Dear Mitchell, no relationship I am aware of even though some relationship (monstrous moonshine) that may first look crazy turn out to be real.

    Your counting is strange - you probably count a whole representation as 1 field? Strange, anyway. An N=8 SUGRA multiplet itself has 256 physical states and they're just physical states.

    ReplyDelete