Counting the Microstates of a Kerr Black Hole

Counting the Microstates of a Kerr Black Hole

Gary Horowitz

UCSB

(with M. Roberts, 2007)

Gary Horowitz

UCSB

(with M. Roberts, 2007)

Black hole entropy

Bekenstein and Hawking showed that black holes have an entropy

This is much greater than the entropy of matter that collapsed to form the black hole.

What are the microstates responsible for this entropy?

String theory has made great progress in finding microstates of black holes

• SUSY extremal black holes• Some non-SUSY extremal black holes• Some near extremal solutions

Relatively little is known about neutral black holes (but see Susskind, and Polchinski, G.H.).

All of these require four dimensional black holes to have four different types of charge.

Strominger, Vafa, …

Microstates of a Kerr Black Hole

Kerr black holes satisfy J M2. The entropy of an extremal Kerr black hole is simply

SBH = 2 J

Since J is quantized (and SBH is independent of G), it suggests that one can count microstates at weak coupling.

There is a bound on perturbative string states

J M2 similar to the bound on rotating black holes

J M2

But in fact: J G M2 (black holes)

J ls2 M2 (strings)

So extremal Kerr is not a maximally spinning string.

Kaluza-Klein theory

GR in 5D with one direction S1 is equivalent to gravity and electromagnetism (and a scalar) in 4D.

Electric charge is momentum along the S1. Magnetic charge is a topological twisting of the S1.

Both of these charges are quantized.

To count the entropy of an extremal Kerr black hole in string theory, we first lift to five dimensions by taking the product of the solution with a circle.

2R

This does not change the entropy since

A5 = A4 (2R), G5 = G4 (2R)

so SBH = A5/4G5 = A4/4G4

We now add one unit of momentum around the circle. In four dimensions, this is one unit of electric charge.

There is a symmetry in string theory which converts this to one unit of magnetic charge.

Geometrically, the five dimensional description of a (4D) magnetic charge looks like:

2R

A Kerr black hole with magnetic charge looks like a five dimensional black hole (Myers, Perry) sitting at the tip. In 5D, a black hole can have two independent angular momenta, J1 and J2.

Kaluza-Klein monopole

monopole with large R

Geometrically, there is no difference between a five dimensional black hole with J1 = - J2 and one with J1 = J2. They are related by a simple reflection.

J1 = - J2

But gluing in a black hole with J1 = J2 corresponds to a four dimensional solution with J=0 and Q 0 (Q denotes electric charge).

So we can exchange J and Q!

J1 = J2

Previous example where J Q

BTZ black hole is 3D solution of gravity with < 0. Described by two parameters (M, J).

If you apply T-duality along the direction, the result is a charged black string described by (M, Q) (Welch, G.H.).

Q is directly related to J.

Recipe for turning J into Q

J=0 Q 0

J0 Q =0

Increase R Decrease R Reflect J2 -J2

Ingredients:

- One black hole with nonzero J

- One unit of magnetic charge

We have now reduced the problem of counting the microstates of extremal Kerr to counting the microstates of a nonrotating charged black hole in Kaluza-Klein theory (Emparan and G.H., 2006).

Quantization of charges

Q is momentum in fifth direction y, and P describes twisting of y circle over S2. If y = y + 2R and N0, N6 are integers, then

In the extremal limit, the entropy is simply

independent of R.

SBH = N0 N6

Embed in M Theory

Take product of KK black hole with T6 and view KK circle as the M theory circle, so R = g ls.

N0 is the number of D0-branes.

N6 is the number of D6-branes.

A symmetry of string theory (T-duality on T6) interchanges N0 and N6.

Recall Entropy Counting for SUSY Black Holes

Need four charges in four dimensions. At weak coupling, one can take four stacks of D3-branes. If T6 = T2 x T2 x T2, the branes wrap one cycle of each T2. All four branes intersect in one point.

At strong coupling, with N branes in each stack, the solution is the product of T6 and an extreme Reissner-Nordstrom black hole.

The entropy of the branes is

S = 2N2

This agrees exactly the black hole entropy.

Since entropy of the branes is independent of the size (and shape) of the torus, it is natural to assume that it is associated with the common intersection point of the branes.

D0-D6 bound states

D0 and D6 branes repel each other, but Taylor (1997) showed that there are still perturbatively stable bound states with N0 = 4 and N6 =4.

After three T-dualities, these states can be described by four D3-branes wrapping the diagonals of the torus. x2

x1

This looks just like a rotated version of the supersymmetric configuration.

Even though there are four D3-branes, the net 3-brane charge vanishes over most three-cycles. It is nonzero only over (135) and (246), each with charge four. This is exactly what should arise after three T-dualities of a D0-D6 system.

With N branes in each stack, N0 = 4N = N6, so the black hole entropy is:

SBH = N0N6 = 16 N2

The branes have two intersection points on each T2, so there are eight intersection points on T6. Each intersection contributes an entropy equal to the SUSY case so

Sbranes = 8 (2 N2) = 16 N2

This is a precise counting of the entropy of a neutral black hole. It can be extended to N0 N6, and by earlier argument, can be applied to extremal Kerr.

Topology of the event horizon is not well defined in M theory

Consider a 4D black hole with N0 =1, N6 =0. This is equivalent to a black hole with N0 =0, N6 =1.

At strong coupling, when R is large, the first has horizon topology S2 x S1 (ignoring the T6), while the second has topology S3.

Yet these are equivalent descriptions of the black hole!

Summary

• One can reproduce the entropy of an extremal Kerr black hole exactly by counting microstates in string theory.

• One can reproduce the entropy of neutral 5D black holes.

• There is a symmetry which exchanges J and Q.

• The topology of the event horizon is not well defined in M theory.

Open Questions

• Why is this working?

• Can one count microstates of nonextremal Kerr and Kaluza-Klein black holes?

• Can one replace the T6 with more general compactifications?