going into the mathematical details, let us explain qualitatively how this

comes about.

Since the four-dimensional black hole has P0 = 1, there is one D6-brane.

In Chapter 8 it was explained that a D6-brane of the type IIA theory is a

higher-dimensional analog of a Kaluza“Klein monopole. This means that,

11.3 Black holes in string theory 579

from the 11-dimensional M-theory perspective, the four dimensions trans-

verse to the brane form the Taub“NUT geometry

R

ds2 = dr2 + r2 d„¦2

1+

TN 2

2r

’1

R 2

dy + R sin2 (θ/2) dφ

+ 1+ . (11.93)

2r

This geometry can be visualized as analogous to a cigar with the D6-brane

localized to the region near the tip. Far from the tip of the cigar, the

geometry looks like 3 —S 1 , where the circle is the M-theory circle, which in

¡

type IIA units has radius R = gs s . The fact that the number of D0-branes

is Q0 means that there are Q0 units of momentum around the M-theory

circle. On the other hand, near the tip of the cigar the geometry would be

nonsingular and look like 4 if there were no other branes in the problem.

¡

However, their presence makes the story more complicated.

Now consider the strong-coupling limit of the previous four-dimensional

picture. In this limit the radius of the M-theory circle approaches in¬nity,

and the Taub“NUT geometry approaches ¬‚at 4 far from the origin. How-

¡

ever, near the origin there is a ¬ve-dimensional black hole. The Q0 units of

momentum around the M-theory circle are still present, but now as angular

momentum J12 = J34 = Q0 /2 about the origin, which was the tip of the

cigar. In the limit one is left with a ¬ve-dimensional black hole with M2-

brane charges Q1 , Q2 , Q3 and J = J12 = J34 . Since the entropy does not

depend on the string coupling constant, which controls the size of the M-

theory circle, its value must be the same for the four- and ¬ve-dimensional

black holes, which is what we found.

EXERCISES

EXERCISE 11.6

Verify that the D0-D4-F1 realization of the black hole discussed in Sec-

tion 11.3 has entropy given by Eq. (11.55), with the charges replaced by the

charges of the dual con¬guration.

SOLUTION

Consider Q0 D0-branes, Q4 D4-branes wrapping the T 4 , which has a volume

580 Black holes in string theory

(2π)4 V , and Q1 F1-branes wrapping the y 1 circle, which has radius R. This

leads to the masses

Q0 Q4 Q1

(2π)4 V,

M1 = , M2 = M3 = 2πR.

(2π)4 gs 5 2π 2

gs s s s

Inserting this into the expression for the entropy Eq. (11.54) gives

2πgs 4

A

=√ s

S= M1 M2 M3 = 2π Q0 Q4 Q1 .

4G5 RV

Comparison of this formula with Eq. (11.55) shows that the D0-D4-F1 sys-

tem gives the same entropy for Q0 = Q1 , Q4 = Q5 and Q1 = n, which is

what we wanted to show.

Note that the various dualities that relate the di¬erent brane descriptions

of the black hole do not change the ¬ve-dimensional metric except by an

overall constant factor. Such a factor has no bearing on the computation of

the entropy, which is dimensionless.

2

EXERCISE 11.7

Compute the area of the horizon of the rotating black hole described in

Section 11.3 and deduce its entropy.

SOLUTION

In the near-horizon limit r ≈ 0 and constant t the metric Eq. (11.74) reduces

to

ds2 = R2 d„¦2 ’ (a/R2 )2 (cos2 θdψ ’ sin2 θdφ)2

3

= R2 dθ2 + R2 (cos θ sin θ)2 (dφ + dψ)2 + (R2 ’ (a/R2 )2 )(cos2 θdψ ’ sin2 θdφ)2 ,

where

R2 = (r1 r2 r3 )2/3 .

The easiest way to compute the area of the horizon described by this metric

is to de¬ne the orthonormal one-forms

e1 = Rdθ,

e2 = R cos θ sin θ(dφ + dψ),

R2 ’ (a/R2 )2 (cos2 θdψ ’ sin2 θdφ).

e3 =

11.3 Black holes in string theory 581

Then the area of the horizon obtained from this metric is given by

e1 § e 2 § e 3 = R 2 R2 ’ (a/R2 )2 cos θ sin θdθ § dφ § dψ

A=

= 2π 2 (r1 r2 r3 )2 ’ a2 .

Using this result,

A

Q1 Q5 n ’ J 2 ,

S= = 2π

4G5

where the angular momentum J is related to the parameter a by

πa

J= .

4G5

If a > r1 r2 r3 , the black hole is over-rotating, and the geometry has a naked

singularity, at least in the supergravity approximation. 2

EXERCISE 11.8

Consider the dual con¬guration of the D = 4 extremal four-charge black

described in Section 11.3. Show that this gives the entropy in Eq. (11.84),

with the charges replaced by the charges of the dual con¬guration.

SOLUTION

The dual con¬guration has three sets of D2-branes and one set of D6-branes.

The associated masses are

R2 R3 R4 R5

M1 = (2πR2 )(2πR3 )TD2 = Q1 , M2 = Q2 ,

gs 3 gs 3

s s

R1 · · · R6