G4 ≥ |Q|.

M (11.27)

If this bound is not satis¬ed, then the metric has a naked singularity at r = 0

that is unshielded by a horizon. According to the cosmic censorship conjec-

ture, naked singularities should never be produced in physical processes, so

that these solutions would be unphysical.

Extremal Reissner“Nordstr¨m black hole for D = 4

o

In the limiting case

G4 = |Q|

r± = M G 4 or M (11.28)

the black hole is called extremal, and it has the maximal charge that is al-

lowed given its mass, as follows from the bound (11.27). When the Reissner“

Nordstr¨m solution arises as a solution of a supersymmetric theory, the sat-

o

uration of this bound is often equivalent to the saturation of a BPS bound,

which then implies that the extremal Reissner“Nordstr¨m black-hole solu-

o

tion has some unbroken supersymmetry.

The metric of an extremal Reissner“Nordstr¨m black hole takes the form

o

r0 r0 ’2

2

ds2 = ’ 1 ’ dt2 + 1 ’ dr2 + r2 d„¦2 , (11.29)

2

r r

where r0 = M G4 . Let us shift the de¬nition of r by letting r = r ’ r0 and

˜

11.1 Black holes in general relativity 559

then dropping the tilde. After some simple algebra this leaves

r0 ’2 2 r0 2

ds2 = ’ 1 + dr2 + r2 d„¦2 .

dt + 1 + (11.30)

2

r r

This is a convenient form of the extremal Reissner“Nordstr¨m metric in

o

which the horizon is located at r = 0. As in the Schwarzschild case, the

space-time is regular at the horizon, which is again only a coordinate sin-

gularity. In the near-horizon limit, where r ≈ 0, the geometry approaches

AdS2 — S 2 , as is shown in Exercise 11.3.

Extremal Reissner“Nordstr¨m black hole for D = 5

o

Reissner“Nordstr¨m black holes have straightforward generalizations to other

o

space-time dimensions. As pointed out in the introduction, an extremal

Reissner“Nordstr¨m black hole in D = 5 is of interest in connection with

o

the microscopic derivation of the black-hole entropy. Its metric can be writ-

ten in a form similar to Eq. (11.29)

2 ’2

22

r0 r0

2 2

dr2 + r2 d„¦2 .

ds = ’ 1 ’ dt + 1 ’ (11.31)

3

r r

2

r2 ’ r0 and then drop the tilde to obtain

Alternatively, one can de¬ne r =

˜

a form analogous to Eq. (11.30)

2 ’2

r0 r0 2

ds2 2

dr2 + r2 d„¦2 .

=’ 1+ dt + 1 + (11.32)

5 3

r r

Using this expression, it is easy to see that the horizon at r = 0 has radius

r0 , and therefore and its area is

A = „¦3 r0 = 2π 2 r0 .

3 3

(11.33)

Comparing with Eq. (11.11), the mass and charge (suitably normalized) of

this black hole are

2

Q 3πr0

M=√ = . (11.34)

4G5

G5

EXERCISES

EXERCISE 11.1

The Reissner“Nordstr¨m black hole discussed in Section 11.1 is a solution

o

560 Black holes in string theory

of Einstein“Maxwell theory

√ 1 1

d4 x ’g R ’ Fµν F µν

S= .

2κ2 4

The equation of motion and Bianchi identity for the gauge ¬eld are

√

1

F µν = √ ‚µ ’gF µν = 0,

µ

’g

µνρσ

‚ν Fρσ = 0.

Find the most general solution for the gauge ¬eld that solves these equations

for the spherically symmetric background

ds2 = ’e2A(r) dt2 + e2B(r) dr2 + r2 d„¦2 .

2

SOLUTION

Since Fµν is static and spherically symmetric, there are only two independent

nonvanishing components for the ¬eld strength, Ftr (r, θ, φ) and Fθφ (r, θ, φ).

For the particular metric of this exercise

√

’g = eA+B r2 sin θ.

The nontrivial equation of motion for the electric ¬eld is

√

’gF rt = ‚r eA+B r2 sin θ · (’e’2A’2B Frt )

‚r

= ‚r e’A’B r2 sin θFtr = 0.

The most general static solution of this equation is

q(θ, φ)

Ftr = eA+B .

r2

The Bianchi identity µµνρσ ‚ν Fρσ = 0, leads to additional constraints, ‚θ Ftr =

0 = ‚φ Ftr , so that

q

Ftr = eA+B ,

r2

where q is constant. For the values of A and B given in Eq. (11.21), this

reduces to Eq. (11.25). These values also solve the Einstein equation (11.23).

The equations of motion for the magnetic ¬eld takes the form

‚θ eA+B r2 sin θF θφ = 0,

‚φ eA+B r2 sin θF φθ = 0.

11.1 Black holes in general relativity 561