String Theory and M-Theory

The solution to these equations is
11.1 Black holes in general relativity 561
Fθφ = p(r, t) sin θ.
Taking into account the Bianchi identity, ∂rFθφ = ∂tFθφ = 0, one obtains
Fθφ = p sin θ,
where p is a constant. This field can then be inserted in the Einstein equation
to determine the functions A and B. ✷
EXERCISE 11.2
Show that the parameters q and p in the previous exercise are electric and
magnetic charges.
SOLUTION
As discussed in Chapter 8, magnetic and electric charge are given by
Qmag = 1
F =
4π
1
π 2π
dθ dφ Fθφ
4π
and
Qel = 1
4π
0
⋆F = 1
π 2π
dθ dφ (⋆F )θφ.
4π 0 0
Inserting Fθφ = p sin θ in the first integral gives Qmag = p. To evaluate the
electric charge it is necessary to compute the dual of the electric field:
Thus Qel = q.
(⋆F )θφ = √ −gF rt = e A+B r 2 sin θe −2(A+B) Ftr = q sin θ.
EXERCISE 11.3
Show that the near-horizon geometry of a D = 4 extremal Reissner–Nordström
black hole is AdS2 × S 2 .
SOLUTION
Near the horizon r ≈ 0. In this limit Eq. (11.30) becomes
Setting ˜r = r 2 0
ds 2 = −
r0
r
−2
dt 2 +
r0
r
0
2
dr 2 + r 2 0dΩ 2 2.
/r, and dropping the tilde,
ds 2
r0
2 −dt2 2
=
+ dr
r
+ r 2 0dΩ 2 2.