Gravity and Strings

572 The extended objects of string theory
of these generalizations had already been obtained in certain cases in [254, 255, 274]. As
usual, in the extremal limit W = 1 the Hi are arbitrary independent harmonic functions of
the overall transverse coordinates.
The most interesting relation that we obtain is the one in the last line, that among the ais,
the dimensionality of the branes, and r:
r = (p1 + 1)(p2 + 1)
d − 2
− a1a2
− 1. (19.151)
d − 2
This equation contains the intersection rules and we can apply it to some basic examples,
using the values for the ai constants appropriate for each kind of brane (see page 531).
As an example, let us consider the case of d = 11 SUGRA. 29 Since there is no scalar,
a1 = a2 = 0, which implies r = r0. Equation (19.151) immediately gives the three intersections
M2 ⊥ M2(0), M2⊥ M5(1), andM5 ⊥ M5(3) (but not the overlap M5 ⊥ M5(1),
which requires a different Ansatz, see e.g. [417]). For example, the solution corresponding
to a black intersection M2 ⊥ M2(0) in which brane 1 lies in the directions y2 = (y 1 , y 2 )
and brane 2 lies in the directions z2 = (z 1 , z 2 ) is given by
d ˆs 2 = H − 2 3
1 H − 2 3
2 Wdt2 − H − 2 3
1 H 1 3 2
2 d y 2 − H 1 3
1 H − 2 3
2 dz 2 2
−H 1 3
1 H 1
3 2
2 d w q + W −1dρ2 + ρ2d 2
(5−q) .
Ĉ ty 1 y 2 = α1(H −1
1
− 1),
Ĉ tz 1 z 2 = α2(H −1
2
− 1),
Hi = 1 + hi
ρ 4−q , W = 1 + ω
ρ 4−q , ω= hi
2 1 − αi .
(19.152)
On reducing in a relative transverse dimension, we obtain a black intersecting solution,
F1A ⊥ D2(0).Onreducing in one of the extra isometric transverse directions wq,weobtain
D2 ⊥ D2(0) (this is the reason why the extra isometric wqs are introduced into the Ansatz).
The solution corresponding to a black intersection M2 ⊥ M5(1) in which the M2 lies in
the directions y2 = (y 1 , y 2 ) and the M5 in y 1 , z4 = (z 1 ,...,z 4 ) is
d ˆs 2 = H − 2 3
1 H − 1 3
2 [Wdt 2 − (dy 1 ) 2 ] − H − 2 3
1 H 2 3
2 (dy2 ) 2 − H 1 3
1 H − 2 3
2 dz 2 4
− H 1 3
1 H 2 3
2
Ĉ ty 1 y 2 = α1(H −1
1
Hi = 1 + hi
ρ
−1 2 2 2 W dρ + ρ d(2) .
− 1),
ˆ˜C ty1z1 ···z4 = α2(H −1
2 − 1),
ω
2 , W = 1 + , ω= hi 1 − α ρ i .
(19.153)
29 Examples in d = 10 can be found in the next chapter, where they are used to construct BH solutions: the
black D1 D5 is given in Eqs. (20.14) and the black D2 S5 D6 is given in Eqs. (20.37).