Gravity and Strings

19.2 String-theory extended objects from effective-theory solutions 537
which, in the weak-coupling limit g → 0, with ℓs fixed, goes quickly to zero, giving a flat
spacetime metric. The F1 solution can then be understood as the long-range fields produced
by a fundamental string in the strong-coupling limit. In the weak-coupling limit, the string
decouples from the supergravity fields.
The event horizon of the black solution becomes singular in the extreme limit both in the
Einstein and in the string frame, and in that limit the dilaton also diverges at the horizon
as ˆφ ∼ ln |x8|. Inthe dual string frame (the frame in which the S5-brane is fundamental
and there is no dilaton factor in its worldvolume action, which is related to the string
frame by Eqs. (18.50) and (18.51)), ignoring the constant in HF1 leads to the solution with
metric
d ˆs 2 ρ4
S5 =
h 2 2 2
dt − dy
3
F1
− h 1 dρ 3
F1
2
ρ2 − h 1 3
F1d2 (7) , (19.58)
which is the direct product of the round S7 metric with radius h 1 6
F1 and the metric of a 1brane
in three dimensions (i.e. a domain wall), which is singular. This near-horizon limit
is well defined in this frame, even if it leads to a metric with singularities, but, unlike the
M2 and M5 cases, it is not a maximally supersymmetric vacuum. These geometries play
roles analogous to the AdSn × Sm geometries in non-conformal versions of the AdS/CFT
correspondence [170].
19.2.5 The S5 solution
The ten-dimensional solitonic 5-brane solution is given by
d ˜ ˆs 2 E = H − 1 4
S5
d ˆs 2 s = Wdt2 − d y 2
5
e −2 ˆφ = e −2 ˆφ0 H −1
S5 ,
3
2 2 4
Wdt − d y 5 − HS5 − HS5
−1 2 2 2 W dρ + ρ d(3) ,
−1 2 2 2 W dρ + ρ d(3) ,
−1
HS5 − 1 ,
2 1 − α ,
ˆB (6) ty 1 ···y 5 = αe−2 ˆφ0
HS5 = 1 + hS5
ω
, W = 1 + , ω= hS5
ρ2 ρ2 (19.59)
and, in the extreme limit ω = 0,α=±1inwhich it is usually known as the solitonic
5-brane solution [206, 207] (also known as the NS 5-brane), it takes the form
d ˜ ˆs 2 E = H − 1 4
S5
3
2 2 4 2
dt − d y 5 − HS5 d x 4 ,
d ˆs 2 s = dt2 − d y 2
5 − HS5 d x 2
4 ,
e −2 ˆφ = e −2 ˆφ0 H −1
S5 ,
HS5 = 1 + hS5
,
|x4| 2
˜ ˆB (6) ty 1 ···y 5 =±e−2 ˆφ0
H −1
S5 − 1 ,
(19.60)