Gravity and Strings

19.1 String-theory extended objects from duality 525
Masses from d = 10 string dualities. We are going to apply these rules to the transformation
of the (almost) only object whose mass we know: the (fundamental) string F1 wound
once around a compact coordinate (x 9 ∈ [0, 2π R9], say). Its mass is just the mass of a
winding mode with w = 1inthe mass formula Eq. (14.60) (which is valid for superstrings
if we include left- and right-moving fermionic oscillators),
MF1w = R9
ℓ2 , (19.5)
s
which is just the string tension times the volume of the compact space.
As a warm-up exercise, let us first perform aTduality in the x 9 direction, in which we
know we should obtain an F1 with minimal momentum in the compact direction:
MF1m = M ′ F1w
= R9
ℓ 2 s
= 1
R ′ , (19.6)
9
in agreement with Eq. (14.60) with n = 1. In this example, it did not matter whether we
were dealing with the IIA or IIB fundamental string. Let us now assume that it is the IIB
one F1B. An S-duality transformation should take us to the D-string wound once around
x 9 .Using Eqs. (19.2) and (19.3), we find
MD1 = M ′ F1Bw = g 1 2
B MF1Bw = g 1 R9 2
B
ℓ2 s
= R′ 9
g ′ Bℓ2 .
s
(19.7)
This mass should be equal to the D-string tension times the volume of the circle, so
TD1 = MD1
2π R ′ 1
=
9 (2πℓs)g ′ Bℓs . (19.8)
We can now perform successive T-duality transformations to find the masses and tensions
of all the Dp-branes. A T duality in the direction x 9 takes us to the D0-brane, whose mass
and tension are
TD0 = MD0 = M ′ D1
= R9
gBℓ 2 s
ℓ
=
2 s /R′ 9
g ′ Aℓs/(R ′ 9ℓ2 1
=
s ) g ′ Aℓs . (19.9)
If we T-dualize the D-string in a transverse direction (x 8 ), we obtain instead the
D2-brane:
MD2 = M ′ R9
D1 =
gBℓ2 R9
=
s g ′ Aℓs/(R ′ 8ℓ2s ) = R8 R9
g ′ Aℓ3 1
, ⇒ TD2 =
s
(2πℓs) 2g ′ Aℓs . (19.10)
By repeating this procedure, we obtain the mass of the Dp-brane wrapped around a
p-torus and its tension (removing the primes):
MDp = R10−p ···R9
gℓ p+1 , TDp =
s
The S5B-brane is the S-dual of the D5-brane:
MS5B = g 1 2 M ′ D5 = g′− 1 R
2
′ 5 /g′ 1 2 ···R ′ 9 /g′ 1 2
g ′−1ℓ6 s
TS5B =
1
(2πℓs) 5g2 ,
ℓs
1
(2πℓs) p . (19.11)
gℓs
= R5 ···R9
g2ℓ6 ,
s
(19.12)