String Theory Demystified

CHAPTER 13 D-Branes 233
D-brane 1 and ends on D-brane 2. Now, let’s see how we specify that the string
starts on the fi rst D-brane. We quantify this by writing
a a
X ( , τ ) = x
(13.19)
0 1
To specify that the string ends on the second D-brane, we have:
X ( , ) x
a a
πτ = (13.20)
2
The oscillator expansions for the NN coordinates are unchanged. However, we need
to incorporate the new boundary condition into the oscillator expansion for the DD
coordinates. It is now written as:
( ) + ′
n≠0
α τ
X x x x i
n e
i
a a 1 a a n −in
( στ , ) = + − σ 2α ∑ sin(
nσ ) (13.21)
1 2 1
π
It is easy to see that this gives the correct boundary conditions by setting σ = 0, π.
You might compare this modal expansion to the modal expansion we got for the DD
coordinates earlier, and to the modal expansion for an open string when no D-brane
µ
is present. When there is no D-brane, we have a momentum term p0 τ which is
related to the zeroth mode α µ
0. In the expansion given here, we have a momentum-like
a a
term given by 1/π( x − x σ
2 1 ) . We use this to describe the zeroth mode:
α
1
= ( x − x )
π 2α′
(13.22)
a a a
0 2 1
Notice that this mode does not multiply the timelike coordinate τ , rather it
multiplies σ . This tells us that the mode is like a winding mode of the string, but it’s
really from the stretching of the string from D-brane 1 to D-brane 2. We have to add
a term to the expression for the mass to refl ect the presence of this additional energy.
This is done using:
1 ⎛ −
0 0 =
2α′ ⎝
⎜
2 ′
αα
a a x x
πα
a a
2 1
Previously, with only a single D-brane the mass was given by
p
d
2 1 ⎛ ∞
∞
i † i
a †
m = ∑∑nanan+ ∑ ∑ mama α ′ ⎝
⎜
n=
1 i=
2
m=
1 a= p+
1
⎞
⎠
⎟
2
a
m
⎞
−1
⎠
⎟