String Theory Notes Page 1 of 31
Chapter 9 – Light-Cone Relativistic Strings
1. Choices for t
We previously used the static gauge, in which the world-sheet time is
identified with the spacetime coordinate X0 by X 0 (t, s) = c t . We can,
however, choose all kinds of different gauges. We choose those in which t
is set equal to a linear combination of the string coordinates:
n mX m (t, s) = lt
To understand what this means, consider two points x1 and x2 with the
same fixed value of t . We then have
( )
n m x 1m - x 2m = 0
( )
The vector x 1m - x 2m is clearly on a hyperplane perpendicular to n m . If we
define the string as the set of points X with constant t , then we see that
the string with world-sheet time t is the intersection of the world-sheet
with the hyperplane n ⋅ x = lt
We want the interval DX m between any two points on the string to be
spacelike. Now, consider
( )
o We know that n m x 1m - x 2m = 0 n ⋅ Dx = 0 .
o If n m is timelike, we can anlyse this condition in a frame in which
only the time compoenent of n is non-zerio. In that case, Dx
clearly has a 0 time compoenent, and is therefore spacelike.
It turns out that this also works for n m null.
Now – for open strings, p m is a conserved quantity. We incorporate this in
our Gauge condition and write
n ⋅ X (t, s) = l (n ⋅ p ) t
For open strings attached to D-branes, some components of p m are not
conserved, but we assume that n is chosen so that n ⋅ p is conserved – for
this to happen, we need n ⋅ s = 0 at the string endpoints. Analysing
units and working in natural units then gives
n ⋅ X (t, s) = 2a ¢(n ⋅ p)t (open strings)
© Daniel Guetta, 2009
, String Theory Notes Page 2 of 31
Not quite sure about the comment that says that gauge isn’t Lorentz
invariant for all choices of n. I also don’t understand how n ⋅ s = 0 at
the endpoints is any requirement – surely we already have s for all
endpoints.
2. The Associated s parameterization for open strings
In the static gauge, we required constant energy density over the string –
in other words, constant t 0 . We now require constancy of n m tm = n ⋅ t ,
as well as s Î éêë0, p ùúû .
I don’t understand this range condition on sigma?
From our expression for tm , we have
ds tm ds
tm (t, s) =
(t, s) n ⋅ m (t, s) = n ⋅ t (t, s)
ds ds
Thus, we can always find a parameterisation in which n ⋅ t (t, s) = a(t )
(ie: does not depend on s ) by adjusting ds / ds accordingly. Further, we
note that
p
ò
0
n ⋅ t ds = n ⋅ p = na(t )
And so
n⋅p
n ⋅ t = (open string world-sheet constant)
p
In this parameterisation, s for a point is therefore proportional to the
amount of n ⋅ p momentum carried by the portion of the string between
é 0, s ù .
êë úû
Now, consider the equations of motion
¶ t mt + ¶ s ms = 0
Dotting this with n m , we get
¶ ¶
¶t
(
n ⋅ t +
¶s
)
n ⋅ s = 0( )
¶
¶s
(
n ⋅ s = 0 )
Which implies that n ⋅ s is independent of s .
© Daniel Guetta, 2009
, String Theory Notes Page 3 of 31
We have already seen that for open strings, n ⋅ s = 0 at endpoints, which
implies that this is the case everywhere.
3. The Associated s parameterization for closed strings
In this case, we want s Î éêë 0,2p ùúû and so
n⋅p
n ⋅ t = (closed string world-sheet constant)
2p
Because of this factor of two, we write the gauge condition without the
factor of two, as
n ⋅ X = a ¢ (n ⋅ p ) t
I don’t understand this range condition on sigma?
We can still show that n ⋅ s is independent of s , but it’s now not
possible to set it to 0 at any given point. Furthermore, it’s unclear what
point is s = 0 . We solve this by setting a certain point on a certain string
to have these properties. The proof this can be done is in the book.
There is, however, an obvious ambiguity – our whole parameterisation can
be rigidly moved along the string without affecting anything.
4. Summary
In summary, we have
n ⋅ s = 0
n ⋅ X (t, s) = ba ¢ (n ⋅ p ) t
2p
n⋅p = n ⋅ t
b
Where b = 1 for closed strings, and b = 2 for open strings.
The first condition above, along with an expression for s immediately
gives us X ⋅ X ¢ = 0 . This allows us to simplify our expression for tm ,
and we obtain X 2 + X ¢2 = 0 . This is best summarised, together with the
first condition above, as
(X X ¢)
2
=0
We then get the following simplified expressions
© Daniel Guetta, 2009
, String Theory Notes Page 4 of 31
1 m 1 ¢
tm =
2pa ¢
X sm = -
2pa ¢
Xm ( )
Feeding into the equations of motion, we get
¢¢
X m - X m = 0( )
These are simply wave equations!
When the string is open, we have the additional requirement that the sm
and therefore the X m ( )¢ vanish at the endpoints.
4. Solving the wave equation
Assuming we have a space-filling D-brane and therefore free-boundary
conditions at the endpoints, the most general solution to the wave
equation is
1 m
X m (t, s) =
2
(
f (t + s) + g m (t - s) )
¢
( )
Bearing in mind the relation sm = - X m / 2pa ¢ and the boundary
sm
conditions = 0 , we get
¶X m
=0 s = 0, p
¶s
The boundary condition at 0 informs us that f and g differ at most by a
constant, which can be absorbed into f.
1 m
X m (t, s) =
2
(
f (t + s) + f m (t - s) )
The boundary condition at s = p gives
¶X m
¶s
1
(
(t, p) = f m ¢ (t + p) - f m ¢ (t - p) = 0
2
)
Since this must be true for all t , this implies that f m ¢ is periodic with
period 2p .
We can therefore write
¥
(
f m ¢ (u ) = f1m + å anm cos nu + bnm sin nu )
n =1
¥
(
f m (u ) = f0m + f1mu + å Anm cos nu + Bnm sin nu )
n =1
© Daniel Guetta, 2009
Chapter 9 – Light-Cone Relativistic Strings
1. Choices for t
We previously used the static gauge, in which the world-sheet time is
identified with the spacetime coordinate X0 by X 0 (t, s) = c t . We can,
however, choose all kinds of different gauges. We choose those in which t
is set equal to a linear combination of the string coordinates:
n mX m (t, s) = lt
To understand what this means, consider two points x1 and x2 with the
same fixed value of t . We then have
( )
n m x 1m - x 2m = 0
( )
The vector x 1m - x 2m is clearly on a hyperplane perpendicular to n m . If we
define the string as the set of points X with constant t , then we see that
the string with world-sheet time t is the intersection of the world-sheet
with the hyperplane n ⋅ x = lt
We want the interval DX m between any two points on the string to be
spacelike. Now, consider
( )
o We know that n m x 1m - x 2m = 0 n ⋅ Dx = 0 .
o If n m is timelike, we can anlyse this condition in a frame in which
only the time compoenent of n is non-zerio. In that case, Dx
clearly has a 0 time compoenent, and is therefore spacelike.
It turns out that this also works for n m null.
Now – for open strings, p m is a conserved quantity. We incorporate this in
our Gauge condition and write
n ⋅ X (t, s) = l (n ⋅ p ) t
For open strings attached to D-branes, some components of p m are not
conserved, but we assume that n is chosen so that n ⋅ p is conserved – for
this to happen, we need n ⋅ s = 0 at the string endpoints. Analysing
units and working in natural units then gives
n ⋅ X (t, s) = 2a ¢(n ⋅ p)t (open strings)
© Daniel Guetta, 2009
, String Theory Notes Page 2 of 31
Not quite sure about the comment that says that gauge isn’t Lorentz
invariant for all choices of n. I also don’t understand how n ⋅ s = 0 at
the endpoints is any requirement – surely we already have s for all
endpoints.
2. The Associated s parameterization for open strings
In the static gauge, we required constant energy density over the string –
in other words, constant t 0 . We now require constancy of n m tm = n ⋅ t ,
as well as s Î éêë0, p ùúû .
I don’t understand this range condition on sigma?
From our expression for tm , we have
ds tm ds
tm (t, s) =
(t, s) n ⋅ m (t, s) = n ⋅ t (t, s)
ds ds
Thus, we can always find a parameterisation in which n ⋅ t (t, s) = a(t )
(ie: does not depend on s ) by adjusting ds / ds accordingly. Further, we
note that
p
ò
0
n ⋅ t ds = n ⋅ p = na(t )
And so
n⋅p
n ⋅ t = (open string world-sheet constant)
p
In this parameterisation, s for a point is therefore proportional to the
amount of n ⋅ p momentum carried by the portion of the string between
é 0, s ù .
êë úû
Now, consider the equations of motion
¶ t mt + ¶ s ms = 0
Dotting this with n m , we get
¶ ¶
¶t
(
n ⋅ t +
¶s
)
n ⋅ s = 0( )
¶
¶s
(
n ⋅ s = 0 )
Which implies that n ⋅ s is independent of s .
© Daniel Guetta, 2009
, String Theory Notes Page 3 of 31
We have already seen that for open strings, n ⋅ s = 0 at endpoints, which
implies that this is the case everywhere.
3. The Associated s parameterization for closed strings
In this case, we want s Î éêë 0,2p ùúû and so
n⋅p
n ⋅ t = (closed string world-sheet constant)
2p
Because of this factor of two, we write the gauge condition without the
factor of two, as
n ⋅ X = a ¢ (n ⋅ p ) t
I don’t understand this range condition on sigma?
We can still show that n ⋅ s is independent of s , but it’s now not
possible to set it to 0 at any given point. Furthermore, it’s unclear what
point is s = 0 . We solve this by setting a certain point on a certain string
to have these properties. The proof this can be done is in the book.
There is, however, an obvious ambiguity – our whole parameterisation can
be rigidly moved along the string without affecting anything.
4. Summary
In summary, we have
n ⋅ s = 0
n ⋅ X (t, s) = ba ¢ (n ⋅ p ) t
2p
n⋅p = n ⋅ t
b
Where b = 1 for closed strings, and b = 2 for open strings.
The first condition above, along with an expression for s immediately
gives us X ⋅ X ¢ = 0 . This allows us to simplify our expression for tm ,
and we obtain X 2 + X ¢2 = 0 . This is best summarised, together with the
first condition above, as
(X X ¢)
2
=0
We then get the following simplified expressions
© Daniel Guetta, 2009
, String Theory Notes Page 4 of 31
1 m 1 ¢
tm =
2pa ¢
X sm = -
2pa ¢
Xm ( )
Feeding into the equations of motion, we get
¢¢
X m - X m = 0( )
These are simply wave equations!
When the string is open, we have the additional requirement that the sm
and therefore the X m ( )¢ vanish at the endpoints.
4. Solving the wave equation
Assuming we have a space-filling D-brane and therefore free-boundary
conditions at the endpoints, the most general solution to the wave
equation is
1 m
X m (t, s) =
2
(
f (t + s) + g m (t - s) )
¢
( )
Bearing in mind the relation sm = - X m / 2pa ¢ and the boundary
sm
conditions = 0 , we get
¶X m
=0 s = 0, p
¶s
The boundary condition at 0 informs us that f and g differ at most by a
constant, which can be absorbed into f.
1 m
X m (t, s) =
2
(
f (t + s) + f m (t - s) )
The boundary condition at s = p gives
¶X m
¶s
1
(
(t, p) = f m ¢ (t + p) - f m ¢ (t - p) = 0
2
)
Since this must be true for all t , this implies that f m ¢ is periodic with
period 2p .
We can therefore write
¥
(
f m ¢ (u ) = f1m + å anm cos nu + bnm sin nu )
n =1
¥
(
f m (u ) = f0m + f1mu + å Anm cos nu + Bnm sin nu )
n =1
© Daniel Guetta, 2009
