Summary String Theory for Undergraduates (8.251)_2 - 2009 Lecture Notes
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String Theory for Undergraduates (8.251): miscellaneous partial notes (1, 2, 3) taken during the class, unlikely to be anywhere near as good as the course book by Prof Barton Zweibach who taught it.
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)
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 .
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
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
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