String Theory Demystified

48 String Theory Demystifi ed
EXAMPLE 2.4
What is the length of a string that has both endpoints fi xed?
SOLUTION
If both endpoints are fi xed, then we must also satisfy the boundary condition
In this case, we have
µ
X X
µ
( τσ , = π) + ( τσ , = π)
=0
L R
2
s s
X ( τσ , = π) = p + α e
L
∑ k
2 2 k≠0
µ µ µ − ik(
τ+ π)
2
s s
X ( τσ , = π) = p + α e
R
∑ k
2 2 k≠0
µ µ µ −ik( τ−π) The boundary condition can only be satisfi ed if k is an integer. The overall solution
in this case can be written as
µ
µ µ 2 µ αn
−inτ
X = x + p σ − 2 e nσ
s s∑sin(
) (2.65)
n
Here we applied the conditions in Eqs. (2.63) and (2.64). This expression includes
the winding term
n≠0
w= p s
2 µ (2.66)
Now let’s compute the string coordinates at the endpoints. We have
Hence, the length of the string is
µ µ
X ( τ,
0 ) = x
µ µ
X ( τπ , ) = x + wπ
µ µ
X ( τπ , ) − X ( τ, 0)
= wπ