String Theory Demystified

CHAPTER 2 Equations of Motion 43
In our case, we get
∫ µν ∫
2 µ ν
2
µ ν
δSP =−T d σ ( δX ) ∂+ ∂−X η −T d σ∂− ∂+
X ( δX ) η νν
We’ve dropped the boundary terms, which must vanish for Neumann boundary
conditions in the case of open strings or for the requirement of periodicity for closed
strings. Since δ µ
X is arbitrary and δS = 0, it must be the case that
P
µ
∂∂ X = 0
(2.46)
+ −
This is the wave equation for relativistic strings using light-cone coordinates.
In the next chapter, we will consider the hamiltonian and stress-energy tensor and
write down conserved charges and currents for the string. Right now, let’s focus on
fi nding a solution of the wave equation given in Eq. (2.46).
From elementary mechanics, we know that the solution of a wave equation can
be written in terms of a superposition of waves moving to the left on the string and
waves moving to the right on the string. If the motion is in one dimension (call it x),
then we can write down a solution of the form
f(, t x) = f ( x− vt) + f ( x+ vt)
L R
We will write the equations of motion for the relativistic string in the same way.
µ
We have a solution which is a superposition of left-moving components X L ( τ + σ)
µ
and right-moving components X R ( τ − σ)
:
µ µ µ
X ( τ, σ) = X ( τ + σ) + X ( τ − σ)
(2.47)
L R
You should recall from partial differential equations that the most general solution
can be written as an expansion of Fourier modes. Here, we denote these modes as
, and write the left-moving and right-moving components as
α µ
k
X
X
µ
L
µ
R
Solutions of the Wave Equation
x
p i
k e
µ 2
µ
s µ α s k − ik(
τ + σσ )
( τ, σ) = + ( τ + σ)
+ ∑ 2 2 2 k≠0
x
p i
k e
µ 2
µ
s µ α s k −ik(
ττ−σ) ( τ, σ) = + ( τ − σ)
+ ∑
2 2 2 k≠0 µ
(2.48)
(2.49)