String Theory Demystified

34 String Theory Demystifi ed
and we also have a conjugate momentum corresponding to the coordinate τ :
P
τ
µ
L
=
X X
∂
=
∂
∂
∂
µ µ
−T X 2 ( ⋅ X′ − X 2 2
( ) ( ) ( X′
) )
T
−12
/
=− ( 2
) ( 2 2
⎡ X⋅ X′ − X) ( X′ ) ( X 2
X ) X X
2 ⎣
⎤ ⎡
⎦ 2 ⋅ ′ ′ − 2 µ ′ X
⎤
⎣
µ ⎦
2
( X⋅ X′ ) X′ − X X
µ ′ µ
=−T
2 2
( X⋅ X′ ) − ( X)
( X′ )
2
(2.25)
Now, let’s vary the action to obtain the equations of motion for the string. First, if
it’s been awhile since you’ve had fi eld theory, convince yourself that
δ δ
τ τ δ
µ
X
⎛ µ ∂X
⎞ ∂ µ
=
X
⎝
⎜ ∂ ⎠
⎟ = ( )
∂
µ ⎛ µ ∂X
⎞
δX′ = δ
⎝
⎜ ∂σ
⎠
⎟ = ∂ µ
( δ X )
∂σ
Then, we can vary the action, and using the conjugate momenta we get
δ τ σ
τ δ
σ δ
τ f ∂L
∂ µ L
S =−T
d d
X + X
τ
µ
µ
i ∂X
∂ X
∂ ∂ µµ
∫ ∫ ( ) (
0
∂ ′ ∂
τ
τ µ σ
τ σ
τ
µ
µ
τ δ
σ δ
⎡
⎤
⎢
) ⎥
⎣
⎦
f
∂
∂
=−T∫
d ∫ d Π ( X ) + Π ( XX
i 0 ∂
∂
µ ⎡
⎤
⎢
) ⎥
⎣
⎦
We can rewrite this expression so that we can get terms multiplied by δ µ
X by using
the product rule from calculus. For example,
∂
∂ ( ) τ µ ∂
P δ X = P δX + δX
µ
µ
τ
∂τ
⇒
∂
∂τ
δ P X
µ
τ µ
= ∂
∂ ( ) P δX − δX
µ
τ
∂P
τ µ µ
τ
µ
∂τ
∂P
τ µ µ
τ
µ
∂τ