String Theory Demystified

134 String Theory Demystifi ed
SOLUTION
Let’s write down the lagrangian, which is
T µ α
µ α
L =− ( ∂ X ∂ X −iψ ρ ∂ ψ )
α
µ
α µ
2
µ µ µ
To examine translational invariance, we let X → X + a where aµ is an
infi nitesimal parameter. A key insight into the fact that a µ is infi nitesimal is that we
can drop terms that are second order in a µ µ µ µ
. Taking X → X + a changes the
lagrangian as follows:
T µ µ α
µ α
L →− ⎡
⎣
∂ ( X + a ) ∂ ( X + a ) −iψ ρ ∂ ψ ⎤
α
µ µ
α µ
2
⎦
T
=− ⎡(
∂ +∂ )( ∂ +∂ ) ⎤
2 ⎣
⎦ +
µ µ α α
X a X a L
α α
µ µ F
T µ α
µ α
µ α
µ α
=− ⎡∂
X ∂ X +∂ ∂ +∂ ∂ +∂ ∂ ⎤
α
µµ α µ α µ α µ
2 ⎣
⎦ +
X a a X a a LF
T µ α
µ α
µ α
=− ⎡∂
X ∂ X +∂ X ∂ a +∂ a ∂ X ⎤ L αα
µ α µ α µ F
2 ⎣
⎦ +
µ α
(drop second-order term ∂ a ∂ a )
µ α
= L − ∂ ∂ α µ
T
µ α
⎡ X a +∂ ∂ ⎤
2 ⎣
a X
α µ ⎦
(add in ∂
µ α
X ∂ X to L to get total lagrangian)
F
α
µ α
µ µ µ
Note that the term iψ ρ ∂ ψ ∝ L is unaffected by α µ F
X → X + a . Now, we
focus on the leftover extra term:
L T
⎡
2 ⎣
µ α
µ α
δ = ∂ X ∂ a +∂ a ∂ X
α µ α µ
We will manipulate this expression to get our conserved current, which will be a
term multiplying ∂ α
a . We can fi x up this expression doing some index gymnastics,
µ
which is a good exercise for us to go through given the level of this book. For more
practice doing this, consult Relativity Demystifi ed.
We want to fi x up the second term so that it looks like the fi rst term. We do this
by raising and lowering indices with the metric and using the fact that
µν
µ αβ
η η = δ h h
= δ
νλ λ
µ
α
βγ γ
⎤
⎦
α
µ