String Theory Demystified

CHAPTER 7 RNS Superstrings 135
Now, the fi rst thing to notice is that the order of the derivatives doesn’t matter.
So, the fi rst step is to write
µ α α
∂ a ∂ X =∂ X ∂ a
α
µ
Next, let’s raise the space-time index on X µ and lower it on a µ . Since these are
space-time indices, we use the Minkowski metric to do this:
µ α
α µ α ν µλ
∂ X ∂ a =∂ ( η X ) ∂ ( η a )
µ α
The metric is not space-time dependent (in the fl at space or Minkowski spacetime
we are considering here), so we can pull the metric terms outside of the
derivatives. You might recognize that this is actually true in general—because the
derivatives are with respect to worldsheet coordinates, but the metric, if it depends
on coordinates, is space-time dependent. So,
α ν µλ
µλ α ν
∂ ( η X ) ∂ ( η a ) = η η ∂ X ∂ a
µν
α
µν
λ µν
α
λ α ν
α ν
= δ ∂ X ∂ a =∂ X ∂ a
The index ν is a repeated or dummy index, so we can call it what we like. Let’s
µ α
µ α
change it to match the fi rst term in δL = T/ 2 [ ∂ X ∂ a +∂ a ∂ X ] :
α µ α µ
α ν
α µ
∂ X ∂ a =∂ X ∂ a
α ν
Now, we repeat the process for the indices on the derivatives. This time, the
indices are worldsheet indices. So, we obtain
ν
α
µ
λ
α µ
α µ
αβ µ γ
∂ X ∂ a = h ∂ X h ∂ a
α µ
αβ
µ γ
= h h ∂ X ∂ a
β
= δ ∂
β
αγ
µ
λ
α λ
αγ β µ
γ
µ γ
X ∂ a
ββ µ
µ
X β
β
aµ µ
X α
α
aµ
=∂ ∂ =∂ ∂
And so, the variation in the lagrangian reduces to
α ν
µ α
µ α
µ α
δ = ( ∂ X ∂ a +∂ a ∂ X )
= T∂ X ∂ a
α µ α µ α µ
L T
2