String Theory Demystified

66 String Theory Demystifi ed
µ
Pτ has an immediate physical interpretation. It is the momentum density of the
string. We integrate along the length of the string fi xing τ to get the total momentum
carried by the string, which we label p µ :
σ
1
pµ = ∫ dσP 0
τ
µ
(3.23)
The other conserved current associated with the global symmetries of the action
comes from invariance under Lorentz transformations. In this case
µ µ
δX = ω X
We can show that the lagrangian in Eq. (3.19) is invariant under a Lorentz
transformation in the following way:
( ) ( ) ( )
µ ν
µ
ν
=∂α( δX ) ∂ X η +∂ X ∂ δ η
β µν α β ( X ) µν
µ ρ ν
=∂ ω α ( X ρ ) ∂ X β
µ
+∂ X ∂
µν α β ( ν λ
X λ )
µ ν
µ ν
µ
ν
δ ∂ X ∂ X η = δ ∂ X ∂ X η +∂ X δ ∂ X η
α β µν α β µν α
β µ νν
µ ρ ν
ν µ λ
= ω ∂ X ∂ X η + ωω ∂ X ∂ X η
ρ α
νρ α
β
β
µν
ν
ν
η ω η
µλ α
λ α
ρ ν
µ λ
= ω ∂ X ∂ X + ω ∂ X ∂ X (lower indices with η )
β
β
ρ ν
ρ λ
= ω ∂ X ∂ X + ω ∂ X ∂ X (relabel dummy indices µ → ρ)
νρ α
νρ α
β
β
ρλ α
β
ρ ν
ρ ν
= ω ∂ X ∂ X + ω ∂ X ∂ X (relabel dummy indices λ →ν)
ρν α
β
ρ ν
ρ ν
= ω ∂ X ∂ X + ω ∂ X ∂ X = 0 (antisymmetry ω =− ω )
νρ α β
νρ α β
αββ
βα
So the lagrangian is invariant under a Lorentz transformation, but what are the
currents? This is easy to fi nd, since
LP
⇒ ∂
∂∂X
L
P
µ ( α )
T
=−
2
αβ µ ν
−hh ∂αX ∂βX
ηµν
T
=−
2
αβ ν T
−hh ∂ β X ηµν
=−
2
αβ
−hh
P
µν α
µ µ
µ µ ν
Using ε Jµν= [ ∂LP/ ∂( ∂αX
)] δX
where δ X = ω ν X together with the antisymmetry
of ω µ
ν , we conclude that the Lorentz current is
µν µ ν ν
J = T X P − X P
α
µν
µν
µ
β
µ
( α α ) (3.24)
µν