Lectures on String Theory

– 15 –
This situation should persist in any Lorentz frame. At any point on a string worldsheet
one should always be able to find two vectors: one is time-like and another is
space-like.
Consider S µ (λ) = ∂Xµ + λ ∂Xµ
∂τ ∂σ
must be space- or time-like as λ varies.
S µ S µ = Ẋ2 + 2λẊX′ + λ 2 X ′2 ≡ y(λ) .
Discriminant must be positive then there are two roots and therefore the regions of
λ with time- and space-like vectors. Discriminant is
(ẊX′ ) 2 − Ẋ2 X ′2 > 0 .
This condition guarantees the causal propagation of string.
Action is invariant under reparametrizations
δX µ = ξ α ∂ α X µ ,
ξ α = 0 on the boundary
Two possibilities:
• Open strings: 0 ≤ σ ≤ π
• Closed strings: 0 ≤ σ ≤ 2π
Equations of motion:
Variation
δS
δX µ = ∫ τ2
= −
+
dτ
τ 1
∫ τ2
τ
∫ 1
π
0
S =
∫ π
dτ
0
∫ π
∫ τ2
τ 1
dτ
( δL
dσ
0
dσ
dσ δL
δẊµ δXµ | τ 2
τ 1
+
• Open string boundary conditions:
• X µ (σ + 2π) = X µ (σ).
Canonical formalism. Momentum
We see that
∫ π
0
dσL
∂ τδX µ + δL )
δẊµ δX ∂ σδX µ
(
′µ δL δL
∂ τ
)
+ ∂
δẊµ
σ
δX ′µ
∫ τ2
τ 1
(3.5)
(3.6)
δL
δX ′µ δXµ | σ=π
σ=0 (3.7)
δL
(τ, σ = π) = δL (τ, σ = 0) = 0
δX ′µ δX ′µ
P µ = δL = −T (ẊX′ )X ′µ − (X ′ ) 2 Ẋ µ
√
δẊµ (ẊX′ ) 2 − Ẋ2 X ′2
P µ X ′ µ = 0 (3.8)
P µ P µ + T 2 X ′2 = 0 (3.9)