String Theory- Relativistic Strings

� � � � Lecture 10 8.251 Spring 2007 Lecture 10 -Topics • Relativistic strings: Nambu-Gotto action, equations of motion and boundaar conditions Reading: Section 6.1 -6.5 S ∝ world-sheet area (A)Units of const. of proportionality[S]=[ � dtL] = Energy × time = ML2 T [World Sheet Area] = L2 [Const. of Proportionality] = M T Recall for tension T0:[T0] = [Force] = ML . T0 = M . So let T0/c be constant T 2 cT of proportionality for S ∝ A. Nambu-Gotto Action � τf � σ1 � S = − Tc 0 τi dτ 0 dσ (X˙· X�)2 − (X˙)2(X�)2X˙µ ∂Xµ ∂Xµwhere = dτ . X� = dσ . The two tangents to the surface.This defines the dynamics of the string.World sheet metric: Let ζ1 = τ , ζ2 = σ.Distance between 2 points on the worldsheet:−dx2 = dXµdXµ = ηµν dXµdXν∂Xµ ∂Xν= ηµν dζαdζβ ∂ζα ζβ= γαβ (ζ)dζαdζβwhere γαβ = ηµν ∂Xµ dXν ∂ζα ∂ζβ X˙2 ˙XX�γαβ =˙XX� (X�)2 String world-sheet metric. All signs from the η are in the X’s. So: 1 �� � Lecture 10 8.251 Spring 2007 T0 dτdσ√−γS = − c where γ = det(γαβ ). Lorentz invar., reparam invar. � τf � σ1 X = dτ dσL(X˙µ,X�µ) τi 0 Lagrangian Density: L(X˙µ,X�µ)= − T0 (X˙X�)2 − (X˙)2(X�)2 c · δS =0 � τf � σi �� ∂L �� ∂(δXµ) �� ∂L �� ∂(δXµ) �� δS = dτ dσ + ∂X˙µ ∂τ ∂X˙µ ∂σ τi 0 δX˙µ = δ∂Xµ ∂τ Define: ∂L = Pµτ = − T0 �(X˙· X�)Xµ�− (X�)2X˙µ ∂X˙µ c (X˙X�)2 − (X˙)2(X�)2 · XX�)˙∂L σ T0 (˙Xµ − (X˙)2Xµ�∂X�µ = Pµ = − c √... Things will simplify soon. So: �� � τσ � δS = dτdσ ∂ (δXµτ )+ ∂ (δXµσ ) − δXµ( ∂Pµ + ∂Pµ )µµ∂τ P∂σ P∂τ ∂σ δS = 0 so ... µµ∂Pτ + ∂Pσ =0 ∂τ ∂σ For open strings: � τf σ dτ [δX0(τ,σ1)Pσ(τ,σ1)−δX0(τ, 0)Pσ (τ,σ1)+δX1(τ,σ1)P σ)(τ,σ1 )−δX1(τ,0)P1 (τ,σ1)+...+δXd ...] 00 1 τi 2D Constraints. 2 Lecture 10 8.251 Spring 2007 For most, get choice as to how the term vanishes since product of 2 terms (so either can be 0). For µ = 0: �1. Dirichlet BCs: Xµ(τ,σ) =constant, δXµ(τ,σ) = 0 (for σ=0 or σ1)∗∗∗ 2. Free BCs: Pµσ(τ,σ)=0 for σ=0 or σ1.∗∗ For µ = 0: 1. ∂X0/∂τ < 0 can’t have Dirichlet. 2. Free BCs. Pσ(τ, 0) = Pτ (τ,σ1) = 0. 00 3