String Theory- Angular Momentum of the Rotating String

� � Lecture 14 8.251 Spring 2007 Lecture 14 -Topics Momentum charges for the string • Lorentz charges for the strings • Angular momentum of the rotating string • Discuss α� and the string length �s• General gauges: Fixing τ and natural units • Reading: Section 8.4-8.6 and 9.1 S = dξ0dξ1 ...dξpL(φa,∂αφa) ξα: coordinates, φa(ξ): fields, ∂α = ∂ξ∂ α α: coord. index α =0, 1,...,p a: field index a =1,...,m i: index for various symmetries δφa(ξ)=i hai (φ(ξ)) Leaves L invar. to first order. 1. ∂L δφa + ∂L ∂α(δφa)=0 ∂φa ∂(∂αφa) 3. Jµ Jα iJiα = ∂L δφa ∂(∂αφa) Similar to mechanics: ∂∂Lq˙δq Claim: Given this transformation leaves L invar. to first order then: → i → 2. ∂αJα =0 ∀i Conserved Current i Check this yourself using 1. and the E − L equations of motion. Done in book as well. Conserved charge too: Qi = JiO(ξ)dξ1dξ2 ...dξp Answer independent of time. 1 � � � Lecture 14 8.251 Spring 2007 dQ = 0 dξO Nambu-Gotta action: S = dξ0 dξ1 L(∂0xµ,∂1xµ)���� dτ dσ µThis means α =0, 1. φa = xa =0,...,d = spatial dimension ⇒Let’s look for asymmetry. A variation of the field that leaves the field invar. δxµ =µ= constant Constant translations of a worldsheet should by asymmetric. Why would Nambu-Gotta action care if rigidally moved worldsheet through time or space?So:δ(∂0xµ)= ∂0(δxµ)=0 δ(∂1xµ)=0 So δxµ =µ indeed asymmetric. Apply (3) µJα ∂L µ = µ ∂(∂αxµ) Jα ∂L= µ ∂(∂αxµ) (J0 ∂L ∂L τσ µ,Jµ1)= ∂x˙µ , ∂x�µ =(P, Pµ )µ Conservation law: ∂αJα = 0 gives us collection of conservation laws for µ. ∂αJα =0= ∂Pµτ + ∂Pµa ∂τ ∂a � σ1Pµ(τ )= 0 Pµτ (τ,σ)dσ conserved quantity indexed by spacetime index µ. Pµ(τ ): conserved momentum for the string not dependent on τ since conserved. Check Pµ is conserved dPµ = � σ1 ∂Pµτ (τ,σ)dσ dτ 0 ∂τ � σ1 ∂Pµσ = − dσ ∂σ 0 =[−Pµσ]σ0 1 2 Lecture 14 8.251 Spring 2007 This yields the free BCs.This is the hardest part of the course. After this, it gets easier.A momentum is in general a variation of a Lagrangian with respect to a velocityeg ∂∂x˙Lµ conserved, has units of momentum. We will see this is indeed the relativemomentum of a piece of string.When we had (ρ,J�):Q[ρ]=3L[J�]=Q TL2 Now we have (P]= Pµ L , Pmomentum density, and Pσµτµ): [Pτµ[P]= Pµ Call PT σµτµσµmomentum current. Okay, we have: dPµ =0 dτ But would like: dPµ =0 dt Conserved for Lorentz observer. Is this the case? (Yes). dPµSure, could work in static gauge. τ = t dt =0 ⇒ But what about an arbitrary τ curve on worldsheet?Look for a generalization formula (clue from divergence theorem)A =flux of vector field3 � � � � � � � � � � � � � � Lecture 14 8.251 Spring 2007 ∂Ax ∂Ay (Axdy − Aydx) = + dxdy R ∂x ∂y dτ]=R ∂Pτ + ∂Pσ dτdσ =0 [Pµτ dσ −Pµσ ∂τ ∂σ Γ Given an arbitrary curve γ, claim momentum given by: Pµ(γ)= [Pµτ dσ −Pµσdτ] γ γ = α → γ2 → β →−γ1 ( γ2 + α + −γ1 + β )(�Pµτ � µσdτ �)=0 dσ −Pκ κ = κ =0 αβ Pµ(γ1)= Pµ(γ2) � σ1Usually will use Pµ(τ)= 0 Pµτ (τ,σ)dτ with constant τ, but nice to have this general formulation. 4 Lecture 14 8.251 Spring 2007 Lorentz Transformation xµ = Lνµx ν Leaves ηµν xµxν invar. Vary xµ subject to xν δxµ =µα xα µx ν )=2ηµα νδ(ηµν xµν xαx =2µαxαxµ where ηµν xν = xµ µIf we want δ(ηµν xxν ) = 0, we make antisymmetric. µν νµ = −Claim: δ(ηµν ∂αxµ∂β x ν )=0 So Nambu-Gotta action invar and get new set of symmetries: δxµ(τ,σ)=µν xν (τ,σ) µν Jα ∂L α αµν µν = ∂(∂αxµ) δxµ = Pµ δxµ = Pµ xν 1 µν µν Jα αα µν = − 2(xµPν − xν Pµ ) No physical relevance to − 1 2 So define: m α (τ,σ)= xµPαα µν ν − xν Pµ αConserved currents: ∂αmµν =0 � σ1 Mµν = mµν 0 dσ 0 Conserved Charge � σ � σ1 Mij = mij dσ =(xiPjτ − xj Piτ )dσ = ijk Lk 00 123 = +1, totally antisymmetric. eg: 5 Lecture 14 8.251 Spring 2007 � σ1 M12 =(x1P2 τ − x2P1 τ )dσ = 12l Lk = L3 0 L�= �r × p�So Mij =angular momentum, conserved. Angular momentum of rotating string: � σ1 M12 = L3 = J =(x1P2 τ − x2P1 τ )dσ 0 ����� ��� �x(t,σ)= σ1 cos πσ cos πct , sin πct πσ1 σ1 σ1 Parametrized String � �� ��� �τ = T0 ∂�x = T0 cos πσ − sin πct , cos πct Pc2 ∂t c σ1 σ1 σ1 ��2 �� σ1 T02 πσ x1P2 − x2P1 = cosπc σ1 1 J = E2 2πT0c E = σ1T0 6