String Theory- String Action contd..

� �� � � � � � � � Lecture 12 8.251 Spring 2007 Lecture 12 -Topics • The σ-parameterization • Equations of motion and Virasoro constraints • General motion for open strings • Rotating open string Reading: Chapter 7 So far: X0(τ,σ)= ct = cτ ∂X�∂X�∂X�∂�x2 v = ⊥ ∂t − ∂t · ∂s ∂s � ds v2 (˙xx�)2 x2x�2 = c ⊥·− ˙dσ 1 − c2 ∂�x ∂�x x˙x� = · ∂t · ∂s ��2 ��2 (x�)2 = ∂�x = ds ∂σ dσ ��2 (˙x)2 = −c 2 + ∂�x ∂t |d�x| = ds ∂�x ∂�x =0 ∂t · ∂σ x˙x� =0 · ∂�x v= ⊥ ∂t T0 −(x�)2 ∂xµ T0 ds/dσ ∂xµ PTµ = − c √... ∂τ = c � 1 − v2/c2 ∂τ x)2 ∂xµ σµ T0 −(˙∂σ T0 1 − v2/c2 ∂xµ �σ)P= − c √... = ds/dσ ∂σ = (0, P∂Pτµ + ∂Pσµ =0 ∂τ ∂σ 1 � � � � � � � � � Lecture 12 8.251 Spring 2007 µ = 0: ∂Pτ 0 =0 (Pσ0 = 0) ∂t ∂T0 ds/dσ� =0 ∂τ c 1 − v2/c2 Consider a constant, fixed dσ dT0ds � =0 dt 1 − v2/c2 ∂Pτµ + ∂Pσµ =0 ∂t ∂σ T0 � ds/dσ ∂2�x ∂ 1 − v2/c2 ∂�x =0 c 1 − v2/c2 ∂t2 − T0 ∂σ ds/dσ ∂σ Wouldn’t it be nice if the 1 − v2/c2 disappeared? Then we would have a nice wave equation. Let’s fix magnitude of σ s.t. ds/dσ/1 − v2/c2 =1 ds dσ = � 1 − vc2 2 1 T0ds 1 dσ = T0 √... = T0 dEnergy 2 � � Lecture 12 8.251 Spring 2007 E σ 0,σ1 = T0 Note σ not equal to the length -more convenient this way, proportional to enerrgy Our cleverness so far: Static gauge Time on world = time on worldsheet Keep lines orthogonal Set σ proportional to energy ��2 ��2 ��2ds v2 ∂x 1 ∂�x=1 − + =1 dσ c2 ⇒ ∂σ c2 ∂t Recall: ∂�x ∂�x =0 ∂t · ∂σ These two boxed equations are the parameterization conditions. Now wave equation is simple: ∂2�x 1 ∂2�x ∂σ2 − c2 ∂t2 =0 For normal non-rel. string, get wave equation. For new rel. string, get wave equation and 2 parameterization conditions. Combine the equations: ��2∂�x 1 ∂�x ∂σ ± c ∂t =1 Now: T0 ∂xµ Pτµ = c2 ∂t ∂xµ Pσµ = −T0 ∂σ Nice and simple. 3 Lecture 12 8.251 Spring 2007 Open String Motion Totally Free 1 X(t,σ)= (F�(ct + σ)+ G�(ct − σ))2 This is all the wave equation tell you. x: position of string. Now BCs: ∂�x =0 ∂σ σ =0,σ1 ∂�x 1 =(F��(ct + σ)+ G��(ct − σ))∂σ 2 Primes indicate derivative with respect to σ BC 1: ∂�x =0 ∂σ σ=0 F��(ct)+ G��(ct)=0 F��(u)= G��(u) G�(u)+ �a0⇒⇒ Back to X�= 1 (F�(ct + σ)+ F�(ct − σ)+ �a0). Absorb �a0 into F�.2 1 X�=(F�(ct + σ)+ F�(ct − σ))2 �x(t, 0) = F�(ct). F tells you the motion of one endpoint. BC 2: ∂�x =0 ∂σ σ=σ1 F��(ct + σ1)= F��(ct − σ1) F��(u +2σ1)= F��(u) F � periodic. F�(u +2σ1)= F�(u)+ 2σ1 �v0 c ∂�x c(t,σ)= (F��(ct + σ)+ F��(ct − σ))∂t 2 Let tt + 2σc 1 then velocity doesn’t change! (since F��(u +2σ1)= F��(u)))→ 4 Lecture 12 8.251 Spring 2007 X(t +2σ1 ,σ)= 1(F (ct +2σ1 + σ)+ F�(ct +2σ1 − σ)) = �x(t,σ)+ 2σ1 �v0 c 2 c So �v0 = average velocity of any fixed-σ point on the string. This explanation is a bit different than the book’s. ∂�x 1 ∂�x += F��(ct + σ)∂σ c∂t ∂�x 1 ∂�x ∂σ − c ∂t = −F��(ct − σ) These yield: ∂x 1 ∂x∂σ ± c ∂t = ±F �(ct ± σ)|F��(u)| =1 | dF�du (u) | =1 u: length parameter on curve |dF�| = du Example: Most famous example. Open string doing circular motion. 5 Lecture 12 8.251 Spring 2007 l: length of string X�(t,σ = 0) = l (cos ωt, sin ωt)2 Recall: X�(t, 0) = F�(ct) ⇒ F (u)= 2 l (cos(ωu/c), sin(ωu/c)) F��(u)= lω (− sin(ωu/c), cos(ωu/c))2 c ωl Unit vector: ωl =1 = c c 2 ⇒ 2 String endpoints move at speed of light! Periodicity of F��: ω(2cσ1) = m(2π) m = 1: 1 l x(0,σ)= 2(F (σ)+ F (−σ)) = 2(cos(πmσ/σ1), 0) ω2σ1 =2π c c σl = = ω π 2 E π σ1 = E =(lT0)T0 ⇒ 2 String has π 2 more energy since rotating. 6