� �� � � � � � � � 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 ( ˙x x�)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 PT µ = − 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∂Pτ µ + ∂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 d T0ds � = 0 dt 1 − v2/c2 ∂Pτ µ + ∂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 en-ergy. 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 Pτ µ = c2 ∂t ∂xµ Pσµ = −T0 ∂σ Nice and simple. 3Lecture 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 t t + 2σc 1 then velocity doesn’t change! (since F��(u + 2σ1) = F��(u)))→ 4Lecture 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. 5Lecture 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) = l ω (− sin(ωu/c), cos(ωu/c))2 c ωl Unit vector: ω l = 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 σ l = = ω π 2 E π σ1 = E = (lT0)T0 ⇒ 2 String has π 2 more energy since rotating.