Lecture 15 8.251 Spring 2007 Lecture 15 - Topics • Solution of the open string motion in the light-cone gauge Reading: Sections 9.2-9.4 x 0(τ0, σ) = cτ0 τ = τ0 is a line, goes to intersection of the worldsheet with the x = cτ0 hyper-plane. nµxµ = λτ0 nµx1 µ = λτ0 nµx2 µ = λτ0 nµ(x1 µ − x2 µ) = 0 nµ = (1,�0), λ = c, recover static gauge If nµ to be timelike: nµΔxµ = 0 Same for ηµ = (a,�0). Δxµ = (0,�v). 1Lecture 15 8.251 Spring 2007 Set λ usefully (aim at τ, σ dimensionless) n x(τ, σ) = λ�(n p)τ· ���· � const � σ1 P µ τ µdσ= P0 �σ1dP µ σµ dτ = −P0 Ask for n · Pσ = 0 at endpoints so that d (n p) = 0. dτ · Reminder of Units: J: Angular momentum of rotating string J = α�E2 ¯h [α�] = [E1]2 since Jh¯is dimensionless. Let’s use natural units (as opposed to Planck units), set c = 1, ¯h = 1. Thus: L = 1 T ML2 = 1 ML = 1 ⇒ Thus everything can be written in terms of units of length (sometimes people use mass instead) 1 1So in natural units, [α�] = [E]2 = M2 = L2. So string length ls = √α� in natural units (to get actual numbers, must replace the c’s and ¯h’s) ls = ¯hc√α� In natural units: T0 1 = πα� c 2 To remember: ls = √α� T0 1 = πα� c 2 Back to I. L = [�] L 1 ⇒ [λ�] = L2 ⇒ λ�∝ α�. As it turns out, n x = 2α�(n p)τ (the 2 will be convenient) · · 2= � �� Lecture 15 8.251 Spring 2007 σ parameterization Static gauge: oT0 (x�)2x˙τ o =Pc √. . . T0 (∂�x/∂σ)2 c ds/dσ 1 − v2 /c2 ⊥� �2 � �2∂�x ds = ∂σ dσ T0 ds/dστ 0 2P = c � 1 − v⊥/c2 1. Try to make n ·Pτ constant along the parameterized string. 2. Get a range σ ∈ [0, π] Imagine had some parameter σ�,P�τ µ(τ, σ). If change parameter, how does it transform? Claim transformation law: Pτ µ(τ, σ) = dσdσ�P�τ µ(τ, σ�) Makes sense that Pτ µdσ is reparam. invar. Multiply by n: n · Pτ (τ, σ) = dσdσ�n · P(τ, σ�) Can set to be A is constant with respect to σ, might be a function of τ � σ1 n · Pτ (τ, σ) = σ1A(τ ) 0 Also = n P (momentum). So, A(τ ) = n p/σ. A not τ dependent! · · n · Pτ (τ, σ) = nσ· 1 p n · Pτ (τ, σ) = nπ · p σ ∈ [0, π] Recall eq. of motion of string: 3Lecture 15 8.251 Spring 2007 ∂Pσµ ∂Pσµ + = 0 ∂τ ∂σ Dot with nµ ∂ ∂ ∂τ (n · Pτ ) + ∂σ (n · Pσ) = 0 ∂ ∂σ (n · Pσ) = 0 We had n · Pσ = 0 at string boundaries (σ = 0, σ = σ1 = π) and since ∂ ∂σ (n · Pσ ) = 0, n · Pσ = 0 ∀σ and ∀ times. Closed Strings n x = 2α�(n p)τ· · For closed strings, more convenient to remove 2: n x = α�(n p)τ· · n · Pτ (τ, σ) = n 2· πp n·Pσ = 0??? Do have ∂ (n·Pσ) = 0, but don’t have endpoints having n·Pσ = 0. ∂σ Open strings give rise to E&M. Closed strings give rise to gravity (harder! more subtle). For a closed string, know how to put σ param. on strings at different times, but we don’t know how to correlate these “σ ticks”. No special points on closed string like endpoints on open string. Compute: σ 1 ( ˙x x�) − x˙2∂σ(n�x) n · P =2πα� ·√. . . · n x ∝ τ , so ∂σ (n x) = 0, so to get n ·Pσ = 0, make ˙x x� = 0. So in spacetime · · · sense want ˙x⊥x�. x�: tangent to string x˙: line of constant σ If given space vector x� and ∃ timelike vector, then ∃ unique vector orthogonal to x� prop to x˙ . So we lock the params. on the string, but still remains the ambiguity of translation of string (where do we set σ = 0 No one knows.) 4� � Lecture 15 8.251 Spring 2007 Summary 1. n x = βα�(n p)τ· · where β = 2 if open string, or 1 if closed string. 2. n · Pτ = npβ 2π 3. 2π σ ∈ [0, ]β 4. n · Pσ = 0 everywhere ⇒ x˙ x� = 0 · 1 x�2x˙µτ µP =2πα� √−x˙2x�2 Dot by n 1 x�2 (n p)β n · Pτ =2πα √−x˙2x�2 βα�(n · p) = 2· π x�2 1 = −x˙2x�2 (x�2)2 = −( ˙x 2)(x�2), (x�)2 = 0 �x�2 = x˙2 x˙2 + x�2 = 0 Using (4), get: ( ˙x ± x�)2 = 0 In static gauge, got: ∂�x 1 ∂�x ∂σ ± c ∂t = 1 1 ∂xµ Pτ µ =2πα� ∂τ 5Lecture 15 8.251 Spring 2007 Eq. of Motion: 1 ∂xµ Pσµ = − 2πα� ∂σ x¨µ − xµ�� = 0 Wave equation for everyone!