String Theory- Open String Motion in Light Cone Gauge

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 hyperplaane 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). 1 Lecture 15 8.251 Spring 2007 Set λ usefully (aim at τ, σ dimensionless) nx(τ,σ)= λ�(np)τ· �� · const � σ1 P µ τµdσ= P0 �σ1dP µ σµ dτ = −P0 Ask for n ·Pσ = 0 at endpoints so that d (np) = 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) 11So 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, nx =2α�(np)τ (the 2 will be convenient)·· 2 = � � � Lecture 15 8.251 Spring 2007 σ parameterization Static gauge: oT0 (x�)2x˙τo =Pc √...T0 (∂�x/∂σ)2c 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�τµ(τ,σ). If change parameter, how does ittransform?Claim transformation law:Pτµ(τ,σ)= dσdσ�P�τµ(τ, σ�) Makes sense that Pτµdσ 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 = nP (momentum). So, A(τ )= n p/σ. A not τ dependent! ··n ·Pτ (τ,σ)= nσ· 1 pn ·Pτ (τ,σ)= nπ · p σ ∈ [0,π]Recall eq. of motion of string:3 Lecture 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 nx =2α�(np)τ·· For closed strings, more convenient to remove 2: nx = α�(np)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 (˙xx�) − x˙2∂σ(n�x) n ·P=2πα� ·√... · nx ∝ τ, so ∂σ (nx) = 0, so to get n ·Pσ = 0, make ˙xx� = 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. nx = βα�(np)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 (np)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 Pτµ =2πα� ∂τ 5 Lecture 15 8.251 Spring 2007 Eq. of Motion: 1 ∂xµ Pσµ = − 2πα� ∂σ x¨µ − xµ�� =0 Wave equation for everyone! 6