String Theory-Nonrelativistic Strings

� ���� ���� Lecture 5 8.251 Spring 2007 Lecture 5 -Topics • Nonrelativistic strings • Lagrangian mechanics Reading: Zwiebach, Chapter 4 Non-Relativistic Strings Study nonrelativistic strings first to develop intuition and math notation before moving to the relativistic strings that we actually care about. Non-relativistic string: Characterized by:Tension, T0:[T0] = [Force] = [Energy/Length] = M [v2]L Mass/Length: µ0 2T0 ≈ µ0vNatural velocity: v =T0/µ0 Transverse Oscillation: Mark point P on string and see it moving up and down: y(P,t),x(P,t)= x(P )(x not dependent on t) Small Oscillation: << 1 ∂x ∂y (t,x) Consider small section of string: 1 � � � Lecture 5 8.251 Spring 2007 Approximate tensions on endpoints as equal (good for transverse waves, terrible for longitudinal) dFν = T0 ∂y (t,x + dx) − T0 ∂y (t,x)∂x ∂x∂2y= T0 (t,x)dx ∂x2∂2y≈ µ0dx ∂t2 ∂2y 1 ∂2y ∂x2 − T0/µ0 ∂t2 =0 The Wave Equation! t,x are parameters. Motion described by y(t,x). (If had motion in more than 1 dimension �y(t,x)) Stretching of string: Δl = dx2 + dy2 − dx = dx( 1+(dy/dx)2 − 1)1= dx(dy/dx)2 ((small))2 General form of wave equation: ∂2f 1 ∂2f ∂x2 − v2 ∂t2 =0 v: velocity of wave, v = T0/µ0 General Solution: y(x,t)= h+(x − v0t)+ h(x + v0t)−Note: the h’s are function of 1 variable (x ± v0t) not 2 variables x and t independeently Boundary Conditions: Behavior of endpoints at all times (special points at all times) Open string: y(t,x = 0) = 0 (Dirichlet condition -for fixed end point) ∂y (t,x = 0) = 0 (Free BD, Neumann condition) ∂x 2 Lecture 5 8.251 Spring 2007 For free endpoint (hoop on string), means string must be perp. here Initial Conditions: All points on string at some t0 (all points at special time) y(λ,t = 0) ∂y (x,t = 0) ∂x Example: Fixed Endpoints: y(t, 0) = h+(−v0t)+ h(v0t)=0 Let u = v0t−= h+(−u)+ h−(u) h−(u)= −h+(−u) y(t,x = a)=0= h+(a − v0t)+ h−(a + v0t) h+(a − v0t)= −h−(a + v0t)= h+(−a − v0t) Let u = −a − v0t h+(u +2a)= h+(u) Variational Principle Consider point mass m doing 1D motion x(t).Assume x(ti)= xi, x(tf )= xf . Under the influence of potential V (x)Know: 3 Lecture 5 8.251 Spring 2007 Possible motions: Not possible: Given a path: 4 � � � � � � � � � � � � � � � � � � � Lecture 5 8.251 Spring 2007 Functional: S : x(t) ⇒� (not a function of time) Hamilton’s Principle: Principal path makes S stationary. Call true path x(t). Consider new path x(t)+ δx(t) S[x(t)+ δx(t)] = S[x]+ θ[(δx)2] Assume δx(ti) = 0, δx(tf )=0 Lagrangian: L(t) = Kinetic Energy -Potential Energy t2 t2 1 S = L(t)dt = m(˙x(t))2 − V (x(t)) dt2t1 t1 tf 1 S[x + δx]= m(˙x + δx˙)2 − V (x + δx) dt2ti tf ∂V tf 11 = S[x]+ m ˙x − ∂x (x(t)δx(t)) dt +2 m(δ ˙− 2 V ��(δx)2xδ ˙x(t))2 ti ti θ(δx2) Need to eliminate second term. � tf xδ ˙∂V ti [m ˙x − ∂x (x(t)δ(x(t)))]dt must go away for S[x + δx]= S[x]+ θ[(δx)2] to be true. Call this the variation δS tf d δS = dt xδx) − m¨(m ˙xδx − V �(x(t))δ(x(t))dtti Integrate by parts tf dS = mx˙(tf )δx(tf ) − mx˙(ti)δx(ti)+ dtδx(t)[−mx¨− V �(x(t))] ti δx(tf )= δ(ti) = 0 from before. The integral tf dtδx(t)[−mx¨− V �(x(t))] must be 0 too, so: ti mx¨= −V �(x(t)) 5 � � � � � � � � � � � � Lecture 5 8.251 Spring 2007 String Lagrangian��2 T : Kinetic energy = 1 µ0dx ∂y 2 ∂t ��2 Potential Energy = � ΔlT0 = � a 1 dx ∂y T0string 0 2 ∂x a 11 L = dx 2 µ0(∂y/∂t)2 − 2 T0(∂y/∂t)2 0 tf S = L(t)dt ti Call L: Lagrangian Density 1 µ0(∂y )2 1( ∂y )L =2 ∂t − 2 ∂t So: tf a ∂y ∂y S = dt dxL ,∂t ∂x ti 0 δy(ti,x)=0 δy(tf ,x)=0 Don’t know δy(x =0,t) or δy(x = a,t) δS = tf dt a dx ∂L δy˙+ ∂L δy� ti 0 ∂y˙ ∂y� 6 � � � � � � Lecture 5 8.251 Spring 2007 Let: tP= ∂L/∂y˙ Px = ∂L/∂y� tf a ∂(δy) ∂(δy)δS = Pt ∂t + Px ∂x ti 0 ���� ��� � δS = tf dt a dx −δy(x,t) ∂Pt + ∂Px + a dxPt[δy]tf + tf x[δy]x=a ti x=0 ti 00 ti ∂t ∂x Pδy(ti)= δy(tf )=0 Must have: ∂Pt + ∂Px =0= µ0 ∂2y − T0 ∂2y ∂t ∂x ∂t2 ∂x2 Some kind of conservation law like ∂µJµ =0 tf tf dtPx[δy]x=a = dt[Px(t,x = a)δy(t,x = a) −Px(t,x = 0)δy(t,x = 0)] x=0ti tiFor ∗∈ 0,a: Px(t,x∗)δy(t,x∗) Dirichlet condition: y(t,x) = fixed, δy(t,x)=0 ∗∗Free boundary condition:Px(t,x∗) = 0, ∂y/∂x = 0 (Neumann condition)7