� ���� ����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 ≈ µ0v Natural 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 inde-pendently. 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 2Lecture 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: 3Lecture 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 1 1 = 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 1 1 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 0 0 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=0 ti ti For ∗ ∈ 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)