� � � � Lecture 6 8.251 Spring 2007 Lecture 6 - Topics • The relativistic point particle: Action, reparametrizations, and equations of motion Reading: Zwiebach, Chapter 5 Continued from last time. ∂Pt + ∂Px = 0 ∂t ∂x Pt = µ0∂y/∂t Px = −T0∂y/∂x Similar to ∂µJµ = 0, ∂ρ/∂t + � · J�= 0, Q = dxρ Free BC (Neumann BC): � Px(t, x∗) = 0 �a Py = µ0dx(∂y/∂t) = dxPt � 0� 0a a a ∂Py/∂t = dx∂Pt/∂t = − dx∂Px/∂x = −[Px(t, x = a) − Px(t, x = 0)] 0 0 Conservation of momentum? Free Relativistic Particle Non-relativistic Action: 1 S = dt mv 2 2 Calculation: dv/dt = 0 Relativistic Particles: 1� � � � � � � �� � = � Lecture 6 8.251 Spring 2007 Everyone should agree on action. It’s a Lorentz invar. −ds2 = −ηµν dxµdxν ds = cdt 1 − v2/c2 = cdτ ds s = −mc 2 = −mc ds cP P So: s = −mc2 tf dt 1 − v2/c2 ti Check: Lagrangian: v2 L = −mc 2 1 − c2 1 v2 = −mc 2(1 − − . . .) Taylor Expansion 2 c2 1 = −mc 2 + mv 2 � �� � 2 rest energy kinetic energy Momentum: ∂L P�= ∂�v 1 −2�v = −mc 2 �2 c2 1 − c2 · v2 2mv1 − vc2 2 Hamiltonian: 2mcH = p� �v − L = . . . = � 1 − c2 · v2 Parameterization Have parameterization xµ(τ ) (the xµ’s are functions of τ) ds2 = −ηµν dxµdxν 2� � � � Lecture 6 8.251 Spring 2007 � � dxµ �� dxν � ds = −ηµν dτ dτ dτ � tf � dxµ dxν s = −mc ti −ηµν dτ dτ dτ τ�(τ ): dxµ dxµ dτ� = dτ dτ � dτ s = −mc � tf ti = −mc � tf ti � −ηµν � dxµ dτ � � −ηµν � dxµ dτ � dxν dτ� � dτ� dτ dτ dxν dτ� � dτ� So using a different parameter, τ � (instead of τ ) gets same action s. s is reparameterization-invariant. � � Quick calculation to find equation of motion from s = −mc 1 − vc22 dt. Should get derivative of rel. momentum with respect to time = 0. S = −mc dS δS = −mc δ(dS) dS2 = −ηµν dxµdxν dxµ dxν (dS)2 = −ηµν (dτ )2 dτ dτ 2(dS) · δ(dS) = −2ηµν δ dxµ dτ dxν dτ (dτ)2 3�� � � � Lecture 6 8.251 Spring 2007 d dxν δ(dS) = −ηµν (δxµ) dτ dτ ds Must vary with dxµ/dτ and dxν /dτ , but since ηµν is symmetric sufficient to vary just dxµ/dτ and multiply by 2. δ(dS) = − d (δxµ) dxµ dτ dτ dS � τf � � d(δxµ) dxµ δS = mc dτ dτ ds �τi τf � � = d (δxµPµ) − δxµ dPµ dτ dτ dττi δxµ(τi) = δxµ(τf ) = 0 � τf � � dPµdS = − δxµ(τ) dτ dττi Equation of Motion: dPµ = 0 dτ This means that Pµ constant on world-line. Constant as a function of any pa-rameter! dPµ = dPµ dτ · dt�dt��� �dτ��� ���� 0 0=0 d dxµ d2Therefore: dτ ds = 0, ds2 (dxµ) = 0 (if τ = s. Okay because τ is arbitrary.) But can’t assign s = τ : d2xµ/dτ 2 == 0. �d dxµ = 0 ds dτ �4� � Lecture 6 8.251 Spring 2007 Coupling to Electromagnetism Lorentz Force Equation: dPµ q dxν dS = c · Fµν ds dPµ q dxν dτ = c · Fµν dτ S = −mc dS + qAµ(x(τ)) dxµ dτ c dτP P A: Nevitz-Schwartz Tensor