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Relativistic Solutions 11.1 Free Particle Motion

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Relativistic Solutions 11.1 Free Particle Motion Physics 411 Lecture 11 Relativistic Solutions Lecture 11 Physics 411 Classical Mechanics II September 21st, 2007 With our relativistic equations of motion, we can study the solutions for x(t) under a variety of different forces. The hallmark of a relativistic solution, as compared with a classical one, is the bound on velocity for massive particles. We shall see this in the context of a constant force, a spring force, and a one-dimensional Coulomb force. This tour of potential interactions leads us to the question of what types of force are even allowed { we will see changes in the dynamics of a particle, but what about the relativistic viability of the mechanism causing the forces? A real spring, for example, would break long before a mass on the end of it was accelerated to speeds nearing c. So we are thinking of the spring force, for example, as an approximation to a well in a more realistic potential. To the extent that effective potentials are uninteresting (or even allowed in general), we really only have one classical, relativistic force { the Lorentz force of electrodynamics. 11.1 Free Particle Motion For the equations of motion in proper time parametrization, we have x¨µ = 0 −! xµ = Aµ τ + Bµ: (11.1) Suppose we rewrite this solution in terms of t { inverting the x0 = c t equation with the initial conditions that t(τ = 0) = 0, we have c t τ = (11.2) A0 and then the rest of the equations read: c t c t c t x = A1 + B1 y = A2 + B2 z = A3 + B3; (11.3) A0 A0 A0 1 of 8 11.2. MOTION UNDER A CONSTANT FORCE Lecture 11 and we can identify the velocities: A1 c A2 c A3 c vx = vy = vz = : (11.4) A0 A0 A0 Remember that we always have the constraint that defines proper time: dt 1 = (11.5) dτ q v2 1 − c2 and this can be used to relate A0 to the rest of the constants: c A0 = : (11.6) q v2 1 − c2 The \lab" measurements, which proceed with a clock reading coordinate time, give back constant velocity motion with velocity v and we pick up a relation between the coordinate time and the proper time of the particle (in its own rest frame). 11.2 Motion under a Constant Force dp Constant force, according to Newton's second law, is one for which dt = F with constant F . In one spatial dimension, we can describe relativistic constant forcing by interpreting p as the spatial, relativistic momentum, i.e. p = p m v . From this point of view, the solution is straightforward { to 1−v2=c2 0 dx(t) avoid confusion, let v(t) = x (t) ≡ dt , then 0 ! d m x (t) 0 F t p = F −! x (t) = q (11.7) dt 1 − x0(t)2=c2 F 2 t2 m 1 + m2 c2 which can be integrated again { if we set x(t = 0) = 0, then 0s 1 m c2 F t 2 x(t) = 1 + − 1 ; (11.8) F @ m c A the usual hyperbolic motion. 2 of 8 11.2. MOTION UNDER A CONSTANT FORCE Lecture 11 If we start with the relativistic free particle Lagrangian in proper time p µ ν parametrization, L = −m c −x_ ηµν x_ , then we'd like to add a \poten- tial" that gives us the constant force case. We know such a potential will be . strange { after all, it must be a scalar, meaning here that it will have to be constructed out of vectors (specifically, combinations of xµ(τ) and its τ derivatives). In this two-dimensional setting, about the only thing that suggests itself is the cross product of xµ andx _ µ. As we shall see, the Lagrangian: 1 F L = −m c p−x_ µ η x_ ν + x_ µ xν (11.9) µν 2 c µν corresponds to the constant force as defined above. Let's work out the equations of motion { written out, the Lagrangian is p L = −m c c2 t_2 − x_ 2 + F (x t_ − t x_): (11.10) Remember, from the definition of proper time, we have c2 = c2 t_2 −x_ 2, which we can solve forx _, for example { then p c t_t¨ _2 x_ = c t − 1x ¨ = p : (11.11) t_2 − 1 Using the proper time relation, we get equations of motion as follows: d @L @L F − = 0 −! m c t¨= x_ _ dτ @t @t c (11.12) d @L @L − = 0 −! m x¨ = F t:_ dτ @x_ @x Upon introducing the equalities (11.11), these degenerate to the single equa- tion: F p t¨= t_2 − 1 (11.13) m c with solution: m c F τ F τ t(τ) = α + cosh(β) sinh + sinh(β) cosh ; (11.14) F m c m c setting α = β = 0 (our choice) to get t(τ = 0) = 0, we have m c F τ m c F t t(τ) = sinh −! τ(t) = sinh−1 : (11.15) F m c F m c 3 of 8 11.3. MOTION UNDER HOOKEAN POTENTIAL Lecture 11 Meanwhile, we can solvex _ from (11.11): m c2 F τ x(τ) = α + cosh ; (11.16) F m c this is the spatial coordinate in proper time parametrization { to find x(t), we can use the inverse relation τ(t) and just replace: r m c2 F 2 t2 x(t) = α + 1 + (11.17) F m2 c2 and our lone integration constant α can be used to set x(t = 0) = 0, the physical boundary condition. When all is said and done, we recover (11.8). 11.3 Motion Under Hookean Potential We can also consider a relativistic spring { here the difficulty is that for an arbitrary displacement, the maximum velocity can be arbitrarily large classically. We expect our relativistic mechanics to take care of this, pro- viding a maximum speed < c. This time, we will start from the relativistic Lagrangian written in coordinate time parametrization: r x0(t)2 1 L = −m c2 1 − − k x(t)2 (11.18) c2 2 and the potential is the usual one with spring constant k. As a check, this gives the correct Newton's second law form upon variation: 0 1 d @L @L d m x0(t) − = 0 −! @ A = −k x(t): (11.19) dt @x0(t) @x(t) dt q x0(t)2 1 − c2 We could integrate this directly (not easy). Instead, we can use the Hamil- tonian associated with L: r ! @L m x0(t)2 x0(t)2 1 H = x0(t) − L = − −m c2 1 − − k x(t)2 @x0(t) q x0(t)2 c2 2 1 − c2 m c2 1 = + k x(t)2; q x0(t)2 2 1 − c2 (11.20) 4 of 8 11.3. MOTION UNDER HOOKEAN POTENTIAL Lecture 11 no surprise. But it does tell us that numerically, this particular combination is a constant, we'll call it the total energy (notice that it contains the rest energy as well as the kinetic and potential energies). Setting H = E gives us a first integral of the motion, and we could solve it1 { instead, we can take the total time derivative to recover a second order ODE analagous to the equation of motion itself: m x00(t) = −k x(t): (11.21) 3=2 x0(t)2 1 − c2 From here, it is easiest to find the solution numerically { we'll use initial 0 conditions x(t = 0) = x0, x (t = 0) = 0, so we're starting at maximum am- plitude, then the position and velocity for a few different starting locations are shown in Figure 11.1 and Figure 11.2. Notice that the \low-velocity" case, corresponding to the smallest starting position looks effectively just like a classical spring, both in position and velocity. As the initial ampli- tude is increased, the solution for x(t) begins to look more like a sawtooth as the velocity turns into a step function close to c. x(t) x0 = 4 x0 = 2 x0 = 1 t x0 = .5 4 8 12 16 20 2 − 4 − Figure 11.1: Position for a relativistic spring (this is the result of numerical integration, with k = 1, c = 1 and x0 shown). 1To get the asymptotic behavior, solve H = E for v in terms of x and send E ! 1 { the relation you will recover is v = ±c, i.e. the sawtooth motion is predictable from the Hamiltonian 5 of 8 11.4. INFALL Lecture 11 x0 = 4 1 x0 = 2 x0 = 1 ) c .5 of x0 = .5 s nit u 0 t 4 8 12 16 20 (in ) t ( ! .5 x − 1 − Figure 11.2: Velocity for the relativistic spring potential (here, k = 1, c = 1, and we start at a variety of maximum amplitudes as shown). 11.4 Infall Consider the one-dimensional infall problem for a test charge moving under the Coulomb field coming from a point charge at the origin. The total energy of the test charge is 1 q2 E = m x_ 2 − : (11.22) 2 x If we start a particle from rest at spatial infinity at time t = −∞, then the energy is E = 0, and this is conserved along the trajectory. We can easily solve the above with E = 0 and integration constant written in terms of t∗, the time it takes to reach the origin (starting from t = 0): !2=3 32=3 r 2 x(t) = q (t∗ − t) : (11.23) 2 m The problem, as always, with this classical solution is that the velocity grows arbitrarily large as t −! t∗: dx v(t) = ∼ (t∗ − t)−1=3; (11.24) dt allowing the test particle to move faster than light.
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