Chapter 5 the Relativistic Point Particle
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Chapter 5 The Relativistic Point Particle To formulate the dynamics of a system we can write either the equations of motion, or alternatively, an action. In the case of the relativistic point par- ticle, it is rather easy to write the equations of motion. But the action is so physical and geometrical that it is worth pursuing in its own right. More importantly, while it is difficult to guess the equations of motion for the rela- tivistic string, the action is a natural generalization of the relativistic particle action that we will study in this chapter. We conclude with a discussion of the charged relativistic particle. 5.1 Action for a relativistic point particle How can we find the action S that governs the dynamics of a free relativis- tic particle? To get started we first think about units. The action is the Lagrangian integrated over time, so the units of action are just the units of the Lagrangian multiplied by the units of time. The Lagrangian has units of energy, so the units of action are L2 ML2 [S]=M T = . (5.1.1) T 2 T Recall that the action Snr for a free non-relativistic particle is given by the time integral of the kinetic energy: 1 dx S = mv2(t) dt , v2 ≡ v · v, v = . (5.1.2) nr 2 dt 105 106 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE The equation of motion following by Hamilton’s principle is dv =0. (5.1.3) dt The free particle moves with constant velocity and that is the end of the story. Since even a free relativistic particle must move with constant veloc- ity, how do we know that the action Snr is not correct in relativity? Perhaps the simplest answer is that this action allows the particle to move with any constant velocity, even one that exceeds that of light. The velocity of light does not even appear in this action. Snr cannot be the action for a relativistic point particle. We now construct a relativistic action S for the free point particle. We will do this by making an educated guess and then showing that it works properly. But first, how should we describe the motion of the particle? Since we are interested in relativistic physics, it is convenient to represent the motion in spacetime. The path traced out in spacetime by the motion of a particle is called its world-line.Inspacetime even a static particle traces a line, since time always flows. Akey physical requirement is that the action must yield Lorentz invariant equations of motion. Let’s elaborate. Suppose a particular Lorentz observer tells you that the particle is moving according to its equations of motion, that is, that the particle is performing physical motion. Then, you should expect that any other Lorentz observer will tell you that the particle is doing physical motion. It would be inconsistent for one observer to state that a certain motion is allowed and for another to state that the same motion is forbidden. If the equations of motion hold in a fixed Lorentz frame, they must hold in all Lorentz frames. This is what is meant by Lorentz invariance of the equations of motion. We are going to write an action, and we are going to take our time to find the equations of motion. Is there any way to impose a constraint on the action that will result in the Lorentz invariance of the equations of motion? Yes, there is. We require the action to be a Lorentz scalar: for any particle world-line all Lorentz observers must compute the same value for the action. Since the action has no spacetime indices, it is indeed reasonable to demand that it be a Lorentz scalar. If the action is a Lorentz scalar, the equations of motion will be Lorentz invariant. The reason is simple and neat. Suppose one Lorentz observer states that, for a given world-line, the action is stationary 5.1. ACTION FOR A RELATIVISTIC POINT PARTICLE 107 against all variations. Since all Lorentz observers agree on the values of the action for all world-lines, they will all agree that the action is stationary about the world-line in question. By Hamilton’s principle, the world-line that makes the action stationary satisfies the equations of motion, and therefore all Lorentz observers will agree that the equations of motion are satisfied for the world-line in question. Asking the action to be a Lorentz scalar is a very strong constraint, and there are actually valid grounds for suspecting that it may be too strong. The action in (5.1.2), for example, is not invariant under a Galilean boost v → v + v0 with constant v0. Such a boost is a symmetry of the theory, since the equation of motion (5.1.3) is invariant. We will find, however, that this complication does not arise in relativity. We can indeed find fully Lorentz-invariant actions. Quick Calculation 5.1. Show that the variation of the action Snr under a boost is a total time derivative. Figure 5.1: A spacetime diagram with a series of world-lines connecting the origin to the spacetime point (ctf ,xf ). We know that the action is a functional – it takes as input a set of functions describing the world-line and outputs a number S.Nowlet us imagine a particle that is moving relativistically in spacetime, starting at the initial position (0, 0) and ending at (ctf ,xf ). There are many possible world-lines between the starting and ending points, as shown in Figure 5.1 (our use of just one spatial dimension is only for ease of representation). 108 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE We would like our action to assign a number to each of these world-lines, a number which all Lorentz observers agree on. Let P denote one world-line. What quantity related to P do all Lorentz observers agree on? The answer is simple: the elapsed proper time! All Lorentz observers agree on the amount of time that elapsed on a clock carried by the moving particle. So let’s take the action of the world-line P to be the proper time associated to it. To formulate this idea quantitatively, we recall that −ds2 = −c2dt2 +(dx1)2 +(dx2)2 +(dx3)2 , (5.1.4) where the infinitesimal proper time is equal to ds/c.Ofcourse, if we simply take the integral of the proper time, then we will not have the correct units for an action. Suppose instead that we integrate ds, which has the units of length. To get the units of action we need a multiplicative factor with units of mass times velocity. This factor should be Lorentz invariant, to preserve the Lorentz invariance of our partial guess ds.For the mass we can use m, the rest mass of the particle, and for velocity we can use c, the fundamental velocity in relativity. If we had instead used the velocity of the particle, the Lorentz invariance would have been spoiled. Therefore, our guess for the action is the integral of (mc ds). Of course, there is still the possibility that a dimensionless coefficient is missing. It turns out that there should be a minus sign, but the unit coefficient is correct. We therefore claim that S = −mc ds , (5.1.5) P is the correct action. This action is so simple that it may be baffling! It probably looks nothing like the actions you have seen before. We can make its content more familiar by choosing a particular Lorentz observer and ex- pressing the action as an integral over time. With the help of (5.1.4) we can relate ds to dt as v2 ds = cdt 1 − . (5.1.6) c2 This allows us to write the action in (5.1.5) as an integral over time: tf v2 S = −mc2 dt 1 − , (5.1.7) 2 ti c 5.1. ACTION FOR A RELATIVISTIC POINT PARTICLE 109 where ti and tf are the values of time at the initial and final points of the world-line P, respectively. From this version of the action, we see that the relativistic Lagrangian for the point particle is given by v2 L = −mc2 1 − . (5.1.8) c2 The Lagrangian makes no sense when |v| >csince it ceases to be real. The constraint of maximal velocity is therefore implemented. This could have been anticipated: proper time is only defined for motion where the velocity does not exceed the velocity of light. The paths shown in Figure 5.1 all rep- resent motion where the velocity of the particle never exceeds the velocity of light. Only for such paths the action is defined. At any point in any of those paths, the tangent vector to the path is a timelike vector. To show that this Lagrangian gives the familiar physics in the limit of small velocities, we expand the square root assuming |v| << c . Keeping just the first term in the expansion gives 1 v2 1 L −mc2 1 − = −mc2 + mv2 . (5.1.9) 2 c2 2 Constant terms in a Lagrangian do not affect the equations of motion, so when velocities are small the relativistic Lagrangian gives the same physics as the the non-relativistic Lagrangian in (5.1.2). This also confirms that we normalized our relativistic Lagrangian correctly. The canonical momentum is the derivative of the Lagrangian with respect to the velocity. Using (5.1.8) we find ∂L v 1 mv p = = −mc2 − = . (5.1.10) ∂v c2 1 − v2/c2 1 − v2/c2 This is just the relativistic momentum of the point particle.