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Chapter 5

The Relativistic

To formulate the of a system we can write either the of , or alternatively, an action. In the case of the relativistic point par- ticle, it is rather easy to write the . 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 , so the units of action are just the units of the Lagrangian multiplied by the units of time. The Lagrangian has units of , 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 of the : 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 of motion following by Hamilton’s principle is dv =0. (5.1.3) dt The free particle moves with constant 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 . 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 , it is convenient to represent the motion in . The path traced out in spacetime by the motion of a particle is called its -line.Inspacetime even a static particle traces a line, since time always flows. Akey physical requirement is that the action must yield Lorentz 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 : 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 . 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 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 with a series of world-lines connecting the origin to the spacetime point (ctf ,xf ).

We know that the action is a – it takes as input a 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 (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 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 ! 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 . To get the units of action we need a multiplicative factor with units of 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 , 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 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. What about the Hamiltonian? It is given by

mv2 H = p · v − L = + mc2 1 − v2/c2 1 − v2/c2 mc2 = . (5.1.11) 1 − v2/c2 110 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE

This coincides with the relativistic energy (2.4.2) of the point particle. We have therefore recovered the familiar physics of a relativistic particle from the rather remarkable action (5.1.5). This action is very elegant: it is briefly written in terms of the geometrical quantity ds,ithas a clear physical interpretation as total proper time, and it manifestly guarantees the Lorentz invariance of the physics it describes.

5.2 Reparametrization invariance

In this section we will explore an important property of the point particle action (5.1.5). This property is called reparametrization invariance. To eval- uate the integral in the action, an observer will find it useful to parametrize the particle world-line. Reparametrization invariance of the action means that the value of the action is independent of parametrization chosen to cal- culate it. It should be so, since the action is defined independently of any parametrization. We now investigate this point in detail and learn how to rewrite the action in such a way that this property is manifest.

Figure 5.2: Aworld-line fully parametrized by τ. All spacetime coordinates xµ are functions of τ.

To integrate ds,weparameterize the world-line P using a parameter τ, as shown in Figure 5.2. This parameter must be strictly increasing as the µ µ world-line goes from the initial point xi to the final point xf , but is otherwise arbitrary. As τ ranges in the interval [τi,τf ]itdescribes the motion of the 5.2. REPARAMETRIZATION INVARIANCE 111 particle. To have a parametrization of the world-line means that we have expressions for the coordinates xµ as functions of τ: xµ = xµ(τ) . (5.2.1) We also require µ µ µ µ xi = x (τi) ,xf = x (τf ) . (5.2.2) Note that even the time coordinate x0 is parameterized. Normally, we use time as a parameter and describe position as a of time. This is what we did in section 5.1. But if we want to treat and time coordinates on the same footing, we must parameterize both in terms of an additional parameter τ. We now reexpress the integrand ds using the parametrized world-line. To 2 µ ν this end, we use ds = −ηµνdx dx to write dxµ dxν ds2 = −η (dτ)2 . (5.2.3) µν dτ dτ Forany motion where the velocity does not exceed the velocity of light ds2 = (ds)2, and therefore the action (5.1.5) takes the form τf dxµ dxν S = −mc −ηµν dτ . (5.2.4) τi dτ dτ This is the explicit form of the action when the path has been specified by a parametrization xµ(τ) with parameter τ. We have already seen that the value of the action is the same for all Lorentz observers. We have now fixed an observer, who has calculated the action using some parameter τ.Does the value of the action depend on the choice of parameter? It does not. The observer can reparametrize the world-line, and the value of the action will be the same. Thus S is reparam- eterization invariant. To see this, suppose we change the parameter from τ to τ . Then, by the chain rule, dxµ dxµ dτ = . (5.2.5) dτ dτ dτ Substituting back into (5.2.4), we get τ τ  f dxµ dxν dτ f dxµ dxν − − − − S = mc ηµν dτ = mc ηµν dτ , dτ dτ dτ  dτ dτ τi τi (5.2.6) 112 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE which has the same form as (5.2.4), thus establishing the reparametrization invariance. Because the verification of this property is quite simple, we say that the action is manifestly reparameterization invariant.

5.3 Equation of motion

We now move on to the equations of motion. For this we must calculate the variation δS of the action (5.1.5), when the world-line of the particle is varied byasmall amount δxµ. The variation is simply given by δS = −mc δ(ds) . (5.3.1)

The variation of ds can be found from the simpler variation of ds2. Since 2 2 µ ν (ds) = ds = −ηµνdx dx ,wefind

2 µ ν µ ν δ(ds) =2ds δ(ds)=−ηµνδ(dx dx )=−2ηµνδ(dx )dx . (5.3.2)

The factor of two on the right-hand side arises because, by symmetry, the variations of dxν and dxν give the same result. Simplifying a little, dxν δ(ds)=−η δ(dxµ) . (5.3.3) µν ds We now wish to understand why

δ(dxµ)=d(δxµ) . (5.3.4)

Youmay already be familiar with this. A simpler version of this result states that the variation of a velocity is just the derivative with respect to time of the variation of the coordinate. Equation (5.3.4) becomes quite clear when we spell out the meaning of d.Forany τ-dependent quantity A(τ)wehave dA = A(τ + dτ) − A(τ). Using this on both sides of equation (5.3.4) we find

δ [xµ(τ + dτ) − xµ(τ)] = δxµ(τ + dτ) − δxµ(τ) . (5.3.5)

Since δ is linear, the two sides are indeed equal. We can understand the equality (5.3.4) a little more geometrically by referring to Figure 5.3. The original world-line of the particle is the lower curve, and the world-line after variation is the upper curve. Shown in the lower curve are the points xµ(τ) 5.3. EQUATION OF MOTION 113 and xµ(τ + dτ). By definition, they are separated by the infinitesimal vector dxµ. The variations δxµ(τ) and δxµ(τ + dτ) are also shown as vectors in the figure, with their ends defining the varied points. In the upper curve, the sep- aration between the varied points is the new dx: the old dx plus the variation δ(dx). The little quadrilateral on the left-side of the figure is shown to the right with labels a, a,b,b for the vectors that represent the corresponding sides. We see that δ(dxµ) corresponds to a − a.Onthe other hand d(δxµ) corresponds to b − b. The equality of these two quantities follows from the vector identity a + b = b + a.

Figure 5.3: Relating δ(dxµ) and d(δxµ)inthe variation of a world-line.

Using (5.3.4) back in (5.3.3), we write the final expression for δ(ds):

d(δxµ) dxµ δ(ds)=−η ds . (5.3.6) µν ds ds We can now go ahead and vary the action using (5.3.1): sf d(δxµ) dxν δS = mc ηµν ds . (5.3.7) si ds ds

We introduced here explicit limits to the integration: si and sf denote the values of the proper time parameter at the initial and final points of the world-line, respectively. To get an equation of motion we need to have δxµ multiplying an object under the integral – the equation of motion is then 114 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE simply the vanishing of that object. Since there are still derivatives acting on δxµ,wemust rewrite the integrand as a total derivative plus additional terms where δxµ appears multiplicatively : sf d dxν sf d2xν δS = mc ds η δxµ − ds δxµ mc η . (5.3.8) µν µν 2 si ds ds si ds The first integral just gives some expression evaluated at the boundaries of the world-line. But we fix the coordinates on the boundaries, so the first term vanishes. The variation of the action then reduces to sf d2xν δS = − ds δxµ(s) mc η . (5.3.9) µν 2 si ds Since δxµ(s)isarbitrary, everything multiplying it must vanish in order for the variation of the action to be zero. We thus find d2xν mc η =0. (5.3.10) µν ds2 The constants are nonvanishing and can be removed. Even the Minkowski ν metric can be removed: an equation of the form 0 = ηµνb , implies that ν ρµ ρµ ν b =0. Indeed, multiplying of the equation by η you find 0 = η ηµνb = ρ ν ρ δν b = b . Therefore, in its simplest form, the equation of motion is

d2xµ =0. (5.3.11) ds2 This is the equation of motion for a free particle. We obtained this equation by varying the relativistic action for the point particle using fully-relativistic notation. Equation (5.3.11) is formulated using the proper time parameter s. It tells you that the four-velocity uµ is constant along the world-line: d dxµ duµ dxµ =0 → =0,uµ = . (5.3.12) ds ds ds ds This means that if the path is marked by equal intervals of proper time, the change in xµ between any successive pair of marks is the same. It also follows from (5.3.12) that the relativistic momentum pµ = muµ is constant along the world-line: dpµ =0. (5.3.13) ds 5.4. RELATIVISTIC PARTICLE WITH ELECTRIC 115

What if you used some arbitrary parameter τ instead of s? Equation (5.3.11) would not hold with s replaced by τ. This is reasonable: changing arbitrarily the parameter is like changing arbitrarily the marks, and as a consequence, the change in xµ between any successive pair of new marks will not be the same. It is actually possible to write a slightly more complicated version of the equation of motion (5.3.11). That version uses an arbitrary parameter, and is manifestly reparametrization invariant (see Problem 5.2). Quick Calculation 5.2. Show that equation (5.3.13) implies that

dpµ =0, (5.3.14) dτ

dτ holds for an arbitrary parameter τ(s). Is there a constraint on ds ? Our goal in this section has been achieved: we have shown how to derive the physically expected equation of motion (5.3.11) starting from the Lorentz invariant action (5.1.5). As we explained earlier, the resulting equation of motion is guaranteed to be Lorentz invariant. Let us check this explicitly. Under a , the coordinates xµ transform as indi- µ µ ν µ cated in equation (2.2.32): x = L ν x , where the constants L ν can be viewed as the entries of an invertible L. Since ds is the same in all Lorentz frames, the equation of motion in primed coordinates is (5.3.11), with xµ replaced by xµ:

d2xµ d2 d2xν 0= = (Lµ xν)=Lµ . (5.3.15) ds2 ds2 ν ν ds2 Since the matrix L is invertible, the above equation implies equation (5.3.11). Namely, if the equation of motion holds in the primed coordinates, it holds in the unprimed coordinates as well. This is the Lorentz invariance of the equations of motion.

5.4 Relativistic particle with

The point particle we studied so far in this chapter is free and it moves with constant four-velocity or four-momentum. If a point particle is electrically charged, and there are nontrivial electromagnetic fields, the particle will ex- perience and the four-momentum will not be constant. You know, in fact, how the momentum of such particle varies in time. Its time derivative 116 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE is governed by the Lorentz equation (3.1.5). The relativistic version of the equation was given in Problem 3.1: dpµ q = F µνu . (5.4.1) ds c2 ν This is an intricate equation. In the spirit of our previous analysis we try to write an action that gives this equation upon variation. The action turns out to be remarkably simple.

Since the Maxwell field couples to the point particle along its world- line, we should add to the action (5.1.5) an integral over P representing the interaction of the particle with the electromagnetic field. The integral must be Lorentz invariant, and must involve the four-velocity of the particle. Since the four-velocity has one spacetime index, to obtain a Lorentz scalar we must multiply it against another object with one index. The natural candidate is the gauge potential Aµ.Iclaim that the interaction term in the action is q dxµ ds Aµ(x(s)) (s) . (5.4.2) c P ds Here q is the electric charge, and the integral is over the world-line P, parametrized with proper time. At each s, the four-velocity (dxµ/ds)is multiplied by the gauge potential Aµ evaluated at the position x(s)ofthe µ particle. The integrand can be written more briefly as Aµdx ,bycancelling the factors of ds.Inthis form, the interaction term is manifestly independent of parametrization. The world-line of the particle is a one-dimensional space, and the natural field that can couple to a particle in a Lorentz invariant way is a field with one index. This will have an interesting generalization when we consider the motion of strings. Since strings are one-dimensional, they trace out two-dimensional world-sheets in spacetime. We will see that they couple naturally to fields with two Lorentz indices! The full action for the electrically charged point particle is obtained by adding the term in (5.4.2) to (5.1.5): q µ S = −mc ds + Aµ(x)dx . (5.4.3) P c P

This Lorentz invariant action is simple and elegant. It is the correct action, and the equation of motion (5.4.1) arises by setting to zero the variation of 5.4. RELATIVISTIC PARTICLE WITH ELECTRIC CHARGE 117

S under a change δxµ of the particle world-line. I do not want to take away from you the satisfaction of deriving this important result. Therefore, I have left to Problem 5.5 the task of varying the action (5.4.3). 118 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE Problems

Problem 5.1. Point particle equation of motion and reparametrizations. If we parametrize the path of a point particle using proper time, the equation of motion is (5.3.11). Consider now a new parameter τ = f(s). What is the most general function f for which (5.3.11) implies d2xµ =0. dτ 2 Problem 5.2. Particle equation of motion with arbitrary parametrization. Vary the point particle action (5.2.4) to find a manifestly reparametriza- tion invariant form of the free particle equation of motion. Problem 5.3. Current of a charged point particle. Consider a point particle with charge q moving in a D = d+1 dimensional spacetime as described by functions xµ(τ)={x0(τ),x(τ)} where τ is param- eter. Recall that the electromagnetic current jµ is defined as jµ =(cρ,j) where ρ is the (charge per unit volume) and j is the (current per unit ). (a) Use functions to write expressions for j0(x, t) and ji(x, t) describ- ing the electromagnetic current associated to the point particle. (b) Show that the expressions you wrote in (a) arise from the following integral representation dxµ(τ) jµ(x, t)=qc dτ δd+1(x − x(τ)) (1) dτ Here δd+1(x) ≡ δ(x0)δ(x1) ···δ(xd). Problem 5.4. Hamiltonian for a non-relativistic charged particle. The action for a non-relativistic particle of mass m and charge q coupled to electromagnetic fields is obtained by replacing the first term in (5.4.3) by the non-relativistic action for a free point particle: 1 q dxµ S = mv2 dt + A (x) dt . (1) 2 c µ dt We have also chosen to use time to parametrize the integral. Problems for Chapter 5 119

(a) Rewrite the above action in term of the potentials (Φ, A) and the or- dinary velocity v. What is the Lagrangian?

(b) Calculate the canonical momentum p conjugate to the position of the particle and show that it is given by q p = mv + A. (2) c

(c) Construct the Hamiltonian for the charged point particle and show it is given by 1 q 2 H = p − A + q Φ . (3) 2m c Problem 5.5. Equations of motion for a charged point particle. Consider the variation of the action (5.4.3) under a variation δxµ(x)ofthe particle . The variation of the first term in the action was obtained in section 5.3. Vary the second term (written more explicitly in (5.4.2)) and show that the equation of motion for the point particle in the presence of an electromagnetic field is (5.4.1). Begin your calculation by explaining why ∂A δA (x(s)) = µ (x(s)) δxν(s) . µ ∂xν Problem 5.6. Electromagnetic field dynamics with charged particle. The action for the dynamics of both acharged point particle and the EM field is given by q µ 1 d+1 µν S = −mc ds + Aµ(x)dx − d x FµνF . P c P 16πc Here dd+1x = dx0dx1 ···dxd. Notice that the total action S is a hybrid. The last term is an integral over spacetime and the first two terms are over the particle world-line. While included for completeness, the first term will play no role here. Vary the action S under a fluctuation δAµ(x) and obtain the equation of motion for the electromagnetic field in the presence of the charged point particle. The answer should be equation (3.3.23), where the current is the one calculated in Problem 5.3. [ Hint: to vary Aµ(x)in the world-line action it is useful to first turn this term into a full spacetime integral with the help of delta functions]. 120 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE

Problem 5.7. Point particle action in

2 µ ν In section 3.6 we considered the invariant interval ds = −gµν(x)dx dx in curved-space. The motion of a point particle of mass m on curved space is studied using the action S = −mc ds .

Show that the equation of motion takes the form

d2xµ dxα dxβ +Γµ =0, (1) ds2 αβ ds ds where 1 ∂g ∂g ∂g Γµ = gµλ λα + λβ − αβ . αβ 2 ∂xβ ∂xα ∂xλ µ µ The Christoffel coefficients Γ are symmetric in the lower indices Γαβ =Γβα, and are built from the metric and its first derivatives. When the metric is constant the Christoffel coefficients vanish and we recover the familiar equation of motion of the free point particle in . Equation (1) is called the equation. It is a rather nontrivial differential equation that follows from the very simple action S.