Relativistic Mechanics

Transcription

1 TAYL I 12/9/02 2:53 PM Page 46 C h a p t e r2 Relativistic Mechanics 2.1 Introduction 2.2 Mass in Relativity 2.3 Relativistic Momentum 2.4 Relativistic Energy 2.5 Two Useful Relations 2.6 Conversion of Mass to Energy 2.7 Force in Relativity 2.8 Massless Particles 2.9 When Is Nonrelativistic Mechanics Good Enough? 2.10 General Relativity 2.11 The Global Positioning System: An Application of Relativity Problems for Chapter 2 These sections can be omitted without serious loss of continuity. 2.1 Introduction We have seen that the laws of classical mechanics hold in a family of inertial frames that are related to one another by the classical, Galilean transformation; in other words, classical mechanics is invariant under the Galilean transformation. But we now know that the correct, relativistic transformation between inertial frames is the Lorentz, not the Galilean, transformation. It follows that the laws of classical mechanics cannot be correct and that we must find a new, relativistic mechanics that is invariant as we pass from one inertial frame to another, using the correct Lorentz transformation. We will find that relativistic mechanics, like classical mechanics, is built around the concepts of mass, momentum, energy, and force. However, the relativistic definitions of these concepts are all a little different from their classical coounterparts. In seeking these new definitions and the laws that connect them, we must be guided by three principles: First, a correct relativistic law must be valid in all inertial frames; it must be invariant under the Lorentz transformation. Second, we would expect the relativistic definitions and laws to reduce to their nonrelativistic counterparts when applied to systems moving much slower than the speed of light. Third, and most important, our relativistic laws must agree with experiment. 2.2 Mass in Relativity 46 We start by considering the mass m of an object an electron, a space rocket, or a star. The most satisfactory definition of m turns out to be remarkably simple. We know that at slow speeds a suitable definition is the classical one (for

2 TAYL I 12/9/02 2:53 PM Page 47 example, m = F>a, where a is the acceleration produced by a standard force F). In relativity we simply agree to use the same, classical definition of m with the proviso that before measuring any mass we bring the object concerned to rest. To emphasize this qualification, we will sometimes refer to m as the rest mass. It can also be called the proper mass since it is the mass measured in the frame where the object is at rest. Observers in different inertial frames all agree on the rest mass of an object. Suppose that observers in a frame S take some object, bring it to rest (in S), and measure its mass m. If they then pass the object to observers in a different frame S, who bring it to rest in S and measure its mass m, it will be found that m =m. That is, rest mass is invariant as we pass from S to S. In fact, this is required by the postulates of relativity: If m were different from m, we could define a preferred frame (the frame where the rest mass of an object was minimum, for example). As we will describe shortly, some physicists use a different definition of mass, called the variable mass. We will not use this concept, and whenever we use the word mass without qualification, we will mean the invariant rest mass defined here. Section 2.3 Relativistic Momentum Relativistic Momentum The classical definition of the momentum of a single body is p = mu (2.1) where m and u are the body s mass and velocity. Since we know how both m and u are measured in relativity, it is natural to start by asking the question: Is it perhaps the case that the classical definition (2.1) is the correct definition in relativity as well? Strictly speaking, this question has no answer. There can be no such thing as a correct or incorrect definition of p since one is at liberty to define things however one pleases. The proper question is rather: Is the definition p = mu a useful definition in relativity? In classical mechanics the concept of momentum has many uses, but its single most useful property is probably the law of conservation of momentum. If we consider n bodies with momenta p 1, Á, p n then, in the absence of external forces, the total momentum a p = p 1 + Á + p n can never change. It would certainly be useful if we could find a definition of momentum such that this important law carried over into relativity. Accordingly, we will seek a definition of relativistic momentum p with the following two properties: 1. p L mu when u V c 2. The total momentum a p of an isolated system is conserved, as measured in every inertial frame.

3 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics 2 2 y (a) x Frame S y (b) 1 x Frame S FIGURE 2.1 Two different views of a collision between two identical balls. (a) In frame S the velocities of the two balls are equal and opposite before and after the impact. (b) The same experiment as seen in frame S, which travels along the x axis at the same rate as ball 1. In this frame ball 1 travels straight up the y axis and back down again. Requirement (1) is just that the relativistic definition should agree with the nonrelativistic one in the nonrelativistic domain. Requirement (2) is that the law of conservation of momentum must hold in all inertial frames if it holds at all. If we were to adopt the classical definition p = mu, requirement (1) would be satisfied automatically. However, it is fairly easy to construct a thought experiment which illustrates that the classical definition p = mu does not meet requirement (2). Consider two identical billiard balls that collide as shown in Fig Relative to the frame S of Fig. 2.1(a), the initial velocities of the two balls are equal and opposite. Further, the collision is arranged symmetrically, such that the x components of the velocities are unchanged by the impact, whereas the y components reverse. It is an experimental fact that such collisions between particles of equal mass do occur. It is easy to check that if we adopt the classical definition, p = mu, then as seen in S this collision conserves momentum, The proof that this is so is shown in Table 2.1, where we have denoted by a and b the x and y components of the initial velocity of ball 1. Notice that, as seen in S, the total classical momentum is actually zero before and after the collision; so it is certainly conserved. TABLE 2.1 a mu(before) = a mu(after) The experiment of Fig. 2.1 as seen in S. Each pair of numbers represents the x and y components of the vector indicated. First Ball 1u 1 2 Second Ball 1u 2 2 mu 1 mu 2 Before: 1a, b2 1-a, -b2 10, 02 After: 1a, -b2 1-a, b2 10, 02 Let us now consider the same collision from a second frame S traveling in the positive x direction of S, at the same rate as ball 1 (that is, the speed v of S relative to S is equal to the x component of the velocity of ball 1). Figure 2.1(b) shows the collision as seen in S, with ball 1 traveling straight up the y axis and bouncing straight down again. If the frames S and S were related by the classical, Galilean transformation (which we know is actually incorrect), the velocities measured in S could be found from those in Table 2.1 using the classical velocity-addition formula, with the results shown in Table 2.2. From the last column, we see that the total classical momentum has the same values before and after impact; that is, the total classical momentum is TABLE 2.2 The experiment of Fig. 2.1 as seen in frame S, assuming that S and S are related by the Galilean transformation. Each velocity here is obtained from that of Table 2.1 by subtracting a from its x component. First Ball 1u œ 1 2 Second Ball 1u œ 2 2 œ mu 1 œ mu 2 Before: 10, b2 1-2a, -b2 1-2ma, 02 After: 10, -b2 1-2a, b2 1-2ma, 02

4 TAYL I 12/9/02 2:53 PM Page 49 conserved in the frame S also. This is just what we should have expected since we know that the laws of classical mechanics are invariant under the Galilean transformation. In fact, however, the Galilean transformation is not the correct transformation between frames S and S, and we must compute the velocities in S using the relativistic transformations (1.43) and (1.44). The results of this rather tedious calculation (Problem 2.5) are shown in Table 2.3. The details of this table are not especially interesting, but there are two important points: First, because the transformation of u y depends on u x, the y components of the two balls velocities transform differently, and, as seen in S, they are no longer equal in magnitude. (Compare the y components in the first and second columns.) Consequently, the y component of the total classical momentum (final column of Table 2.3) is positive before the collision and negative after. Thus, even though the total classical momentum is conserved in frame S, it is not conserved in frame S. Therefore, the law of conservation of classical momentum (defined as p = mu for each body) is incompatible with the postulates of relativity, and the classical definition of momentum does not satisfy our requirement (2) listed above. TABLE 2.3 The experiment of Fig. 2.1 as seen in S, based on the correct relativistic velocity transformation from S to S. The relative velocity of the two frames is v = a; accordingly, b denotes a>c and g = 11 - a 2 >c 2 2-1>2. Section 2.3 Relativistic Momentum 49 First Ball 1u œ 1 2 Second Ball 1u œ 2 2 mu 1 œ mu 2 œ Before: 0, b g11 - b 2 2-2a 1 + b 2, -b g11 + b 2 2-2ma 1 + b 2, 2mbb 2 g11 - b 4 2 After: 0, -b g11 - b 2 2-2a 1 + b 2, b g11 + b 2 2-2ma 1 + b 2, -2mbb2 g(1 - b 4 ) If there is a law of momentum conservation in relativity, the relativistic definition of momentum must be different from the classical one, p = mu. If we rewrite this classical definition as dr p = m dt [classical] (2.2) we get a useful clue for a better definition. The difficulty in our thought experiment originated in the complicated transformation of the velocity dr>dt (particularly the y component). These complications arose because both dr and dt change as we transform from S to S. We can avoid some of this problem if we replace the derivative dr>dt in (2.2) by the derivative dr>dt 0 with respect to the proper time, t 0, of the moving body: dr p = m dt 0 [relativistic] (2.3) In (2.2), dr is the vector joining two neighboring points on the body s path, and dt is the time for the body to move between them both as measured in any one inertial frame S. In (2.3) dr is the same as before, but dt 0 is the proper time

5 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics between the two points; that is, the time as measured in the body s rest frame. From its definition, the proper time dt 0 (just like the proper mass m) has the same value for all observers in all frames. Thus the vector defined in (2.3) transforms more simply than the classical momentum (2.2) since only the numerator dr changes as we move from one frame to another. In particular, the y component, p y = m dy>dt 0, does not change at all as we pass from S to S, and the difficulty encountered in our thought experiment would not occur if we were to adopt the definition (2.3). (For details, see Problem 2.6.) At slow speeds, dt and dt 0 are indistinguishable and the new definition (2.3) agrees with the classical one; that is, the definition (2.3) meets requirement (1). Further, one can show that it always meets requirement (2); specifically, if the total momentum a p, as defined by (2.3), is constant in one inertial frame S, the same is true in all inertial frames.* Since the proof is rather long, although reasonably straightforward, we leave it as a problem (Problem 2.14) at the end of this chapter. With the definition (2.3) of momentum, the law of conservation of momentum would be logically consistent with the postulates of relativity. Whether momentum defined in this way is conserved must be decided by experiment. The unanimous verdict of innumerable experiments involving collisions of atomic and subatomic particles is that it is: If we adopt the definition(2.3) for the momentum of a body, the total momentum a p of an isolated system is conserved. Under the circumstances, we naturally adopt (2.3) as our definition of momentum. It is convenient to express the definition (2.3) a little differently: The time-dilation formula implies that dt = g dt where g = 11 - u 2 >c 2 2-1>2 0,. Therefore, (2.3) is the same as dr dr p = m = mg dt 0 dt = gmu Thus we adopt as the final form of our definition: The momentum of a single body of mass m and velocity u is p = mu 31 - u 2 >c 2 = gmu (2.4) An important consequence of the factor g in the momentum (2.4) is that no object can be accelerated past the speed of light: We will find that in relativity, just as in classical mechanics, a constant force on a body increases its momentum p at a constant rate; if the force acts for long enough, we can make p as large as we please. In classical mechanics, where p = mu, this means that a constant force steadily increases u and can eventually make u as large as we please. In relativity, an increase in p = gmu is reflected by increases in u and g. Now, as u : c we know that g increases without limit. Thus, as u : c the * The proposition as stated is a little oversimplified. What one can actually prove is this: If both momentum, as defined by (2.3), and energy (whose definition we give in Section 2.4) are conserved in one inertial frame, they are both automatically conserved in all inertial frames.

6 TAYL I 12/9/02 2:53 PM Page 51 Section 2.4 Relativistic Energy 51 constant force keeps increasing g without u ever reaching c. This difference between classical and relativistic mechanics is illustrated in Fig. 2.2, which shows a plot of u against p for both cases. Example 2.1 A 1-kg lump of metal is observed traveling with speed 0.4c. What is its momentum? What would its momentum be if we doubled its speed? Compare with the corresponding classical values. When b = 0.4, the factor g = 11 - b 2 2-1>2 is easily calculated to be g = 1.09, and p = gmu = 1.09 * 11 kg2 * 10.4 * 3 * 10 8 m>s2 = 1.31 * 10 8 kg # m>s (2.5) If we double the speed, b = 0.8 and g = Thus the momentum becomes p = gmu = 1.67 * 11 kg2 * 10.8 * 3 * 10 8 m>s2 = 4.01 * 10 8 kg # m>s (2.6) c O u Classical Relativistic FIGURE 2.2 The speed of a body as a function of its momentum in classical and relativistic mechanics. At low speeds the two curves merge. In classical mechanics u grows indefinitely as p increases. In relativity u never exceeds c, however large p becomes; instead, u is asymptotic to c as p : q. p which is more than three times the previous answer. The classical answers are found by omitting the factors of g: If b = 0.4, then p = 1.20 * 10 8 kg # m>s, just a little less than the correct answer (2.5); if b = 0.8, then p = 2.40 * 10 8 kg # m>s, significantly less than the correct answer (2.6). Some physicists like to think of the relativistic momentum as the product of gm and u, which they write as p = m var u (2.7) where m var = gm = m 31 - u 2 >c 2. (2.8) m var The quantity is called the variable mass since, unlike the rest mass m, it varies with the body s speed u.the form (2.7) has the advantage of making the relativistic momentum look more like its nonrelativistic counterpart p = mu. On the other hand, it is not always a good idea to give two ideas the appearance of similarity when they are in truth different. Further, the introduction of the variable mass does not achieve a complete parallel with classical mechanics. For example, we will see that the quantity 1 2 m var u 2 is not the correct expression for the relativistic kinetic energy and the equation F = m var a is not the correct relativistic form of Newton s second law (Problems 2.13 and 2.35). For these reasons, we will not use the notion of variable mass in this book. 2.4 Relativistic Energy Having found a suitable relativistic definition for momentum p, our next task is to do the same for the energy E of a body. Just as with momentum, one is in principle free to define E however one pleases. But the hope of finding a useful definition suggests two requirements analogous to those used for momentum:

7 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics 1. When applied to slowly moving bodies, the new definition of E should reproduce as closely as possible the classical definition. 2. The total energy a E of an isolated system of bodies should be conserved in all inertial frames. The definition that fits these requirements turns out to be this: The energy of a single body of mass m, moving with speed u,is E = mc u 2 >c 2 = gmc 2 (2.9) It is important to note that this applies to any single body an elementary particle, like an electron; an assembly of particles, like an atom; or an assembly of atoms, like a baseball, a space rocket, or a star. Although we will not do so here (but see Problem 2.14), one can prove that with the definition (2.9), a law of conservation of energy would be logically consistent with the postulates of relativity: If a E were constant as measured in one inertial frame, the same would be true in all inertial frames. Furthermore, experiment shows that the quantity a E is conserved for any isolated system. Thus the definition (2.9) meets requirement (2). Just because the quantity a E is conserved, we are not yet justified in giving E, as defined by (2.9), the name energy. The main reason for doing so will emerge when we establish the connection of (2.9) with the classical definition of energy, that is, when we check requirement (1). Before we do so, we mention two other important points. First, since g is dimensionless and mc 2 has the dimensions of energy, our definition E = gmc 2 at least has the correct dimensions for an energy. Second, although we have not yet defined the concept of force in relativity, when we do so in Section 2.7, we will prove the following important theorem: If a total force F acts on a body as it moves through a small displacement dr, the resulting change in the energy E, as defined by (2.9), is de = F # dr (2.10) You should recognize the product on the right as the work done by the force F, and the equation (2.10) as the work-energy theorem: The change in a body s energy is the work done on it.the fact that this theorem applies to our new definition of E is strong reason for regarding E as the relativistic generalization of the classical notion of energy. Let us now evaluate the relativistic energy (2.9) for a slowly moving body. With u V c, we can use the binomial approximation to write the factor g as g = 1 - u2 c 2-1>2 L u 2 c 2 Therefore, u 2 E = gmc 2 L c 2 mc2 = mc mu2 [when u V c] (2.11)

8 TAYL I 12/9/02 2:53 PM Page 53 We see that for a slowly moving body, the relativistic energy is the sum of two 1 terms: a constant term mc 2 that is independent of u and a second term 2 mu2 that is precisely the classical kinetic energy of a body of mass m and speed u. In classical physics it was believed that mass was always conserved, and the term mc 2 in (2.11) would therefore have been an immutable constant. Further, you will recall that one was always at liberty to add or subtract an overall constant from the energy since the zero of energy was arbitrary. Thus, in the classical context, (2.11) implies that when u V c the relativistic energy E of a body is just the classical kinetic energy plus an irrelevant constant mc 2. Therefore, the relativistic definition (2.9) meets both our requirements (1) and (2), and our identification of (2.9) as the appropriate generalization of the classical notion of energy is complete. We will find that in relativity the irrelevant constant mc 2 in (2.11) is actually extremely important. The reason is that the classical law of conservation of mass turns out to be wrong. This law was based mainly on nineteenthcentury measurements of masses in chemical reactions, where no change of mass was ever detected. In this century, however, nuclear processes have been discovered in which large changes of mass occur, and even where the rest mass of certain particles disappears entirely. Further, we now know that even in chemical reactions the total rest mass of the participating atoms and molecules does change, although the changes are much too small (1 part in 10 9 or so) to be detected directly. Given that rest masses can change, it should be clear that the term mc 2 in (2.11) is important. In Section 2.6 we describe some processes in which the rest mass of a system does change, and will see just how important the term mc 2 is. In the remainder of this section we give two more definitions connected with the relativistic energy (2.9) and describe an application of the laws of energy and momentum conservation to processes in which rest masses do not change the so-called elastic processes. It is clear from either the exact equation E = gmc 2 or the nonrelativistic approximation (2.11) that even when a body is at rest, its energy is not zero but is given, instead, by the famous equation Section 2.4 Relativistic Energy 53 E = mc 2 [when u = 0] (2.12) This energy is called the rest energy of the mass m, and we will see in Section 2.6 how it can be converted into other forms, such as the kinetic energy of other bodies. Example 2.2 What is the rest energy of a 1-kg lump of metal? Substituting into (2.12), we find that E = mc 2 = 11 kg2 * 13 * 10 8 m>s2 2 = 9 * joules about the energy generated by a large power plant in one year. This incredible amount of energy would be of no interest if it could not be converted into other forms of energy. In fact, however, such conversion is possible. For example, if the metal is uranium 235, about 1 part in 1000 of the rest energy can be converted into heat by the process of nuclear fission. Thus 1 kg of 235 U can yield a fantastic 9 * joules of heat.

9 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics When a body is not at rest, we can think of its total energy E = gmc 2 as the sum of its rest energy mc 2 plus the additional energy 1E - mc 2 2 that it has by virtue of its motion.this second term we naturally call the kinetic energy K, and we write where E = mc 2 + K K = E - mc 2 = 1g - 12mc 2 (2.13) At slow speeds we have seen that K L 1 2 mu2, but in general the relativistic kinetic energy is different from 1 2 mu2. In particular, as u : c, the kinetic energy approaches infinity. However much energy we give a body, its speed can therefore never reach the speed of light a conclusion we reached before by considering the relativistic momentum. Notice that, since g Ú 1, the relativistic 1 kinetic energy is always positive (like its nonrelativistic counterpart 2 mu2 ). In classical mechanics a surprising number of interesting problems can be solved using just the laws of energy and momentum conservation. In relativity also, there are many such problems, and we conclude this section with an example of one. Example 2.3 Two particles with rest masses m 1 and m 2 collide head-on, as shown in Fig Particle 1 has initial velocity u 1, while particle 2 is at rest 1u 2 = 02. Assuming that the collision is elastic (that is, the rest masses are unchanged), use conservation of energy and momentum to find the velocity u 3 of particle 1 after the collision.apply the result to the case that a pion (a subatomic particle with mass m 1 = 2.49 * kg) traveling at 0.9c makes an elastic head-on collision with a stationary proton 1m 2 = 1.67 * kg2. Since the solution of this problem is very similar to that of the corresponding nonrelativistic problem, let us first review the latter. Conservation of energy implies that E 1 + E 2 = E 3 + E 4 If the rest masses of the two particles are unchanged, this can be rewritten as 1m 1 c 2 + K m 2 c 2 + K 2 2 = 1m 1 c 2 + K m 2 c 2 + K 4 2 and, canceling the mass terms, we see that kinetic energy is conserved, K 1 + K 2 = K 3 + K 4 which is the usual definition of an elastic collision in classical mechanics. The conservation of momentum and kinetic energy imply (in nonrelativistic mechanics) m 1 u 1 = m 1 u 3 + m 2 u 4 (2.14) FIGURE 2.3 An elastic, head-on collision. Before After u 1 u 3 u 4 u 2 0

10 TAYL I 12/9/02 2:53 PM Page 55 Section 2.4 Relativistic Energy 55 and 1 2 m 1 u 1 2 = 1 2 m 1 u m 2 u 4 2. (2.15) These two equations can be solved for the two unknowns u 3 and u 4. Since one of the equations is quadratic, we get two solutions. The first solution, u 3 = u 1 and u 4 = 0, gives the initial velocities before the collision occurred. The second solution is the interesting one and gives (Problem 2.11) u 3 = m 1 - m 2 m 1 + m 2 u 1 (2.16) and u 4 = 2m 1 m 1 + m 2 u 1 (2.17) Several features of these answers deserve comment. If m 1 7 m 2, u 3 is positive and particle 1 continues in its original direction; if m 1 6 m 2, u 3 is negative and particle 1 bounces back in the opposite direction. If m 1 = m 2, u 3 = 0 and u 4 = u 1 ; that is, particle 1 comes to a dead stop, giving all of its momentum and kinetic energy to particle 2. If m 1 V m 2 (1 is much lighter than 2), u 3 L -u 1 and particle 1 bounces back off the much heavier target with its speed barely changed. The solution of the corresponding relativistic problem is very similar, although considerably messier because of the square roots involved in the factors g. Because there are several different velocities in the problem, we must be careful to distinguish the corresponding factors of g. We therefore write g 1 for 11 - u 2 and g for 11 - u 2 2 >c 2 2-1>2 1 >c 2 2-1>2 2 (which is equal to 1 in this problem), and so on. The conservation of relativistic momentum implies that g 1 m 1 u 1 = g 3 m 1 u 3 + g 4 m 2 u 4 while conservation of relativistic energy implies that g 1 m 1 c 2 + m 2 c 2 = g 3 m 1 c 2 + g 4 m 2 c 2 (2.18) These are two equations for the two unknowns, u 3 and u 4. If, for example, we eliminate u 4, some fairly messy algebra leads us to a quadratic equation for u 3. One solution of this equation is u 3 = u 1 (the original velocity before the collision) and the other is m m 2 u 3 = u 1 (2.19) m m m 1 m u 2 1 >c 2 The answer (2.19) has much in common with the nonrelativistic (2.16). When m 1 7 m 2, particle 1 continues in its original direction ( u 3 positive); if m 1 6 m 2, particle 1 bounces back ( u 3 negative). If m 1 = m 2, particle 1 comes to a dead stop. If particle 1 is moving nonrelativistically, the square root in the denominator can be replaced by 1 to give u 3 L m m 2 2 1m 1 + m u 1 = m 1 - m 2 m 1 + m 2 u 1 which is precisely the nonrelativistic answer (2.16).

11 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics For the case of a pion, with u 1 = 0.9c, colliding with a stationary proton we can substitute the given numbers into (2.19) to give u 3 = -0.76c [relativistic] (Notice that this is negative, indicating that the light pion bounces backwards.) If we were to put the same numbers into the nonrelativistic result (2.16), we would find that u 3 = -0.62c [nonrelativistic] The difference between these two answers is large enough to be easily detected, and it is the relativistic answer that proves to be correct. More generally, in all collisions involving atomic and subatomic particles, one finds perfect agreement between the experimental observations and the predictions based on conservation of relativistic energy and momentum. This is, in fact, the principal evidence that these quantities are conserved. 2.5 Two Useful Relations We have introduced four parameters, m, u, p, and E, that characterize the motion of a body. Only two of these are independent since p and E were defined in terms of m and u as p = mu 31 - u 2 >c 2 (2.20) and E = mc u 2 >c 2 (2.21) We can, of course, rearrange these definitions to give an expression for any one of our parameters in terms of any two others. There are two such expressions that are especially useful, as we discuss now. First, dividing (2.20) by (2.21), we find that or p E = u c 2 B K u c = pc E (2.22) which gives the dimensionless velocity B = u>c in terms of p and E. (Since pc has the dimensions of energy, the right side is dimensionless, as it must be.)

12 TAYL I 12/9/02 2:53 PM Page 57 Section 2.5 Two Useful Relations 57 Second, by squaring both (2.20) and (2.21) it is easy to verify that (Problem 2.16) E pc E 2 = 1pc mc (2.23) This useful expression for E in terms of p and m shows that the three quantities E, pc, and mc 2 are related like the sides of a right-angled triangle, with E as the hypotenuse, as illustrated in Fig [At this stage there is no deep geometrical significance to this statement; we mention it only as a convenient way to remember the relation (2.23).] In our applications of relativistic mechanics, we will frequently use both the result (2.22) and the Pythagorean relation (2.23). Now is a convenient point to mention some of the units used to measure the parameters u, E, m, and p. First, as we have seen repeatedly, relativistic velocities are best expressed as fractions of c or by using the dimensionless b = u>c. For energy, the SI unit is the joule (J). However, most of our applications of relativity will be in atomic and subatomic physics, where the joule is an inconveniently large unit of energy, and a much more popular unit is the electron volt or ev. This is defined as the work needed to move an electron (of charge q = -e = * coulomb) through a voltage drop of 1 volt ( V = -1 volt); thus 1 ev = q V = * coulomb2 * 1-1 volt2 = 1.60 * J. (2.24) mc 2 (a) E mc 2 K E mc 2 K mc 2 mc 2 (b) K K pc pc We shall find that typical energies in atomic physics are of the order of 1 ev or so; those in nuclear physics are of order 10 6 ev, or 1 MeV. Like the joule, the kilogram is inconveniently large as a unit for atomic and subatomic physics. For example, the mass of the electron is 9.11 * kg. In fact, in most applications of relativity one is not so much concerned with the mass m as with the corresponding rest energy mc 2. [For instance, in (2.23) it is mc 2, rather than m itself, that appears.] Thus in many relativistic problems, masses are given implicitly by stating the rest energy mc 2, usually in ev. Another way to say this is that mass is measured in units of ev>c 2, as we discuss in the following example. Example 2.4 Given that the electron has mass m = * kg, what is its rest energy in ev, and what is its mass in ev>c 2? The rest energy is or* mc 2 = * kg2 * * 10 8 m>s2 2 = * J2 * = 5.11 * 10 5 ev 1 ev * J mc 2 (c) mc 2 FIGURE 2.4 (a) The Pythagorean relation (2.23) means that the three variables E, pc, and mc 2 form the sides of a right triangle, with E as the hypotenuse. (b) Since E = mc 2 + K, the hypotenuse can be divided as shown. When pc is much smaller than mc 2, the energy is mostly rest energy. (c) When pc is larger than mc 2, the energy is mostly kinetic. mc 2 = MeV (2.25) *Note that to get this answer, which is correct to three significant figures, we used four significant figures in all input numbers. These were taken from Appendix A, which lists the best known values of the fundamental constants.

13 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics Thus a convenient way to specify (and remember) the electron s mass is to say that its rest energy mc 2 is roughly half an MeV. An alternative way to put this is to divide both sides of (2.25) by c 2 and say that m = MeV>c 2 If you have not met ev>c 2 before as a unit of mass, it will probably seem a bit odd at first. The important thing to remember is that the statement m = 0.5 MeV>c 2 is precisely equivalent to the more transparent mc 2 = 0.5 MeV. In most applications of relativity we will be less interested in the momentum p than in the product pc. [For example, both of the relations (2.22) and (2.23) involve pc rather than p.] The quantity pc has the dimensions of energy and so is often measured in ev or MeV. This is equivalent to measuring p itself in ev>c or MeV>c. Example 2.5 An electron (rest mass about 0.5 MeV>c 2 ) is moving with total energy E = 1.3 MeV. Find its momentum (in MeV>c. and in SI units) and its speed. Given the energy and mass, we can immediately find the momentum from the Pythagorean relation (2.23) or pc = 3E 2-1mc = MeV MeV2 2 = 1.2 MeV p = 1.2 MeV>c (Notice how easily the units work out when we measure E in MeV, p in MeV>c, and m in MeV>c 2.) This is easily converted into SI units if necessary: The required conversion is MeV 1 c = * J * 10 8 m>s = 5.34 * kg # m>s (2.26) (This and several other conversion factors are listed inside the front cover; more exact values are given in Appendix A.) Therefore, p = 11.2 MeV>c2 * 5.3 * kg # m>s 1 MeV>c L 6 * kg # m>s. Once we know the energy and momentum, the speed follows immediately from (2.22). b = pc E = 1.2 MeV 1.3 MeV = 0.92 or u L 0.9 c

14 TAYL I 12/9/02 2:53 PM Page 59 Section 2.6 Conversion of Mass to Energy Conversion of Mass to Energy According to the relativistic definition of energy, even a mass at rest has an energy, equal to mc 2. As we have emphasized, this statement is meaningless unless there is some way in which the supposed rest energy mc 2 can be converted, or at least partly converted, into other more familiar forms of energy, such as kinetic energy a process that is often loosely called conversion of mass to energy. Such conversion would require the classical law of conservation of mass to be violated. In this section we argue that if the relativistic mechanics we have developed is correct, the nonconservation of mass is logically necessary; and we describe some processes in which mass is not conserved. Let us consider two bodies that can come together to form a single composite body.the two bodies could be two atoms that can bind together to form a molecule, or two atomic nuclei that can fuse to form a larger nucleus. Because it is probably easier to think about everyday objects, we will imagine two macroscopic (that is, nonmicroscopic) blocks. When the bodies are far apart, we take for granted that the forces between them are negligible. But when they are close together, there will be forces, and we distinguish two cases: If the forces are predominantly repulsive, we will have to supply energy to push the two bodies together; if the forces are predominantly attractive, energy will be released as they move together. As a model of the repulsive case, we imagine two blocks with a compressible spring attached to one of them, as shown in Fig. 2.5(a). To hold the blocks together, we attach a pivoted catch to the second block, as shown. Suppose now that we push the two blocks together until the catch closes, as in Fig. 2.5(b).This will require us to do work, which is stored by the system as potential energy of the spring. Because of this potential energy, the bound state of the system is unstable: If we release the catch, the blocks will fly apart and the stored energy will reappear as kinetic energy, as shown in Fig. 2.5(c).What we will now argue is that the potential energy stored in the bound system manifests itself as an increase in the mass of the system; that is, the rest mass M of the unstable bound state is greater than the sum of the separate rest masses m 1 and m 2 of the blocks when far apart. (In classical mechanics, M = m 1 + m 2 of course.) 1 2 (a) 1 2 Rest mass M (b) 1 2 m 1 m 2 (c) FIGURE 2.5 (a) Two blocks repel one another at close range because of the spring attached to block 1. (b) If the blocks are pushed together, the catch on block 2 can hold them in a bound state of rest mass M. (c) The bound state is unstable in the sense that the blocks fly apart when the catch is released.

15 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics As long as the two blocks remain together, we can treat them as a single body, whose rest mass we denote by M. If the body is at rest, then according to the definition (2.9) of relativistic energy its total energy is total energy = Mc 2 (2.27) [Notice that we are assuming that the definition (2.9) can be applied to a composite system like our two blocks locked together; in the final analysis this assumption must be and is justified by experiment.] If we now gently release the catch, the two blocks will fly apart. Once they are well separated, we can treat them as two separate bodies with rest masses and and total energy. m 1 m 2 total energy = E 1 + E 2 = K 1 + m 1 c 2 + K 2 + m 2 c 2 (2.28) By conservation of energy, (2.27) and (2.28) must be equal. Therefore, Mc 2 = 1K 1 + K m 1 + m 2 2c 2 (2.29) K 1 K 2 Since both and are greater than zero, we are forced to the conclusion that M is greater than the sum of the separate masses, m 1 + m 2. This conclusion is independent of the mechanical details of our example and applies to any unstable system of rest mass M that can fly apart into two (or more) pieces. We can solve (2.29) to give the amount by which the rest mass decreases as the blocks fly apart. If we denote this by then (2.29) implies that M = M - 1m 1 + m 2 2 Mc 2 = K 1 + K 2 = energy released as bodies fly apart (2.30) We can say that a mass M has been converted into kinetic energy, the rate of exchange between mass and energy being given by the familiar energy = mass * c 2. The kinetic energy released as the bodies move apart is the same as the work done to bring them together in the first place. Thus we can rephrase our result to say that when we push our two blocks together, the work done results in an increase in rest mass, M, given by Mc 2 = work done to push bodies together (2.31) Example 2.6 The nuclei of certain atoms are naturally unstable, or radioactive, and spontaneously fly apart, tearing the whole atom into two pieces. For example, the atom called thorium 232 splits spontaneously into two offspring atoms, radium 228 and helium 4, 232 Th 228 Ra + 4 He (2.32)

16 TAYL I 12/9/02 2:53 PM Page 61 The combined kinetic energy of the two offspring is 4 MeV. By how much should the rest mass of the parent 232 Th differ from the combined rest mass of its offspring? Compare this with the difference in the measured masses listed in Appendix D. The reaction (2.32) is analogous to the example of the two blocks just discussed. (In fact, the analogy is remarkably good. The repulsive force is the electrostatic repulsion of the offspring nuclei, both of which are positively charged; the catch that holds the offspring together is their nuclear attraction, which is not quite strong enough and eventually releases, letting the offspring fly apart.) The required mass difference M is given by (2.30) as Section 2.6 Conversion of Mass to Energy 61 M = K 1 + K 2 c 2 = 4 MeV>c 2 Therefore, 232 Th should be heavier than the combined mass of 228 Ra and 4 He by 4 MeV>c 2. We can convert this to kilograms if we wish. MeV M = 4 c 2 = 7 * kg = 4 * 106 * 1.6 * J 13 * 10 8 m>s2 2 (2.33) This mass difference is very small, even compared to the masses of the atoms involved. ( 232 Th has a mass of about 4 * kg; so the mass difference is of order 1 part in 10 5 of the total mass.) Nevertheless, it is large enough to be measured directly. To check our predicted mass difference against the measured masses, we refer to Appendix D, which lists the masses concerned as follows: Atom Mass (in u) 262 Th Initial 228 Ra He = Final Total Difference These masses are given in atomic mass units (denoted u). We will discuss this unit in Chapter 3, but for now we need to know only that 1 u = 1.66 * kg Thus the measured mass difference is M = u * 1.7 * kg 1 u = 7 * kg in agreement with our prediction (2.33).

17 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics We next consider briefly the case that the force between the bodies is predominantly attractive, so that work is required to pull them apart. For example, if our two bodies are an electron and a proton, they attract one another because of their opposite electric charges; if they come close together, they can form the stable bound state that we call the hydrogen atom and an external agent must do work (13.6 ev, in fact) to pull them apart again. The work needed to pull a bound state apart (leaving the pieces well separated and at rest) is called the binding energy and is denoted by B. As before, we denote by M the rest mass of the bound state and by m 1 and m 2 the separate masses of the two bodies. By conservation of energy Mc 2 + B = m 1 c 2 + m 2 c 2 (2.34) In this case we see that M is less than m 1 + m 2, the difference M = m 1 + m 2 - M being given by Mc 2 = binding energy, B = work to pull bodies apart (2.35) If we now released our two bodies from rest, they would accelerate back together and could reenter their bound state, with the release of energy B. (In the example of the electron and proton, the energy is released as light when they form into a hydrogen atom.) Thus we can rephrase (2.35) to say that as the two bodies come together and form the stable bound state, there is a release of energy and corresponding loss* of mass, M, given by Mc 2 = energy released as bodies come together to form bound state (2.36) Example 2.7 It is known that two oxygen atoms attract one another and can unite to form an O 2 molecule, with the release of energy E out L 5 ev (in the form of light if the reaction takes place in isolation). By how much is the O 2 molecule lighter than two O atoms? If one formed 1 gram of O 2 in this way, what would be the total loss of rest mass and what is the total energy released? (The O molecule has a mass of about 5.3 * kg. ) By (2.36) the mass of one O 2 molecule is less than that of two O atoms by an amount M = E out c 2 L 5 ev>c 2 L 5 * 1.6 * J 13 * 10 8 m>s2 2 L 9 * kg *It is quite easy to confuse the direction of the mass change (gain or loss) in a given process. If this happens, just go back to the fact of energy conservation, which you can easily write in a form like (2.29) or (2.34). From this you can see immediately which mass is greater.

18 TAYL I 12/9/02 2:53 PM Page 63 Section 2.6 Conversion of Mass to Energy 63 Dividing this by the mass 5.3 * kg of a single O 2 molecule, we see that the fractional loss of mass is about 2 parts in Thus if we were to form 1 gram (g) of O 2 this way, the total loss of mass would be M L 2 * g (2.37) which is much too small to be measured directly. This is fairly typical of the mass changes in chemical reactions and explains why nonconservation of mass does not show up in chemistry. While the mass change (2.37) is exceedingly small, the total energy released, E out = Mc 2 L 12 * kg2 * 13 * 10 8 m>s2 2 L 2 * 10 4 J is large much more than would be needed to boil a gram of water, for example. This is because the conversion factor, c 2, from mass to energy is so large. So far in this section we have focused on the conservation of relativistic energy and the concomitant nonconservation of mass. For an isolated system, momentum is also conserved, and in many problems conservation of energy and momentum give enough information to determine all that one needs to know. We conclude this section with an example. Example 2.8 The particle is a subatomic particle that (as mentioned in Example 1.2) can decay spontaneously into a proton and a negatively charged pion. : p + p - (This immediately tells us that the rest mass of the is greater than the total rest mass of the proton and pion.) In a certain experiment the outgoing proton and pion were observed, both traveling in the same direction along the positive x axis with momenta p p = 581 MeV>c and p p = 256 MeV>c Given that their rest masses are known to be m p = 938 MeV>c 2 and m p = 140 MeV>c 2 find the rest mass m of the. We can solve this problem in three steps: First, knowing p and m for the proton and pion, we can calculate their energies using the Pythagorean relation E 2 = 1pc mc (2.38) This gives E p = 1103 MeV and E p = 292 MeV

19 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics Second, using conservation of energy and momentum, we can reconstruct the energy and momentum of the original : E = E p + E p = 1395 MeV (2.39) and p = p p + p p = 837 MeV>c (along the positive x axis) (2.40) Finally, knowing E and p for the, we can use the relation (2.38) again to give the mass m = 4E 2-1p c2 2 c 2 = 1116 MeV>c 2 This is, in fact, how the masses of many unstable subatomic particles are measured. As we discuss in the next section, it is fairly easy to measure the momenta of the decay products as long as they are charged. If the decay masses are known, we can then calculate their energies and hence reconstruct the parent particle s energy and momentum. From these, one can calculate its mass. 2.7 Force in Relativity We have come a surprisingly long way in relativistic mechanics without defining the notion of force. This reflects correctly the comparative unimportance of forces in relativity. Nonetheless, forces are what change the momentum of a body and we must now discuss them. Just as with momentum and energy, our first task is to decide on a suitable definition of force. In classical mechanics two equivalent definitions of the force acting on a body are F = ma (2.41) and F = dp dt (2.42) These are equivalent since p = mu, with m constant, so dp>dt = ma. In relativity we have defined p as gmu, and it is no longer true that dp>dt = ma. Therefore, we certainly cannot carry over both definitions, (2.41) and (2.42), into relativity. In fact, for most purposes the convenient definition of force in relativity is the second of the classical definitions, (2.42). The total force F acting on a body with momentum p is defined as F = dp dt (2.43)

20 TAYL I 12/9/02 2:53 PM Page 65 Clearly, the definition (2.43) reduces to the nonrelativistic definition if the body concerned is moving nonrelativistically. Further, with this definition of F, the work-energy theorem carries over into relativity, as we prove in the following example. Example 2.9 Prove that if a mass m, acted on by a total force F, moves a small distance dr, the change in its energy, de, equals the work done by F. Section 2.7 Force in Relativity 65 de = F # dr (2.44) To prove this, we replace F by dp>dt and dr by u dt on the right to give F # dr = dp dt # u dt (2.45) Now, we will prove in a moment that dp # u = de dt dt (2.46) Thus from (2.45) it follows that F # de dr = dt = de dt which is the work-energy theorem. It remains only to prove the identity (2.46), as follows: From the Pythagorean relation, we know that E = 1p 2 c 2 + m 2 c 4 2 1>2 Using the chain rule to differentiate this, we find (Problem 2.36) de dt = 1 2 1p2 c 2 + m 2 c 4 2-1>2 2pc 2 # dp dt = pc2 # dp E dt = u # dp dt (2.47) which is (2.46). The most important force in most applications of relativistic mechanics is the electromagnetic force. It is found experimentally that when a charge q is placed in electric and magnetic fields E and B, its relativistic momentum changes at a rate equal to q1e + u * B2. Thus, having defined F as dp>dt, we find that the electromagnetic force is given by the classical formula, often called the Lorentz force, F = q1e + u * B2 Even when the force F is known, the equation dp>dt = F is usually hard to solve for a body s position as a function of time. One important case where it is easily solved is for a charged body in a uniform magnetic field B. In this case the force F = q1u * B2 is perpendicular to u, and the body s energy is

21 TAYL I 12/9/02 2:53 PM Page Chapter 2 Relativistic Mechanics therefore constant. [This follows from the work-energy theorem (2.44) because F is perpendicular to dr as the body moves along.] Therefore, the velocity is constant in magnitude and changes only its direction. In particular, if the body is moving in a plane perpendicular to B, it moves in a circular path, whose radius R can be found as follows: Since g is constant, the equation dp dt = qu * B becomes Since du>dt = a and u is perpendicular to B, this implies that For motion in a circle, we know that a is the centripetal acceleration u 2 >R. (This purely kinematic result is true in relativity for exactly the same reasons as in classical mechanics.) Therefore, whence, since gmu = p, du gm dt = qu * B gma = qub u 2 gm R = qub R = p qb (2.48) This result provides a convenient way to measure the momentum of a particle of known charge q. If we send the particle into a known magnetic field B, then by measuring the radius R of its curved path we can find its momentum p from (2.48). Example 2.10 A proton of unknown momentum p is sent through a uniform magnetic field B = 1.0 tesla (T), perpendicular to p, and is found to move in a circle of radius R = 1.4 m. What are the proton s momentum in MeV> c and its energy in MeV? The proton s charge is known to be q = e = 1.6 * coulomb (C). Thus from (2.48) its momentum is p = qbr = 11.6 * C2 * 11.0 T2 * 11.4 m2 = 2.24 * kg # m>s From the list of conversion factors inside the front cover, we find 1 MeV>c = 5.34 * kg # m>s. [This was derived in (2.26).] Therefore, p = * kg # m>s2 * 1 MeV>c 5.34 * kg # m>s = 420 MeV>c

22 TAYL I 12/9/02 2:53 PM Page 67 Section 2.8 Massless Particles 67 Since the proton s rest mass is known to be m = 938 MeV>c 2, we can find its energy from the Pythagorean relation (2.38) to be E = 31pc mc = 1030 MeV 2.8 Massless Particles In this section we consider a question that will probably strike you as peculiar if you have never met it before: In the framework of relativistic mechanics, is it possible to have particles of zero mass, m = 0? In classical mechanics the answer to this question is undoubtedly no. 1 The classical momentum and kinetic energy of a particle are mu and 2 mu2. If m = 0, both of these are also zero, and a particle whose momentum and kinetic energy are always zero is presumably nothing at all. In relativity the answer is not as clear cut. Our definitions of energy and momentum were p = gmu (2.49) and E = gmc 2 (2.50) and from these we derived the two important relations E 2 = 1pc mc (2.51) and b = u c = pc E (2.52) Let us consider the last two relations first. If there were a particle with m = 0 [and if the relations (2.51) and (2.52) applied to this particle], then (2.51) would imply that E = pc (if m = 0) (2.53) and (2.52) would then tell us that b = 1 or u = c (if m = 0) That is, the massless particle would always have speed u = c. The converse of this statement is also true. If we discovered a particle that traveled with speed c, by (2.52) it would satisfy pc = E, and (2.51) would then require that m = 0. If we turn to our original definitions of p and E, (2.49) and (2.50), a superficial glance suggests the same result as in classical mechanics. With m = 0, (2.49) and (2.50) appear at first to imply that p and E are zero. However, a closer look shows that this is not so. We have already seen that a massless particle would have to travel at speed c; and if u = c, then g = q. Thus both