Exam (elaborations) TEST BANK FOR A Unified Grand Tour of Theoretical Physics 3RD Edition By Lawrie I. (Solution Manual)

Exam (elaborations) TEST BANK FOR A Unified Grand Tour of Theoretical Physics 3rd Edition By Lawrie I. (Solution Manual) A Unied Grand Tour of Theoretical Physics Solution Manual Ian D Lawrie September 2012 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 Chapter 2 Geometry Exercise 2.1 Consider a Cartesian coordinate system S and and a second one, S′, which is obtained by giving S a velocity v, without rotating its axes. Then the origin of S′ moves with constant velocity v relative to S, and we take the two origins to coincide at t = t′ = 0. Assume that the relation between the two sets of coordinates is linear and that space is isotropic. The most general form of the transformation law can then be written as x′ = α [ (1 − λv2)x + (λv · x − βt)v ] t′ = γ [ t − (δ/c2)v · x ] where α, β, γ, δ and λ are functions of v2. For the case that v is in the positive x direction, write out the transformations for the four coordinates. Write down the trajectory of the S′ origin as seen in S and that of the S origin as seen in S′ and show that β = 1 and α = γ.Write down the trajectories seen in S and S′ of a light ray emitted from the origin at t = t′ = 0 that travels in the positive x direction, assuming that it is observed to travel with speed c in each case. Show that δ = 1. The transformation from S′ to S should be the same as the transformation from S to S′, except for the replacement of v by −v. Use this to complete the derivation of the Lorentz transformation [2.2] by finding γ and λ. Solution First, a few words about the way this question is set up. Once we have learned enough about the geometry of Minkowski spacetime, the best way of arriving 2 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 at the Lorentz transformation is to ask about coordinate transformations that preserve the metric or, equivalently, the form of the proper time interval [2.6]. Thus, an inertial Cartesian frame of reference is a set of coordinates such that c2(dτ )2 = c2(dt)2−(dx)2−(dy)2−(dz)2 and, using xμ for (ct, x, y, z) and xμ′ for (ct′, x′, y′, z′), we look for a constant matrix Λμ′ μ such that, if xμ′ = Λμ′ μxμ, then c2(dt′)2 − (dx′)2 − (dy′)2 − (dz′)2 = c2(dt)2 − (dx)2 − (dy)2 − (dz)2. By studying the most general matrix that satisfies this requirement, we find that the change of coordinates can be interpreted as a combination of a relative velocity of the origins and a rotation of the spatial axes. (There is some further discussion in §3.5.) For the purposes of this question, we are taking a more primitive point of view, by simply trying to find a transformation rule that works, without any insight into its geometrical meaning. For that reason, the terminology needs to be considered carefully in the light of the later theory. In particular, a ‘rotation of spatial axes’ turns out to mean different things in two frames of reference that are in relative motion. In setting up the problem, I assumed that space (more accurately, spacetime) is isotropic. That means that there is no naturally-occurring vector that distinguishes one direction from any other direction. Consequently, the new 3-dimensional vector x′ must be constructed from the only vectors we have to hand, namely x and v. That is, x′ = Ax + Bv. The coefficients A and B can depend only on scalar quantities that are unchanged by spatial rotations, namely t and the dot products of vectors, x · x = |x|2, v · v = v2 and v · x. Since we also assume that the transformation is linear in x and t, we find that A can be a function only of v2, while B can only have the form B = B1(v2)v · x + B2(v2)t. Similarly, t′, which is a scalar from the 3- dimensional point of view, can only have the form t′ = C1(v2)t+C2(v2)v ·x. I traded in the five functions A, B1, B2, C1 and C2 for five other functions α, β, γ, δ and λ because I happen to know that this will simplify the algebra. Now for the problem itself. Say that v = (v, 0, 0). Then the Lorentz transformation given above reads explicitly x′ = α[(1 − λv2)x + (λvx − βt)v] = α(x − βvt) (2.1) y′ = α(1 − λv2)y (2.2) z′ = α(1 − λv2)z (2.3) t′ = γ(t − δvx/c2). (2.4) 3 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 The origin of the S′ system is at x′ = y′ = z′ = 0, so its coordinates in S are at x = βvt and y = z = 0. It is supposed to be moving along the x axis with speed v, so we must have β = 1 . The origin of S is at x = y = z = 0, so with β = 1, its coordinates in S′ are y′ = z′ = 0 and x′ = −αvt = −(αv/γ)t′. It must be moving in the negative x′ direction with speed v, so we find α = γ . A light ray that sets out from x = 0 at t = 0 finds itself at x = ct at time t relative to S. Substituting x = ct in (2.1) and (2.4), we find that its position in S′ when t′ = γ(1−δv/c) is x′ = γ(c−v)t = [(c−v)/(1−δv/c)]t′. If the light ray also travels with speed c relative to S′, this position must be x′ = ct′, so we conclude that δ = 1 . At this point, two of the transformation equations read x′ = γ(x − vt) and t′ = γ(t − vx/c2), and these equations can be solved to give x = x′ + vt′ γ(1 − v2/c2) and t = t′ + vx′/c2 γ(1 − v2/c2) . (2.5) This transformation from S′ to S should have the same form as the original transformation from S to S′, if we replace v with −v, so we conclude that γ = (1 − v2/c2)−1/2 . For the same reason, we conclude from (2.2) and (2.3), with α = γ, that γ(1 − λv2) = 1, or λ = (γ − 1)/γv2 . Clearly, these results give the special form of the Lorentz transformation [2.2], which applies when v is in the x direction. But since the functions α, β, . . . depend only on the magnitude of v, they remain valid when v is in any direction, and we get the more general result x′ = x + (γ − 1)(v · x) v2 v − γvt (2.6) t′ = γ(t − v · x/c2), (2.7) with γ = (1 − v2/c2)−1/2. 4 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 Exercise 2.2 Two coordinate frames are related by the Lorentz transformation (2.2). A particle moving in the x direction passes their common origin at t = t′ = 0 with velocity u and acceleration a as measured in S. Show that its velocity and acceleration as measured in S′ are u′ = u − v 1 − uv/c2 , a′ = (1 − v2/c2)3/2 (1 − uv/c2)3 a . Solution The particle’s trajectory as seen in S is x = ut+ 1 2at2. Substituting this into the Lorentz transformation equations gives x′ = γ[(u − v)t + 1 2at2] (2.8) t′ = γ[(1 − uv/c2)t − (av/2c2)t2]. (2.9) One way of proceeding would be to solve (2.9) for t as a function of t′ and substitute this into (2.8) to get the trajectory x′(t′) as seen in S′. We would then differentiate to find u′ = dx′/dt′ and a′ = d2x′/dt′2. A neater way is to treat these two equations as a parametric form of the trajectory. Then we can calculate the velocity and acceleration as u′ = dx′ dt′ = dx′/dt dt′/dt (2.10) a′ = du′ dt′ = du′/dt dt′/dt = d2x′/dt2 (dt′/dt)2 − (dx′/dt)(d2t′/dt2) (dt′/dt)3 . (2.11) Evaluating all the derivatives at t = 0, we get dx′ dt = γ(u − v), d2x′ dt2 = γa, dt′ dt = γ(1 − uv/c2), d2t′ dt2 = −γav/c2, (2.12) and substituting these results into the two previous equations gives the advertised answers. 5 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 Exercise 2.3 A rigid rod of length L is at rest in S′, with one end at x′ = 0 and the other at x′ = L. Find the trajectories of the two ends of the rod as seen in S and show that the length of the rod as measured in S is L/γ, where γ = (1 − v2/c2)−1/2. This is the Fitzgerald contraction. If the rod lies along the y′ axis of S′, what is its apparent length in S? A clock is at rest at the origin of S′. It ticks at t′ = 0 and again at t′ = τ . Show that the interval between these ticks as measured in S is γτ . This is time dilation. Solution Clearly, the rod points in the x′ direction, and we might as well take it to be on the x′ axis. Then its two ends are at (x′, y′, z′) = (0, 0, 0) and (x′, y′, z′) = (L, 0, 0). Substituting these values into the Lorentz transformation equations, we find that at time t as seen in S, the two ends are at (x, y, z) = (vt, 0, 0) and (x, y, z) = (L/γ + vt, 0, 0). Thus, as seen in S at time t, the differences in coordinates of the two ends of the rod are (Δx,Δy,Δz) = (L/γ, 0, 0) and the length of the rod is L/γ. This is always ≤ L because γ is always ≥ 1. The transformation equation t′ = γ(t − vx/c2) is irrelevant to the above calculation, but it tells us something that is worth noting. For example, the right-hand end of the rod (the one at x′ = L) is seen by an observer in S to pass the point x = L/γ at t = 0. This is an event that occurs at one definite point in space and time. For an observer in S′, this event happens at t′ = −vL/c2. Now suppose that the two ends of the rod are at (x′, y′, z′) = (0, 0, 0) and (x′, y′, z′) = (0, L, 0). As seen in S, the corresponding positions are (x, y, z) = (vt, 0, 0) and (vt, L, 0). We get (Δx,Δy,Δz) = (0, L, 0), so the observer in S sees a rod of length L. The first tick of the clock occurs at (x′, y′, z′, t′) = (0, 0, 0, 0) as seen in S′. Substituting these values into the transformation equations gives four simultaneous equations to solve for the corresponding coordinates of this event as seen in S, and the solution is (x, y, z, t) = (0, 0, 0, 0). The second tick occurs at (x′, y′, z′, t′) = (0, 0, 0, τ ), and this again gives four simultaneous equations. Two of them are 0 = y and 0 = z, which are quite easy to solve. The other two are 0 = γ(x−vt) and τ = γ(t−vx/c2). Using the first one to eliminate x, we get τ = γ(t − v2t/c2) = γ(1 − v2/c2)t = t/γ, so the solution is t = γτ . Then we also have x = vt = γvτ . Thus, for the observer in S, 6 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 the time interval between the two ticks is γτ . This is a longer time than τ , so it is often said that a moving clock appears to run slow. Obviously, the observer in S sees the clock tick for the first time when it is at x = 0 and for the second time when it is at x = γvτ . 7 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter 2 Exercise 2.4 As seen in S, a signal is emitted from the origin at t = 0, travels along the x axis with speed u, and is received at time τ at x = uτ. Show that, if u c2/v then, as seen in S′, the signal is received before being sent. Show that if such paradoxes are to be avoided, no signal can travel faster than light. Solution The y and z coordinates are irrelevant to this problem, so I will ignore them. As seen in S, the signal is emitted at (x, t) = (0, 0) and received at (x, t) = (uτ, τ ). As seen in S′, the Lorentz transformation tells us that it is emitted at (x′, t′) = (0, 0) and received at x′ = γ(u − v)τ and t′ = γ(1 − uv/c2)τ . So if uv c2, an observer in S′ sees the signal being received before it is sent. Since this seems to defy our usual expectation that a cause should precede its effect, we suspect that there must be some maximum speed, say umax with which any signal can travel. Now, the relative speed v of S and S′ cannot be greater than c, because this would lead to an imaginary value of γ, and hence imaginary values of x′ and t′. So the maximum value of uv is umaxc, and since this maximum value is supposed to be c2, we find that umax = c. It is important to check that this maximum speed applies equally to any frame of reference, and we can do this by using the result of exercise 2.2. If the signal has velocity u relative to S, then its velocity relative to S′ is u′ = (u − v)/(1 − uv/c2). With a short calculation, we can work out that u′2 − c2 = (u2 − c2)(1 − v2/c2) (1 − uv/c2)2 . (2.13) Clearly, if the signal travels with velocity u = ±c relative to some frame S, it also travels with velocity u′ = ±c relative to any other frame. This is just as well, since the constancy of the speed of light was a basic assumption of the theory. We also see, though, that if |u| c, then the right-hand side of (2.13) is negative, and thus |u′| c. Therefore, if a signal travels with speed less than c relative to any one inertial frame, it also travels with speed less than c in any other frame; the statement that a signal cannot travel faster than c is independent of which frame of reference we use. 8 This document can be downloaded from . c⃝ Ian D Lawrie 2012 It may be freely shared, but may not be altered or sold. Chapter