Set 1 solutions 4-11 In the ring's inertial frame they meet again at T = pR/V, so the time elapsed for both Alice and Bob is
 Dt = pR V ų 1-V²
and their clocks will be synchronized. The time dilation argument does not apply because they are not in inertial frames.

4-14. In the satellites' frame let one be at the origin, the next at x' = L, y' = 0. Let the signals be emitted at tóA = tóB = 0. In that frame the oberver is in the "middle", at x'=L/2, y' = y = h, when he receives both signals together. Since he is at a distance (h² + L²/4)½ from the origin, that is also the time t' of the reception event. So that event has coordinates (t', x', y') = ((h² + L²/4)½, L/2, h).
Transform to the observer's system,

x = gxó+ bgtó = g[L/2 + b(h² + /4)½]
To first order in b, g = 1, so deviation from L/2 is
dx = b(h² + /4)½.

4-19 In a snapshot all the light arrives at the camera at the same time, so if the object has depth, it originated from the more distant parts at their earlier positions. Objects moving in a transverse plane are Lorentz contracted by the normal amount, all their parts are "seen" simultaneously. Hence the object "really" looks distorted, and this is equivalent to a rotation only in the two-dimensional image on the film. (A stereo view would show the difference; or, consider a railroad car on tracks as the object. The wheels have to stay on the tracks, so the whole car + wheels cannot consistently look rigidly rotated.)
So the rectangular cross section of a box car "really" becomes a parallelogram, the sides (size a) "really" look rotated, but not the front and back walls (proper size b). One should show that the Lorentz contraction of the walls is consistent with the foreshortening we would get from a rigid rotation:
We see the back wall a time a/c earlier than the front wall, so it lags behind by a distance av/c, and the angle q between the side wall (as "really" seen) and the line of sight, satsfies tanq = (av/c)/a = v/c. But then it would also be seen as longer than it would be if rigidly rotated. If we want to interpret what we see as a rigid rotation, where the side wall still has length a, we have to say sinq = v/c The front wall is forshortened by the Lorentz contraction factor (1-(v/c)²)½ = (1 - sin2q)½ = cosq and that is the correct foreshortening factor for a rigid rotation of a transverse length, rotated by the same angle q.

5-7 Differentiating the normalization condition for the four-velocity
 u·u = -1
(1)
we find that generally
 a·u = 0
(2)
where a is the four-acceleration a = du/dt

In the rest frame of the particle therefore a = (0, g). The invariant a·a is thus
 a·a = g2
(3)
When written out in a general frame where the particle is moving, equatiuons (1), (2), and (3) become
 -(ut)2 + (ux)2 = -1
(4)

 -atut + axux = 0
(5)

 -(at)2 + (ax)2 = g2
(6)
These three equations can be solved for at and ax in terms of g and the components of u. One finds
 at ║ dut/dt = gux
(7)

 ax ║ dux/dt = g ut
(8)
The solution where ux is zero at t = 0 is
 ux = Asinh(gt)
(11)

 ut = Acosh(gt)
(12)
but the normalization condition (4) implies A = 1. These equations are easy to integrate with the given boundary conditions to find
 x(t) = x0 + 1/g [cosh(gt) - 1]
(13)

 t(t) = 1/g sinh(gt)
(14) The world line is the hyperbola t2 = (x-x0)2 + (2/g)(x-x0). The image was taken from an answer to a problem in Schutz. a is the present g, l is the present t.

5-15 Symmetry arguments can show that the detection frequency is the same as the emission frequency. For example, any deviation in frequency should not depend on the direction of rotation. Also a mirror image of the apparatus should not change the frequency deviation. But the mirror image reverses the rotation and the angle f, exchanging emitter and receiver. Thus neither can observe, for example, a higher frequency than the other.
But symmetry does not tell you what is "really" going on: Think of the events of emission and reception in the inertial frame. They happen somewhere on the circle, the light is in a straight line between them, so the angle between ray direction and emitter velocity is essentially the same as between ray and detector velocity (actually p minus that angle). Therefore the frequency shift factor for one is the inverse of the other - if the rotation is toward the emitter, the red shift on emission is canceled by the blue shift on detection (Eq 5.73).

17. a) everyone got this part, since the answer was known: a photon making angle a with the x-axis has four-momentum

p =(p, pcosa, psina, 0)       p = (h/2p)w.
Transform to the observer's frame, evaluate cosaó = póx/pót.

b) 2p sinaó daó dtó fó(aó) is the number of photons emitted at an angle aó into an annulus of angular width daó in time dtó. This must be the same as the corresponding number in the rest frame, i.e. the formula with primes omitted. With the result from part a) and the transformation of dt we find
 fó(aó) = f / g [g(1-V cosaó)]2

c) We have (inverse of (5.73))     (h/2p) w = Eó g (1 - V cosaó). Solve for E', use result of part b):
 Ló(aó) = Eó(aó) fó(aó) = L / g [g(1 - V cosaó)]3

d) The ratio of intensity forward/backward direction is
 Ló(0) Ló(p) = µĶ 1+V 1-V ÷° 3
which as V« 1 becomes very large, meaning most of the radiation is beamed forward.

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On 26 Sep 2003, 16:31.