Problem 1:
A possible clock is shown in the figure below. It consists
of a flashtube F and a photocell P shielded so that each views only the mirror
M, located a distance d away, and mounted rigidly with respect to the
flashtube-photocell assembly. The electronic innards of the box are such that,
when the photocell responds to a light flash from the
mirror, the flashtube is triggered with a negligible delay and emits a short
flash towards the mirror. The clock thus "ticks" once every (2d/c) seconds
when at rest.
(a) Suppose that the clock moves with a uniform velocity v,
perpendicular to the line from PF to M, relative to an observer. Using the
second postulate of relativity, show by explicit geometrical or algebraic
construction that the observer sees the relativistic time dilatation as the
clock moves by.
(b) Suppose that the clock moves with a velocity v parallel to the line from PF
to M. Verify that here, too, the clock is observed to tick more slowly, by
the same time dilatation factor.
Solution:
 | Concepts, principles, relations that apply to the
problem: The postulates of relativity. |
| Why do they apply? In vacuum, light propagates with respect to any inertial frame and in all directions with the universal speed c. This speed is a constant of nature. We use this postulate to calculate the time it takes a light pulse to move from F to P. |
| How do they apply?
 d2 = (c2 - v2)t2 (a) The observer sees the pulse move with speed c. But the flash is not moving a distance 2d, but a distance 2(d2+v2t2)1/2. The time between ticks is Dt = 2(d2+v2t2)1/2/c = 2(d2+v2d2/(c2-v2))1/2/c = (2d/c)(1+v2/(c2-v2))1/2 = (2d/c)(c2/(c2-v2))1/2 = g(2d/c). We have Dt = gDt. |
| Details of the calculation: (b) The observer sees the pulse move with speed c. It travels a distance d1 + d2. d1 = d' + vt1 = ct1, t1 = d'/(c-v). d2 = d' - vt2 = ct2, t2 = d'/(c+v). The clock therefore ticks every t1 + t2 seconds. The time interval between ticks is Dt = d'/(c-v) + d'/(c+v) = [d'(c+v) + d'(c-v)]/(c2-v2) = 2d'c/(c2-v2) = g2(2d'/c). But d' = d(1-v2/c2)1/2 = d/g, (Lorentz contraction), therefore Dt = g(2d/c) = gDt. |
Problem 2:
How far will a muon (whose lifetime is 2.2 µsec in its rest frame) move before it decays in the lab if it moving at 0.999c?
Solution:
| Concepts, principles, relations
that apply to the problem: Time dilation |
| Why do they apply? The proper time interval between two events (creation and decay) is given. We are asked to find the time interval in a reference frame moving with speed v = 0.999c with respect to the frame in which the two events have the same space coordinates. How do they apply? dt = dt/g. dt = 2.2 µsec/(1 - 0.9992)1/2 = 49.2 µsec. |
| Details of the calculation: None |
Problem 3:
A rod of length L0 is inclined at angle q from the x-axis in its rest frame. Find the inclination angle of the rod as measured by an observer moving with relativistic speed “v” in the x-direction.
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic Kinematics, Lorentz contraction |
| Why do they apply? To an observer in motion relative to an object, the dimensions of the object are contracted by a factor of 1/g in the direction of motion. |
| How do they apply? In the rods rest frame: Lx = L0cosq, Ly = L0sinq. In the observers rest frame Lx’ = Lx/g, Ly’ = Ly. Here g = (1 - v2/c2)-1/2. tanq’ = Ly’/Lx’ = g tanq. |
| Details of the calculation: None |
Problem 4:
If Bill, who is on a train moving with speed 0.6c, waves to Julie at four second intervals as measured in Bill's frame, how much time will have passed between waves in Julies rest frame?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic kinematics, the proper time interval |
| Why do they apply? We are given the proper time interval between two events in frame 1 and are asked to find the time interval in another inertial frame moving with speed v with respect to frame 1. |
| How do they apply? gDt = Dt, g = (1 - 0.36)-1/2 = 1.25, Dt = 5s. |
| Details of the calculation: None |
Problem 5:
For two inertial coordinate systems K and K˘
in relative motion at speed “v” along their x1 axes, the
Lorentz transformations provide that
x1˘ =
g(x1 – vt)
x2˘ = x2
x3˘ = x3 and
t˘
= g(t – vx1/c2),
where g = [1 – v2/c2]-1/2
.
(a) Derive the velocity transformations relating speeds uj
(in K) and uj˘ (in K˘),
with j = 1, 2, 3.
(b) Now assume Kurt (in system K) fires an energetic photon straight up, along
his x2 (i.e., y) axis. If Paula’s system K˘
travels at v = 0.6c relative to Kurt, calculate the components of the photon’s
velocity in Paula’s reference frame.
(c) From your results in part (b), calculate the magnitude of the photon's
velocity in Paula's frame. Explain why this result is to be expected.
Solution:
|
Concepts, principles, relations that apply to the
problem: |
|
Why do they apply? |
|
How do they apply? |
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Details of the calculation: |
Problem 6:
Consider the following diagram where a distant source at point B is moving at a velocity v ~ c toward the point B'.
At time t1 the source at point B emits a light signal that is detected at time t'1 by an observer at point A. When the source reaches the point B' at time t2, it emits another light signal that is detected by observer A at time t'2. You may assume the distance from A to B to be very large compared with the distances associated with transverse motion.
(a) Derive an expression for the time observed at A for the source to appear to move from B to B' and an expression for the apparent transverse velocity bT (apparent velocity perpendicular to the line of sight) for the source observed at A in terms of the angle q and the true velocity b = v/c.
(b) Show that this apparent velocity is maximized for an angle qmax = cos-1b, find the expression for the maximal apparent velocity, and show that it has no upper bound, even though the actual velocity satisfies b < 1 [you may find sin(cos-1b) = (1 – b2)1/2 to be useful in this regard].
(c) Calculate the apparent transverse velocity if q = 10° and the actual velocity from B to B' is b = v/c = 0.995.
This optical illusion of an apparent transverse velocity exceeding that of light is observed frequently in radio astronomy where it is called superluminal motion.
Solution:
| Concepts, principles, relations that apply to the
problem: Postulate: In vacuum, light propagates with respect to any inertial frame and in all directions with the universal speed c. |
| Why do they apply? Assume A and B' are at rest in the inertial frame K. B is moving with velocity v in K and is approaching B'. The times t1, t2, t1', and t2' are measured in K and the diagram is drawn in K, i.e. the angles Df and q are measured in K. |
| How do they apply? (a) Pulse 1, emitted at t1 travels a distance d + v(t2 - t1)cosq before reaching A. Its travel time is d/c + v(t2 - t1)cosq/c. Pulse 2, emitted at t2 travels a distance d before reaching A. Its travel time is d/c. The difference in arrival times of the two pulses at A, t2' - t1' is t2' - t1' = t2 + d/c - (t1 + d/c + v(t2 - t1)cosq/c) = (t2 - t1)(1 - vcosq/c). The pulses arriving a time interval Dt = t2' - t1' apart come from directions separated by an angle Df, and the transverse distance an object a distance d from A would have to cover to produce the two pulses separated by Dt is dDf @ v(t2 - t1)sinq. The apparent transverse velocity therefore is vtrans = v(t2 - t1)sinq/((t2 - t1)(1 - vcosq/c)) = (cbsinq)/(1 - bcosq). |
| Details of the calculation: (b) dvtrans/dq = cbcosq/(1 - bcosq) - cb2sinq2/(1 - bcosq)2. Set dvtrans/dq = 0 to find the angle q for which vtrans is maximized. cbcosq(1 - bcosq) - cb2sinqsinq = 0, cbcosq - cb2 = 0, cosq = b. qmax = cos-1b. vtrans_max = (cbsin(cos-1b))/(1 - bcos(cos-1b)) = cb(1 - b2)1/2/(1 - b2) = cb/(1 - b2)1/2. As b --> 1, (1 - b2) --> 0, and vtrans_max increases without limit. (c) vtrans = (0.995c*sin10o)/(1 - 0.995*cos10o) = 8.589c. |