Question

Two inertial frames of reference O and O’ are in the standard configuration, frame O’ having velocity v with respect to frame O. At the instant when their spatial origins O and O’ coincide, a light beam is emitted from O and O’ along the positive x- and x’- axis. The beam is reflected by a mirror M fixed in O at a distance d from O and with its plane perpendicular to the x-axis. Consider the following events:
(a) light beam reaches M,
(b) reflected beam returns to O’,
(c) reflected beam returns to O.
Sketch and calculate the times of these events as measured by observers in frames O and O'

Answer 1

You gotta use lorenz transformations.. i guess.. i m just learning relativity.. this is interesting question for me :D..

Answer 2

in Os frame of reference, the light beam travels with speed c ,
distance d so c = d/t hence
t =d/c

in O's frame of reference , the light beam ALSO travels with speed c
then the frame O is traveling with speed -v. Hence for him, the distance between the mirror and origin is length contracted by a factor gamma.. so you can calculate time again?

Answer 3

@Vincent-Lyon.Fr

Answer 4

There are two approaches you can take to this problem.

One is the more conventional of using distance and dividing by the velocity. However this requires the Lorentz velocity transformation.

The second, as @Mashy suggests, is the Lorentz transformation which is much cleaner but requires you to be aware of what frame you're in and what frame you're looking for.

Answer 5

For the first part, the light travels half the distance in the stationary frame so yes it is d/c, however, in part b, the light hits the mirror then returns and this is in fact 2d/c

Both of these times are measured in the stationary frame O.

The Lorentz transformation for time is
\[\Large t'=\gamma(t-{vx\over c^2})\]

However, since you are measuring a period of time, Lorentz transformation uses "events" rather than an actual length of time.

So event 1
\[\Large t_1'=\gamma(t_1-{vx_1\over c^2})\]
Event 2
\[\Large t_1'=\gamma(t_1-{vx_1\over c^2})\]

The time it takes:
\[\Large\Delta t'=\gamma(\Delta t-{v\Delta x\over c^2})\]

Answer 6

For the return beam, in the stationary frame, there is no length contraction so delta x is simply d, the distance between the origin and the mirror. The time it takes is 2d/c.

Using some algebra, you can then calculate the time it takes in the O' frame.