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I am at Point A, and I have a laser pointer. I point it at a mirror, one mile west of me, at point B. The beam travels to the mirror and comes back to make a nice round dot on my shirt at point C. In my frame, Points A and C are the same.

Bob zooms over me in his spaceship, going half the speed of light due north. Since he is going half the speed of light, he travels one mile in the time it takes my laser to go two miles. In his frame, he sees my laser fire at point A, hit the mirror at B, and reflect to point C, which is a long ways south of Point A.

As we all know, the Lorentz transformation arranges everything so that both of us see the light beam traveling at speed $c$. But the Lorentz transformation does not change the fact that my laser is pointed due west, in my frame and in Bob's frame. Yet Bob sees the beam travel at an angle, and reflect at the opposite angle, to arrive well south of where it originated.

In my frame, the laser is stationary, and the beam it emits travels parallel to its axis. In Bob's frame, the laser is moving south at half the speed of light. How does that motion cause the beam to be produced at an angle to the laser's axis?

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    $\begingroup$ "But the Lorentz transformation does not change the fact that my laser is pointed due west, in my frame and in Bob's frame." <-- I think that assumption is your problem. $\endgroup$
    – DanielSank
    Commented Oct 21, 2015 at 18:32
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    $\begingroup$ Don't you also have this paradox with galilean relativity? $\endgroup$
    – Brian Moths
    Commented Oct 21, 2015 at 18:37
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    $\begingroup$ For example consider a pipe carrying water from A to B and back to A, so the pipe makes a very long narrow "U" shape. Here you have water instead of light, but you still have your paradox, and you can treat it with galileon space time or minkowski space time. $\endgroup$
    – Brian Moths
    Commented Oct 21, 2015 at 18:42
  • $\begingroup$ Daniel - The laser is pointing West in my frame, by definition. Are you suggesting that a Lorentz transformation rotates it? Which way, CW or CCW? And why that way, and not the other? $\endgroup$
    – Jerome Berryhill
    Commented Oct 22, 2015 at 0:58
  • $\begingroup$ NowI - I'm not following you. If there is a pipe, it has a shape, which is independent of the frame. It either starts and stops at the same point, or else it does not. $\endgroup$
    – Jerome Berryhill
    Commented Oct 22, 2015 at 1:08

2 Answers 2

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Let's have a closer look at your laser projector in your rest frame. In particular let's concentrate on a single photon inside the projector:

Photon motion

Actually I've just realised I've drawn my diagram with the laser moving East not West - oh well. Anyhow in this frame the laser beam moves due East in a straight line. So far so good.

Now let's look at the situation in Bob's frame:

Bob's frame

In Bob's frame the laser projector is moving with time at 0.5$c$. The projector still points due east, but the photon making its way along the projector towards the exit window is now not travelling due east but instead is travelling at a diagonal.

So the point is that the laser projector does not need to be rotated for the beam to emerge at an angle.

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  • $\begingroup$ John - That sounds a little better. Basically, you're saying that the photon has momentum, and in Bob's frame, that momentum has a Southward component. Which means, that while the velocity of light does not depend upon the motion of the source, the momentum does. $\endgroup$
    – Jerome Berryhill
    Commented Oct 22, 2015 at 18:39
  • $\begingroup$ @JeromeBerryhill: what I am saying is that the direction that the photon is moving looks different in different frames - the speed of the photon is $c$ in all frames. The magnitude of the momentum is $h\nu/c$ in all frames, and the frequency $\nu$ can change due to the Doppler shift. However in this case there is no Doppler shift and the magnitude of the momentum doesn't change (though its direction does). $\endgroup$
    – John Rennie
    Commented Oct 23, 2015 at 5:06
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Angle of emission $\theta_s$ and angle of reception $\theta_o$ are tied by relativistic aberration formula.

$$ \cos {\theta_0} = \frac {\cos {\theta_s} - \frac v c} {1- \frac v c \cos \theta_s} $$

This simple animation demonstrates path of laser beam in different frames:

https://www.youtube.com/watch?v=FQKp3FU8vR8

By the way. Angles are different, and observations are different. What about provisions of Special Relativity?

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