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I'm struggling to understand a hopefully silly implication about light clocks on trains. The train is in motion. SR and all that. Bob is on the side of the tracks watching. We're going to get super specific and say that the emitter for the light clock is an LED with a collimator.

Problem 1: The light beam leaves the LED, passes through the collimator, strikes the reflector on the ceiling of the train, and returns to the clock. Bob sees the clock tick. Bob must conclude that the beam of light traveled at an angle relative to his reference frame. Yet the laws of optics must still be obeyed.

Proposed solution to problem 1: Bob figures out that the LED emitted light in many directions. While the light was traveling to the collimator, the collimator was traveling with the train. So the light that made it through the collimator was moving at the proper angle relative to Bob's coordinates for it to hit the reflector on the far side of the train. The optics works from Bob's perspective, and the beam travels at the "correct angle". Do I have that right? (I think a similar argument could be made for a laser by recognizing that the lasing cavity is moving. The photons that are coherent are the ones that are moving "correctly" to the cavity.)

Problem 2: We replace the mirror at the top of the train with a retroreflector. Now Bob is completely baffled (and so am I). The clock still ticks, but that implies that the angle of reflection was equal to the angle of incidence. Yet, that's a retroreflector at the top of the train. The light should have reflected back along the incident beam. How did it make it back to the light clock? The laws of optics must still be obeyed.

The retroreflector can be considered (in the classic 2D case) to be a pair of mirrors at right angels to each other. All I can imagine is that if I were to carefully trace out a beam striking one of the mirrors and work out the time and locations of the reflections from the two mirrors in Bob's frame, it wouldn't behave like a retroreflector anymore, but rather like a plane mirror. Is that really the result? Or is there something else going on that I'm missing?

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  • $\begingroup$ Do I get it correctly that the LED is located at the floor of the train, shining upwards, vertically (as seen from inside the train)? $\endgroup$ – md2perpe Jun 24 at 11:29
  • $\begingroup$ @md2perpe, yes, the LED is on the floor with the collimator arranged so that it shines a collimated beam straight up with respect to the moving reference frame of the train. $\endgroup$ – Philip Freeman Jun 24 at 19:56
  • $\begingroup$ @Phillip Freeman I am not certain that the laws of optics must be obeyed in the moving frame. Newton’s laws are not relativistic, nor are the laws of circuit theory. You can identify the laws that must hold in all frames by the fact that they can be written in tensor form, like Maxwell’s equations. I am not sure if the laws of optics qualify $\endgroup$ – Dale Jun 25 at 20:23
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Imagine that instead of a beam of light, a ball is rolled across the floor of the railway carriage from one side to the other. Upon hitting the opposite wall of the carriage, it bounces back to where it started.

In the frame of the train, the ball has travelled directly back and forth at right angles to the wall. In the frame of Bob at the trackside, the ball has taken a diagonal path toward the far wall and been reflected back on a diagonal path, with the angle of incidence equalling the angle of reflection. So far, so good.

Now, let us introduce a retroreflector at the far wall of the carriage which will ensure that the ball bounces back in the same direction in which it arrives (which we can make with two rigid surfaces joined at right angles, just as we could make a retroreflector for light with two mirrors).

On the train, the ball hits the retroreflector and bounces back as before. From the frame of reference of Bob, the ball is moving on a diagonal path, so following your argument the retroreflector should send the ball back on the diagonal path on which it arrived. It does no such thing, of course. The reason is that the retroreflector only reverses the path of the ball in the frame in which it is stationary.

From Bob's frame, both the ball and the retroreflector are moving. When the ball strikes the moving surfaces of the retroreflector, the angles of incidence and reflection are not equal, and the resulting effect is that, in Bob's frame, the retroreflector does not reverse the path of the ball.

The paper available at this link (Am. J. Phys. 80, 680 (2012)) explains the underlying principles of the reflection of light from moving mirrors. It contains the case of a moving mirror angled at 45 degrees to its direction of motion, from which it is straightforward to establish that a moving retroreflector made of two mirrors each angled at 45 degrees to the direction of motion will not reverse the path of the incident light in Bob's frame, but will return the light to the source.

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  • $\begingroup$ "When the ball strikes the moving surfaces of the retroreflector, the angles of incidence and reflection are not equal..." Then you've stated a condition in which the rules of optics are violated. I don't think SR changes the rule in optics (or kinetic bouncing) that the angle of incidence equals the angle of reflection. Am I wrong on that? $\endgroup$ – Philip Freeman Jun 24 at 19:59
  • $\begingroup$ No, the angle of incidence is not necessarily equal to the angle of reflection for a moving mirror. Google velocity aberration. $\endgroup$ – Marco Ocram Jun 24 at 21:32
  • $\begingroup$ Hmmm.... Doesn't seem to explain what you contend it does. It seems that the apparent position would be displaced, but that doesn't seem to indicate that a reflection of that apparent ray wouldn't still have angle of incidence equal to angle of reflection in Bob's frame. $\endgroup$ – Philip Freeman Jun 24 at 22:51
  • $\begingroup$ Check this link. The formula for the relationship between the angles of incidence and reflection is too complicated to include in a comment... physics.weber.edu/galli/RelativisticReflection.pdf $\endgroup$ – Marco Ocram Jun 25 at 6:13
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The retroflector is moving. The mirrors in it will change their angles (as seen by Bob) because of length contraction. This makes the retroflector reflect the light to another angle than from where it entered.

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    $\begingroup$ This effect was actually considered in the retro-reflectors placed on the moon by astronauts (they were not intentionally made at non-right angles, but it was one term in the error budget): link.springer.com/article/10.12942/lrr-2010-7 $\endgroup$ – Mike Jun 24 at 20:31
  • $\begingroup$ That totally makes sense! I'm going to play with the numbers a bit and see if the change in angle is enough. Or, do I also need to consider the "velocity aberration" mentioned below? $\endgroup$ – Philip Freeman Jun 24 at 23:01
  • $\begingroup$ @PhilipFreeman. I think that this is enough explanation for the retroflector. $\endgroup$ – md2perpe Jun 25 at 7:02
  • $\begingroup$ So the question is "is this what it is?" and the answer is "yes it is". We need a calculation. And being a retroreflector, it needs to work for all angles of incidence. $\endgroup$ – JEB Jun 25 at 14:19
  • $\begingroup$ It's not enough. The example by Troposphere below is an excellent reason why. If the retroreflector was aligned so that one mirror was perpendicular to the direction of travel and the other parallel, then the angle would remain pi/2 between the surfaces in Bob's frame. Yet, the reflected beam clearly isn't returning along the incident path in Bob's frame. $\endgroup$ – Philip Freeman Jun 25 at 22:19
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This is a common rebuttal to the light clock, or variant thereof. The solution lays in the relativity of simultaneity.

Of course, a numerical solution can always be found by using a photons 4-momentum:

$$ p^{\mu} = (\frac E c, {\bf p})= (||{\bf p}||, {\bf p})$$

which transforms as a four vector:

$$ p'^{\mu} = \Lambda^{\mu}_{\ \nu}p^{\nu}$$

If you need to consider frequency and wave number, then use the 4-wave number:

$$ k^{\mu}=\frac 1{\hbar} p^{\mu}=(\frac{\omega}c, {\bf k})$$

But the problem here is conceptual: how can a laser shoot a beam sideways, or how can a retroreflector reflect off 90 degrees?

In both these problems, a laser beam from an aperture, or a retroreflector, there is a plane of constant phase:

$$ \phi'(x', t') $$

(where the source extends to $x'\in(-L/2, +L/2)$) that leads to a light ray orthogonal to the plane-of-constant phase (in the train's primed-frame$^1$)).

Here, we are talking about a light ray in the vertical direction:

$$ E(x',y',z') = E_0e^{I\phi'(x', z', t')} = E_0e^{i(k'z'-\omega' t')}$$

Note that the phase can be written as:

$$\phi' = (k'z'-\omega' t')$$ $$ \phi'= ({\bf k}'\cdot {\vec r}' - \omega' t')$$ $$ \phi'= -k'^{\mu}r'_{\mu}$$

which show it is a manifestly invariant Lorentz scalar. It is the same in all frames.

When you boost to Bob's (who should be Alice) unprimed frame (a boost along $-x'$):

$$ \phi(x, t) = \phi'(x', t') $$

but:

$$ t = \gamma\big(t'+\frac{vx'}{c^2}\big)$$ $$ x = \gamma\big(x'+vt'\big)$$

Because $t$ and $t'$ cannot be synchronized across the aperture, what appears as constant phase on the train is a phase ramp from the train station.

A phase ramp works the same way here as it does in an electronically steered radar, it caused the beam to diverge from normal.

[1] When discussing relativity, it customary to have the so-called at-rest frame be unprimed, and the moving frame be prime. The train moves, hence it is primed. Naming is alphabetical, moving outward. We start at rest with Alice in the train station and Bob on the train. It may seem nit-picky, but the number-one aspect of thought experiments must be clarity. And consistency.

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  • $\begingroup$ I skimmed over this briefly. Are you saying that in the "other" frame the beam's phase fronts are not perpendicular to the propagation vector? $\endgroup$ – garyp Jun 24 at 17:14
  • $\begingroup$ The answer by @md2perpe seems plausible. Is your analysis consistent with length contraction of the mirrors? $\endgroup$ – garyp Jun 24 at 18:02
  • $\begingroup$ This is nonsense, isn't it? $\endgroup$ – TonyK Jun 24 at 18:57
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    $\begingroup$ @TonyK. It's not nonsense. It explains how the direction of the light transforms. But it doesn't really explain why the moving retroreflector doesn't send the light back to the source. $\endgroup$ – md2perpe Jun 24 at 20:29
  • $\begingroup$ @garyp no, the I'm saying the phase fronts aren't parallel to the surface (it's an imaginary surface in a retro-reflector, OP didn't include a diagram which would make easier to discuss). This is the standard explanation of stellar aberrations, I'm not sure why there is push back. $\endgroup$ – JEB Jun 25 at 5:17
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The angles of the mirrors in the retroreflector don't necessarily change due to length contraction -- for all we know, a corner reflector might be mounted such that each mirror is either exactly perpendicular to the direction of the train's motion, or parallel to it. (We'll move the LED a bit off center, such that its beam still gets to bounce off all three mirrors -- that shouldn't change the underlying point of your question).

However, an unavoidable complication (from the trackside point of view) is that one of the mirrors is moving in a direction perpendicular to the mirror itself. Such a moving mirror does not satisfy the law that the angle of incidence equals the angle of reflection -- so the retroreflector is not exactly "retro" as observed from the trackside.

To see this qualitatively, imagine that there's a mirror hanging in the middle of the back wall of the carriage. We emit a light signal from the right front corner of the carriage, bounce it off the mirror, and detect it in the left front corner.

Seen from the train, the path of this light signal forms an isosceles triangle together with the front wall of the carriage.

However, the trackside observer can see that the "emission" and "detection" events happen at different positions along the track, and therefore at different distances (still along the track) from the "bounce" event. Both observers agree that emission and detection happen at equal (but opposite) lateral distances from the center of the track. But that means that the trackside observers must conclude that the incoming and reflected light rays make different angles to the track centerline (and therefore different angles to the plane of the mirror, too).

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  • $\begingroup$ That was an excellent and simple example! I also found this more complete description that explains a bit more about reflections in SR. $\endgroup$ – Philip Freeman Jun 25 at 14:36

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