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In a thought experiment, there are two parallel mirrors which are very long which also face each other. One has a hole in it and one flash from a strobe light shines into that hole.

One observer in the same frame of reference as the mirrors sees the light bouncing between the mirrors at about a 30 degrees to the perpendicular. To this observer the light appears to be making a zig-zag pattern between the mirrors.

Another observer traveling parallel to the mirrors at half the speed of light observes the light from the strobe bouncing between the two mirrors in a line perpendicular to the mirrors. To that observer the light appears to be like a standing wave.

Since both observers are seeing the same light move between the mirrors, does that mean that the up direction and the down direction are different when associated with the same dimension perpendicular to the mirror for the two frames of reference?

= = = = = = = =

Question: Does the dimension perpendicular to the velocity vector between two inertial frames of reference have a different association with the in-bound and out-bound directions in the two frames of reference?

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    $\begingroup$ I don't buy that the light will form a standing wave for the observer moving by. That would mean that you zig zag route woudld map on the same space-time points in the moving coordinate system. Transfromations in special relativity does not work that way. $\endgroup$ – Mikael Fremling Dec 7 '15 at 20:02
  • $\begingroup$ That is why the "about" adjective is on the 30 degree angle for the first observer. The point is that the moving frame is at a sufficient speed for the light to be a standing wave. The paradox could be setup as a standing wave for the observer in the mirror's frame of reference. That would mean that the moving observer sees the light follow a zig-zag. The only real difference is which frame are the mirrors in, but they do not affect the question about the relationship between the perpendicular dimension and the up and down directions. $\endgroup$ – Allyn Shell Dec 7 '15 at 20:57
  • $\begingroup$ I think I see what you're getting at, but let me see if I understand: take the second observer (whom we'll call B) as stationary and the mirror/strobe/first observer (A) as moving toward B at 0.5_c_ (along what we'll call the x axis). Then the beam of light, being pointed toward B at 30°, will appear to move toward B with an x-component of $c \sin 30° = 0.5c$. However, the mirrors are also moving at 0.5c in the same direction, so the beam should be hitting the mirrors in the same spot each time it reflects, thus creating a standing wave. Is that an accurate rephrasing of the question? $\endgroup$ – iamnotmaynard Dec 7 '15 at 21:22
  • $\begingroup$ Allyn, the answer is yes. One of your guys says the light moves like this /\/\/\ and the other guy claims it moves like this | . See this question for an example involving a mast. $\endgroup$ – John Duffield Dec 7 '15 at 21:24
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    $\begingroup$ @JohnDuffield, Please, I really am not trying to be difficult. Dimensions fascinate me. I want to know all sorts of things about them like "why is a dimension stiff?," and "why do the orthogonal dimensions stay oriented as they do?," and "is the geodetic path followed by light actually following a dimension?," and "do dimensions reorganize themselves in different frames of reference?," and "in the beginning did the time dimension originally have two directions?," etc. $\endgroup$ – Allyn Shell Dec 7 '15 at 22:09
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Given the comments I think I can now give a coherent answer to why no paradox arises. The bottom line is that no standing wave is formed, and the simple reason is that the emitter of the light will look as if it is moving backward, and this will create a temporal zig-zag pattern instead of a spacial one.

More formally it goes like this: A light source is emitting light at an angle of $30^\circ$ such that the $x$-component of the light is $v=\frac c2$. For a stationary observer the bouncing light beam will look like a zig-zag pattern.

The path of the pattern will be $$\vec r=(\frac c2(t-t_0),\mathrm{zz}(\omega (t-t_0)))$$

Here $\mathrm{zz}$ is the zig-zag, bouncing pattern. $\omega$ is the frequency of the bounce, and depends on the distance between the two mirrors. $t_0$ is the time the light pulse was emmited. I'm including it here as it will become important soon.

A moving observer now flies by at $v=\frac c2$. It concludes that the light beam is bouncing up an down between the same two positions. However, before it concludes that the light is forming a standing wave, it also notices that the emitter now seems to be moving at a speed $v_e=-\frac c2$.

The position of the beam will now be given by $$ x'=\frac{x-\frac c2t}{\sqrt{\gamma}}=\frac{\frac c2(t-t_0)-\frac c2t}{\sqrt{\gamma}}=-\frac{\frac c2t_0}{\sqrt{\gamma}}$$

The $y$ coordinate does not transform, but time itself gets dilated as

$$ t=\frac{t'+\frac c2\frac{x'}{c^2}}{\sqrt{\gamma}}=\frac{t'+\frac c2\frac{-\frac{\frac c2t_0}{\sqrt{\gamma}}}{c^2}}{\sqrt{\gamma}}=\frac{t'-\frac 14\frac{t_0}{\sqrt{\gamma}}}{\sqrt{\gamma}}$$

So the position of the light beam is now

$$\vec r'=(-\frac{\frac c2t_0}{\sqrt{\gamma}},\mathrm{zz}(\omega (\frac{t'}{\sqrt{\gamma}}-\frac 14\frac{t_0}{\gamma}-t_0))).$$

We now conclude that the light beam appears to be bouncing up and down, as $x'$ in constant. The frequency of the bounce is a bit different because of time dilation though.

However, the effect of the moving emitter is that although the light is going up and down, no light beams will ever overlap.

So there is not paradox here, both observer see a zig-zag pattern. It's just that for the stationary observer the pattern is stationary and for the moving observer the patterns (compressed by length contraction) seems to be moving backwards.

NOTE: The same phenomena happens also in without mixing in relativity. To see this, put $\gamma=1$ and let $t=t'$.

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  • $\begingroup$ and what is the relationship between the perpendicular dimension and the up and down directions? $\endgroup$ – Allyn Shell Dec 9 '15 at 14:10
  • $\begingroup$ @AllynShell There is no distinction between the perpendicular outgoing (up?) and the incoming (down?) directions. You mistakenly assume there will be a standing wave, but that is not the case. $\endgroup$ – Mikael Fremling Dec 9 '15 at 14:55
  • $\begingroup$ Please ignore the term "standing wave." If (for the moving observer) the light's motion is aligned with the dimension perpendicular to the velocity of that inertial frame of reference, then that motion is staying on this dimension's axis. But there is about a 60 degree angle between the up and down motion for the stationary observer. My question is about the (change? of) nature of the dimension perpendicular to the velocity vector of an inertial reference frame. Is that truely different for the two reference frames? $\endgroup$ – Allyn Shell Dec 9 '15 at 19:51
  • $\begingroup$ @AllynShell I think the standing wave part was cause of much confusion. I refined my answer and gave some mathematical details. $\endgroup$ – Mikael Fremling Dec 10 '15 at 10:01
  • $\begingroup$ Yes I see that (and you have explained it well), but my question only relates to the nature of a dimension. The light itself is irrelevant to my question. I apologize that it introduce confusion. I have changed the title of this question to move the emphasis of the paradox to the dimension itself. $\endgroup$ – Allyn Shell Dec 10 '15 at 15:02
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The two mirrors will appear to tilt to the moving observer. If he is moving at the rate of speed to make the light appear to bounce back and forth, the angle will be such that the incoming angle from the source will equal the tilt angle. I should edit to say the incoming ray will appear to tilt from the original angle.

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