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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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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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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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