Special relativity paradox involving two light clocks

Question

I came up with this thought experiment that seems to exemplify a paradox. In this paradox, one clock is ticking unevenly for one observer and evenly for another.

Essentially, the clocks record a “tick” each time the light reaches a mirror. There is a light clock in the standard arrangement perpendicular to the motion and a second light clock arranged parallel to the motion. In the parallel light clock it takes more time for the light to go forward than backward. So it seems like one light clock ticks evenly and the other ticks unevenly.

I created a diagram to explain the paradox. I can't figure out the error in my logic. Can anyone help?

Answer 1

Typically we would not consider each mirror to constitute a “tick” but rather a full round trip. However, that is rather minor and is just a small semantic issue.

In principle your analysis is correct. The effect you have noted is due to the relativity of simultaneity. Indeed, the time between the rear and the front tick is different than the time between the front tick and the rear tick in the frame where the horizontal clock is moving. (With a “tick” defined your way)

The relativity of simultaneity is the most challenging concept for new students of relativity to grasp. Hence it is the source of most of the relativistic paradoxes that you will find. In the frame where the light clock is stationary the ticks are even, and in the frame where it is moving the ticks are not. This requires the relativity of simultaneity as neither time dilation nor length contraction can produce this effect.

Answer 2

Dale's answer is right. Here is a little more detail:

Let's set the clocks right next to each other, so that the bottom of Clock 1 coincides with the left end of Clock 2. Let's conveniently make each clock one light-second long.

When the clocks start, light leaves the origin in both the up and rightward directions. In the train frame, the first tick of the vertical clock occurs at $(t=0,x=1,y=0)$ and the first tick of the second clock occurs at $(t=1,x=0,y=1)$. Lorentz transform these and you'll find that in the platform frame, these ticks occur at $t'=(1-xv)/\sqrt{1-v^2}$ and $t'=1/\sqrt{1-v^2}$. These are not the same so the ticks in the platform frame are not simultaneous.

The second ticks both occur at $(t=2,x=0,y=0)$. Because these are the same point in spacetime, you don't have to do any arithmetic to know the coordinates will transform identically to the platform frame.

The fact that the first (and third and fifth and seventh....) ticks are out of synch in the platform frame generally makes them inconvenient for illustrating various simple points. The fact that the second (and fourth and sixth and eighth...) ticks are simultaneous in both frames (together with the fact that they obviously have to be simultaneous even before you get into any details) makes them very convenient illustrating those same points.

Answer 3

The size of your clock is important. If the size of the clock is effectively zero, there is no problem. A clock of size zero is one based on local events, such as detecting light at a fixed location (the clock in which the light moves back and forth qualifies as a zero size clock). If your clock ticks in the way you defined it, then it is not based on local events, but on events happening at different locations. This happens in the "every other tick" that you defined, your clock will have an error equal to the lack of synchronicity between clocks located at these two different positions.

The reason is that each tick in your clock will measure the time of a different zero sized clock, these different clocks are the zero-size clocks located at the two ends of the finite sized clock. So it is not surprising that your clock does not tick uniformly. It is effectively ticking in synchrony with clocks located at different locations on alternating ticks. Because these clocks are out of sync, it looks as "non-uniform" ticks in your clock. Thus any finite clock will have an error accounting for this lack of synchrony of ideal clocks located at its different parts.

Answer 4

I can't be certain with my answer as I am not a scientist, but in the scenario the light only becomes uneven when the train increases or decreases speed. Person A would not see this affect as they are part of the increase or decrease in speed, however the onlooker person B would see this because they are traveling at a different speed. This scenario describes the idea that your speed determines your time.

Farther information on this topic can be found at https://www.space.com/36273-theory-special-relativity.html#:~:text=One%20of%20the%20many%20implications,when%20it%20is%20at%20rest.

Answer 5

Congratulations, you have discovered an example that the speed of light is not isotropic by Einstein's theory, despite the claim .
Another point is that the light clocks in motion, oriented perpendicular to motion (y- and z-axis, in both positive and negative directions), all run synchronously, and that non-simultaneity only exists in one direction. Usually this is glossed over by assuming that the ticks are evened out between the clocks, despite the obvious difference in this description to the mathematics. Maybe you would be less likely to be confused about this if you knew that these clocks only tick at the same rate when a mirror is added, and that clocks aligned perpendicular to each other tick at the same rate, eventually, because they invented length contraction. Without length contraction clocks will never run at the same rate, and since assuming just the second postulate contradicts the first postulate, since the clocks tick at different rates, you can combine your acute observation, with the fudge factor of invented length contraction, and realize the theory is worthless as a description of reality. Without length contraction, the two postulates of special relativity contradict each other, all other weirdness aside.