Is the Light Clock Calibration the Key to Understanding Relativity?

[DAC]

Good question! The answer is, both clocks tick at the same rate as each other, regardless of which frame you choose (but that rate will be different in different frames, illustrating time dilation).

This is a good illustration of the difficulties you run into if you try to apply Newtonian mechanics while also trying to keep the speed of light constant in all frames. In the frame in which both light clocks are at rest (the train frame), obviously both tick at the same rate; the vertical light makes the round trip in the same time as the horizontal light, so both light beams, if they start out together, finish each round trip (each "tick") together. That means the two light beams coincide (are at the same point in space) once per "tick".

But if the two light beams coincide in one frame, they must coincide in all frames; the coincidence of the two light beams, each time it happens, is a "marker" in spacetime (the same kind I was talking about in an earlier post), and you can't change such markers by changing frames. So the two light clocks must tick at the same rate as each other in all frames.

But how is that possible, since the vertical clock is not length contracted (light moving perpendicular to direction of motion), but the horizontal one is (light moving parallel to direction of motion)? As DaleSpam recommended, you should work out the math for this explicitly; but here's a quick way to see how it could work.

We already know that the spatial path of the vertical clock's light, in the frame in which the train is moving, is not vertical but diagonal--so it's longer (and the light therefore takes longer to traverse it). The spatial path of the horizontal clock's light is length contracted, yes--but the front mirror is moving away from the light on the forward leg (if we assume the light starts from the rear mirror) while the rear mirror is moving towards the light on the return leg. It turns out that the effect of these two things on the light's travel distance, in the frame in which the train is moving, is not symmetrical: the forward leg is lengthened (by the front mirror moving away from the light) more than the return leg is shortened (by the rear mirror moving toward the light). This just compensates enough for length contraction so the end result is that the horizontal light travels the same spatial distance, in the frame in which the train is moving, as the vertical light does.

Does the trip out and back total the same each return trip, or is the trip out getting increasingly longer each time?
Does the length contraction affect both trips equally?
Does the horizontal clock time dilate and or length contract?
Thanks.

 

[Dale]

Does the trip out and back total the same each return trip, or is the trip out getting increasingly longer each time?

The distance the light travels is the same each trip. What do you conclude from that?

Does the length contraction affect both trips equally?

Yes, insofar as length contraction can be said to affect a trip at all.

Does the horizontal clock time dilate and or length contract?

Both.

 

[DAC]

From what you said, I will assume both clocks in the moving frame, experiencing the same motion of the train, are both time dilated. But that motion also causes length contraction, but that only applies to the horizontal clock. So on the one hand the clock ticks more slowly, yet on the other it ticks faster, the mirrors being closer together.
You would think if both are time dilated and only one is length contracted, they couldn't tick at the same rate, but this SR. How about a clue.

 

[PeterDonis]

You would think if both are time dilated and only one is length contracted, they couldn't tick at the same rate

You might, if you didn't actually do the math. We have already advised you several times to do the math. My post #27 gives good hints at how to proceed with that. If you get into it and you are stuck, post what you're able to do and we can help with it. But it's no good just continuing to say "I don't see how..." without actually working through the math. This isn't something that's just going to come by intuition; if your intuition were already trained to deal with relativity we wouldn't have to have this discussion.

 

[Dale]

How about a clue.

Calculate how long it takes for a horizontal clock to tick given its length L and speed v.

 

[DAC]

Calculate how long it takes for a horizontal clock to tick given its length L and speed v.

Calculate how long it takes for a horizontal clock to tick given its length L and speed v.

Time is distance over c. And distance in this case is the length of the clock plus the distance it has moved. Or for the return leg, less the distance moved.
So if the distance between mirrors is, say, one metre, and the train/clock advances 0.1 metres, the distance the light travels is 1.1 metres, divided by c gives the clock tick over time. Or 0.9metres divided by c for the return leg. I don't see what this proves other than the there and back leg is 2 metres, which is the same as the stationary frame

 

[Dale]

Time is distance over c. And distance in this case is the length of the clock plus the distance it has moved. Or for the return leg, less the distance moved.

Write it out algebraically, in terms of $L$ and $v$. What does $L$ have to be in order for the horizontal clock to tick in sync with a vertical clock of length $L_0$ ?

So if the distance between mirrors is, say, one metre, and the train/clock advances 0.1 metres, the distance the light travels is 1.1 metres, divided by c gives the clock tick over time. Or 0.9metres divided by c for the return leg. I don't see what this proves other than the there and back leg is 2 metres, which is the same as the stationary frame

Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?

 

[DAC]

Write it out algebraically, in terms of $L$ and $v$. What does $L$ have to be in order for the horizontal clock to tick in sync with a vertical clock of length $L_0$ ?Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?

Write it out algebraically, in terms of $L$ and $v$. What does $L$ have to be in order for the horizontal clock to tick in sync with a vertical clock of length $L_0$ ?Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?

OK you win, anyway don't you guys provide answers rather than questions. Keep up the good work.

 

[Nugatory]

And on that note I think we can close the thread