The forum discussion centers on the calibration of light clocks in the context of Einstein's theory of relativity. Participants debate whether calibrating a light clock to tick every time light travels one meter affects the perceived passage of time in different frames of reference. It is established that while one meter is a consistent measure, the path lengths differ in moving frames, leading to time dilation. The conclusion emphasizes that the definition of a "tick" must remain invariant across frames to accurately illustrate time dilation.
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PeterDonis said: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.
The distance the light travels is the same each trip. What do you conclude from that?DAC said:Does the trip out and back total the same each return trip, or is the trip out getting increasingly longer each time?
Yes, insofar as length contraction can be said to affect a trip at all.DAC said:Does the length contraction affect both trips equally?
Both.DAC said:Does the horizontal clock time dilate and or length contract?
DAC said:You would think if both are time dilated and only one is length contracted, they couldn't tick at the same rate
Calculate how long it takes for a horizontal clock to tick given its length L and speed v.DAC said:How about a clue.
DaleSpam said:Calculate how long it takes for a horizontal clock to tick given its length L and speed v.
DaleSpam said:Calculate how long it takes for a horizontal clock to tick given its length L and speed v.
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## ?DAC said: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.
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 said: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
DaleSpam said: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.DaleSpam said: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?