Trying to understand relativity of lengths and times

[JesseM]

i understand. let me try it this way: the stick is stationary in the "rest frame". the axis of the stick lies along the x-axis of the "rest frame". the end A of the stick coincides with the origin of the "rest frame". L is the distance between the end A of the stick and the end B of the stick (the length of the stick). the length L is measured by a meterstick which is also stationary. a constant velocity v relatively to the origin of the "rest frame" in the direction of increasing x is imparted to the stick at TA (i will explain later what i mean by TA). the observer moving together with the meterstick and the stick to be measured, measures the length of the stick again directly by superposing the meterstick, in just the the same way as if all three were at rest.

clocks are set up in the "rest frame". these clocks are in sync. a clock is attached to the end A of the stick. another clock is attached to the end B of the stick. the clock attached to the end A of the stick and the clock attached to the end B of the stick are in sync with the clocks set up in the "rest frame", that is to say that their indications correspond at any instant to the "time of the 'rest frame'" at the places where they happened to be. these clocks are therefore in sync in the "rest frame".

a ray of light departs from the end A of the stick at the time TA (keep in mind that at the time TA the end A of the stick coincides with the origin of the "rest frame"). (keep in mind also that TA is the position of the hands of the clock set up at the origin of the "rest frame" and the position of the clock attached to the end A of the stick.) the ray of light is reflected at the end B of the stick at the time TB , and at the time TB the end B of the stick coincides with the point x1 on the x-axis of the "rest frame".

We are still assuming the stick is moving, correct? i.e. the B end of the stick was only at x1 at time TB, at earlier or later times (including the time TA) it would be at a different point on the x-axis?

(keep in mind that TB is the position of the hands of the clock set up at the point x1 in the x-axis of the "rest frame" and the position of the hands of the clock attached to the end B of the stick.) the ray of light arrives at the end A again of the stick at the time T'A (the position of the clock attached to the end A of the stick).

OK.

the two clocks are in sync if TB - TA = T'A - TB.

This is incorrect. You said earlier that the stick was moving in the "rest frame", but that the clocks at the A end and the B end would match the time in the rest frame. But in the rest frame, when the light departs from the A end, the B end is moving away from the starting point of the light, whereas when the light is reflected from the B end, the A end is moving towards the point where the light was reflected. Since the rest frame says the light itself moves at c in both directions, it should take more time for the light to go from A to B than it does for the light to go from B back to A.

L' is the distance between the origin and the point x1 on the x-axis of the "rest frame".

OK, x1 is the point on the x-axis where the B-end of the stick is at time TB. And you said earlier that at time TA, the A-end of the stick was at the origin x=0. Since the stick is moving, you're clear that according to this definition L' would not be the length of the stick in the "rest frame", correct?

taking into consideration the principle of the constancy of the velocity of light we find that

L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA.

False, given your definition of L'. Where did you get this formula? If you derived it yourself, can you explain the derivation? And if you just copied it from somewhere, I feel confident that the author, whoever they were, intended the symbol L' to represent the length of the stick in the "rest frame", which would be equal to the position of the B-end at TA (assuming that the A-end was at the origin at TA), not to the position of the B-end at TB. Again, if you assume the B-end is at position x = L' at time TA, and it's moving along the x-axis at velocity v, then at time TB it will have moved an additional distance of v*(TB - TA) along the x-axis, so its new position will be x = L' + v*(TB - TA). And if the light beam started at the origin at time TA, and reached the B end of the stick at time TB, then we have [distance]/[time] = c which gives:
[L' + v*(TB - TA)] / [(TB - TA)] = c
L'/(TB - TA) + v = c
L'/(TB - TA) = c - v
(TB - TA)/L' = 1/(c - v)
TB - TA = L'/(c - v)

So you can see it is easy to derive the formula you give if we assume L' represents the length of the stick in the "rest frame", so that if the A-end was at x=0 at time TA, the B-end would be at x=L' at time TA, and would later be at x=L' + v*(TB - TA) at time TB.

On the other hand, if you assume that the B-end was at position x=L' at time TB, so that L' is not the length of the stick, then I have no idea how you would derive the formula L'/(c - v) = TB - TA. If you think it is possible that whoever wrote this formula meant L' in the way that you mean it, can you show the derivation of the formula?

Also, note that the two formulas L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA contradict your earlier assertion that "the two clocks are in sync if TB - TA = T'A - TB." TB - TA cannot equal T'A - TB according to these formulas, because L'/(c - v) will not equal L'/(c + v) for any nonzero value of v.

 

[myoho.renge.kyo]

We are still assuming the stick is moving, correct? i.e. the B end of the stick was only at x1 at time TB, at earlier or later times (including the time TA) it would be at a different point on the x-axis? OK. This is incorrect. You said earlier that the stick was moving in the "rest frame", but that the clocks at the A end and the B end would match the time in the rest frame. But in the rest frame, when the light departs from the A end, the B end is moving away from the starting point of the light, whereas when the light is reflected from the B end, the A end is moving towards the point where the light was reflected. Since the rest frame says the light itself moves at c in both directions, it should take more time for the light to go from A to B than it does for the light to go from B back to A. OK, x1 is the point on the x-axis where the B-end of the stick is at time TB. And you said earlier that at time TA, the A-end of the stick was at the origin x=0. Since the stick is moving, you're clear that according to this definition L' would not be the length of the stick in the "rest frame", correct? False, given your definition of L'. Where did you get this formula? If you derived it yourself, can you explain the derivation? And if you just copied it from somewhere, I feel confident that the author, whoever they were, intended the symbol L' to represent the length of the stick in the "rest frame", which would be equal to the position of the B-end at TA (assuming that the A-end was at the origin at TA), not to the position of the B-end at TB. Again, if you assume the B-end is at position x = L' at time TA, and it's moving along the x-axis at velocity v, then at time TB it will have moved an additional distance of v*(TB - TA) along the x-axis, so its new position will be x = L' + v*(TB - TA). And if the light beam started at the origin at time TA, and reached the B end of the stick at time TB, then we have [distance]/[time] = c which gives:
[L' + v*(TB - TA)] / [(TB - TA)] = c
L'/(TB - TA) + v = c
L'/(TB - TA) = c - v
(TB - TA)/L' = 1/(c - v)
TB - TA = L'/(c - v)

So you can see it is easy to derive the formula you give if we assume L' represents the length of the stick in the "rest frame", so that if the A-end was at x=0 at time TA, the B-end would be at x=L' at time TA, and would later be at x=L' + v*(TB - TA) at time TB.

On the other hand, if you assume that the B-end was at position x=L' at time TB, so that L' is not the length of the stick, then I have no idea how you would derive the formula L'/(c - v) = TB - TA. If you think it is possible that whoever wrote this formula meant L' in the way that you mean it, can you show the derivation of the formula?

Also, note that the two formulas L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA contradict your earlier assertion that "the two clocks are in sync if TB - TA = T'A - TB." TB - TA cannot equal T'A - TB according to these formulas, because L'/(c - v) will not equal L'/(c + v) for any nonzero value of v.

i am sorry. i understand now where i am having difficulties. i will set up the problem differently in the near future in a different thread. thanks!