I Question about a Square Light Clock

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TL;DR Summary
Light clock with 4 plates
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At each vertex of a square, there is a mirror with its plane perpendicular to the diagonal containing the vertex. The reflecting side faces the center of the square.

A beam of light is successively reflected off each of the mirrors, such that it travels around a square circuit. Each time it completes a circuit, we count a unit of time.

Does the unit of time defined in this way depend on speed?
 
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Yes, all clocks depend on speed in the same way.
 
south said:
Does the unit of time defined in this way depend on speed?
Speed relative to what?
Imagine two of these light clocks, identically constructed and moving relative to one another. Both will find that the other one is running slow (and length-contracted in the direction of motion, and Terrell-rotated).
 
"Speed" of what?

That clock you've designed and constructed in your lab is currently moving relativistically (as observed by someone in, say, Andromeda).

Do you observe your own clock in your lab behaving weirdly?
 
Consider this: Instead of just that light clock, you have second standard Light clock next to it, with the mirrors spaced apart such that the back and forth trip is the same length as the square light clock path. These clocks are fixed relative to each other. Anyone at rest with respect to them would see the two clocks ticking in perfect sync. Now add an observer who in in motion with respect to both clocks and measures the clocks undergoing time dilation. If the two clocks didn't maintain the same tick rate, they'd go out of sync and you'd have the two observers recording two physically contradicting results.
 
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south said:
TL;DR Summary: Light clock with 4 plates

Does the unit of time defined in this way depend on speed?
@Janus' analysis shows that this clock (and indeed any other) must time dilate the same way, or else there must be an absolute sense of rest. No further analysis is needed. However, the analysis from first principles of this clock is a trivial extension to the usual light clock as long as velocity is parallel to one side (it's messier for a velocity that is not parallel to a side, but still not particularly difficult). You should carry it out (edit: note that there's a completely maths-free way of doing it if you think a bit). The only thing to remember is the length contraction factor of the side parallel to the motion.
 
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Length contraction: ##a' = a\sqrt{1-v^2/c^2}##

##T' = {a \over \sqrt{c^2-v^2}} + {a' \over c-v} + {a \over \sqrt{c^2-v^2}} + {a' \over c+v}##

##T' = {a \over \sqrt{c^2-v^2}} + {a' (c+v)\over c^2-v^2} + {a \over \sqrt{c^2-v^2}} + {a' (c-v)\over c^2-v^2}={4a \over c}\gamma##
 
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Janus said:
If the two clocks didn't maintain the same tick rate, they'd go out of sync and you'd have the two observers recording two physically contradicting results.
You'd have a way to distinguish between a state of rest and a state of uniform motion, violating the 1st Postulate.
 
Thanks to the persons wich help me. Best regards.
 
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