|Jan16-06, 08:42 AM||#1|
clock synchonization in relativity
I'm a bit confused about a statement in my course on relativity. I have a system S' that is moving with a direction v (in the x'-direction) compared to the system S, so the following transformation formula holds:
t = gamma*(t'+vx'/c^2)
The textbook now goes on to say (translated from dutch.. possibly poorly):
"Clocks that are synchronized in S (t=constant) will be delayed more (t'=smaller) as their distance along the x' axis increases."
Indeed, I agree that t' will be smaller, as x' increases. However, I was wondering the following. If at a time t in S all the clocks give a short lightflash or something, to show for instance that it's twelve o'clock, then in the S' system this flash will be seen at a time t' at a location x'. Now, in S' the lightflash will be seen sooner (t' smaller) as x' increases, so an observer in S' will conclude that as x' increases the clock actually runs more and more ahead, knowing that the flash means '12 o'clock' and that that '12 o'clock'-signal happens first for larger x'.
Is the statement in my course-book therefore wrong or am I messing something up? (probably the second, but I've thought this through a number of times now and I can't seem to figure it out)
|Jan16-06, 09:58 AM||#2|
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The question really is: For events that occur simultaneously in S (the S clocks all strike 12 noon, for example) when do they occur according to S'?
Start with the appropriate Lorentz transformation:
[tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]
Since [itex]\Delta t = 0[/itex], this tells you that according to S' a clock at position x (x > 0) in S will strike noon before a clock at the origin (x = 0). ([itex]\Delta t'[/itex] will be negative.)
A way to summarize the clock desynchronization effect is: Moving clocks, synchronized in their rest frame but separated by a distance D along their direction of motion, are not synchronized in the stationary frame; The front clock lags the rear clock by an amount:
[tex]T = Dv/c^2[/tex].
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