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ghwellsjr
#28
Mar27-12, 01:22 PM
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Quote Quote by JM View Post
Quote Quote by ghwellsjr View Post
OK, fine. Now do you accept Einstein's calculation for τ (tau), the rate at which a moving clock ticks in your stationary frame as a function of t, the rate at which the stationary coordinate clocks tick in your stationary frame and v, the velocity of the moving clock?
τ = t√(1-v2/c2)
George: This formula comes directly from the transform equation t = ( T-vX/c2)/√(1-v2/c2), with the assumption that X=vT.
But is the formula intended as an example, or as a universal truth? Consider.....
All theclocks are synchronized, so that all clocks of K read T, not just the clock at X=vT,and all the clocks of k read t, not just the one at x=0. If X=0, thus 'pointing at the origin of K', t = T/√ ( 1- v2/c2), and t>T. Because of synch. this result applies also to the theclocks at the origin of k. The conclusion, seemingly, is that the moving clock at the origin of k can run slow or fast depending on the value of X.
So is 'slow clocks' a universal truth? I'm looking for help, yes or no, and why.
JM
Yes, slow clocks is a universal truth in Special Relativity.

Einstein's derivation of the Proper Time on a clock moving at speed v as a function of t, Coordinate Time, in a frame comes from section 4 of his 1905 paper. Remember, Coordinate clocks always remain fixed at the locations at which they were synchronized within a particular Frame of Reference. If you look at his derivation, he starts off talking about "one of the clocks which are qualified to mark time t when at rest relatively to the stationary system". What he means is that there is a second synchronized clock located at the spatial origin of one reference frame prior to time zero which then becomes stationary in a second reference frame moving at v with respect to the first reference frame after their mutual time zero. He asks the question, "What is the rate of this clock, when viewed from the stationary system?"

So τ is the Proper Time of a single clock put in motion at time zero compared to the infinite number of Coordinate Clocks that remain stationary. We are comparing the time on this moving clock to the times on the adjacent clocks as it passes by them. The moving clock will always run slower than the stationary clocks. But remember, we are comparing one clock to a bunch of different clocks that have been previously synchronized.

So we always know the tick rate of a clock moving in a Frame of Reference by the simple formula expressed above.