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Special Relativity Clocks 
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#19
Mar2712, 10:29 AM

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JM 


#20
Mar2712, 11:07 AM

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DaleSpam:
I hope this clarifies my understanding. JM 


#21
Mar2712, 11:25 AM

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JM 


#22
Mar2712, 11:30 AM

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#23
Mar2712, 11:31 AM

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JM 


#24
Mar2712, 11:38 AM

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JM 


#25
Mar2712, 12:14 PM

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#26
Mar2712, 12:18 PM

P: 231

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 v^{2}/c^{2}), 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 


#27
Mar2712, 01:18 PM

PF Gold
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#28
Mar2712, 01:22 PM

PF Gold
P: 4,695

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. 


#29
Mar2712, 01:49 PM

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#30
Mar2712, 02:29 PM

Mentor
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#31
Mar2712, 03:24 PM

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With an illegal copy of a notsoperfect translation linked in post #2. 


#32
Mar3112, 08:16 PM

P: 231

Einsteins mathematical derivation of the transform of time begins with the equation defining the synchronization of the moving clocks. The light postulate is then used with geometry to develop the transform. The only feature of the moving clocks used in this analysis is the synch. relation. He later speaks of clocks qualified to mark the time of the moving frame. Since the time of the moving frame is defined by the transform, doesnt 'qualified' mean the clocks display the time given by the transform? At this point there seems to be a choice. Either the clocks originally placed in the moving frame are 'programed' to display t, or additional clocks are supplied ( as we would use stop watches) to display t. JM 


#33
Mar3112, 09:04 PM

P: 231

Einstein refers to a clock qualified to mark the time t (my notation) when at rest relativily to the moving system and so adjusted that it marks the time t. This adjustment seems to mean that the moving clock displays the time t given by the transform, doesn't it? Then he says "Between the quantities x,t,and τ, which refer to the position of the clock,..." (his notation), x and t being the coordinates of the stationary frame and τ being the time of the moving frame. By what justification does x refer to the position of the clock? In the transforms, as they are usually viewed, x and t are independent variables allowed over the range ∞ to +∞. If slow clocks is universal then x must be permanently restricted to the values x=vt. If x is an independent variable then there is no significance to where x is 'pointing' because all clocks read the same value wherever located. Thanks for your participation. JM 


#34
Apr112, 09:37 PM

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#35
Apr712, 11:25 AM

P: 231

George and DaleSpam,
Are we done here? The discussion here leads to the suggestion that the 'slow clock' idea, with τ<t, is only one possible result for the moving clocks. Choices of x other than x=vt lead to different relations, such as τ>t for x=0. Synchronization means that the result applies to all the moving clocks, including the one at the origin of the moving frame. Do you all accept this idea, or do we have some more to talk about? JM 


#36
Apr712, 01:18 PM

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P: 16,989

If you have a clock which is moving in an arbitrary fashion (including, but not limited to, x=vt) you use the following formula to calculate the time displayed on the clock: [tex]\tau = \int \sqrt{1v(t)^2/c^2} dt[/tex] http://en.wikipedia.org/wiki/Proper_...ial_relativity The integrand is always less than or equal to 1, so you never get [itex]d\tau>dt[/itex] where t is the time coordinate in an inertial frame and v is the clock's velocity in that frame. 


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