synchronized clocks with respect to rest frameby mananvpanchal Tags: clocks, frame, respect, rest, synchronized 

#55
Mar312, 06:38 AM

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If you want to generalize it then you are certainly welcome to. You would need to use an arbitrary velocity, v, instead of 0.6 c as I used here. Then your more general expression would reduce to these equations for v = 0.6 c and would reduce to the equations for the train going the opposite direction for v = 0.6 c. I have proven beyond any doubt that the clocks are desynchronized after t=τ=0 for the scenario in the OP. Do you understand and agree with that? 



#56
Mar312, 06:42 AM

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P: 16,477

It would help greatly if you would label different specific events on the worldlines and calculate the spacetime coordinates in the given frame and what the clocks read at those events. Then we can communicate clearly about your drawings. 



#57
Mar312, 08:20 AM

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#58
Mar312, 11:15 AM

P: 249

I don't think it is that hard to see that if two locations in an object are seen to travel at the same speed that they would have to expereince the same amount of time dialation, and they are just observed by an outside observer to be out of sync. I just don't know what else to tell ya, besides maybe that if you insist on only using that equation, I agree you shouldn't ever just use the time dialation equation alone without it... 



#59
Mar312, 09:49 PM

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#60
Mar512, 06:21 AM

P: 215

There is another doubt! This may be the case again of generalization. [itex]t'=\tau  0.75 d, \mbox{if } \tau \ge 0[/itex] As [itex]\tau[/itex] increases, desynchronization between two clocks decreases. So, for [itex]\tau < 0[/itex], all clocks are synchronized. At [itex]\tau = 0[/itex] train changes its frame, and starts moving with constant speed 0.6c, the clocks is most desynchronized. For [itex]\tau > 0[/itex], as [itex]\tau[/itex] increases, desynchronization between two clocks becomes less and less. And after much more time, there is very negligible amount of desynchronization remains with respect to O. Please, explain me how can we solve this? 



#61
Mar512, 06:32 AM

P: 215

http://www.fourmilab.ch/documents/Re...fSimultaneity/ Which says that event can be said to be occurred when it is perceived by observer. (Or this is not telling what I have wrote, please correct me.) http://www.pitt.edu/~jdnorton/teachi...ivity_rel_sim/ Please, read last topic "What the Relativity of Simultaneity is NOT" Which says that event is happened at some time and location. event occurring is not depend on when observer persist it. (Or this is not telling what I have wrote, please correct me.) 



#62
Mar512, 06:52 AM

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P: 16,477

For example, consider the clocks [itex]d=0[/itex] and [itex]d=1[/itex]. At t'=100 we have [itex]\tau_0=100[/itex] and [itex]\tau_1=100.75[/itex] so the desynchronization is [itex]\tau_1  \tau_0=0.75[/itex]. At t'=200 we have [itex]\tau_0=200[/itex] and [itex]\tau_1=200.75[/itex] so the desynchronization is [itex]\tau_1  \tau_0=0.75[/itex]. 



#63
Mar512, 07:10 AM

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P: 16,477

The section "What the Relativity of Simultaneity is NOT" is correct. In relativity all of the "appearance" effects due to the finite speed of light are compensated for. I.e. in the Fourmilab page the yellow, blue, and gray observers are not stupid but they realize that the speed of light is finite and they account for the finite speed of light and the different distances to the red and green flashes. They would all determine that the flashes happened simultaneously. 



#64
Mar512, 07:30 AM

P: 215

Can you please explain me how can I get [itex]\tau_a[/itex] and [itex]\tau_b[/itex]? 



#65
Mar512, 07:37 AM

P: 215

Hello John232, DaleSpam
I think there might be some misunderstanding with you guys. 



#66
Mar512, 09:52 AM

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P: 16,477

[tex]r_d=\left( t=\begin{cases} \tau_d & \mbox{if } \tau_d \lt 0 \\ 1.25 \tau_d & \mbox{if } \tau_d \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau_d \lt 0 \\ 0.75 \tau_d+d & \mbox{if } \tau_d \ge 0 \end{cases} ,0,0\right)[/tex] [tex]r'_d=\left( t'=\begin{cases} 1.25 \tau_d  0.75 d & \mbox{if } \tau_d \lt 0 \\ \tau_d  0.75 d & \mbox{if } \tau_d \ge 0 \end{cases}, x'=\begin{cases} 1.25 d  0.75 \tau_d & \mbox{if } \tau_d \lt 0 \\ 1.25 d & \mbox{if } \tau_d \ge 0 \end{cases} ,0,0\right)[/tex] So for [itex]\tau_d \ge 0[/itex] we get [itex]\tau_d=0.8t[/itex] in the unprimed frame and we get [itex]\tau_d=t'+0.75d[/itex] in the primed frame. 



#67
Mar512, 03:09 PM

P: 249

"It might appear possible to overcome all the difficulties attending the definition of “time” by substituting “the position of the small hand of my watch” for “time.” And in fact such a definition is satisfactory when we are concerned with defining a time exclusively for the place where the watch is located; but it is no longer satisfactory when we have to connect in time series of events occurring at different places, or—what comes to the same thing—to evaluate the times of events occurring at places remote from the watch."  Albert Einstein
http://www.fourmilab.ch/etexts/einst...www/#SECTION11 



#68
Mar512, 06:51 PM

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P: 16,477

Yes, good quote and excellent link. You will note that Einstein derives the Lorentz transform as the general equation in section 3 and then derives time dilation as a special case in section 4. This corroborates my earlier claim that the Lorentz transform is more general.




#69
Mar612, 12:38 AM

P: 215

Now, we have two unknown variables [itex]t'[/itex] and [itex]\tau_d[/itex]. 



#70
Mar612, 05:38 AM

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#71
Mar712, 09:29 AM

P: 141

First lets take another situation, lets just say that the two clocks on the train are in sync in the train's frame of reference. Lets say that the train is moving at .5c in the stations frame of reference. Do you understand why the clocks won't by in sync in the stations frame of reference, if they are in sync in the trains?




#72
Mar2912, 11:13 PM

P: 215




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