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Synchronized clocks with respect to rest frame 
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#37
Feb2912, 09:31 PM

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In other inertial frames or in non inertial frames they will not be synchronized. O's frame is non inertial. 


#38
Mar112, 01:16 AM

P: 71

mananvpanchal first post (with quotation marks) and my comments without quotation marks.
"Suppose, A and B is clocks at both end of train. A is at left and B is at right. Observer O is at middle of train at point M. Observer R is on platform.Train is at rest and O synchronize both clock. The clock is synchronized with respect to both observer." clocks are synchronized. "Now, train starts moving to right. It accelerate and after some time it runs with constant speed. Now, the clock is still synchronized with respect to O." At the point that the entire train is again inertial, yes the clocks, tell the same time as O and O confirms this as being true. During acceleration A clock will appear to run behind O and B will appear to run ahead of O WRT the inertial synchronization (also B will appear larger and A will appear smaller very minutely during acceleration). The aberration in time measurement during acceleration corrects itself when the object returns to an inertial state there is no measurable difference in the clocks after the train return to an inertial state. "But what about R? Is clocks synchronized with respect to R? If no, then which clock is ahead A or B?" Clearly The clock of R will be running faster than A and B and O and continue to run at the same faster rate with the train again inertial. From the moment the train started moving WRT R, the clock of R was not synchronized with the clocks on the train. "Thanks." I hope this answers your question. mathal 


#39
Mar112, 01:19 AM

#40
Mar112, 04:56 AM

P: 249

For each velocity, there could only exist one distortion of spacetime that would conclude that there is the same value for the speed of light. For instance, two people traveling together on a train couldn't be seen to measure time differently because then they would each have different values of the speed of light. Their location as being in the front or back of the train doesn't affect how they can make that measurement. The basis of SR is that the speed of light is constant in all frames. I think clocks should only be checked if they are synchronized using an entanglement experiment, something Einstein never would have done because nothing was supposed to travel faster than the speed of light at the time. 


#41
Mar112, 10:48 AM

P: 71

When I posted my responce I edited myself twice and I'm still not satisfied with what I wrote. It all boils down to the measurement problem.
If there are only 2 clocks held by A and B then while the train is inertial O will observe the same time on both clocks. For instance if the train is dark inside and the clocks flash seconds the signals will reach O at the same moment. If O uses a flash of light to see each clock the images will show the same time. During acceleration things are different. Using flashing clocks the signal from A is travelling to O who is accelerating away from this flash and so will receive it later than the flash from B which is approaching him as he accelerates towards it. The impression this gives is that A clock is lagging behind B clock. The observation when O uses flashes of light towards A and B clocks is the opposite. His flash arrives later to B then it does to A. The picture of the clock times gives the impression that A clock is running faster than B clock. (tic) The images of the clocks will arrive back to O at the same moment. If the acceleration is constant there will be a constant lag time. Simultaneous identical acceleration of two clocks does not break the synchronization. mathal 


#42
Mar112, 10:59 AM

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P: 17,541

So, in R's frame the worldline of A, O, and B are: [tex]r_d=\left(t,x=\begin{cases} d & \mbox{if } t \lt 0 \\ 0.6 t+d & \mbox{if } t \ge 0 \end{cases} ,0,0\right)[/tex] where d=1 for A, d=0 for O, and d=1 for B. As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, τ, using the spacetime interval. Solving for t we get: [tex]t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}[/tex] Substituting in to the above we get: [tex]r_d=\left( t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)[/tex] Noting that τ does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame. Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain. [tex]r'_d=\left( t'=\begin{cases} 1.25 \tau  0.75 d & \mbox{if } \tau \lt 0 \\ \tau  0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d  0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)[/tex] Noting that τ does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. If there is some step that you do not follow then please ask for clarification. But it is quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame. 


#43
Mar212, 02:53 AM

P: 215

Thanks for your reply.
From 1st equation the calculation is done considering speed 0.6c. So, you are trying to say is: at t < 0, clocks is synchronized for both R and O. And at t = 0 train changes its frame and travels at 0.6c, and at t=0 the clocks becomes desynchronized for O. Whatever you stated before you state same here with maths. Actually, I like maths, but, equations comes from theory. It state same thing as theory states. I need to understand the theory first. Actually I need be explained that how clocks becomes desynchronized for O, when train changes its frame. Please, shade some light on my post #36 and #39. Please, tell me what is wrong with my understanding. Thanks again DaleSpam. 


#44
Mar212, 02:53 AM

P: 215

Thanks mathal



#45
Mar212, 07:40 AM

P: 249

Looks to me like you rigged the equations to give different values for A, O, and B from the intitial setup of the first equation. Say for instance instead, I used different equations like the time dialation equation. Then I solved for A, O, and B. I would get the same answer for each one that would end up getting me different values than what you got...



#46
Mar212, 07:53 AM

P: 249

This would be because d has different values for A, O, and B. If I simply put the values of A in the time dialation equation it would have the same values for O and B, so then I would know that they should all come out to be the same since they had the same value for t and v. Distance wouldn't be an issue in my setup. I don't think it should since the distance the objects are away from each other wouldn't affect how they measured the speed of light...



#47
Mar212, 10:26 AM

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#48
Mar212, 11:16 AM

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[tex]r'_d=\Lambda \cdot r_d = \left( \begin{array}{cccc} 1.25 & 0.75 & 0 & 0 \\ 0.75 & 1.25 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \cdot \left( \begin{array}{c} t = \begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases} \\ x = \begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} \\ 0 \\ 0 \end{array} \right)[/tex] [tex]r'_d=\left( \begin{array}{c} t'= 0.75 \begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} + 1.25 \begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases} \\ x'= 1.25 \begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases}  0.75 \begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases} \\ 0 \\ 0 \end{array} \right)[/tex] [tex]r'_d=\left( t'=\begin{cases} 1.25 \tau  0.75 d & \mbox{if } \tau \lt 0 \\ \tau  0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d  0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)[/tex] 


#49
Mar212, 11:26 AM

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P: 17,541

What you have not done is indicate what the clocks at A and B read at those events. However, if you look at the length of the line from the bend, when the clocks all read 0, to the events in question you will see that they are different lengths. Therefore A and B will not read the same at those events. They will therefore be desynchronized according to O. 


#50
Mar212, 01:27 PM

P: 249




#51
Mar212, 01:47 PM

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P: 17,541

Alternatively, feel free to actually work through the problem mathematically using whatever approach you prefer, and then post your work. If you get a different result then I will gladly help you see where you went wrong. 


#52
Mar212, 01:49 PM

P: 3,967

Sorry for the layman's sloppy explanation but I think that is the gist of it ;) 


#53
Mar312, 02:17 AM

P: 215

As we get the equation below
[itex]t'_a=\tau + 0.75,[/itex] [itex]t'_o=\tau, [/itex] [itex]t'_b=\tau  0.75[/itex] But suppose, train is going to opposite direction, then the equation cannot distinguish both [itex]t'[/itex]. We get same values for A and B for both direction, if A is ahead in one direction, A will be ahead in other direction too. The equations is depend on [itex]\tau[/itex], and [itex]\tau[/itex] is not depend on [itex]x[/itex]. [itex]\tau[/itex] is just time dilation, in whatever direction the train is going But, there must be meaning of direction of motion. What clock is ahead of other is matter of method used for reading, as mathal and I said before. But using any one method, the result should be opposite if direction of motion is opposite. If A is ahead in one direction, then B must be ahead in opposite direction. So, far We get only one reason for desynchronization is "length contraction". desync by length contraction.bmp The above image shows that train has length of l1 at start X, train accelerated to Y point, at that time length becomes l2. From now Y to Z train maintains its constant velocity, so now length remain same as l2. llxy, lrxy, llyz and lryz is distance traveled by front and end point of train. We can easily see that ratio between llxy and lrxy is more than llyz and lryz. So, as train maintains its constant velocity for longer time the desynchronization created by length contraction decreases. If we guess that one way speed of light is not same in all direction then we have now another reason to explain desynchronization on frame change. Like, light takes more time to reach to observer in direction of motion and less time in opposite direction. But, in this idea direction of motion matters. If we change direction of motion A and B changes its state of desynchronization. But, we cannot say one way speed of light is not same in all direction. And we cannot say change in frame cannot affect synchronization. May be any one we have to pick. 


#54
Mar312, 03:12 AM

P: 215

Ok, please look at second and third image. In second image O is currently in first frame. Second frame is also O's frame, but second frame is future of O. O sees that all events in first frame is simultaneous for him. If somehow O could see second frame, O can conclude that in second frame events will not be simultaneous for him. Now, O changes his frame, if we want to look actual scenario, we have to transform from first frame to second frame. To show transformation I converted second image to third image. Now, please look at third image. Again O sees in second frame all events is simultaneous for him. If somehow O could see his first frame, O can conclude that events would not be simultaneous for him in first frame. So, conclusion is if current frame is inertial then all events would be simultaneous for O. When frame is changed, some desynchronization occurs if this case happens. Suppose, O have fires pulses in both direction. O changes his frame in mean time before the pulses reach to the clockmirror. In that case desynchronization created. But it will be for just that one pulse firing. Now, O again in inertial frame. Now O fires pulses, O feels clocks is synchronized again. 


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