## synchronized clocks with respect to rest frame

 Quote by DaleSpam But O isn't in constant motion, O accelerates. That acceleration is detectable in multiple ways, including the desynchronization of A and B.
I don't recal ever hearing that clocks that accelerated together at the same rate would ever no longer be synchronized. Thinking back to the Einstein in the elevator thought experiment, I thought, what if he had a stepladder. Could he move up the step ladder and find that his light ray bends by a different amount? I think no at each step, but yes as he moved up the steps. Then what if he did the same experiment at Earth? Closer to the Earth the beam would bend more, since it would have a stronger gravitational pull. If he moved farther away from the Earth it would bend less being farther away from the planet.

The trains acceleration, unlike Earths gravitational pull, would act equally on the whole train if it was rigid.
 Hello, The below images show O's accelerating motion with respect to R. As speed changes desynchronization becomes bigger with respect to R. Please, look at first frame of below image. It shows desynchronization of clocks when frame is in motion. But, it is with respect to R. When we transform to first frame, as below image shows there is no longer desynchronization with respect to O. But, second frame is desynchronized with respect to O. But, again we transform to second frame. there is no longer desynchronization with respect to O. But, first frame is desynchronized with respect to O. So, desynchronization is only there when we see other frame. Our own frame cannot have desynchronization with respect to itself. So, as I understand constant speed of frame would not create desynchronization, and changing in frame (acceleration) also would not create desynchronization. Can you explain me how acceleration can create desynchronization with respect to O?

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 Quote by John232 I don't recal ever hearing that clocks that accelerated together at the same rate would ever no longer be synchronized.
In an inertial frame this is correct. Two clocks which are initially synchronized and share the same speed profile will remain synchronized, but only in that inertial frame.

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

There is other doubt too.

 Quote by mananvpanchal Please, look at first frame of below image. It shows desynchronization of clocks when frame is in motion. But, it is with respect to R.
second image of post #36 says clocks is synchronized with respect to O, but not for R. For R clock A is ahead of clock B, because signal reaches lately to B with respect to R because of direction of motion.

Suppose, that train is coming from far left of R, it passes R and goes far right with constant velocity. So below image show how clocks becomes desync with respect to R. It also tells train coming from left and goes right not make changes in desync. A is always ahead of B with respect to R during whole journey.

But, wait if we think how R sees clocks during whole journey. Below image show that

There is matter that train coming from left and goes to right. Coming from left tells R that clock B is ahead of A, and going right tells R that clock A is ahead of B.
Look, at the point where R and O is on perpendicular line of train path way. This is the point where R thinks both clock is synchronized.

From my understanding, first image shows that "R thinks that clocks would be desynchronized like this with respect to O". Neither for R the clocks would be desynchronized like this, nor for O.

What I understand is clocks is perfectly synchronized for O and, clocks is desyhnchonized with respect to R as showed in second image.

Can anyone shade some light on this?

Thanks

 Quote by DaleSpam In an inertial frame this is correct. Two clocks which are initially synchronized and share the same speed profile will remain synchronized, but only in that inertial frame. In other inertial frames or in non inertial frames they will not be synchronized. O's frame is non inertial.
I fail to see how that is true without an explanation. For instance, how does the relation of the distance of two objects make them observered from rest as measuring time differently when they have accelerated the same amount for the same duration? I think the proof in the pudding here would be that if you considered each object separately you would find that they both expereince the same amount of time dialation, but if you considered them together or linked to an action they would be seen to experience different amounts of time dialation. So, then why would an object require different amounts of spacetime dialation if it was seen to travel at the same speed of another object in order to measure the same speed for light? It would make it seem that the origanal answer of the amount of spacetime dialation occured for each object would be false unless you considered your own relation to the object. Then answers that considered your relation to the object would not be interchangeable with answers that didn't consider your relation to the object, since they would give different values. Then if the object is traveling with a velocity then would the amount of spacetime dialation that occured change as it moved to a different location? Then the whole thing falls apart, none of the results would give the same speed of light or the same amount of time dialation if it was considered to be X units away from something else, and that affect the outcome of the amount of spacetime dialation that occured.

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.
 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

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 Quote by John232 I fail to see how that is true without an explanation.
 Quote by mananvpanchal Can you explain me how acceleration can create desynchronization with respect to O?
Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore lets use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame.

So, in R's frame the worldline of A, O, and B are:
$$r_d=\left(t,x=\begin{cases} d & \mbox{if } t \lt 0 \\ 0.6 t+d & \mbox{if } t \ge 0 \end{cases} ,0,0\right)$$
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:
$$t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}$$

Substituting in to the above we get:
$$r_d=\left( t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$

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.
$$r'_d=\left( t'=\begin{cases} 1.25 \tau - 0.75 d & \mbox{if } \tau \lt 0 \\ \tau - 0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d - 0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$

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.

 Quote by DaleSpam Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore lets use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame.
 Quote by DaleSpam 1. $r_d=\left(t,x=\begin{cases} d & \mbox{if } t \lt 0 \\ 0.6 t+d & \mbox{if } t \ge 0 \end{cases} ,0,0\right)$ 2. $t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}$ 3. $r_d=\left( t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$

 Quote by DaleSpam Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain. 4. $r'_d=\left( t'=\begin{cases} 1.25 \tau - 0.75 d & \mbox{if } \tau \lt 0 \\ \tau - 0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d - 0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$
But I don't follow this. I have tried to get it, but I don't succeed. That's fine, don't worry about it.

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.

Thanks mathal

 Quote by mathal 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
That is what I think. Constant speed cannot affect synchronization. And simultaneous identical acceleration creates constant deference between clocks. As soon as the speed becomes constant the clocks again becomes synchronized.
 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...
 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...

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 Quote by John232 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.
There is more to relativity than just time dilation. The Lorentz transform is more general and automatically reduces to the time dilation formula whenever appropriate. I recommend against ever using the time dilation formula. It will cause mistakes, as this example would show.

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 Quote by mananvpanchal But I don't follow this. I have tried to get it, but I don't succeed. That's fine, don't worry about it.
So, using the matrix form of the Lorentz transform we have:

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

$$r'_d=\left( \begin{array}{c} t'= -0.75 \begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} + 1.25 \begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases} \\ x'= 1.25 \begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} - 0.75 \begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases} \\ 0 \\ 0 \end{array} \right)$$

$$r'_d=\left( t'=\begin{cases} 1.25 \tau - 0.75 d & \mbox{if } \tau \lt 0 \\ \tau - 0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d - 0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$

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 Quote by mananvpanchal Please, shade some light on my post #36 and #39. Please, tell me what is wrong with my understanding.
If you go back to your #36 and #39, you can see that you have identified some events on A and B which are identified as being simultaneous for O.

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.

 Quote by DaleSpam There is more to relativity than just time dilation. The Lorentz transform is more general and automatically reduces to the time dilation formula whenever appropriate. I recommend against ever using the time dilation formula. It will cause mistakes, as this example would show.
Thats like saying Lorentz was the real genius and Einstein was a dumb***. I went to the wiki and in the fine print under the first diagram to the right it says that it describes when one event detects another event. So then any result useing that equation would only tell what each observer see's when they detect an event. So that doesn't mean that A and B are no longer in sync it only means that the signal to R is not in sync. I wouldn't use the Lorentz Transform to ever find out any information about what clocks say to determine the amount of time dialation from one frame of reference to another, or you would always be wrong, since your just finding out what an observer detects at a certain moment.

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 Quote by John232 Thats like saying Lorentz was the real genius and Einstein was a dumb***.
Clearly that is an absurd mischaracterization of what I said.

 Quote by John232 I went to the wiki and in the fine print under the first diagram to the right it says that it describes when one event detects another event. So then ....
Please link to the page in question. You have a definite misunderstanding of relativity and the Lorentz transform, but I am not sure why.

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.