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Synchronized clocks with respect to rest frame

by mananvpanchal
Tags: clocks, frame, respect, rest, synchronized
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DaleSpam
#37
Feb29-12, 09:31 PM
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Quote Quote by John232 View Post
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.
mathal
#38
Mar1-12, 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
mananvpanchal
#39
Mar1-12, 01:19 AM
P: 215
There is other doubt too.

Quote 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.

Click image for larger version

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

Click image for larger version

Name:	desync actually from R.jpg
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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
John232
#40
Mar1-12, 04:56 AM
P: 249
Quote Quote by DaleSpam View Post
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.
mathal
#41
Mar1-12, 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
DaleSpam
#42
Mar1-12, 10:59 AM
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Quote Quote by John232 View Post
I fail to see how that is true without an explanation.
Quote Quote by mananvpanchal View Post
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:
[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.
mananvpanchal
#43
Mar2-12, 02:53 AM
P: 215
Thanks for your reply.

Quote 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 Quote by DaleSpam
1. [itex]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)[/itex]

2. [itex]t=\begin{cases}
\tau & \mbox{if } \tau \lt 0 \\
1.25 \tau & \mbox{if } \tau \ge 0
\end{cases}[/itex]

3. [itex]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)[/itex]
I follow this.

Quote Quote by DaleSpam
Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain.
4. [itex]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)[/itex]
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.
mananvpanchal
#44
Mar2-12, 02:53 AM
P: 215
Thanks mathal

Quote Quote by mathal View Post
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.
John232
#45
Mar2-12, 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...
John232
#46
Mar2-12, 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...
DaleSpam
#47
Mar2-12, 10:26 AM
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P: 16,999
Quote Quote by John232 View Post
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.
DaleSpam
#48
Mar2-12, 11:16 AM
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P: 16,999
Quote Quote by mananvpanchal View Post
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:

[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]
DaleSpam
#49
Mar2-12, 11:26 AM
Mentor
P: 16,999
Quote Quote by mananvpanchal View Post
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.
John232
#50
Mar2-12, 01:27 PM
P: 249
Quote Quote by DaleSpam View Post
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.
DaleSpam
#51
Mar2-12, 01:47 PM
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P: 16,999
Quote Quote by John232 View Post
Thats like saying Lorentz was the real genius and Einstein was a dumb***.
Clearly that is an absurd mischaracterization of what I said.

Quote Quote by John232 View Post
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.
yuiop
#52
Mar2-12, 01:49 PM
P: 3,967
Quote Quote by John232 View Post
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?
If we have a long object such as a rocket with initially synchronised clocks at the tail and nose and accelerate the rocket, the clock at the tail travels a further distance than the clock at the front of the rocket in a given time as measured in the original rest frame. This causes the clocks become unsynchronised in the rest frame of the rocket as the rear clock effectively has to travel faster (in the original frame) and time dilates to a greater extent. This is because in order for the rocket to retain its proper length in the rest frame of the accelerating rocket, it is length contracting in the original frame. THis form of acceleration is known as Born rigid acceleration (See http://www.mathpages.com/home/kmath422/kmath422.htm or Google Bell's rocket paradox). The clocks will remain synchronised (in the launch frame) under acceleration if we insist the clocks remain the same distance apart in the launch frame, but under these conditions the the rocket will be increasing its proper length and eventually be torn apart. Even under these conditions, the clocks will be unsychronised in the rest frame of the accelerating rocket because the rocket has a velocity relative to the launch frame and the relativity of simultaneity means that if the clocks appear synchronised in the launch frame they will not be synchronised in the accelerating rocket frame.

Sorry for the layman's sloppy explanation but I think that is the gist of it ;)
mananvpanchal
#53
Mar3-12, 02:17 AM
P: 215
As we get the equation below
Quote Quote by DaleSpam
[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]
We can see that [itex]t'=\tau - 0.75 d, \mbox{if } \tau \ge 0[/itex]. And we know that d=-1 for A, d=0 for O and d=1 for B then, we can get this

[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.
mananvpanchal
#54
Mar3-12, 03:12 AM
P: 215
Quote Quote by DaleSpam
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.
"with respect to O" described in post #36. Actually I have identified all events as being simultaneous for O.
Quote Quote by DaleSpam
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.
Actually, I want to say with this three images that all events is simultaneous for O.
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 clock-mirror. 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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