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Synchronized clocks with respect to rest frame

by mananvpanchal
Tags: clocks, frame, respect, rest, synchronized
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DaleSpam
#55
Mar3-12, 06:38 AM
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Quote Quote by mananvpanchal View Post
But suppose, train is going to opposite direction, then the equation cannot distinguish both [itex]t'[/itex].
The equation was derived specifically for the scenario represented in the OP. It is not a general equation and cannot be used as is to make any conclusions about other scenarios. Any reasoning based on using this equation for a different scenario is inherently flawed.

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?
DaleSpam
#56
Mar3-12, 06:42 AM
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Quote Quote by mananvpanchal View Post
"with respect to O" described in post #36. Actually I have identified all events as being simultaneous for O.
I am not sure what you are trying to say here. Not all events are simultaneous for O, for example no two distinct events on O's worldline are simultaneous.

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.
mananvpanchal
#57
Mar3-12, 08:20 AM
P: 215
Quote Quote by mananvpanchal
So, far We get only one reason for desynchronization is "length contraction".

Attachment 44664

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.
I am very sorry about this. Sometimes mind works like crazy. Discussion is going on here about O and I am talking about "with respect to R" in this paragraph.
John232
#58
Mar3-12, 11:15 AM
P: 249
Quote Quote by DaleSpam View Post
Please link to the page in question. You have a definite misunderstanding of relativity and the Lorentz transform, but I am not sure why.
It was the same page that you gave a link to when you put the equation on the forum to begin with. I think you have a clear misunderstanding of the difference between the actual time dialation and the observed time from a location, and that is why when you end up using the time dialation formula you always end up getting answers that are "wrong".

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...
DaleSpam
#59
Mar3-12, 09:49 PM
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Quote Quote by John232 View Post
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
But due to the relativity of simultaneity that is only the case in R's frame. In all other frames the objects do not travel at the same speed at all times, therefore they do not experience the same amount of time dilation and they become desynchronized.
mananvpanchal
#60
Mar5-12, 06:21 AM
P: 215
Quote Quote by DaleSpam View Post
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.
Yes, you are right. We will get the answer as you said.

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?
mananvpanchal
#61
Mar5-12, 06:32 AM
P: 215
Quote Quote by DaleSpam View Post
I am not sure what you are trying to say here. Not all events are simultaneous for O, for example no two distinct events on O's worldline are simultaneous.

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.
Actually, I am confused about event. I have read this links. Which tell me opposite thing. Please, tell me what do I understand wrong?

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.)
DaleSpam
#62
Mar5-12, 06:52 AM
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Quote Quote by mananvpanchal View Post
Yes, you are right. We will get the answer as you said.

There is another doubt!

This may be the case again of generalization.
My apologies, I just noticed that my notation is unclear. The variable [itex]\tau[/itex] is intended to be a parameter of the worldline, i.e. different values of [itex]\tau[/itex] pick out different events on the worldline which correspond to the reading of the clock at that event. However, since there are three different worldlines there should be three different parameters. I.e. I should have used [itex]\tau_d[/itex] instead of [itex]\tau[/itex]. I am sorry for any confusion that resulted.

Quote Quote by mananvpanchal View Post
[itex]t'=\tau - 0.75 d, \mbox{if } \tau \ge 0[/itex]

As [itex]\tau[/itex] increases, desynchronization between two clocks decreases.
This is an incorrect reading of the expression. The desynchronization between two clocks remains constant as the [itex]\tau_d[/itex] increase (for [itex]\tau_d>0[/itex]).

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].
DaleSpam
#63
Mar5-12, 07:10 AM
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Quote Quote by mananvpanchal View Post
Actually, I am confused about event. I have read this links. Which tell me opposite thing. Please, tell me what do I understand wrong?

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.)
You are reading the two pages correctly as far as I can tell. They are indeed contradictory. Rather embarassingly the fourmilab page has the physics wrong and the history/philosophy page has the physics correct.

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.
mananvpanchal
#64
Mar5-12, 07:30 AM
P: 215
Quote Quote by DaleSpam View Post
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].
As we got [itex]\tau = 0.8t[/itex].

Can you please explain me how can I get [itex]\tau_a[/itex] and [itex]\tau_b[/itex]?
mananvpanchal
#65
Mar5-12, 07:37 AM
P: 215
Hello John232, DaleSpam

I think there might be some misunderstanding with you guys.

Quote Quote by DaleSpam
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 t does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.
Quote 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 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.
You both might saying the same thing that the clocks is in sync with respect to R.
DaleSpam
#66
Mar5-12, 09:52 AM
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Quote Quote by mananvpanchal View Post
As we got [itex]\tau = 0.8t[/itex].

Can you please explain me how can I get [itex]\tau_a[/itex] and [itex]\tau_b[/itex]?
Just solve the first component of the corrected equations listed below for [itex]\tau_d[/itex] and then substitute in the appropriate value for d.

[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.
John232
#67
Mar5-12, 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
DaleSpam
#68
Mar5-12, 06:51 PM
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P: 16,985
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.
mananvpanchal
#69
Mar6-12, 12:38 AM
P: 215
Quote Quote by DaleSpam View Post
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.
I am sorry, but we would get [itex]\tau_d=t'+0.75d[/itex] for primed frame, not [itex]\tau_d=t+0.75d[/itex].
Now, we have two unknown variables [itex]t'[/itex] and [itex]\tau_d[/itex].
DaleSpam
#70
Mar6-12, 05:38 AM
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P: 16,985
Quote Quote by mananvpanchal View Post
I am sorry, but we would get [itex]\tau_d=t'+0.75d[/itex] for primed frame, not [itex]\tau_d=t+0.75d[/itex].
Now, we have two unknown variables [itex]t'[/itex] and [itex]\tau_d[/itex].
Oops, you are correct. I will fix it above.
darkhorror
#71
Mar7-12, 09:29 AM
P: 140
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?
mananvpanchal
#72
Mar29-12, 11:13 PM
P: 215
Quote Quote by DaleSpam View Post
Quote Quote by mananvpanchal View Post
I am sorry, but we would get [itex]\tau_d=t'+0.75d[/itex] for primed frame, not [itex]\tau_d=t+0.75d[/itex].
Now, we have two unknown variables [itex]t'[/itex] and [itex]\tau_d[/itex].
Oops, you are correct. I will fix it above.
Ok, so please clarify the confusion. I still don't get what should be the value of [itex]\tau_d[/itex]


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