Synchronized clocks with respect to rest frame

[Dale]

Thats like saying Lorentz was the real genius and Einstein was a dumb***.

Clearly that is an absurd mischaracterization of what I said.

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]

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]

As we get the equation below

r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau - 0.75 d &amp; \mbox{if } \tau \lt 0 \\<br /> \tau - 0.75 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

We can see that t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0. And we know that d=-1 for A, d=0 for O and d=1 for B then, we can get this

t&#039;_a=\tau + 0.75,
t&#039;_o=\tau,
t&#039;_b=\tau - 0.75

But suppose, train is going to opposite direction, then the equation cannot distinguish both t&#039;. 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 \tau, and \tau is not depend on x. \tau 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 anyone 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".

View attachment 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 anyone we have to pick.

 

[mananvpanchal]

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.

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.

 

[Dale]

But suppose, train is going to opposite direction, then the equation cannot distinguish both t&#039;.

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?

 

[Dale]

"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]

So, far We get only one reason for desynchronization is "length contraction".

View 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]

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 dilation and the observed time from a location, and that is why when you end up using the time dilation 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 dilation equation alone without it...

 

[Dale]

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]

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.

t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0

As \tau increases, desynchronization between two clocks decreases.

So, for \tau &lt; 0, all clocks are synchronized. At \tau = 0 train changes its frame, and starts moving with constant speed 0.6c, the clocks is most desynchronized. For \tau &gt; 0, as \tau 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]

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/RelativityOfSimultaneity/

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/teaching/HPS_0410/chapters/Special_relativity_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.)

 

[Dale]

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 \tau is intended to be a parameter of the worldline, i.e. different values of \tau 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 \tau_d instead of \tau. I am sorry for any confusion that resulted.

t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0

As \tau increases, desynchronization between two clocks decreases.

This is an incorrect reading of the expression. The desynchronization between two clocks remains constant as the \tau_d increase (for \tau_d&gt;0).

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

 

[Dale]

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/RelativityOfSimultaneity/

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/teaching/HPS_0410/chapters/Special_relativity_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]

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

As we got \tau = 0.8t.

Can you please explain me how can I get \tau_a and \tau_b?

 

[mananvpanchal]

Hello John232, DaleSpam

I think there might be some misunderstanding with you guys.

Substituting into the above we get:
r_d=\left(<br /> t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Noting that t does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.

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.

 

[Dale]

As we got \tau = 0.8t.

Can you please explain me how can I get \tau_a and \tau_b?

Just solve the first component of the corrected equations listed below for \tau_d and then substitute in the appropriate value for d.

r_d=\left(<br /> t=\begin{cases}<br /> \tau_d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 1.25 \tau_d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 0.75 \tau_d+d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases}<br /> ,0,0\right)
r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau_d - 0.75 d &amp; \mbox{if } \tau_d \lt 0 \\<br /> \tau_d - 0.75 d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau_d &amp; \mbox{if } \tau_d \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau_d \ge 0 <br /> \end{cases}<br /> ,0,0\right)

So for \tau_d \ge 0 we get \tau_d=0.8t in the unprimed frame and we get \tau_d=t&#039;+0.75d in the primed frame.

 

[John232]

"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/einstein/specrel/www/#SECTION11

 

[Dale]

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]

So for \tau_d \ge 0 we get \tau_d=0.8t in the unprimed frame and we get \tau_d=t+0.75d in the primed frame.

I am sorry, but we would get \tau_d=t&#039;+0.75d for primed frame, not \tau_d=t+0.75d.
Now, we have two unknown variables t&#039; and \tau_d.

 

[Dale]

I am sorry, but we would get \tau_d=t&#039;+0.75d for primed frame, not \tau_d=t+0.75d.
Now, we have two unknown variables t&#039; and \tau_d.

Oops, you are correct. I will fix it above.

 

[darkhorror]

First let's take another situation, let's just say that the two clocks on the train are in sync in the train's frame of reference. Let's 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]

I am sorry, but we would get \tau_d=t&#039;+0.75d for primed frame, not \tau_d=t+0.75d.
Now, we have two unknown variables t&#039; and \tau_d.

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 \tau_d

 

[whosapopstar?]

As much as i see this, we have here four dominant terms: time dilation, length contraction, light speed and rest frame.

my question is this:
Lets assume that we all think and believe that the clock aboard the train returns with time dilation on it, also as a result of constant speed (put aside time dilation as a result of acceleration). There is some sort of effect on the clock, while moving at constant speed, and acceleration has nothing to do with it at all, that can't be denied.

Now, say this guy has some sort of machine that has clocks and light detectors and works in a certain way, that is not effected by length contraction. This machine was calibrated to work in a certain way, before the train was moving. We know and agree that light does not change its speed. We agree that time dilation exists at constant speed. Yet, after the train is moving, that machine is still working as it did, it stays calibrated, after it was calibrated while at bay!

Let us see what we have: a calibrated machine, regardless if the train is moving or not moving at constant speed. Light speed is the same regardless if train is moving or not. Length contraction does not effect that machine.

We are left with only time dilation effecting that machine while at constant speed. and with the word : rest frame.

If time dilation is effecting that machine, and yet it stays calibrated, there must be somthing counter effecting it to stay calibrated. What is this thing called?

How can a choice of a rest frame have this effect? my rest frame is always on the train station and all through this experiment i never bother to exchange any signals with the train. Only when it returns do i ask these question. What does a rest frame have to do with it?

Or should we return to acceleration as a source for all this?

 

[Dale]

Ok, so please clarify the confusion. I still don't get what should be the value of \tau_d

\tau_d is the reading on the clock. The clock doesn't just show one number, it shows a different number at each point in time. The formula relates the coordinate time to the time reading on the clock.

 

[mananvpanchal]

Hello DaleSpam,

First you had came up with this equation

r&#039;_d=\left(<br /> t&#039;=\begin{cases}<br /> 1.25 \tau - 0.75 d &amp; \mbox{if } \tau \lt 0 \\<br /> \tau - 0.75 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x&#039;=\begin{cases}<br /> 1.25 d - 0.75 \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,0,0\right)

Then, I had created the doubt

We can see that t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0. And we know that d=-1 for A, d=0 for O and d=1 for B then, we can get this

t&#039;_a=\tau + 0.75,
t&#039;_o=\tau,
t&#039;_b=\tau - 0.75

But suppose, train is going to opposite direction, then the equation cannot distinguish both t&#039;

And you had solved this by

v = -0.6 c

So, I had to get t&#039;_a, t&#039;_o, t&#039;_b using \tau and d.

After this I had created another doubt

Yes, you are right. We will get the answer as you said.

There is another doubt!

This may be the case again of generalization.

t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0

As \tau increases, desynchronization between two clocks decreases.

And you had came up with the idea of \tau_a, \tau_o, \tau_b

This is an incorrect reading of the expression. The desynchronization between two clocks remains constant as the \tau_d increase (for \tau_d&gt;0).

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

So, I had to get t&#039;_a, t&#039;_o, t&#039;_b using \tau and d. But you had came up with \tau_a, \tau_o, \tau_b.

When, I had asked you

As we got \tau = 0.8t.

Can you please explain me how can I get \tau_a and \tau_b?

You had came up with the idea

So for \tau_d \ge 0 we get \tau_d=0.8t in the unprimed frame and we get \tau_d=t&#039;+0.75d in the primed frame.

Then, I had told you

Now, we have two unknown variables t&#039; and \tau_d.

So, now the problem is we have to derive

t&#039;_a=\tau + 0.75,
t&#039;_o=\tau,
t&#039;_b=\tau - 0.75 using \tau, and d.

We know here d_a=-1, d_o=0, d_b=1 and \tau = 0.6t.

But, To solve "decreasing desync as \tau increases" problem you came up with the idea of \tau_a, \tau_o, \tau_b.

Now, we have the equations.

t&#039;_a=\tau_a - 0.75 d_a,
t&#039;_o=\tau_o - 0.75 d_o,
t&#039;_b=\tau_b - 0.75 d_b.

And as I said before we have now two unknown variables per eqaution (t&#039;_a, \tau_a), (t&#039;_o, \tau_o) and (t&#039;_b, \tau_b).

So, the question is how can we get values of t&#039;_a, \tau_a, t&#039;_o, \tau_o, t&#039;_b and \tau_b using known variables t, d_a, d_o and d_b?

 

[whosapopstar?]

Here is another question:

When the train returns, we can see that somthing happened, e.g. we have time dilation on the clock, and we agree that at least part of that time dilation was produced by constant speed (CS). There is true evidence that somthing happened there.

Now regarding the clocks that are not synchronized, although they are both on the same train (but apart from each other): is there an experiment that can be done, which will show us this difference of de-synchroniztion between them, after the clocks will return to the station, and not by sending signals when the train is on the move? if not, how come one clock can bring back evidence to the station of a phenomenon (CS time dilation on a single clock), while another phenomenon, the de-synchronization of two clocks, is not somthing that can be brought back as evidence? or is such an experiment plausible after all for two clocks? or is this de-synchronization, a result of accelerating and de-accelerating and not of constant speed?

 

[Dale]

Now, we have the equations.

t&#039;_a=\tau_a - 0.75 d_a,
t&#039;_o=\tau_o - 0.75 d_o,
t&#039;_b=\tau_b - 0.75 d_b.

And as I said before we have now two unknown variables per eqaution (t&#039;_a, \tau_a), (t&#039;_o, \tau_o) and (t&#039;_b, \tau_b).

So, the question is how can we get values of t&#039;_a, \tau_a, t&#039;_o, \tau_o, t&#039;_b and \tau_b using known variables t, d_a, d_o and d_b?

I never put in any subscripts for t and t'. You pick a value for t or t' and solve for the various \tau_d.

 

[mananvpanchal]

I never put in any subscripts for t and t'. You pick a value for t or t' and solve for the various \tau_d.

Ok, here are you telling this?

t&#039;=\tau_a - 0.75 d_a
t&#039;=\tau_o - 0.75 d_o
t&#039;=\tau_b - 0.75 d_b
Now, please look at this

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 let's 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}<br /> d &amp; \mbox{if } t \lt 0 \\<br /> 0.6 t+d &amp; \mbox{if } t \ge 0 <br /> \end{cases}<br /> ,0,0\right)
where d=-1 for A, d=0 for O, and d=1 for B.

Here, you simply defined time t in R's clock. So, you can define distance of train clocks from R with x=0.6t + d.

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}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}

Here, you took \gamma = 1.25, and you define t = 1.25 \tau. Here \tau is train clocks' reading for R. If we want synchronized clocks in R's frame, there must be a single value for \tau. We can easily see that there is same value of all three train's clocks' readings \tau exist for R. So we can say that train' clocks remain synchronized in R's frame. And you proved this by

Substituting into the above we get:
r_d=\left(<br /> t=\begin{cases}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases},<br /> x=\begin{cases}<br /> d &amp; \mbox{if } \tau \lt 0 \\<br /> 0.75 \tau+d &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}<br /> ,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.

Clocks is only synchronized in R's frame only when there is only one value of \tau exist. If we get different value of each clock reading for R then we cannot say that clocks is synchronized in R's frame.

But, you said that there are three value \tau_a, \tau_o and \tau_b exist. So, train clocks cannot be remain synchronized for R.

 

[mananvpanchal]

Please, look at this too

t&#039;=\tau - 0.75 d, \mbox{if } \tau \ge 0

As \tau increases, desynchronization between two clocks decreases.

This is an incorrect reading of the expression. The desynchronization between two clocks remains constant as the \tau_d increase (for \tau_d&gt;0).

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

Here, I am talking about t&#039;. I have take here t&#039;_a=\tau - 0.75 d_a, t&#039;_o=\tau - 0.75 d_o and t&#039;_b=\tau - 0.75 d_b. So as \tau increase deference between t&#039;_a, t&#039;_o and t&#039;_b decreases. And that is why I have used subscript with t&#039;.

t is R's clock's reading for R.

\tau is train's clocks (A, O, B)'s reading for R (The reading is same, so train's clocks is synchronized for R) (\tau is not train's clocks reading for O, so it cannot have different values as per d).

And t&#039; is trains clock's reading for O (We have boosted here the train clocks reading for R (\tau) to train's clocks reading for O (t&#039;)) (Train's clocks is not synchronized for O, so t&#039; should have different values as per d). So actually subscript should be used with t&#039; not with \tau.

 

[Dale]

OK, there are so many things incorrect with your previous two posts that I think it will be better to explain the general approach rather than respond point-by-point. If you still have specific questions afterwards then I will be glad to respond point-by-point.

First, when you are doing spacetime diagrams what you are doing is mapping the position of an object as a function of time to the shape of a geometric object. This increases the dimensionality of an object, i.e. a 0D point particle is represented by a 1D worldline (for this analysis we are ignoring the size of the clocks so they are represented by worldlines).

Second, you can always write a 1D worldline in parametric form with a single parameter. So, you can always write it in the form (t(\lambda),x(\lambda),y(\lambda),z(\lambda)) where \lambda is the parameter.
http://en.wikipedia.org/wiki/Parametric_equation

Third, the parameterization is not unique. So if you have any continuous and invertible function \lambda(\zeta) then (t(\zeta),x(\zeta),y(\zeta),z(\zeta)) is also a valid parameterization of the same worldline.

Fourth, the parameter need not have any significance, but it can. For example, one common parameterization is to use the time in a given reference frame, t, another common parameterization is to use the proper time displayed on the clock, τ. The latter parameterization is particularly common since it is frame-invariant and physically measurable.

Fifth, if you have multiple worldlines then each worldline will have its own parameterization and thus its own parameter.

Sixth, for each reference frame there are a series of surfaces of constant coordinate time called hyperplanes of simultaneity, or just planes of simultaneity. There is one such plane for each value of coordinate time.

Seventh, two clocks are considered to be synchronized in a specified reference frame at a given coordinate time iff their displayed proper time is the same at their respective intersections with the plane of simultaneity.

Do you understand these concepts? Do you need further explanation of any? I will try to post the application of these principles to this specific problem later today.

 

[mananvpanchal]

Please, look at this

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}<br /> \tau &amp; \mbox{if } \tau \lt 0 \\<br /> 1.25 \tau &amp; \mbox{if } \tau \ge 0 <br /> \end{cases}

You have defined here \tau is time reading of train's clocks for R.

Suppose, there is only one clock in train. When train is at station the R, O and the clock have same space component in platform co-ordinate system. Now at t=0 train's speed is 0.6c. So we can draw the world line of the clock like this.

We can easily see that when 1.25 reading in R's clock for R, O's clock reading is 1 for R. And when 1.25 reading in O's clock for O, R's clock reading is 1 for O. So moving O's clock slow down with respect to R, and moving R's clock slow down with respect to O. Here you have used simple transformation formula t = \gamma \tau, because space component is 0 at initial stage. We get one value for t for \tau = 1 and x = 0.

Now, if we put another two clocks at train's front and train's end. And you have said that the the clocks would be dilated for R at same rate and would remain synchronized for R. So, as per your saying, the world lines of the three clocks would look like this.

You can see that lines of simultaneity of R's frame says that clocks is dilated with same rate and clocks is synchronized.

But, do you notice what is wrong with this diagram?

We cannot use the t = \gamma \tau formula to calculate readings of front clock and end clock. Because, now the space component is not same for R, front clock and end clock. We have to consider space component in calculation. Here space component is not 0 of the new two clocks at initial stage. Another thing is you can see in above diagram that difference between space component doesn't change after frame changing. But, if we transform some (t, x)=(0, x) point to another co-ordinate system we will get (t', x')=(0, x'), where x ≠ x' surely. So after transformation space component of clock should be changed.

So, we have to use this equation t = \gamma (\tau - 0.6x). We get three values for t by putting (\tau = 1, x=-1), (\tau = 1, x=0) and (\tau = 1, x=1) in equation. So R will see \tau = 1 at different different time in his frame. The above space component problem will also be solved with this equation. This scenario can be described by below image.

We can easily see that lines of simultaneity of O's frame says that train's clocks is synchronized for O, but not for R. So, train's clocks would not be synchronized for R.

 

[mananvpanchal]

OK, there are so many things incorrect with your previous two posts that I think it will be better to explain the general approach rather than respond point-by-point. If you still have specific questions afterwards then I will be glad to respond point-by-point.

I would be happy to see my errors.

First, when you are doing spacetime diagrams what you are doing is mapping the position of an object as a function of time to the shape of a geometric object. This increases the dimensionality of an object, i.e. a 0D point particle is represented by a 1D worldline (for this analysis we are ignoring the size of the clocks so they are represented by worldlines).

Second, you can always write a 1D worldline in parametric form with a single parameter. So, you can always write it in the form (t(\lambda),x(\lambda),y(\lambda),z(\lambda)) where \lambda is the parameter.
http://en.wikipedia.org/wiki/Parametric_equation

Third, the parameterization is not unique. So if you have any continuous and invertible function \lambda(\zeta) then (t(\zeta),x(\zeta),y(\zeta),z(\zeta)) is also a valid parameterization of the same worldline.

Fourth, the parameter need not have any significance, but it can. For example, one common parameterization is to use the time in a given reference frame, t, another common parameterization is to use the proper time displayed on the clock, τ.

I follow this.

The latter parameterization is particularly common since it is frame-invariant and physically measurable.

I don't follow this.

Fifth, if you have multiple worldlines then each worldline will have its own parameterization and thus its own parameter.

Sixth, for each reference frame there are a series of surfaces of constant coordinate time called hyperplanes of simultaneity, or just planes of simultaneity. There is one such plane for each value of coordinate time.

Seventh, two clocks are considered to be synchronized in a specified reference frame at a given coordinate time iff their displayed proper time is the same at their respective intersections with the plane of simultaneity.

I follow this.

I will try to post the application of these principles to this specific problem later today.

Certainly.

 

[Dale]

I don't follow this.

The proper time is a frame invariant quantity, meaning that all reference frames agree on what it is. It is the integral of the spacetime interval along the clock's worldline. The proper time is also measurable, specifically, the reading that a clock displays is the measurement of proper time. At any event along the worldline all reference frames must agree on what the clock actually reads.

 

[Dale]

OK, so to use the above approach in this problem the first thing that I did was to write the worldline of the clocks in parametric form. Since you first described the situation in R's frame, I used that frame for the description and parameterized it in terms of coordinate time. The result is the first equation of post 42.

Then, since the parameterization is not unique and since proper time has the advantages listed above as well as the direct application for determining synchronization, I decided to re-parameterize in terms of proper time instead of coordinate time. The relationship between the two parameters is the second equation in post 42, and the resulting parametric equation of the worldline is given in the third equation.

Once that is done, then the worldline in the primed frame is found simply by using the Lorentz transform, as shown in post 48.

Now, we have a parameterization for each worldline in terms of proper time in both the primed and the unprimed frame. So at this point we make 3 worldlines, corresponding to A, O, and B, by substituting d=-1,0,1 respectively. As mentioned above, each worldline has its own parameter, hence the subscripts on the proper time. I.e. there are three clocks displaying numbers and so there are three proper time values listed.

Now, in order to determine if a pair of clocks are synchronized in some frame we find the proper time at the intersection of the worldline of each clock with the same plane of simultaneity. This is done by setting t or t' equal to some value and solving for the two values of proper time. If the values are equal then the pair of clocks is synchronized and vice versa.

Does that make sense now?

 

[Dale]

You have defined here \tau is time reading of train's clocks for R.
...
We cannot use the t = \gamma \tau formula to calculate readings of front clock and end clock.

Yes, we can. The proper time displayed on the clock is given by the spacetime interval between the event on the worldline where t=0 and any other event on the worldline.

\Delta\tau^2 = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2
For τ>0
(\tau - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau^2 = 0.64 t^2
\tau = 0.8 t

The d cancels out.

 

[mananvpanchal]

OK, so to use the above approach in this problem the first thing that I did was to write the worldline of the clocks in parametric form. Since you first described the situation in R's frame, I used that frame for the description and parameterized it in terms of coordinate time. The result is the first equation of post 42.

This is fine.

Then, since the parameterization is not unique and since proper time has the advantages listed above as well as the direct application for determining synchronization, I decided to re-parameterize in terms of proper time instead of coordinate time. The relationship between the two parameters is the second equation in post 42, and the resulting parametric equation of the worldline is given in the third equation.

This is fine.

Once that is done, then the worldline in the primed frame is found simply by using the Lorentz transform, as shown in post 48.

This is not fine. Because you even don't need to transform at all. \tau is proper time = co-ordinate time in train frame. Because clocks is at rest in train frame. Then why you want to transform it? and to which frame? and for what?

Now, we have a parameterization for each worldline in terms of proper time in both the primed and the unprimed frame. So at this point we make 3 worldlines, corresponding to A, O, and B, by substituting d=-1,0,1 respectively. As mentioned above, each worldline has its own parameter, hence the subscripts on the proper time. I.e. there are three clocks displaying numbers and so there are three proper time values listed.

This is not fine too. How three clocks at rest in train frame can have three different different proper time readings in the same train frame? If the clocks is at rest in some frame then there is only one proper time reading of all clocks. The spatial difference of clocks doesn't make any change in time readings for a frame in which the clocks is at rest. The spatial difference only makes change time readings for a frame in which the clocks are moving. So spatial difference of clocks cannot change \tau value in train frame, because clocks is at rest in train frame and proper time (\tau) = co-ordinate (\tau) time in train frame. But spatial difference of clocks can change t value in R's frame. Because clocks is moving in R's frame and proper time (\tau) ≠ co-ordinate (t) time in R's frame.

Now, in order to determine if a pair of clocks are synchronized in some frame we find the proper time at the intersection of the worldline of each clock with the same plane of simultaneity. This is done by setting t or t' equal to some value and solving for the two values of proper time. If the values are equal then the pair of clocks is synchronized and vice versa.

Above two is not fine, so this is also not fine. Please, first define what is t&#039;?

 

[mananvpanchal]

Yes, we can. The proper time displayed on the clock is given by the spacetime interval between the event on the worldline where t=0 and any other event on the worldline.

\Delta\tau^2 = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2
For t>0
(\tau - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau^2 = 0.64 t^2
\tau = 0.8 t

The d cancels out.

This seems right, but this is not right. This is not the equation to find time readings.

Please, look at this.

\Delta\tau^2 = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2

(\tau_2 - \tau_1)^2 = (t_2-t_1)^2 - ((0.6t_2+d) - (0.6t_1 + d))^2

(\tau_2 - \tau_1)^2 = 0.64 (t_2-t_1)^2

(\tau_2 - \tau_1) = 0.8 (t_2-t_1)

And this will be converted into simple time dilation equation.

\Delta \tau = 0.8 \Delta t

So, do you notice here the problem?

We are taking here \tau_1 = 0 = t_1. And that's why you able to write down the equation like this \tau = 0.8 t. Here we can take \tau_1 = 0 = t_1, because transformation of \tau_1 = 0 gives us t_1 = 0. But we cannot find any value of t_1 (where t_1 ? 0), so from that we can get \tau_1 where \tau_1 = t_1.

You are right that in time dilation equation \Delta \tau = 0.8 \Delta t you always get a constant difference \Delta \tau by taking some constant difference \Delta t. But that doesn't mean that \tau_2 = t_2 and \tau_1 = t_1.

So picking time difference (\tau_2 - \tau_1) in train frame we can get time difference (t_2 - t_1) in R's frame, and piking again the same time difference (\tau_3 - \tau_2) in train frame we get the same before time difference (t_3 - t_2) in R's frame. But if we take here \tau_1 = t_1 it is guaranteed that we have now \tau_2 ? t_2 and \tau_3 ? t_3.

To find t_1, t_2, t_3 from \tau_1, \tau_2, \tau_3 we still wants lorentz transformation equation not time dilation equation. Time dilation equation only can be used to find dilated time difference not dilated time reading.

So, from the above time dilation equation we can write \tau_2 = 0.8 t_2 where \tau_1 = 0 = t_1. But we cannot write \tau_3 = 0.8 t_3 because we never have \tau_2 = t_2.

The above description is just to differentiate time difference and time reading.

We have now LT equation t_n = 1.25 (\tau_n - 0.6d).
Now if we want to transform t_1, t_2, t_3 from \tau_1, \tau_2, \tau_3 we have to take d into account.

So, to find time difference we don't need to take d into account (\Delta \tau = 0.8 \Delta t), but to find time reading we have consider d in LT equation (t_n = 1.25 (\tau_n - 0.6d)).

So...

So, we have to use this equation t = \gamma (\tau - 0.6x). We get three values for t by putting (\tau = 1, x=-1), (\tau = 1, x=0) and (\tau = 1, x=1) in equation. So R will see \tau = 1 at different different time in his frame.

 

[Dale]

This is not fine. Because you even don't need to transform at all. \tau is proper time = co-ordinate time in train frame. Because clocks is at rest in train frame. Then why you want to transform it? and to which frame? and for what?

There are two inertial frames. The unprimed frame which is R's frame, and the primed frame where O is at rest after τ=0.

Of course you need to transform it because you are interested in the synchronization in both frames. You therefore need to be able to determine the planes of simultaneity in both frames.

This is not fine too. How three clocks at rest in train frame can have three different different proper time readings in the same train frame?

If the clocks is at rest in some frame then there is only one proper time reading of all clocks.

They could easily have different readings if they are not synchronized. My watch and my computer clock are both at rest in my frame, one reads 7:37 and the other reads 7:35. So being at rest in the same frame does not imply that they have the same proper time reading.

Furthermore, since you are interested in whether or not they are synchronized you cannot assume that they are synchronized. That is a logical fallacy called "begging the question". You need to assume that they could be synchronized or they could be desynchronized. You do that by allowing each one to vary. Then you prove that the way they vary is such that they are equal (or unequal) at the intersection with a given plane of simultaneity.

The spatial difference of clocks doesn't make any change in time readings for a frame in which the clocks is at rest. The spatial difference only makes change time readings for a frame in which the clocks are moving. So spatial difference of clocks cannot change \tau value in train frame, because clocks is at rest in train frame and proper time (\tau) = co-ordinate (\tau) time in train frame. But spatial difference of clocks can change t value in R's frame. Because clocks is moving in R's frame and proper time (\tau) ≠ co-ordinate (t) time in R's frame.

The math disagrees. You are simply assuming the conclusion you think is right, and thereby committing a logical fallacy. If you do not assume the conclusion you find out that it is a false assumption.

Above two is not fine, so this is also not fine. Please, first define what is t&#039;?

See above, t' is the time coordinate in the inertial frame where O is at rest after τ=0.

 

[Dale]

This seems right, but this is not right. This is not the equation to find time readings.

Do you agree or disagree that, as part of the problem set up, all 3 clocks read τ=0 at t=0? Remember that t is the coordinate time in the unprimed (R) frame where they are initially at rest.

If you disagree, then what is the value of t when τ_d=0 for each clock?

 

[mananvpanchal]

I think I have to again define the whole scenario.

For t &lt; 0 there is three clocks A, O's clock and B is on train in rest frame of R. R also has a clock. All clocks is synchronized.
Now at t = \tau = 0 train is moving with 0.6c speed in R's frame. The three clocks is at rest in O's frame and moving in R's frame.

You are using t = 1.25 \tau to find co-ordinate time (t) from proper time (\tau) of train's clocks.

But I not agree with you. Because t = 1.25 \tau is TD eqaution. It is used to find dilated time deference from proper time deference. It is not used to find co-ordinate time reading from proper time reading. To find out co-ordinate time reading from proper time reading we have to use LT equation t = 1.25 (\tau - 0.6d). In case of O's clock (where d=0) the equation can be simplified to t = 1.25 \tau. But this does not mean that we can use the same simplified equation for A and B clock. We have to use t = 1.25 (\tau - 0.6d) for A and B clocks.

If you disagree, then what is the value of t when τ_d=0 for each clock?

When \tau_a = 0, value of t_a=0.75
When \tau_o = 0, value of t_o=0
When \tau_b = 0, value of t_b=-0.75

Now, we have transformed O's frame's clocks readings into R's frame already. We don't need further transformation. Because, \tau is proper time and t&#039; is co-ordinate time in O's frame. The clocks is at rest in O's frame so in this case proper time (\tau) = co-ordinate time (t&#039;). You cannot transform O's clocks reading into O's clocks reading. Because they both readings are single clock's reading which is at rest in your frame.

See above, t' is the time coordinate in the inertial frame where O is at rest after τ=0.

Here, \tau = t&#039;. Because, clocks is at rest in O's frame. So, we don't need to transform it.

If you want to prove that O's clocks is synchronized for R, and not for O. You would pick (\tau_a = \tau_o = \tau_b) for any value of \tau \ge 0, you must get single value of t. You have tried to achieve this using TD equation, but we cannot use TD equation to get time reading. And we cannot get single value of t for any (\tau_a = \tau_o = \tau_b) using LT equation.

So, train's clocks is synchronized for O (\tau_a = \tau_o = \tau_b), but not for R (t_a \ne t_o \ne t_b).

 

[Dale]

All clocks is synchronized.
Now at t = \tau = 0
...
When \tau_a = 0, value of t_a=0.75
When \tau_o = 0, value of t_o=0
When \tau_b = 0, value of t_b=-0.75

Please make up your mind. These are two mutually contradictory scenarios so you have to pick only one.

All of our discussion to date has been using the first condition. Changing the scenario at this late date is rather irritating, but I can certainly re-work everything with this new scenario. But you must recognized that it is different and contradictory to the previously discussed scenario.

Here, \tau = t&#039;. Because, clocks is at rest in O's frame. So, we don't need to transform it.

I already gave you a concrete example why the simple fact that a clock is at rest is not sufficient to guarantee that it is synchronized. Your logic is wrong.

 

[mananvpanchal]

When \tau_a = 0, value of t_a=0.75
When \tau_o = 0, value of t_o=0
When \tau_b = 0, value of t_b=-0.75

I know that this calculation creates gap between world line and overlapping of world line. And I know that is strange to you. So, please see below image in which I have tried to show smooth acceleration of three clocks.

We can easily see that for t &lt; 0, the clocks is at rest in R's frame and lines of simultaneity says that the three clocks are synchronized in R's frame.

Now at t = 0 clocks starts accelerating. And at some time t = t_1 the clocks achieves 0.6c speed and after this clocks moving with the constant speed.

Now, you can see lines of simultaneity of O's frame are not parallel during acceleration phase (0 \le t &lt; t1). And lines of simultaneity of O's frame are parallel for t &lt; 0 and for t \ge t_1.

Space axis and time axis is slowly skewed during acceleration. And after achieving constant speed it will not be skewed more. Lines of simultaneity are always parallel to space axis, so it is also skewed with space axis.

Lines of simultaneity of any frame indicates synchronized clocks in that moving frame.

So, lines of simultaneity of O's frame indicates synchronized clocks in O's frame. But, lines of simultaneity of O's frame are not parallel to lines of simultaneity of R's frame. So, O's clocks which is synchronized in O's frame is not synchronized in R's frame.

Now, please see below image in which I have tried to show what you are saying.

You can easily see that it solves the purpose "Clocks of O's frame are synchronized in R's frame".
But, to solve the purpose, the diagram creates some problems.
- Only time axis is skewed. Space axis is not skewed.
- Space axis is not skewed, so space scale of O's frame remains equal to space scale of R's frame. Which should not be.
- Clocks is synchronized in R's frame, but the diagram cannot show how the clocks is not synchronized in O's frame.

Now, please tell me. Do you agree that after finishing acceleration the lines of simultaneity of O's frame are skewed for R's frame?
If you find that I couldn't show your idea in second diagram then please provide me your diagram.

 

[mananvpanchal]

Please make up your mind. These are two mutually contradictory scenarios so you have to pick only one.

All of our discussion to date has been using the first condition. Changing the scenario at this late date is rather irritating, but I can certainly re-work everything with this new scenario. But you must recognized that it is different and contradictory to the previously discussed scenario.

From my original post I am telling that clocks is synchronized for O, but not for R. But, when you posted your maths, I thought that you were telling right. But, I found some errors in your maths.

You are using t = 1.25 \tau to prove that clocks is synchronized for R. Here, you assuming single value of \tau for all three clocks, so you can prove that clocks is synchronized for R.

But when "decreasing desync with increasing \tau" problem created, you came up with idea of \tau_a, \tau_o and \tau_b. So, now you can easily prove that clocks is not synchronized for O with these three values of \tau.

So, now how are you going to prove that clocks is still synchronized for R with these three values of \tau using t = 1.25 \tau?

 

[Dale]

But, I found some errors in your maths.

No, you didn't find any errors. You found that my math disproved your assumptions so you went ahead and assumed your conclusion and stated that therefore the math is wrong. That is a logical fallacy known as "begging the question".

You are using t = 1.25 \tau to prove that clocks is synchronized for R.

I showed you how that was derived. It was derived correctly from the spacetime interval formula. My other results were derived from the Lorentz transform. Those two equations, the spacetime interval and the Lorentz transform, are the fundamental equations of special relativity. If you are not using them then you are not doing special relativity, and in special relativity you can always use them. The fact that they are contradictory to your assumptions proves your assumptions wrong.

Here, you assuming single value of \tau for all three clocks, so you can prove that clocks is synchronized for R.

I didn't assume it, I proved it. The d cancels out.

But when "decreasing desync with increasing \tau" problem created, you came up with idea of \tau_a, \tau_o and \tau_b. So, now you can easily prove that clocks is not synchronized for O with these three values of \tau.

Yes, I made a small mistake in notation, which I quickly corrected immediately upon your pointing it out.

So, now how are you going to prove that clocks is still synchronized for R with these three values of \tau using t = 1.25 \tau?

I already proved it, by deriving (not assuming) the fact that the proper time doesn't depend on d.

 

[Dale]

I know that this calculation creates gap between world line and overlapping of world line. And I know that is strange to you.

Not just strange, it is wrong. It is the most obvious of physical nonsense and should have immediately clued you in that you were wrong. You cannot have clocks magically appearing and disappearing, with duplicates and missing clocks. It is physical nonsense and violates all sorts of conservation laws.

Lines of simultaneity of any frame indicates synchronized clocks in that moving frame.

No, the lines of simultaneity indicate lines of constant coordinate time in the frame. What you need to prove is that the clocks display the same proper time at their respective intersections with one of these lines of simultaneity. That you have never done, but simply assumed.

In fact, your very diagram disproves your assumption. I have only added some dots to your diagram, but not changed the lines. I am assuming (according to the original scenario, not your contradictory scenario) that each clock reads τ=0 at the black dots.

We can easily see that the spacetime intervals between the black dots and the corresponding red dots are the same. That means that the clocks read the same time at the red dots. Since the red dots are on a single plane of simultaneity in the unprimed frame that implies that they are synchronized in the unprimed frame.

We can also easily see that the spacetime intervals between the black dots and the corresponding green dots are not the same. That means that the clocks do not read the same time at the green dots. Since the green dots are on a single plane of simultaneity in the primed frame that implies that they are not synchronized in the primed frame.

Your own graphics disprove your assumptions, as does all of the math.

 

[mananvpanchal]

We can easily see that the spacetime intervals between the black dots and the corresponding red dots are the same. That means that the clocks read the same time at the red dots. Since the red dots are on a single plane of simultaneity in the unprimed frame that implies that they are synchronized in the unprimed frame.

We can also easily see that the spacetime intervals between the black dots and the corresponding green dots are not the same. That means that the clocks do not read the same time at the green dots. Since the green dots are on a single plane of simultaneity in the primed frame that implies that they are not synchronized in the primed frame.

Is this not contradictory?

 

[Dale]

Is this not contradictory?

No, it is not contradictory.

 

[mananvpanchal]

Thanks DaleSpam.

 

[Dale]

You are welcome. Do you understand why it is not contradictory?

 

[mananvpanchal]

I want to clear some points here.

- You are using TD equation to find time readings, that is wrong. I agree that time dilation difference between two readings of each three clocks are same for R. But time readings of each three clocks are not same for R. To find time reading you have to use LT equation.

- You are telling \tau is not dependent of d, and you have proved forcefully that clocks is synchronized for R using TD equation and using single independent value of \tau. But then you have a goal to prove that clocks is not synchronized for O. So, you came up with three values of \tau. But, how could you not realize that if \tau is not dependent of d then how could you get three values of \tau? And if there are three values of \tau exist then how could you prove that clocks is synchronized for R using the TD equation?

- You said my modified diagram (modified by you) shows that clocks is not synchronized for O. Then how does the modified diagram shows that clocks is synchronized for R?

- I cannot understand that why you don't want to understand this: If a clock is at rest in a frame then its proper time (what time interval it has covered in the frame) is equal to co-ordinate time of the frame (because covered space interval is zero in the frame). If a clocks is moving in a frame then proper time (what time interval it has covered in the frame) is not equal to co-ordinate time of the frame (because covered space interval is not zero in the frame). So, \tau is a proper time of train's clocks, the clocks is at rest in O's frame then co-ordinate time of O's frame is equal to proper time of the clocks. And, \tau is a proper time of train's clocks, the clocks is moving in R's frame then co-ordinate time of R's frame is not equal to proper time of the clocks.

- I cannot understand that why you don't want to understand this: Line of simultaneity is the line on which, two events occurred, called simultaneous events. So short hand and long hand of clocks on the line meets at 12 must be simultaneous events on the line of simultaneity. So clocks on the line can be called synchronized. If clocks on line of simultaneity is not synchronized then what is other method you have to prove that two events occurred on the line of simultaneity is simultaneous events?

- Accelerating scenario can be made up using too many small inertial frames changing. My diagram shows how lines of simultaneity of O's frame becomes unparallel to space axis of R's frame (during acceleration and after acceleration). So, clocks on the lines of simultaneity becomes desynchronized for R. But, lines of simultaneity is always parallel to space axis of whatever small inertial frame in which O exist at that time. So, clocks on the lines of simultaneity remain synchronized for O.

- Using LT equation if I pick single value of t, I get three values of \tau, and if I pick single value of \tau, I get three values of t. So, train's clocks is not synchronized for R. If we set two another observer at A and B. The thee A, O and B confirms that train's clocks is synchronized for them, but they all see different value in R's clock.

- You guys explained me that how clocks becomes desync on frame changing: when train is at rest light pulse take equal time to reach to O from both clocks. But after frame change light pulse take different time to reach to O from both clocks (front clock's light beam traveling in opposite direction of train travel and rear clock's light beam traveling in same direction of train travel). But, light speed is same in all direction, so clocks is actually desync.

We can think a experiment on the basis of this. Suppose, there are two more observers at A and B. Train is at rest in R's frame. Clocks firing a pulse to all observers. O gets A and B's "0" pulse at same time but after some time of his own clock reads "0" because of light propagation time. O confirms that all clocks is synchronized. A gets "0" pulse first from O and then from B after some time of his own clock reads "0". A confirms that all clocks is synchronized. B gets "0" pulse first from O and then from A after some time of his own clock reads "0". B confirms that all clocks is synchronized.

Now, train has changed its frame. O gets A and B's "0" pulse at different time but after some time of his own clock reads "0". O confirms that clocks is not synchronized. O confirms that B (front) is ahead of A (rear) (O > B > A). A gets "0" pulse first from O and then from B after some time of his own clock reads "0". Time duration in receiving two pulse is obviously less than what was in rest frame. But A cannot confirms that clocks is not synchronized. A can say that O and B is behind of his own clock (A > O = B). B gets "0" pulse first from O and then from A after some time of his own clock reads "0". Time duration in receiving two pulse is obviously more than what was in rest frame. But B cannot confirms that clocks is not synchronized. B can say that O and A is behind of his own clock (B > O = A).

Can you see the problem here? Three observers in inertial frame is not agree with each other. They are not agree on sequence of desync (which is ahead and which is behind). They are not agree even on the clocks is synchronized or not.

- Earth's motion is accelerating motion around Sun, but we know that Earth's clocks remain synchronized for us. But Earth's clocks would not remain synchronized for Sun.

- I have proved that train's clocks is synchronized for O and not synchronized for R using diagram and maths (LT equation).