Synchronized clocks with respect to rest frame

[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'=\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>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).