synchronized clocks with respect to rest frame

 Quote by DaleSpam 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 $\tau_d=t'+0.75d$ for primed frame, not $\tau_d=t+0.75d$.
Now, we have two unknown variables $t'$ and $\tau_d$.

Mentor
 Quote by mananvpanchal I am sorry, but we would get $\tau_d=t'+0.75d$ for primed frame, not $\tau_d=t+0.75d$. Now, we have two unknown variables $t'$ and $\tau_d$.
Oops, you are correct. I will fix it above.
 First lets take another situation, lets just say that the two clocks on the train are in sync in the train's frame of reference. Lets say that the train is moving at .5c in the stations frame of reference. Do you understand why the clocks won't by in sync in the stations frame of reference, if they are in sync in the trains?

Quote by DaleSpam
 Quote by mananvpanchal I am sorry, but we would get $\tau_d=t'+0.75d$ for primed frame, not $\tau_d=t+0.75d$. Now, we have two unknown variables $t'$ 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 $\tau_d$
 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?

Mentor
 Quote by mananvpanchal 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.

Hello DaleSpam,

First you had came up with this equation
 Quote by DaleSpam $$r'_d=\left( t'=\begin{cases} 1.25 \tau - 0.75 d & \mbox{if } \tau \lt 0 \\ \tau - 0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d - 0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$
Then, I had created the doubt
 Quote by mananvpanchal We can see that $t'=\tau - 0.75 d, \mbox{if } \tau \ge 0$. And we know that d=-1 for A, d=0 for O and d=1 for B then, we can get this $t'_a=\tau + 0.75,$ $t'_o=\tau,$ $t'_b=\tau - 0.75$ But suppose, train is going to opposite direction, then the equation cannot distinguish both $t'$
And you had solved this by
 Quote by DaleSpam v = -0.6 c
So, I had to get $t'_a$, $t'_o$, $t'_b$ using $\tau$ and $d$.

After this I had created another doubt
 Quote by mananvpanchal Yes, you are right. We will get the answer as you said. There is another doubt! This may be the case again of generalization. $t'=\tau - 0.75 d, \mbox{if } \tau \ge 0$ As $\tau$ increases, desynchronization between two clocks decreases.
And you had came up with the idea of $\tau_a$, $\tau_o$, $\tau_b$
 Quote by DaleSpam 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 $\tau_1 - \tau_0=0.75$.
So, I had to get $t'_a$, $t'_o$, $t'_b$ using $\tau$ and $d$. But you had came up with $\tau_a$, $\tau_o$, $\tau_b$.

 Quote by mananvpanchal 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
 Quote by DaleSpam 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.
 Quote by mananvpanchal Now, we have two unknown variables $t'$ and $\tau_d$.
So, now the problem is we have to derive

$t'_a=\tau + 0.75,$
$t'_o=\tau,$
$t'_b=\tau - 0.75$ 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.

$t'_a=\tau_a - 0.75 d_a,$
$t'_o=\tau_o - 0.75 d_o,$
$t'_b=\tau_b - 0.75 d_b$.

And as I said before we have now two unknown variables per eqaution ($t'_a$, $\tau_a$), ($t'_o$, $\tau_o$) and ($t'_b$, $\tau_b$).

So, the question is how can we get values of $t'_a$, $\tau_a$, $t'_o$, $\tau_o$, $t'_b$ and $\tau_b$ using known variables $t$, $d_a$, $d_o$ and $d_b$?
 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?

Mentor
 Quote by mananvpanchal Now, we have the equations. $t'_a=\tau_a - 0.75 d_a,$ $t'_o=\tau_o - 0.75 d_o,$ $t'_b=\tau_b - 0.75 d_b$. And as I said before we have now two unknown variables per eqaution ($t'_a$, $\tau_a$), ($t'_o$, $\tau_o$) and ($t'_b$, $\tau_b$). So, the question is how can we get values of $t'_a$, $\tau_a$, $t'_o$, $\tau_o$, $t'_b$ 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 $\tau_d$.

 Quote by DaleSpam 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?

$t'=\tau_a - 0.75 d_a$
$t'=\tau_o - 0.75 d_o$
$t'=\tau_b - 0.75 d_b$

 Quote by DaleSpam Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore lets use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame. So, in R's frame the worldline of A, O, and B are: $$r_d=\left(t,x=\begin{cases} d & \mbox{if } t \lt 0 \\ 0.6 t+d & \mbox{if } t \ge 0 \end{cases} ,0,0\right)$$ where d=-1 for A, d=0 for O, and d=1 for B.
Here, you simply defined time $t$ in R's clock. So, you can define distance of train clocks from R with $x=0.6t + d$.
 Quote by DaleSpam As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, τ, using the spacetime interval. Solving for t we get: $$t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}$$
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
 Quote by DaleSpam Substituting in to the above we get: $$r_d=\left( t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$ Noting that τ does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.
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.

Quote by DaleSpam
 Quote by mananvpanchal $t'=\tau - 0.75 d, \mbox{if } \tau \ge 0$ 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 $\tau_1 - \tau_0=0.75$.
Here, I am talking about $t'$. I have take here $t'_a=\tau - 0.75 d_a$, $t'_o=\tau - 0.75 d_o$ and $t'_b=\tau - 0.75 d_b$. So as $\tau$ increase deference between $t'_a$, $t'_o$ and $t'_b$ decreases. And that is why I have used subscript with $t'$.

$t$ 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 $t'$ 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 ($t'$)) (Train's clocks is not synchronized for O, so $t'$ should have different values as per $d$). So actually subscript should be used with $t'$ not with $\tau$.
 Mentor 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.

 Quote by DaleSpam As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, τ, using the spacetime interval. Solving for t we get: $$t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}$$
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.

 Quote by DaleSpam 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.
 Quote by DaleSpam 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.
 Quote by DaleSpam The latter parameterization is particularly common since it is frame-invariant and physically measurable.
 Quote by DaleSpam 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.
 Quote by DaleSpam I will try to post the application of these principles to this specific problem later today.
Certainly.

Mentor
 Quote by mananvpanchal 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.
 Mentor 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?

Mentor
 Quote by mananvpanchal 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.