Synchronized clocks with respect to rest frame

[Mentz114]

These diagrams might help. They are essentially the same as the one posted by Dalespam in post #95 but show the events from both sets of observers POV.

In the first diagram the stationary frame is the comoving observers A,B,C. The primed observers begin moving away simultaneously in the unprimed observers frame.

The second diagram is the rest frame of the primed observers, and it is clear that their clocks are no longer synchronised because they did not begin their motion simultaneously in their frame.

The diagrams are completely clear and show what is happening. There are no contradictions or paradoxes.

The mistake in your original diagram in post #1 is that you tried to make the moment of departure simultaneous in both frames, which it is not.

 

[mananvpanchal]

@Mentz114

Ok, so A, B, C and A', B', C' is synchronized in ABC's frame (before frame change).
So t_{abc}=t_{a'b'c'} (before frame change).

Now, clocks is changed its frame at t_{abc}=0 simultaneously in ABC's frame and clocks moving with 0.6c speed.

Now, can you calculate for me what is the value of t_{a'}, t_{b'} and t_{c'} at t_{abc}=0? (please, take origin on B's world line)

 

[Mentz114]

@Mentz114

Ok, so A, B, C and A', B', C' is synchronized in ABC's frame (before frame change).
So t_{abc}=t_{a'b'c'} (before frame change).

Now, clocks is changed its frame at t_{abc}=0 simultaneously in ABC's frame and clocks moving with 0.6c speed.

OK. The speed is 0.52c in the diagram below which has the lines of simultaneity. You can read off the answers from this.

Now, can you calculate for me what is the value of t_{a'}, t_{b'} and t_{c'} at t_{abc}=0? (please, take origin on B's world line)

Can you rephrase this in terms of proper times ? t_{abc}=0 is only true in the unprimed frame.

I've run out of time. Try calculating proper times by triangulation from the diagrams using dtau^2 = dt^2-dx^2.

 

[Dale]

- 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.

Not only is my use of the equation correct, I proved that it is correct from first principles, the spacetime interval equation.

The Lorentz transform is used to transform from the unprimed frame to the primed frame, it is not useful within a single frame. It is always used, once you have completely specified the problem in one frame, to determine how the problem appears in any other frame.

The LT cannot be used to specify the problem within a single frame. You should have realized that you were making a mistake when your use of the LT contradicted your initial scenario.

- 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?

This is a valid point. My apologies for the confusion. There are always 3 values of τ in both frames. So I should have written:
for \tau_d>0
(\tau_d - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau_d = 1.25 t
Where \tau_a, \tau_o, \tau_b are found by substituting the appropriate values of d into the equation.

Is that more clear?

- 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?

First, do you agree that the clocks each read 0 at the black dots in your scenario?

If so, then along each worldline simply find points of equal spacetime interval from the black dots and note that they occur simultaneously for R.

If not, then please mark where each clock reads 0 clearly on your diagram.

- 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).

Again, my watch and my desk clock are both at rest in my frame, and they have different readings of proper time. One says 1:19 and the other says 1:21. So it is not possible for them to both be equal to the co-ordinate time. They are not synchronized. This disproves your claim by counterexample.

You are focusing on the fact that they are running at the same rate as coordinate time in my frame. However, that is not relevant to determining synchronization. Two clocks may be running fast or slow and yet be synchronized. They may be running at the correct rate and yet not be synchronized. You are mixing two separate concepts.

- 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?

You are again confusing simultaneity with synchronization. Simply because you can draw a line of simultaneity on a spacetime diagram does not imply that all clocks will read the same on that line of simultaneity unless the clocks are synchronized.

The procedure for synchronization was given by Einstein. In your problem setup you stated that the clocks were initially synchronized in the unprimed frame. Meaning that Einstein's procedure synchronization procedure was carried out initially.

Because the clocks were synchronized in the unprimed frame, and because their velocity profile is the same in the unprimed frame at all times, they remain synchronized in the unprimed frame, as I proved. Because they are synchronized in the unprimed frame, they are not synchronized in the primed frame frame, as I proved.

- 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.

No they don't. Look at the gaps between your "sweeping" lines of simultaneity. Do those gaps look like they are marking equal spacetime intervals for each clock? Please answer this question in bold directly.

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

No you haven't. You first drew physically impossible diagrams, then you misinterpreted your own diagrams. Your math is incorrect because you tried to use a transform to specify a problem rather than to transform it into a new reference frame.

 

[mananvpanchal]

Can you rephrase this in terms of proper times ? t_{abc}=0 is only true in the unprimed frame.

I've run out of time. Try calculating proper times by triangulation from the diagrams using dtau^2 = dt^2-dx^2.

Suppose, that at some t_a=t_b=t_c=-0.001, value of t'_a=t'_b=t'_c=-0.001 too. But, at t_a=t_b=t_c=0, what should be the value of t_a=?, t_b=?, t_c=?

I think I don't need difference of proper time d\tau and difference of co-ordinate time dt. Please, clarify your point if I have misunderstood this.

 

[mananvpanchal]

This is a valid point. My apologies for the confusion. There are always 3 values of τ in both frames. So I should have written:
for \tau_d>0
(\tau_d - 0)^2 = (t-0)^2 - ((0.6t+d) - (0 + d))^2
\tau_d = 1.25 t
Where \tau_a, \tau_o, \tau_b are found by substituting the appropriate values of d into the equation.

Is that more clear?

No, this is more confusing. If there are three values of \tau, then you should have three values of t using the \tau = 1.25 t equation. Which implies that clocks is not synchronized in R's frame.

No they don't. Look at the gaps between your "sweeping" lines of simultaneity. Do those gaps look like they are marking equal spacetime intervals for each clock? Please answer this question in bold directly.

No, they are not marking equal spacetime intervals for each clock in R's frame.

 

[mananvpanchal]

@Mentz114

Yes, I have to agree that changing the frame of three clocks is simultaneous events for unprimed frame, so it should not be simultaneous for primed frame. From primed frame, first C changes its frame, then B and then A. This is easily understandable with two inertial frames that in one this events is simultaneous, but not in other. But, with this frame change scenario this is very hard for me to understand this. Because, the one by one frame changing of clocks require some time. And the time duration is the cause of desynchronization. So, I want to know how do the clocks changes its time readings in primed frame to be desynchronized?

 

[Mentz114]

Yes, I have to agree that changing the frame of three clocks is simultaneous events for unprimed frame, so it should not be simultaneous for primed frame. From primed frame, first C changes its frame, then B and then A.

I think A', then B' then C' change speed.

So, I want to know how do the clocks changes its time readings in primed frame to be desynchronized?

I can tell you how much, but not how or why. It is not difficult to calculate the proper times on the primed clocks when C' changes speed using the second diagram.

 

[mananvpanchal]

I think A', then B' then C' change speed.

Yes, I forgot that labels are in reverse order in diagram.

I can tell you how much, but not how or why.

That is fine

It is not difficult to calculate the proper times on the primed clocks when C' changes speed using the second diagram.

Should I use LT equation to calculate or some other?

 

[Mentz114]

The simultaneity is not straightforward in this scenario. In the diagram you can see that each of the primed observers will see the other two primed observers changing speed after they have done it.

To calculate proper times use the metric d\tau^2 = dt^2 - dx^2. You can use either diagram because (obviously) proper times will be the same between the same events in either.

I'll try the calculation myself later because I'm going to busy for some hours.

 

[Dale]

No, this is more confusing. If there are three values of \tau, then you should have three values of t using the \tau = 1.25 t equation. Which implies that clocks is not synchronized in R's frame.

Why? The only reason that you would need 3 values of something is if it were the parameter of the 3 worldlines. 3 worldlines -> 3 parameters. The t is coordinate time, not a parameter. The 3 values of τ are the parameters for the 3 worldlines. You don't need to have 6 parameters for 3 worldlines, so why would you need 3 values of τ and also 3 values of t?

You can use t as the parameter along the worldline, but then it gets confusing about whether you are talking about t one of the 3 parameters, or t coordinate time. This is another reason most people like to parameterize by proper time instead. Once you have parameterized the worldlines by τ then there is no need for further parameterization and to do so would only lead to confusion.

In any case, the t on the right hand side of the spacetime interval equation is the coordinate time here.

No, they are not marking equal spacetime intervals for each clock in R's frame.

Then clearly the clocks cannot remain synchronized at those lines of simultaneity.

I mentioned this back in post 49, but I guess I didn't do a good job describing the importance. Let me try to describe it now. Suppose you have two clocks undergoing arbitrary motion, and you draw two spacelike lines intersecting the worldlines of the clocks. Suppose further that you are given that at the first line the clocks both read 0. Then, to determine what each clock reads at the intersection with the next line, all you have to do is calculate the spacetime interval along the clock's path. If it is longer for one of the clocks then that clock will read a larger number. If the lines are lines of simultaneity then the clocks read the same number at the first line of simultaneity (synchronized) but read different numbers at the second line of simultaneity (desynchronized).

 

[mananvpanchal]

The simultaneity is not straightforward in this scenario. In the diagram you can see that each of the primed observers will see the other two primed observers changing speed after they have done it.

No, actually I don't. I see that A' changes frame first, then B' and then C'. They all should agree on sequence of events.

In train lightning example if we put three observers in train and three at platform. They all 6 agree that lightning occurs in platform frame simultaneously and in train frame unsimultaneously (with sequence).

In our example all three clocks A', B' and C' have changed frame simultaneously for unprimed frame, but they haven't for primed frame. They all A, B, C, A', B', C' should agree on simultaneous changing of frame for unprimed frame and sequence of unsimultaneous changing of frame for primed frame.

If they all don't agree with each other then I afraid their is very big confusion going on.

To calculate proper times use the metric d\tau^2 = dt^2 - dx^2. You can use either diagram because (obviously) proper times will be the same between the same events in either.

I really don't know how to calculate using that equation that is why I am asking you.

 

[Mentz114]

OK, I've done the calculation. Refer to the diagram. All quantities are in the primed frame.

I've labelled some points on the worldlines of A' and B'. a is the event where A' changes speed. The elapsed time on A' clock between a->b is ab. The elapsed time on B' clock is pq = ac.
So
<br /> \tau^2_{B&#039;}=\tau^2_{A&#039;}-(bc)^2= \tau^2_{A&#039;}(1-\beta^2) = \tau^2_{A&#039;}/\gamma^2<br />
I could have a made a mistake in my haste, but you can easily check my work now.

If they all don't agree with each other then I afraid their is very big confusion going on.

I'm sorry you find it confusing because taking into account light travel time that is what happens. If they calculate back they get the expected result.

 

[mananvpanchal]

<br /> \tau^2_{B&#039;}=\tau^2_{A&#039;}-(bc)^2= \tau^2_{A&#039;}(1-\beta^2) = \tau^2_{A&#039;}/\gamma^2<br />

Ok, this is fine. But...

Where is my confusion lies:

When A', B' and C' changes its frame simultaneously in unprimed frame at some time t_{a&#039;}=t_{b&#039;}=t_{c&#039;}=0.
Now, in primed frame first A' changes its frame at t_{a&#039;}=t&#039;_{a&#039;}=0. After, some time B' changes its frame at t_{b&#039;}=t&#039;_{b&#039;}=0, at this time A's value is reached to t&#039;_{a&#039;}=t&#039;_{a&#039;1} in primed frame. So, A' and B' is no longer synchronized in primed frame because t&#039;_{b&#039;}=0 and t&#039;_{a&#039;}=t&#039;_{a&#039;1}. A' is ahead of B' in primed frame.
So, far this is ok for me.

Now, confusion starts. I afraid here. Because a moment before changing frame (t_{a&#039;}=-0.000001) A' see same value in B' (t_{b&#039;}=-0.000001) in unprimed frame. So A' sees value < -0.000001 in B' after changing its frame. Because, A' reaches to t&#039;_{a&#039;}=t&#039;_{a&#039;1} when B' have t_{b&#039;}=t&#039;_{b&#039;}=0. So A' sees sudden decreasing change in B' clock's reading after changing its frame.

Please, look at your diagram when A' in unprimed frame at point "a". A' sees value in B' clock at point "q" because line of simultaneity is skewed in unprimed frame. But, after changing frame, line of simultaneity is changed for A'. So now A' sees value in B' clock at point "p". Now, value at "p" < value at "q", so A' sees sudden decreasing change in B' clock's reading after changing its frame.

Am I clear so far?

 

[Mentz114]

Ok, this is fine. But...

Where is my confusion lies:

When A', B' and C' changes its frame simultaneously in unprimed frame at some time t_{a&#039;}=t_{b&#039;}=t_{c&#039;}=0.
Now, in primed frame first A' changes its frame at t_{a&#039;}=t&#039;_{a&#039;}=0. After, some time B' changes its frame at t_{b&#039;}=t&#039;_{b&#039;}=0, at this time A's value is reached to t&#039;_{a&#039;}=t&#039;_{a&#039;1} in primed frame. So, A' and B' is no longer synchronized in primed frame because t&#039;_{b&#039;}=0 and t&#039;_{a&#039;}=t&#039;_{a&#039;1}. A' is ahead of B' in primed frame.
So, far this is ok for me.

Now, confusion starts. I afraid here. Because a moment before changing frame (t_{a&#039;}=-0.000001) A' see same value in B' (t_{b&#039;}=-0.000001) in unprimed frame. So A' sees value < -0.000001 in B' after changing its frame. Because, A' reaches to t&#039;_{a&#039;}=t&#039;_{a&#039;1} when B' have t_{b&#039;}=t&#039;_{b&#039;}=0. So A' sees sudden decreasing change in B' clock's reading after changing its frame.

Please, look at your diagram when A' in unprimed frame at point "a". A' sees value in B' clock at point "q" because line of simultaneity is skewed in unprimed frame. But, after changing frame, line of simultaneity is changed for A'. So now A' sees value in B' clock at point "p". Now, value at "p" < value at "q", so A' sees sudden decreasing change in B' clock's reading after changing its frame.

Am I clear so far?

A' observer cannot see anything until the light reaches them from B' worldline. Doing the calculation means nothing because the two points ( events) involved are not in causal contact. They are outside each others light-cone. It is not a physical number you are calculating.

I don't understand your notation. Please don't use 't'. The only thing I will work in is proper times, denoted by \tau.

 

[mananvpanchal]

A' observer cannot see anything until the light reaches them from B' worldline. Doing the calculation means nothing because the two points ( events) involved are not in causal contact. They are outside each others light-cone. It is not a physical number you are calculating.

I don't understand your notation. Please don't use 't'. The only thing I will work in is proper times, denoted by \tau.

Yes, you are right. I know that that A' actually doesn't see decreasing value of B' clock because of light propagation time. This is just calculation on line of simultaneity.

Thanks

 

[mananvpanchal]

Hello Mentz114

I have a doubt here.

I have added two line of simultaneity of unprimed frame in your diagram. Suppose, two events occurred on A's path and B's path exactly after first line of simultaneity in unprimed frame. The event occurred after "p" on B's path emits light pulse toward A'. We don't know when light pulse is reached to A'. But, we are sure that light pulse is already emitted by the event.

Now, A' changes its frame. After changing the frame, A's line of simultaneity of primed frame becomes horizontal line through "p". So, that event occurred before frame change on B's path already emitted light pulse to A'. Now, the line of simultaneity again reach to the same point.

Would that event be reoccurred for A'?
Does A' see two light pulse (one from event occurred before frame change and other from event occurred after frame change)?
Would the all events between "p" and "q" be reoccurred for A'?

I think instant frame change might be the cause of the problem here. Am I right?

 

[Mentz114]

Hello Mentz114

Would that event be reoccurred for A'?
Does A' see two light pulse (one from event occurred before frame change and other from event occurred after frame change)?
Would the all events between "p" and "q" be reoccurred for A'?

I think instant frame change might be the cause of the problem here. Am I right?

No, the event would not reoccurr. That would be a causal paradox.

There's no problem. I've tried to add regular light pulses from B/B' to A/A'. As you can see, everything is regular. A and A' see a Doppler shift when either B' or A' change speed.

I don't think there is anything to be gained by looking for problems in this scenario because there aren't any. You are making problems by doing a calculation for the unprimed frame in the primed frame.

 

[mananvpanchal]

No, the event would not reoccurr. That would be a causal paradox.

I am sorry, if you guys find that I am creating problems here.

Anyway, I have some other doubts related to instant frame change scenario.

Please, look at this diagram.

We cannot find contracted length between clocks of primed frame for unprimed frame in above digram. Because, rest length in unprimed frame is defines in diagram is about 10, and if we try to find contracted length by seeing the length between clocks at same time in unprimed frame we again get 10.

We can easily find here contracted length between clocks of unprimed frame for primed frame in above digram. Here, we can see that rest length in primed frame is about 12, whereas contracted length in primed frame is about 9.

I think the problem here might be:
In second image we have described frame change events unsimultaneously. That is why the length between vertical world lines in primed frame are more than the length between diagonal world lines in unprimed frame. But in first image the events described are simultaneous. That is why the length between vertical world lines in unprimed frame are equal to the length between diagonal world lines in primed frame.

This problems might not so hard for you, but it is actually for me. That is why I am creating problems here...

 

[Mentz114]

Those two diagrams are related by a Lorentz transformation with beta = 0.525. That's the way it is.

The velocity changes don't have to be instantaneous, just taking a small time compared to the grid scale.

I'm sorry you're having problems understanding this, but I can't see your difficulties.

 

[mananvpanchal]

Those two diagrams are related by a Lorentz transformation with beta = 0.525. That's the way it is.

The velocity changes don't have to be instantaneous, just shorter than the grid scale.

I'm sorry you're having problems understanding this, but I can't see your difficulties.

That's ok. Don't worry. Thanks Mentz114.

 

[Mentz114]

That's ok. Don't worry. Thanks Mentz114.

I think your problem is with the idea of 'now'. This concept really only has meaning for a particular observer. If the A' observer asks himself 'what is happening to B' now ?' the question has no meaning and leads to weird results, especially if the observers are well separated. The best example is the 'Andromeda paradox', where two observers on Earth with a very small relative velocity apply the idea of 'now' to a very distant place, like the Andromeda galaxy. Using their lines of simultaneity they come up times that are thousands of years apart on Andromeda. So 'now' on a worldline than your own is a frame dependent quantity and so has no useful interpretation.

 

[pervect]

I think your problem is with the idea of 'now'. This concept really only has meaning for a particular observer. If the A' observer asks himself 'what is happening to B' now ?' the question has no meaning and leads to weird results, especially if the observers are well separated. The best example is the 'Andromeda paradox', where two observers on Earth with a very small relative velocity apply the idea of 'now' to a very distant place, like the Andromeda galaxy. Using their lines of simultaneity they come up times that are thousands of years apart on Andromeda. So 'now' on a worldline than your own is a frame dependent quantity and so has no useful interpretation.

I haven't been following the discussion in detail, but I also suspect the problem of "now" is the issue. The other, related issue that may be arising is what "now" means to someone who is making an instantaneous velocity change. I don't think there's any good way of defining the concept of "now" in such a case, frankly. The sudden jump in frames leaves some events without any valid time coordinate, and then there's also the problem of some events having multiple time coordinates.

It may be clearer and make more physical sense if we consider what "now" looks like for an accelerating observer, and take the limit as the acceleration approaches infinity.

This is rather technical, but the end result is rather well known as the "rindler" observer. I'll present the results without explaining how they were arrived at, though time permitting I could probably be convinced to take the effort to draw up the diagrams and post more if there is interest. I should add that I'm sure Mentz already knows this, and he might also be induced to explain it if it's of interest to the OP. I'm not POSITVE this is actually the issue or an issue of interest to the OP, so I'm reluctant to take too much time on it currently.

Anyway, the end result is that there is only a certain region of space-time that can be given valid coordinates, a certain region that the concept of "now" applies to. The edge of this region is bounded by the so-called Rindler horizon, which is an event horizon similar to that formed by a black hole.

The Rindler horizon is at a distance of c^2/g "behind" the accelerating observer. As g approaches infinity, the Rindler horizon gets closer and closer, until in the limit of infinite acceleration, the Rindler coordinates can't cover anything behind the accelerating observer at all.

Since I regard "now" as entirely a mental construct, without much physical significance, this doesn't bother me a whole lot. It's basically mostly a matter of convention - though if you want to set up a coordinate system that obeys Newton's laws, it's an important convention, i.e. you can't both use an arbitrary definition of "now" AND also apply Newton's laws even in a limiting sense.

 

[Dale]

This problems might not so hard for you, but it is actually for me. That is why I am creating problems here...

That and you prefer to argue and come up with really strange ways to tell people that they are wrong rather than try to ask questions and learn. It is not productive, particularly when you know that you are learning a topic.

Btw, you never responded to my post 111.

In general, if you want to describe the behavior of some clock you need to write down a parametric expression for its worldline in some given reference frame. Then, to determine the reading on the clock you use the second formula on this page if the reference frame is inertial:
http://en.wikipedia.org/wiki/Proper_time#In_special_relativity

Or, if it is a non-inertial frame or gravity is involved:
http://en.wikipedia.org/wiki/Proper_time#In_general_relativity

That will give you a complete description of the clock in that frame. If you want to then see the description of the clock in some other frame then you use the Lorentz transform if both frames are inertial or the appropriate coordinate transform otherwise.

That is the general approach for these types of problems.

If you have multiple clocks simply parameterize each one and follow the above, and in order to determine if they are synchronized in some frame simply compare their proper times at some specific coordinate time.

 

[mananvpanchal]

I think your problem is with the idea of 'now'.

Please, understand "see" as "calculate" here.

We might describe this situation like this:
Line of simultaneity of A', B' and C' in unprimed frame moving from bottom to up. There is only one line of simultaneity exist for all three clocks. Now, in unprimed frame all clocks simultaneously change its frame. But, how can we display the simultaneous change in primed frame? We might imagine that three lines of simultaneity created simultaneously in primed frame like three parallel world. This lines of simultaneity is far from each other by time distance. Now, the three lines of simultaneously moving from bottom to up remaining parallel to each other. The events (B' frame change and C' frame change) is already occurred in future of A', but A' cannot see it. The events leaves footprint of own self in spacetime. So, when A's line of simultaneity come to the footprint of the events, A' can see that the events is reoccurring for A'. Here we should not use word "reoccurring", because A' has not seen the events before, it was occurred in future. So, now in primed frame A' still sees in own world that B' has not changed its frame yet. But, B' sees in own world that A' has already changed its frame before himself. The situation is fine uptill now. All events might be occurred already in future of A'. But the events occurs only for one time in A's world. But the symmetry breaks where two frames meet each other. A' might see that events are reoccurring for him in this region (between "p" and "q"). (Possibly reoccurring event is defined as black dot in diagram)

There might be many errors exits in my description, and I expect it to be corrected.

 

[mananvpanchal]

I'll present the results without explaining how they were arrived at, though time permitting I could probably be convinced to take the effort to draw up the diagrams and post more if there is interest.

It would be great help. Thanks.

 

[mananvpanchal]

That and you prefer to argue and come up with really strange ways to tell people that they are wrong rather than try to ask questions and learn. It is not productive, particularly when you know that you are learning a topic.

At starting of this post I had a result in my mind about the scenario. I knew that it might be wrong. So, I have posted my doubt here. I asked that "Am I right? If not then please, give me explanation". I got the answer that "You are wrong", but no explanation. After, this you came with maths. I appreciated your work and was trying to be convinced by your maths that "clocks in synchronized for R, but not for O". But, I found some errors in your maths. I asked about you, and you also came up with solutions.

Btw, you never responded to my post 111.

Can you re-explain your post #42 using subscript with \tau?
But, please this time define clearly:
What is t? What is \tau_a, \tau_o, \tau_b? What is t&#039;?

 

[Mentz114]

mananvpanchal , I'm sorry but I don't understand any of your post#125 except this bit

But, how can we display the simultaneous change in primed frame?

You cannot because the change of velocities is not simultaneous in that frame.
The LT tells me that those 3 events cannot be simultaneous in the primed frame. It is difficult to accept because it is counter-intuitive but it leads to no paradoxes, discontinuities or other unpleasantness.

When you start talking about events 'reoccurring' you've lost me ( and physics). That doesn't happen or someone would have noticed. Similarly there is no point in discussing what goes on in other worldlines because we can only ever see the past of spatially separated observers.

For all I know, the OP might have something important to say, but I can't see what it is and I'm pretty tired of this whole thing.

DaleSpam and Pervect - your posts are right on the button and I hope they help the OP to understand what he/she is trying to fathom.

 

[Dale]

I found some errors in your maths

You found a small inconsistency in notation, not an error. The math is all correct.

Can you re-explain your post #42 using subscript with \tau?
But, please this time define clearly:
What is t? What is \tau_a, \tau_o, \tau_b? What is t&#039;?

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. R's frame is the unprimed frame and t is the coordinate time in the unprimed 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.

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, \tau_d, using the spacetime interval. Solving for t we get:
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}

Substituting into the above we get an expression for the worldline parameterized by the proper time on each clock:
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)

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

Now, we use the Lorentz transform to boost to the primed frame where O is at rest for \tau_d=t&gt;0. This frame is the unprimed frame and t' is the coordinate time in this frame. We obtain the following expression for the worldline of the clocks in the primed frame, again parameterized by the proper time.
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)

Noting that \tau_d does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. It is therefore quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.

 

[mananvpanchal]

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

Noting that \tau_d does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. It is therefore quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.

Thanks for re-explaining.

Ok, so in unprimed frame \tau_a = \tau_o = \tau_b and in primed frame \tau_a \ne \tau_o \ne \tau_b. Right?

 

[Dale]

Yes.