Events for frame changing clocks

[mananvpanchal]

Hello,

The three clocks A, B and C is at rest in S frame. The clocks unsimultaneously change its frame from S to S'. The frame change events is simultaneous in S' frame. Now, clocks is moving in S frame and at rest in S' frame.

I am struggling with this. The events on A clock displayed by red dots seems re-occurring for C clock (The events occurring two times, before frame changing of C and after frame changing of C). The events on C clock displayed by green dots seems never occurring for A clock (The events have no chance to occur, because line of simultaneity of A instantly becomes skewed). Please, shade some light on this.

Manan

 

[ghwellsjr]

Hello,

The three clocks A, B and C is at rest in S frame. The clocks unsimultaneously change its frame from S to S'. The frame change events is simultaneous in S' frame. Now, clocks is moving in S frame and at rest in S' frame.

View attachment 46458

I am struggling with this. The events on A clock displayed by red dots seems re-occurring for C clock (The events occurring two times, before frame changing of C and after frame changing of C). The events on C clock displayed by green dots seems never occurring for A clock (The events have no chance to occur, because line of simultaneity of A instantly becomes skewed). Please, shade some light on this.

Manan

I think you're struggling because you don't have a clear and proper understanding of what an event is. You've drawn some red and green dots depicting totally unique events. The red events on A clock have nothing to do with the green events on C clock.

An event is the set of four co-ordinates, one of time and three of space, for a point in spacetime. If you change anyone of the co-ordinates, you have a different event. So why would you be concerned that the four red events happening where the A clock is located at four different times have any connections with the four green events happening where the C clock is located?

 

[mananvpanchal]

I think you're struggling because you don't have a clear and proper understanding of what an event is. You've drawn some red and green dots depicting totally unique events. The red events on A clock have nothing to do with the green events on C clock.

An event is the set of four co-ordinates, one of time and three of space, for a point in spacetime. If you change anyone of the co-ordinates, you have a different event. So why would you be concerned that the four red events happening where the A clock is located at four different times have any connections with the four green events happening where the C clock is located?

Yes, I know that a event can be defined using three spatial coordinates and one time coordinate. And I know that there is no connection between red events and green events.

The three clocks is changing S frame turn by turn. First of all A changes the frame, then B and then C. The three events is unsimultaneous in S frame, but it is simultaneous in S' frame. If some observer is at rest already in S' frame, he sees that all three clocks have changed its frame simultaneously.

Now, please look closely. I have showed horizontal line of simultaneity (LoS) of C in S frame. A has already changed his frame for C, but C is still in S frame. The red events is occurring for C on A clock (like flashing a light pulse per time unit) during its journey between two horizontal LoS, please note that C is still in S frame where A is in S' frame. After second horizontal LoS C changes its frame. Now C is in S' frame. The LoS of C becomes skewed and the red events seems eligible for re-occurring for C.

I have showed LoS of A in S frame. First A changes its frame for S frame observer. The LoS of A becomes skewed. And the green events on C don't seems eligible to occur for A.

Please, let me know if the scenario is still unclear.

 

[ghwellsjr]

Yes, I know that a event can be defined using three spatial coordinates and one time coordinate. And I know that there is no connection between red events and green events.

The three clocks is changing S frame turn by turn. First of all A changes the frame, then B and then C. The three events is unsimultaneous in S frame, but it is simultaneous in S' frame. If some observer is at rest already in S' frame, he sees that all three clocks have changed its frame simultaneously.

Now, please look closely. I have showed horizontal line of simultaneity (LoS) of C in S frame. A has already changed his frame for C, but C is still in S frame. The red events is occurring for C on A clock (like flashing a light pulse per time unit) during its journey between two horizontal LoS, please note that C is still in S frame where A is in S' frame. After second horizontal LoS C changes its frame. Now C is in S' frame. The LoS of C becomes skewed and the red events seems eligible for re-occurring for C.

I have showed LoS of A in S frame. First A changes its frame for S frame observer. The LoS of A becomes skewed. And the green events on C don't seems eligible to occur for A.

Please, let me know if the scenario is still unclear.

You should not think of the clocks as changing frames. What they are doing is changing speed. You can convert the co-ordinates for an event in one frame into the co-ordinates of the other frame.

Pick a red event. I presume its co-ordinates will be defined in the S' frame. Convert its co-ordinates to the S frame. Keep the time co-ordinate and swap in the spatial co-ordinates for the C clock. Since it is at rest in the S frame, this will be very easy to do. Now you have an event for the A clock that is simultaneous with an event for the C clock in the S frame.

Or, if you want to go the other way, pick a green event defined with co-ordinates in the S frame. Convert it to the S' frame. Keep the time co-ordinate and swap in the spatial co-ordinates for the A clock. Since it is at rest in the S' frame, this will be very easy to do. Now you have an event for the C clock that is simultaneous with an event for the A clock in the S' frame.

Really, this is a trivial issue.

 

[mananvpanchal]

You should not think of the clocks as changing frames. What they are doing is changing speed. You can convert the co-ordinates for an event in one frame into the co-ordinates of the other frame.

Pick a red event. I presume its co-ordinates will be defined in the S' frame. Convert its co-ordinates to the S frame. Keep the time co-ordinate and swap in the spatial co-ordinates for the C clock. Since it is at rest in the S frame, this will be very easy to do. Now you have an event for the A clock that is simultaneous with an event for the C clock in the S frame.

Or, if you want to go the other way, pick a green event defined with co-ordinates in the S frame. Convert it to the S' frame. Keep the time co-ordinate and swap in the spatial co-ordinates for the A clock. Since it is at rest in the S' frame, this will be very easy to do. Now you have an event for the C clock that is simultaneous with an event for the A clock in the S' frame.

Really, this is a trivial issue.

Please, try to understand. I have defined some events in diagram (red events and green events). A can define red events in its S' frame and C can define the same events in its S frame. Same for green events, A can define green events in its S' frame and C can define the same events in its S frame. The all events is defined by two co-ordinate system for A and C.

The issue here is not to define the events in any frame or transforming to one frame to another.

The issue is:

at t1: LoS is horzontal for all three clock.
at t2: A changes its frame and LoS of its become skewed. A misses all green events on C.
at t3: B changes its frame.
at t4: C changes its frame.
between t2 and t4: C is in S frame, and all red events occur for C.
we take t4 of S=t'1 of S'
so, after t'1: LoS of C become skewed. The all red events seems eligible to re-occur for C.

 

[Dale]

Events are geometric objects that exist in the manifold independently of any coordinate chart. A given coordinate chart may or may not cover the entire manifold. Whether or not a given chart covers a given event has nothing to do with the existence of the event.

 

[yuiop]

I am reposting this because there was a major error in original attached drawing.

In the attached drawing one second time intervals measured by the clocks are marked as a0 to a5, b0 to b5 and c0 to c5 on clocks A, B and C respectively. Light blues lines connect equal times indicated on neighbouring clocks. These times are the times actually measured on clocks A, B and C and are not coordinate times in S or S'. As in the OP, the lines labelled S and S' indicates lines of simultaneity in S and S' respectively.

The thing to note is that the light blue lines are not parallel to the LoS in S' even after all the clocks attain the same velocity relative to S and are all at rest in S'. Clocks do not naturally resynchronize themselves after being accelerated to a new rest frame. The same is true even if we use the Born rigid acceleration method, which retains the proper distance between clocks during acceleration.

After the acceleration phase the clocks have to be manually resynchronized in their new rest frame, for example by retarding the B and C clocks relative to the A clock, or advancing the A and B clocks relative to the C clock. It is arbitrary whether we manually adjust the leading or trailing clocks. When we manually adjust a clock we cannot claim we have moved into the future or past of other observers in other reference frames. That is the myth of theories like block time.

 

[ghwellsjr]

Please, try to understand. I have defined some events in diagram (red events and green events). A can define red events in its S' frame and C can define the same events in its S frame. Same for green events, A can define green events in its S' frame and C can define the same events in its S frame. The all events is defined by two co-ordinate system for A and C.

The issue here is not to define the events in any frame or transforming to one frame to another.

First off, let me point out that your scenario has nothing to do with the fact that A, B and C are clocks--they could just as easily be rocks. You haven't asked about the times on the clocks.

The issue is:

at t1: LoS is horzontal for all three clock.

In frame S, LoS is horizontal all the time for all events. You don't have to narrow it down to any specific time.

at t2: A changes its frame and LoS of its become skewed.

A didn't change its frame. A has always been in frame S and in frame S'. In frame S', LoS is skewed all the time for all events. You don't have to narrow it down to any specific time.

However, the event at which A changes its speed has two different sets of co-ordinates, one set for frame S and another set for frame S'. You just can't say at t2, something happened to A without also saying which frame t2 applies to.

A misses all green events on C.

This statement reveals that you don't have any idea what an event is. The green events occur in both frames, just with different co-ordinates. Why do you think that A misses any events? This just doesn't make any sense at all.

at t3: B changes its frame.

B changes its speed, not its frame, and you need to say which frame t3 applies to. B is always in both frames, just at rest prior to a certain time in frame S and after a different certain time is S'. You need to do a Lorentz Transform to determine those two times in both frames, you cannot just say that t3 applies to both frames.

at t4: C changes its frame.

Likewise for C and t4.

between t2 and t4: C is in S frame, and all red events occur for C.

At all times, C is in both frames. Each red event has two different sets of co-ordinates, one for frame S and one for frame S'.

we take t4 of S=t'1 of S'

You cannot compare times in frame S with times in S'. It is meaningless to say that the time co-ordinate in one frame is equal to the time co-ordinate in a different frame (except, of course, for the origins). What you can do is say that t'4=t'1 (if it is, who knows?).

so, after t'1: LoS of C become skewed.

LoS applies to frames, not to rocks or clocks. LoS is skewed in your drawing for frame S' all the time, not just for the one example you drew. LoS is horizontal in your drawing for frame S all the time, not just for the three examples you drew.

The all red events seems eligible to re-occur for C.

This last statement reveals that you have no idea what an event is.

Really, this is a trivial issue. As long as you continue to ignore what I'm telling you and insist that I don't understand, then you will continue to struggle with this very simple issue.

 

[mananvpanchal]

Events are geometric objects that exist in the manifold independently of any coordinate chart. A given coordinate chart may or may not cover the entire manifold. Whether or not a given chart covers a given event has nothing to do with the existence of the event.

This seems logical.

 

[mananvpanchal]

I am reposting this because there was a major error in original attached drawing.

The confusion is created because I have used here clocks. But, please don't go in detail of like time reading, synchronization, co-ordinate time or proper time. I have only used clock because I want to define some (light pulse) events. We can imagine the clocks as some other objects which can generate some light pulse events.

 

[mananvpanchal]

First off, let me point out that your scenario has nothing to do with the fact that A, B and C are clocks--they could just as easily be rocks. You haven't asked about the times on the clocks.

Yes, you are right. I am sorry for misunderstanding. The scenario has nothing to do with clocks.

In frame S, LoS is horizontal all the time for all events. You don't have to narrow it down to any specific time.

Yes, you are right. But, I was just trying to describe some situation at some time.

A didn't change its frame. A has always been in frame S and in frame S'. In frame S', LoS is skewed all the time for all events. You don't have to narrow it down to any specific time.

This is also fine.

However, the event at which A changes its speed has two different sets of co-ordinates, one set for frame S and another set for frame S'. You just can't say at t2, something happened to A without also saying which frame t2 applies to.

Again I was just defining some situation at some time. t2 is in S frame.

This statement reveals that you don't have any idea what an event is. The green events occur in both frames, just with different co-ordinates. Why do you think that A misses any events? This just doesn't make any sense at all.

This last statement reveals that you have no idea what an event is.

What I understand about an event is: The event is a point in 4D spacetime. We can define the same point with some other co-ordinate system too. The point is different than 3D point. Because 4D point has time component and it is there at some time, it is not there for all the time.
If my understanding is limited then please, explain me about it with some detail.

 

[ghwellsjr]

What I understand about an event is: The event is a point in 4D spacetime. We can define the same point with some other co-ordinate system too. The point is different than 3D point. Because 4D point has time component and it is there at some time, it is not there for all the time.
If my understanding is limited then please, explain me about it with some detail.

There are two important issues with regard to events in Special Relativity which you may have missed:

1) The fact that a time co-ordinate is "added" to the three spatial co-ordinates may imply that it is independent as is true in Galilean co-ordinate systems. But in Special Relativity, the time co-ordinate is inextricably linked with the three spatial co-ordinates so that you cannot have it under some conditions and not have it under others.

Think about a 3D Galilean co-ordinate system. What if someone tried to persuade you that the Z-component might not be present some of the time--it just wouldn't make sense.

2) When you define or specify an event in one inertial co-ordinate system, you have already defined and specified it for all other inertial co-ordinate systems and you must use the Lorentz Transformation as the means to obtain the co-ordinates in those other systems. It always produces a time component, as well as three spatial co-ordinates in any other system you want. There are no exceptions.

Why don't you add some numbers to your diagram in your first post, including the co-ordinates of the red and green events and then use the Lorentz Transform to see that all the events occur in both frames without any problem?

 

[mananvpanchal]

There are two important issues with regard to events in Special Relativity which you may have missed:

1) The fact that a time co-ordinate is "added" to the three spatial co-ordinates may imply that it is independent as is true in Galilean co-ordinate systems. But in Special Relativity, the time co-ordinate is inextricably linked with the three spatial co-ordinates so that you cannot have it under some conditions and not have it under others.

Think about a 3D Galilean co-ordinate system. What if someone tried to persuade you that the Z-component might not be present some of the time--it just wouldn't make sense.

2) When you define or specify an event in one inertial co-ordinate system, you have already defined and specified it for all other inertial co-ordinate systems and you must use the Lorentz Transformation as the means to obtain the co-ordinates in those other systems. It always produces a time component, as well as three spatial co-ordinates in any other system you want. There are no exceptions.

Why don't you add some numbers to your diagram in your first post, including the co-ordinates of the red and green events and then use the Lorentz Transform to see that all the events occur in both frames without any problem?

All are fine for me.

 

[mananvpanchal]

What I understand with line of simultaneity is: The line on which all events happened to be considered as simultaneous events.

Please, look at figure 4 in this link http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html.

The "time gap" is created by turn around. The same situation is displayed with green dots. A changes its speed so A becomes at rest in S' frame, and the green events lies in the time gap for A.

The outbound leg is symmetrical for terence and stella, same is true for inbound leg. But, the age difference is created at turn around. Till turn around terence is agging less for stella and stella is aging less for terence. But at turn around terence's age suddenly increased for stella. After turn around the situation again becomes symmetrical.

So, events occurring on terence's location between last line of simultaneity of outbound leg and first line of simultaneity of inbound leg might not get any chance to occur for stella. Terence is suddenly more aged for stella.

The opposite case also to be considered where events get chance to reoccur for C clock.

 

[Dale]

You may want to read this paper. The authors discuss the problem you have noticed and propose my favorite solution to it.

http://arxiv.org/abs/gr-qc/0104077

 

[mananvpanchal]

You may want to read this paper. The authors discuss the problem you have noticed and propose my favorite solution to it.

http://arxiv.org/abs/gr-qc/0104077

Thanks DaleSpam.
This paper perfectly solves the both problem. The method to define line of simultaneity is the key of the solution. Now, red events would not be reoccurred and green events would not be skipped for C and A respectively.

 

[ghwellsjr]

What I understand with line of simultaneity is: The line on which all events happened to be considered as simultaneous events.

You left out, "according to a given Frame of Reference". In Special Relativity, "event" has a particular meaning and along with it, so does "simultaneous". This is really a trivial issue and you are refusing to pay attention to the process you must follow in Special Relativity.

Please, look at figure 4 in this link http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html.

Now you're changing the subject. We agreed that your scenario has nothing to do with the times on clocks:

First off, let me point out that your scenario has nothing to do with the fact that A, B and C are clocks--they could just as easily be rocks. You haven't asked about the times on the clocks.

Yes, you are right. I am sorry for misunderstanding. The scenario has nothing to do with clocks.

and now you're discussing the twin paradox which is all about the times on clocks and "where did the missing time go?".

The "time gap" is created by turn around. The same situation is displayed with green dots. A changes its speed so A becomes at rest in S' frame, and the green events lies in the time gap for A.

At least you are now starting to use the correct terminology, thank you for that. Yes, A changes its speed and therefore the frames change the states of motion and of rest for A. But you continue to miss the point that Frame S and Frame S' (along with the infinity of other inertial frames) continue to exist all the time and over all of space. A has different states of motion in different frames but there is no inertial frame in which A remains at rest all the time.

As I have stated before, A was at rest in Frame S prior to the event of its speed changing and then it is at rest in Frame S' and, by the same token, A was in motion prior to the event of its speed changing in Frame S' and then it was in motion in Frame S. The event of its speed changing is the same event in both frames but it can have different co-ordinates (depending on how you choose to define the origins of the two frames).

The outbound leg is symmetrical for terence and stella, same is true for inbound leg. But, the age difference is created at turn around. Till turn around terence is agging less for stella and stella is aging less for terence. But at turn around terence's age suddenly increased for stella. After turn around the situation again becomes symmetrical.

Once again, this is a different subject than what you agreed on for this thread.

So, events occurring on terence's location between last line of simultaneity of outbound leg and first line of simultaneity of inbound leg might not get any chance to occur for stella.

No, all events occur in all frames. You would never say that an event has no chance to occur in one frame that occurred in another frame if you understood what an event is and how it only has meaning within a Frame of Reference and how you convert the co-ordinates of events between frames using the Lorentz Transform. All events occur in all inertial frames--there are no exceptions.

Terence is suddenly more aged for stella.

You agreed that this thread is not about aging, so this issue doesn't apply.

The opposite case also to be considered where events get chance to reoccur for C clock.

Again, your statement that events can or do reoccur shows you don't understand what an event is. I don't even know what this statement can possibly mean. I don't understand why you are confused over such a trivially simple issue.

And since you agreed that I had a fine idea:

Why don't you add some numbers to your diagram in your first post, including the co-ordinates of the red and green events and then use the Lorentz Transform to see that all the events occur in both frames without any problem?

All are fine for me.

Why don't you actually do it? It may help you understand what an event is and why all events occur in all frames.

 

[Dale]

Thanks DaleSpam.
This paper perfectly solves the both problem. The method to define line of simultaneity is the key of the solution. Now, red events would not be reoccurred and green events would not be skipped for C and A respectively.

Yes, that and it's simplicity are why it is my favorite method for defining a non inertial frame.

 

[darkhorror]

I am trying to understand what you are saying and what you think the problem is.

It sounds like you are saying that you have 3 events happen. they happen at certain times in one frame of reference, then you switch to another frame of reference, then those events don't ever happen? Is that what you are saying?

 

[mananvpanchal]

The below diagram shows that how the paper solves the both problems. I have drawn the diagram in reference of B clock, so both case can be covered in single diagram.

We can see that there is no problem remaining of intersecting LoS and going apart LoS. All LoS of both inertial frames becomes parallel to each other out of light cone. This looks good.

The paper shows twin paradox in case of same outbound and inbound speed. I have drawn a diagram in which outbound and inbound speed is not same just for information purpose.

Thanks.

 

[mananvpanchal]

I am trying to understand what you are saying and what you think the problem is.

It sounds like you are saying that you have 3 events happen. they happen at certain times in one frame of reference, then you switch to another frame of reference, then those events don't ever happen? Is that what you are saying?

I am not switching to other frame of reference. This is not the issue of transformation. The events is exist in all FoR.

There are three clocks at rest with each other in S frame.
There is only time component is changing for the clocks, not space component.
Now, the three clocks decides to come into moving state in S frame.
They come into moving state one by one such that its looks like in S' frame that they have changed its frame simultaneously.
After all clocks came into moving state they are now again at rest with each other in S' frame.

Now, see the #1 post's diagram.
First A clock changes its frame from S to S'. LoS of A becomes skewed.
C clock has not changed its frame yet. So LoS of C is horizontal.
As time component increases some red events occurs on A clock.
As time component increases LoS of C clock goes upward in diagram, the red events is occurred for C.
Now, C clock changes its frame (in S' frame clocks changing its frame is simultaneous events).
So, LoS of C clock becomes skewed. The LoS is below all the red events so, as time component increases the red events is again eligible to occur for C clock.

On the other side, when A changes its frame, the LoS of A becomes skewed. The LoS is above the all green events on C clock. As time component increases the green events never come into the path of the skewed LoS. The green events on C clock is eligible to skip for A clock.

So, the reoccurring and skipping problem is solved using that paper. Please, see post #20 for solution.

 

[mananvpanchal]

Hello ghwellsjr,

Please, see the below diagram.

The LoS of A's world line is red and LoS of C's world line is green.
We can see that red LoS is never intersect to green events. And green LoS intersect two times to red events.

Post #20 provides solution of this problem. We can see in diagram of post #20 that there are no intersecting and moving apart LoS out of light cone.

 

[darkhorror]

Looks to me as if you are just not drawing what you are trying to explain correctly. Seems to me if you tried to draw the "twin paradox" the way you drew that you would get the same problems.

 

[ghwellsjr]

Hello ghwellsjr,

Please, see the below diagram.

The LoS of A's world line is red and LoS of C's world line is green.

World lines don't own Lines of Simultaneity, frames do. You have two sets of LoS, one for frame S which are horizontal and one for frame S' which are skewed. The world lines for A, B and C cross all the Lines of Simultaneity for both frames (as well as for all other frames). You need to stop thinking that frame S quits being a frame when the objects change their speed and frame S' starts being a frame. Frame S continues all the way to the top of your diagram and along with it, horizontal LoS. And frame S' goes all the way to the bottom of your diagram, and along with it, skewed LoS. You have shown an overlap of horizontal and skewed LoS for the red events but not for the green events. Why not?

We can see that red LoS is never intersect to green events.

What you mean is that in frame S' none of the red events are simultaneous with any of the green events. Isn't that trivially obvious? You said that all three clocks simultaneously change their speed in the S' frame and since all the red events happen after the change in speed and all the green events happen before the change in speed, how would you expect any of the events to be simultaneous in the S' frame?

And green LoS intersect two times to red events.

What you mean is that in frame S, all the green events are simultaneous with all the red events. Isn't that trivially obvious? You placed all the green events after clock A has changed speed and you placed all the red events before clock C has changed speed.

If you had drawn more skewed lines below where you drew the existing ones, it would be trivially obvious that in frame S', all the green events happen before any of the red events.

All you are doing is illustrating Relativity of Simultaneity which is that events that are simultaneous in one frame may not be simultaneous in another frame, or to put it another way, simultaneity is frame dependent.

Post #20 provides solution of this problem.

There's no problem.

We can see in diagram of post #20 that there are no intersecting and moving apart LoS out of light cone.

I have no idea what you are trying to convey in that diagram. I have no idea how you drew that diagram. I have no idea why you feel compelled to draw such a complicated diagram when your simple diagram, especially if you would extend it, illustrates perfectly well that events can be simultaneous in one frame and not simultaneous in another.

 

[mananvpanchal]

World lines don't own Lines of Simultaneity, frames do. You have two sets of LoS, one for frame S which are horizontal and one for frame S' which are skewed. The world lines for A, B and C cross all the Lines of Simultaneity for both frames (as well as for all other frames). You need to stop thinking that frame S quits being a frame when the objects change their speed and frame S' starts being a frame. Frame S continues all the way to the top of your diagram and along with it, horizontal LoS. And frame S' goes all the way to the bottom of your diagram, and along with it, skewed LoS.

All the frames exist all the time. But, what is the meaning of a frame if there is no observer to observer from that frame?

You have shown an overlap of horizontal and skewed LoS for the red events but not for the green events. Why not?

Because, at that time no clock exist in S' frame.

What you mean is that in frame S' none of the red events are simultaneous with any of the green events. Isn't that trivially obvious? You said that all three clocks simultaneously change their speed in the S' frame and since all the red events happen after the change in speed and all the green events happen before the change in speed, how would you expect any of the events to be simultaneous in the S' frame?

What you mean is that in frame S, all the green events are simultaneous with all the red events. Isn't that trivially obvious? You placed all the green events after clock A has changed speed and you placed all the red events before clock C has changed speed.

I am talking about here two types of events.
1. Event defined by the point at which clocks changes its speed.
2. Event defined by the point at which clocks emits light pulse.

1st type of events are simultaneous in S' frame, but not in S frame.
2nd type of events are simultaneous in S frame, but not in S' frame.

If you had drawn more skewed lines below where you drew the existing ones, it would be trivially obvious that in frame S', all the green events happen before any of the red events.

Yes, because green events are not simultaneous with red events in S' frame.

 

[ghwellsjr]

World lines don't own Lines of Simultaneity, frames do. You have two sets of LoS, one for frame S which are horizontal and one for frame S' which are skewed. The world lines for A, B and C cross all the Lines of Simultaneity for both frames (as well as for all other frames). You need to stop thinking that frame S quits being a frame when the objects change their speed and frame S' starts being a frame. Frame S continues all the way to the top of your diagram and along with it, horizontal LoS. And frame S' goes all the way to the bottom of your diagram, and along with it, skewed LoS.

All the frames exist all the time. But, what is the meaning of a frame if there is no observer to observer from that frame?

You and I are the observers that can see everything that is going on from start to finish over all of space as defined by a Frame of Reference because we are not in the frame. If we decide to put any observers in the frame, we can use the frame to determine what they will actually see as time progresses. We can have multiple observers at different locations in the frame and all having different states of motion along with other objects which they can observe. But they are not aware of what is happening until light from each object propagates to them. And it doesn't matter which frame we use to analyze what the observers in the scenario will see. All frames will come to exactly the same conclusion about what the observers observe. No frame provides any observers with any additional insight into what they can observe.

Besides, you didn't even specify any observers in your scenario and the three clocks that you did specify were all described from the standpoint of just two frames, not three. And I'm trying to show you that you only need one frame but that you can then transform the events to any other frame. In fact, you said the clocks could have been rocks.

You have shown an overlap of horizontal and skewed LoS for the red events but not for the green events. Why not?

Because, at that time no clock exist in S' frame.

That could only be true if there was a time when your clocks were brought into existence and you didn't make any such claim.

What you mean is that in frame S' none of the red events are simultaneous with any of the green events. Isn't that trivially obvious? You said that all three clocks simultaneously change their speed in the S' frame and since all the red events happen after the change in speed and all the green events happen before the change in speed, how would you expect any of the events to be simultaneous in the S' frame?

What you mean is that in frame S, all the green events are simultaneous with all the red events. Isn't that trivially obvious? You placed all the green events after clock A has changed speed and you placed all the red events before clock C has changed speed.

I am talking about here two types of events.
1. Event defined by the point at which clocks changes its speed.
2. Event defined by the point at which clocks emits light pulse.

1st type of events are simultaneous in S' frame, but not in S frame.
2nd type of events are simultaneous in S frame, but not in S' frame.

In Special Relativity, there aren't two types of events. An event is just the four co-ordinates specifying a time at a location in a Frame of Reference. It doesn't matter if anything actually happens there at that time or even if anything is located there at that time.

Think about an ordinary two-dimensional graph with X and Y co-ordinates. Even without specifying anything in the graph itself, I could talk about the point at X=3 and Y=4, couldn't I? Well, in a FoR, what corresponds to a point in an X-Y graph is called an event because it also includes time. So I could talk about the event at T=5, X=-6, Y=3 and Z=-2, couldn't I, even if there wasn't actually anything there at that time?

In any given FoR, any two or more events that have the same T co-ordinates are simultaneous. The same two or more events, when transformed to a different FoR may or may not be simultaneous. You have to look at their T co-ordinates to see if they are the same in the new FoR. It's a trivially simple concept.

You have classified your two types of events in an arbitrary way. You specified the first type of event to be simultaneous in S' and the second type to be simultaneous in S. That's why they are the way they are, not because there is some intrinsic difference between them. You could just as easily have had all the clocks continue to emit light pulses simultaneously at the moment they changed speed or thereafter in the S' frame.

If you had drawn more skewed lines below where you drew the existing ones, it would be trivially obvious that in frame S', all the green events happen before any of the red events.

Yes, because green events are not simultaneous with red events in S' frame.

Neither are they simultaneous after the change in speed where you did draw them to show that red events happen after green events. I'm just asking why you think there is any difference between the two sets of skewed lines (the set you did draw and the set you didn't draw)?

 

[mananvpanchal]

Hello ghwellsjr,

I cannot understand what you are trying to prove. I have no confusion about which events is simultaneous with what. I have no confusion about transformation. The problem was not related with the simultaneity or transformation. There was a problem which is solved. I cannot accept that you still don't understand that what was the problem.

Please, look at the image. I have drawn red lines as LoS of S' and green lines as LoS of S.

So, please tell me this.
What is your problem with this diagram?
What have I concluded wrong about the diagram?

 

[ghwellsjr]

Hello ghwellsjr,

I cannot understand what you are trying to prove. I have no confusion about which events is simultaneous with what. I have no confusion about transformation. The problem was not related with the simultaneity or transformation. There was a problem which is solved. I cannot accept that you still don't understand that what was the problem.

Please, look at the image. I have drawn red lines as LoS of S' and green lines as LoS of S.

So, please tell me this.
What is your problem with this diagram?
What have I concluded wrong about the diagram?

I have no problem with this diagram. It shows very well that pairs of the green and red events are simultaneous in the S frame and not simultaneous in the S' frame.

But let me ask you something: Why did you draw the skewed lines in red and the horizontal lines in green? Would it have been just as meaningful to you if the skewed lines were in green (or yellow or orange) and the horizontal lines were in red (or blue or purple)?

 

[mananvpanchal]

Why did you draw the skewed lines in red and the horizontal lines in green? Would it have been just as meaningful to you if the skewed lines were in green (or yellow or orange) and the horizontal lines were in red (or blue or purple)?

Yes, It is meaningful to me well. We can use any color we want.

 

[ghwellsjr]

Yes, It is meaningful to me well. We can use any color we want.

Then why do you think there is a problem that needs to be solved?

 

[mananvpanchal]

Please look at below diagram.

The clocks changes its state from "rest" to "moving" unsimultaneously in S frame, and simultaneously in S' frame.

Now, we look the situation from the perspective of A clock.
Before changing state in S frame, blue events occurs on A clock, and orange events occurs on C clock simultaneously.
After changing state in S frame, red events occurs on A clock, and pink events occurs on C clock simultaneously.
Did you see green events on C clock is skipped for A clock?

Now, we look the situation from the perspective of C clock.
Before changing state in S frame, green events occurs on C clock, and red events occurs on A clock simultaneously.
After changing state in S frame, pink events occurs on C clock, and red events occurs on A clock simultaneously.
Did you see red events on A clock is reoccurred for C clock?

To solve the reoccurring and skipping problem, I have drawn the diagrams in post #20.
Please, read paper provided by DaleSpam in post #15 for detailed information.

 

[Mentz114]

re #31 : mananvpanchal, you were told many posts ago in an earlier thread that there was no problem because the events are not in causal contact and not directly observable.

But I'm glad you've now understood what your problem was.

 

[Dale]

Hi mananvpanchal, perhaps it would help if I gave you some explanation of various terminology that are used to describe these concepts.

The first concept is a manifold. That is the mathematical object which represents all of the events in spacetime. The manifold does not have a coordinate system, but it does have a metric. The metric allows you to measure geometric features like distances and times and angles and relative speeds, even without a coordinate system. Because we are not dealing with gravity, the manifold is an infinite flat 4D plane.

The second concept is a worldline, but you already know that concept so I won't go into detail.

The third concept is an inertial worldline. An inertial worldline is one which is a straight line.

The fourth concept is a coordinate chart. A chart is a smooth mapping of some open subset of the manifold onto an open subset of R4. You can have multiple charts associated with a manifold. A chart does not need to cover the entire manifold nor the entire R4. If two charts overlap in some region then there is a function which maps from one chart to the other. This function is called a coordinate transform. If two charts do not overlap in some region then the mapping is not defined in that region.

The fifth concept is an inertial frame*. An inertial frame is a chart which maps straight lines in the manifold (inertial worldlines) to straight lines in R4. When the laws of physics are expressed in terms of the coordinates of an inertial frame then they take their textbook form. Also, since straight lines go on forever in a flat manifold, so does an inertial frame in SR. In GR straight lines become geodesics and inertial frames become local.

So, let's take these concepts and apply them to your scenario. You have drawn the worldlines for the clocks A, B, and C, and they are non-inertial since they have a bend. All of the red dots and all of the green dots are events that exist in the manifold irrespective of whatever coordinate charts we may draw later.

Next, we can draw some coordinate charts, starting with two obvious inertial frames, S and S'. These are inertial frames in flat spacetime, so by definition they go on forever and overlap everywhere. The coordinate transform between them is the Lorentz transform. All of the red dots and all of the green dots have coordinates in S and S'.

Now, we can make some other coordinate charts which are non-inertial.

The first chart will be a subset of S, let's call it S-. S- is specifically the subset of S before clock C accelerates. Assuming that C accelerates at t=T, the coordinate transform between S and S- is the identity transform on the region t<T and is undefined elsewhere.

The second chart will be a subset of S', let's call it S'-. S'- is specifically the subset of S' after clock C accelerates. Assuming that C accelerates at t'=T', the coordinate transform between S' and S'- is the identity transform on the region t'>T' and is undefined elsewhere.

Note that S- and S'- are two different non-inertial charts. Some events in the manifold are covered by both S- and S'-, and some events in the manifold are not covered by either S- or S'-. Both S- and S'- are considered non-inertial simply because they don't cover the whole manifold, but within their respective subsets of the manifold the laws of physics take their standard form, and in their region of overlap they are related by the Lorentz transform.

Now, suppose that we don't like having multiple non-inertial coordinate charts with gaps and overlaps, but we want a single chart which covers the whole spacetime and which maps one of the bent lines to a straight line. Then, we can use the radar-time method suggested by Dolby and Gull. Even though it covers the whole spacetime, this chart is clearly non-inertial since it maps a bent line in the manifold (the clock worldline) to a straight line in R4. The laws of physics become more complicated in this frame, but it is a perfectly valid coordinate chart.

Does this help?

*The technical definition of a frame is a little subtle, but in SR you will be OK with this.

 

[mananvpanchal]

Does this help?

Very well. This clears the idea greatly. Thanks DaleSpam.

 

[mananvpanchal]

re #31 : mananvpanchal, you were told many posts ago in an earlier thread that there was no problem because the events are not in causal contact and not directly observable.

But I'm glad you've now understood what your problem was.

I couldn't understood the term "casual contact" before. And I still can't understand the term. What is meaning of "casual contact" or "directly observable"?

 

[ghwellsjr]

Please look at below diagram.

The clocks changes its state from "rest" to "moving" unsimultaneously in S frame, and simultaneously in S' frame.

Now, we look the situation from the perspective of A clock.
Before changing state in S frame, blue events occurs on A clock, and orange events occurs on C clock simultaneously.
After changing state in S frame, red events occurs on A clock, and pink events occurs on C clock simultaneously.
Did you see green events on C clock is skipped for A clock?

Now, we look the situation from the perspective of C clock.
Before changing state in S frame, green events occurs on C clock, and red events occurs on A clock simultaneously.
After changing state in S frame, pink events occurs on C clock, and red events occurs on A clock simultaneously.
Did you see red events on A clock is reoccurred for C clock?

To solve the reoccurring and skipping problem, I have drawn the diagrams in post #20.
Please, read paper provided by DaleSpam in post #15 for detailed information.

Throughout this thread, A, B, and C were just clocks, not observers. Now you want to talk about their perspectives but this is an ambiguous term. Are you asking about what they can actually see and observe or something different?

In post #10, you introduced the idea that what you had previously called red and green events were actually light pulses. And now you have more different colored light pulses so are you now asking about what A sees of C's different colored light pulses and what C sees of A's different colored light pulses? There is only one answer to this question.

Or are you asking something different? If it is something different, then you need to specify exactly what that something different is because there are an infinite number of different questions with an infinite number of different answers and I have no way of knowing what you have in mind.

 

[Mentz114]

I couldn't understood the term "casual contact" before. And I still can't understand the term. What is meaning of "casual contact" or "directly observable"?

Two events along two worldlines are not causally connected if the proper interval between them is spacelike. This means that one of them is outside the others past light cone. In the 'now' of either event the other event will not be observable until 'now + later' of the observing WL.

 

[mananvpanchal]

Throughout this thread, A, B, and C were just clocks, not observers. Now you want to talk about their perspectives but this is an ambiguous term. Are you asking about what they can actually see and observe or something different?

In post #10, you introduced the idea that what you had previously called red and green events were actually light pulses. And now you have more different colored light pulses so are you now asking about what A sees of C's different colored light pulses and what C sees of A's different colored light pulses? There is only one answer to this question.

Or are you asking something different? If it is something different, then you need to specify exactly what that something different is because there are an infinite number of different questions with an infinite number of different answers and I have no way of knowing what you have in mind.

This is the confusion: I am not asking you. I just telling you.
I am just trying to explain you the problem which is solved. But, I cannot understand why you are not understanding.

 

[mananvpanchal]

Two events along two worldlines are not causally connected if the proper interval between them is spacelike. This means that one of them is outside the others past light cone. In the 'now' of either event the other event will not be observable until 'now + later' of the observing WL.

Thanks Mentz114.

 

[ghwellsjr]

Throughout this thread, A, B, and C were just clocks, not observers. Now you want to talk about their perspectives but this is an ambiguous term. Are you asking about what they can actually see and observe or something different?

In post #10, you introduced the idea that what you had previously called red and green events were actually light pulses. And now you have more different colored light pulses so are you now asking about what A sees of C's different colored light pulses and what C sees of A's different colored light pulses? There is only one answer to this question.

Or are you asking something different? If it is something different, then you need to specify exactly what that something different is because there are an infinite number of different questions with an infinite number of different answers and I have no way of knowing what you have in mind.

This is the confusion: I am not asking you. I just telling you.
I am just trying to explain you the problem which is solved. But, I cannot understand why you are not understanding.

You're right, you weren't asking. You said you wanted me to look at the situation from the perspective of clock A and then from the perspective of clock C. But I don't know what you mean by "perspective". Do you mean what A and C can actually see or do you mean something different?

 

[mananvpanchal]

You're right, you weren't asking. You said you wanted me to look at the situation from the perspective of clock A and then from the perspective of clock C. But I don't know what you mean by "perspective". Do you mean what A and C can actually see or do you mean something different?

Please, don't get tensed too much. I know that clocks cannot observe. But, this is simple way to tell something. Let's think about three observes holding that clocks. Now, is this fine?

 

[ghwellsjr]

You're right, you weren't asking. You said you wanted me to look at the situation from the perspective of clock A and then from the perspective of clock C. But I don't know what you mean by "perspective". Do you mean what A and C can actually see or do you mean something different?

Please, don't get tensed too much. I know that clocks cannot observe. But, this is simple way to tell something. Let's think about three observes holding that clocks. Now, is this fine?

That's fine.

Now when you say you want me to look at the situation from the perspectives of each of the observers holding the different clocks, do you mean what the observers can actually see or do you mean something different?

 

[mananvpanchal]

That's fine.

Now when you say you want me to look at the situation from the perspectives of each of the observers holding the different clocks, do you mean what the observers can actually see or do you mean something different?

No, I don't mean what observer see. When we are talking about some simultaneous events we don't worry about when the events will be observed. I am just telling that some events is simultaneous for some observer. that's it.

The blue and orange events are simultaneous for A observer.
The red and pink events are simultaneous for A observer.
The green and red events are simultaneous for C observer.
The pink and red events are simultaneous for C observer.

 

[ghwellsjr]

No, I don't mean what observer see. When we are talking about some simultaneous events we don't worry about when the events will be observed. I am just telling that some events is simultaneous for some observer. that's it.

The blue and orange events are simultaneous for A observer.
The red and pink events are simultaneous for A observer.
The green and red events are simultaneous for C observer.
The pink and red events are simultaneous for C observer.

What you are doing is implicitly selecting the frame in which the observer is at rest. This can work unambiguously for three of the four cases you presented above because both the observer and the remote clock are at rest in the same frame and therefore we can talk about light pulse events being simultaneous. But it doesn't work unambiguously for the third case because C observer is not at rest with respect to A clock during the time interval of the green events. If you say that the pink and red events are simultaneous in S frame, then, of course they are simultaneous for C observer in his implied rest S frame but now you are forcing the red and green light pulses to occur at different time intervals in their own rest frames without any way to co-ordinate the timing intervals of the two sets of clocks.

You stated in post #3 that you wanted the clocks to emit light pulses at regular intervals of time and your drawing in post #31 appears to show that. The problem is that as soon as A clock changes speed and before C observer and C clock change their speed, there can not be any more simultaneous light pulse events between clock A and observer C with his clock C because the clocks are ticking at different rates. And the only way that we can say that the light pulse events after C observer/clock change their speed are simultaneous is if they happen to be simultaneous with A clock and we ignore the fact that the times on the clocks are no longer ticking synchronously in their common rest frame.

But you said earlier that you weren't concerned about these issues, you were only concerned about the simultaneity of events. Please go back and look at yuiop's post #7 where he illustrated these issues perfectly and your discount of his post in your post #10.

You can't have it both ways, you either care about the timing on the clocks which cause the light pulse events to be simultaneous or you don't care about the timing on the clocks and simply define arbitrary events to be simultaneous in the two frames.

One more thing: you set up your scenario in such a way that the three clocks started out at rest in the S frame and then you stated that they change speed simultaneously in the S' frame to become at rest in the S' frame. How did each clock know when to fire its rocket (or whatever) to cause this simultaneous change in speed in the S' frame?

 

[mananvpanchal]

What you are doing is implicitly selecting the frame in which the observer is at rest. This can work unambiguously for three of the four cases you presented above because both the observer and the remote clock are at rest in the same frame and therefore we can talk about light pulse events being simultaneous. But it doesn't work unambiguously for the third case because C observer is not at rest with respect to A clock during the time interval of the green events. If you say that the pink and red events are simultaneous in S frame, then, of course they are simultaneous for C observer in his implied rest S frame but now you are forcing the red and green light pulses to occur at different time intervals in their own rest frames without any way to co-ordinate the timing intervals of the two sets of clocks.

You stated in post #3 that you wanted the clocks to emit light pulses at regular intervals of time and your drawing in post #31 appears to show that. The problem is that as soon as A clock changes speed and before C observer and C clock change their speed, there can not be any more simultaneous light pulse events between clock A and observer C with his clock C because the clocks are ticking at different rates. And the only way that we can say that the light pulse events after C observer/clock change their speed are simultaneous is if they happen to be simultaneous with A clock and we ignore the fact that the times on the clocks are no longer ticking synchronously in their common rest frame.

The red and pink events is occurred simultaneously in S' frame not in S frame.

Yes, I have stated that the events is occurred at some unit time interval. But, that was over requirement to explain the problem. To explain the problem I actually don't need events occurred at some unit time interval, I don't need readings of clocks, I don't need to bother about synchronization of the clocks.

But you said earlier that you weren't concerned about these issues, you were only concerned about the simultaneity of events. Please go back and look at yuiop's post #7 where he illustrated these issues perfectly and your discount of his post in your post #10.

You can't have it both ways, you either care about the timing on the clocks which cause the light pulse events to be simultaneous or you don't care about the timing on the clocks and simply define arbitrary events to be simultaneous in the two frames.

As I said before events has nothing to do with timings of clocks.

One more thing: you set up your scenario in such a way that the three clocks started out at rest in the S frame and then you stated that they change speed simultaneously in the S' frame to become at rest in the S' frame. How did each clock know when to fire its rocket (or whatever) to cause this simultaneous change in speed in the S' frame?

Clocks don't know, I know that when to fire its rocket. I have showed this in diagram that when I have fired its rocket.

See, I had a problem which is solved. But, I cannot understand what is your problem.
Do you not agree with problem?
or
Do you not agree with solution?

 

[ghwellsjr]

You presented a scenario in your first post that you believed had a problem that you did not know how to solve using the normal methods of Special Relativity. People do this all the time on this forum. They usually call these problems paradoxes because when viewed by two different Frames of Reference, they believe there is an inconsistency or a conflict in Special Relativity.

The reason why these people believe there is a problem is because they don't understand Special Relativity and so we try to help them learn Special Relativity until they can understand why the problem they presented is not a problem. Sometimes our explanations help but other times, despite our best efforts, these people continue to struggle with the concepts of Special Relativity and never arrive at the point where they realize there was no problem.

You are different from most people. In spite of the fact that you continue to believe that your scenario presents a problem that cannot be handled by the normal methods of Special Relativity, you believe that there is an extraordinary solution that does solve the problem that you presented. Whether or not that solution actually works is immaterial if you continue to believe that it is the only way to solve the problem you presented. That's what concerns me.

You have made a lot of statements in this thread that indicate to me that you do not understand the basics of Special Relativity. Here's one from post #5:

I have defined some events in diagram (red events and green events). A can define red events in its S' frame and C can define the same events in its S frame. Same for green events, A can define green events in its S' frame and C can define the same events in its S frame. The all events is defined by two co-ordinate system for A and C.

I explained what was wrong with your idea in post #12:

2) When you define or specify an event in one inertial co-ordinate system, you have already defined and specified it for all other inertial co-ordinate systems and you must use the Lorentz Transformation as the means to obtain the co-ordinates in those other systems. It always produces a time component, as well as three spatial co-ordinates in any other system you want. There are no exceptions.

And you said that was fine but then you continued to express the same problem. What you don't seem to realize is that you can formulate, analyze, demonstrate, present and understand any scenario in Special Relativity (which excludes scenarios involving gravity) using any single inertial Frame of Reference, you don't need two co-ordinate systems for A and C as you stated and as you continue to state.

Your perceived problem is that A, B, and C change frames when they change speeds and since they each do it at different times, this leads you to believe during the transition, there are some events which "disappear" for C and are "duplicated" or "repeated" for A and this is a problem that you believe cannot be handled by the normal single-frame approach but can only be handled by using a very complicated non-linear definition of a frame.

I'm not saying that the complicated non-linear definition of a frame is wrong or that it doesn't also provide an alternate "solution" to your "problem", I'm just saying that it is unnecessary to "solve" your "problem" in that very complicated way and that it is wrong and a misunderstanding of Special Relativity to believe that it is the "only" way or even a "better" way than the normal linear single-frame approach.

It would be one thing if you presented evidence and understanding that your scenario could be explained by any single linear Frame of Reference such as the one you called S or the one you called S' or any other linear frame but you haven't provided any. You have consistently indicated that these won't work but rather it takes two frames, S for the first part and S' for the second part and then you see a problem with that particular two-frame approach. Yes, that is a problem for you to believe that you need two frames but the solution is not to insist on a single non-linear frame but rather to understand that either the S frame all by itself will work with no problem or the S' frame all by itself will also work with no problem or any other single linear frame, they are all equally valid and none is preferred, certainly not a non-linear frame.

 

[yuiop]

The red and pink events is occurred simultaneously in S' frame not in S frame.

Yes, I have stated that the events is occurred at some unit time interval. But, that was over requirement to explain the problem. To explain the problem I actually don't need events occurred at some unit time interval, I don't need readings of clocks, I don't need to bother about synchronization of the clocks.

I find it odd that you think you don't need to bother with synchronisation of the clocks when considering whether events occurred simultaneously or not. You seem to think that simultaneity is a natural phenomena and you liked the explanation that the new line of simultaneity spreads out at the speed of light. However, you do not seem to be aware that simultaneity for spatially separated objects is just a convention that is determined by the method you use to synchronise the clocks. You cannot separate simultaneity from synchronisation of clocks. When we use the Einstein synchronisation method we assume that the one way speed of light is isotropic in all directions, but that is just an assumption and the one way speed of light has never been measured and cannot be measured even in principle. It is inherently unknowable. If we do not assume the speed of light is equal in different directions, then the LoS is not necessarily horizontal in frame S in the first place. I demonstrated in a diagram in #7 that when the clocks accelerate to a new rest frame S' they are no longer (Einstein) synchronised in frame S or S'. They are however still synchronised if we use a synchronisation convention that does not assume that the one way speed of light is the same in all directions. The change in LoS spreading out at the speed of light is just an artificial construct and is just a representation of how fast we can artificially re-sychronise the clocks because that is how long it takes to realize a change has occurred. It is especially artificial when we accelerate more than one spatially separated clock (as in your scenario) because we arbitrarily have to select one master clock

To show how artificial it is, consider some clocks on the rim of a large disc. Let us say the disc is initially not rotating and we sychronise the clocks using the Einstein synchronisation method that assumes isotopic one way speed of light. Now we spin the disc clockwise to relativistic speeds without adjusting any clocks. Observers on the disc would now measure the one way speed of light to be different in different directions. If they now try and resychronise the clocks using the Einstein convention, they find it impossible to do all the way around the disc and they end up with something similar to the international date line where there is a discontinuous jump in time at some arbitrary location on the rim of the disc. Even if they do use this artificial method of resynchronisation, they will find that if they use mirrors to measure the speed of light all the way around the disc that it takes more time to go all the way around the disc clockwise than it takes to go in the other direction when they use any single clock and they cannot get away from the isotropic speed of light in an accelerating reference frame, no matter what convention they use.

 

[mananvpanchal]

Hello ghwellsjr,

You says that we can observe the scenario from any frame, and we get solution. But, here we are not talking about two inertial frame. Here is a non-inertial frame which is made by two sub part of two inertial frame.

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html

Please, see figure 4 in above link. In this case there are one inertial frame in which terence is at rest and one non-inertial frame in which stella is at rest.

The time gap is created in terence's frame for stella. Stella misses all the events of terence's world line which is lies in the time gap. We know that this is the cause of age difference between stella and terence. Actually the time gap helps us in solving the paradox. But, if we look closely the time gap is a big problem in its own. So, to remove the skipping events problem we have to take help from the new method for defining LoS.

 

[ghwellsjr]

Hello ghwellsjr,

You says that we can observe the scenario from any frame, and we get solution.

I'm not the only one saying that, I'm just repeating what Einstein said and what your paper says near the top:

Relativity ... says to us, "Pick whichever frame you like to describe your results. They're all equivalent."

Do you disagree?

But, here we are not talking about two inertial frame. Here is a non-inertial frame which is made by two sub part of two inertial frame.

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html

Please, see figure 4 in above link. In this case there are one inertial frame in which terence is at rest and one non-inertial frame in which stella is at rest.

In a frame in which terence is at rest, stella is moving during the entire scenario. You can analyze everything about both terence and stella from this one frame. Do you disagree?

There is another inertial frame in which stella is at rest during the outbound leg of her trip and moving during the inbound leg of her trip and in which terence is moving during the entire scenario. You can analyze everything about both terence and stella from this one frame. Do you disagree?

There is another inertial frame in which stella is moving during the outbound leg of her trip and at rest during the inbound leg of her trip and in which terence is moving during the entire scenario. You can analyze everything about both terence and stella from this one frame. Do you disagree?

There are an infinite number of additional inertial frames in which both terence and stella are both moving for the entire scenario. You can analyze everything about both terence and stella from any of these inertial frames. Do you disagree?

All of these frames employ the same well-defined method of description and analysis, and they all produce the same result in terms of the final age difference between terence and stella. None of these frames are preferred over any of the others and none of them have any "problem" with regard to missing or duplicated events. Do you disagree?

There are an infinite number of additional non-inertial frames in which you can contort the analysis just about any way you choose. There is no well-defined method of how to do this. Every example is unique. Some may introduce missing events or duplicated events but if they are done correctly, they will still also produce the same result in terms of the final age difference between terence and stella. None of these non-inertial frames are preferred over any of the other non-inertial frames or inertial frames. Do you disagree?

The time gap is created in terence's frame for stella.

That's not what the article says. It says the gap is in terence's world line, not in his reference frame.

Stella misses all the events of terence's world line which is lies in the time gap.

That's not what the article says. It says the gap is devoid of blue lines, meaning it's devoid of events.

We know that this is the cause of age difference between stella and terence.

No it isn't the cause of the age difference. The age difference can be analyzed from any inertial frame in which there are no gaps, no missing events, no duplicated events, no non-linearities, no non-inertial frames, no convoluted random stitching of other disjointed sub-frames, etc. Just simple application of Special Relativity. Clocks moving in any inertial frame tick at a slower rate the faster they move. Therefore, a moving clock will accumulate less time than a stationary clock. Real simple. But as your article pointed out, "there are some pitfalls for the unwary" and I think the author fell into one of the pits when he said below Figure 3, "From the viewpoint of Stella, her clock is running much faster than Terence's during the turnaround." Seems to me that it's the other way around.

Actually the time gap helps us in solving the paradox.

No it doesn't, it just creates a new problem.

But, if we look closely the time gap is a big problem in its own.

You're right about that.

So, to remove the skipping events problem we have to take help from the new method for defining LoS.

If you had only stuck with a single inertial frame, you wouldn't have any skipping events problem that needed to be solved.

Please understand, I'm not saying that any of the analyses in the article are wrong, not even the one that you see a problem with, or that the "new method for defining LoS" is wrong, they just are unnecessary. You misunderstand Special Relativity if you think that a single inertial frame is inadequate to solve any problem (devoid of gravity) or that any non-inertial frame is better at solving any problem (devoid of gravity).

After you answer all my questions in bold, I would like you to explain to me the "new method for defining LoS", please.

 

[mananvpanchal]

Ok, if any single inertial frame can explain everything then pick up S and/or S' and please explain me my original post. The requirement is there should not be any duplicating and skipping events.