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"?