Events for frame changing clocks

1. Apr 21, 2012

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

2. Apr 21, 2012

ghwellsjr

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 any one 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?

3. Apr 21, 2012

mananvpanchal

Yes, I know that a event can be defined using three spacial 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.

4. Apr 21, 2012

ghwellsjr

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.

5. Apr 21, 2012

mananvpanchal

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.

6. Apr 21, 2012

Staff: Mentor

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.

7. Apr 21, 2012

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.

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Last edited: Apr 21, 2012
8. Apr 21, 2012

ghwellsjr

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

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.
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.
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.
Likewise for C and t4.
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'.
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?).
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.
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.

Last edited: Apr 21, 2012
9. Apr 23, 2012

mananvpanchal

This seems logical.

10. Apr 23, 2012

mananvpanchal

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.

11. Apr 23, 2012

mananvpanchal

Yes, you are right. I am sorry for misunderstanding. The scenario has nothing to do with clocks.
Yes, you are right. But, I was just trying to describe some situation at some time.
This is also fine.
Again I was just defining some situation at some time. t2 is in S frame.
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.

Last edited: Apr 23, 2012
12. Apr 23, 2012

ghwellsjr

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?

13. Apr 23, 2012

mananvpanchal

All are fine for me.

14. Apr 23, 2012

mananvpanchal

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

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.

15. Apr 23, 2012

Staff: Mentor

16. Apr 23, 2012

mananvpanchal

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.

Last edited: Apr 23, 2012
17. Apr 23, 2012

ghwellsjr

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.
Now you're changing the subject. We agreed that your scenario has nothing to do with the times on clocks:
and now you're discussing the twin paradox which is all about the times on clocks and "where did the missing time go?".
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).
Once again, this is a different subject than what you agreed on for this thread.
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.
You agreed that this thread is not about aging, so this issue doesn't apply.
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 actually do it? It may help you understand what an event is and why all events occur in all frames.

18. Apr 23, 2012

Staff: Mentor

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

19. Apr 23, 2012

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?

20. Apr 24, 2012

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.

Last edited: Apr 24, 2012
21. Apr 24, 2012

mananvpanchal

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.

22. Apr 24, 2012

mananvpanchal

Hello ghwellsjr,

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.

23. Apr 24, 2012

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.

24. Apr 25, 2012

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

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.
There's no problem.
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.

25. Apr 25, 2012

mananvpanchal

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?
Because, at that time no clock exist in S' frame.
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.
Yes, because green events are not simultaneous with red events in S' frame.