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I am not really sure if this is ATM or not. If not, please close the thread.

Anyway,there are various interpretations of the twin's paradox and various so called explanations for the twin's paradox.

I propose the following interpretation.
O and O' are in the same inertial frame.

O applies a constant acceleration and so does O' in opposite linear directions but for different time periods such that their relative motion is v.

O apples the acceleration for the longer period.

They agree to proceed with this relative v for a long period of agreed upon time.

Then O applies a constant acceleration in the opposite direction of the prior acceleration for the shorter time period.

O' applies a constant acceleration in the opposite direction of the prior acceleration for the longer time period such that they are forced back to the original inertial frame but at a much greater distance.

They can then apply Einstein's clock synchronization procedure since they are now in the same inertial frame.

What will the clocks reveal?

Of course, SR claims they will both be dilated by the same value.

2. Originally Posted by abcdefg
I am not really sure if this is ATM or not. If not, please close the thread.

Anyway,there are various interpretations of the twin's paradox and various so called explanations for the twin's paradox.

I propose the following interpretation.
O and O' are in the same inertial frame.

O applies a constant acceleration and so does O' in opposite linear directions but for different time periods such that their relative motion is v.

O apples the acceleration for the longer period.

They agree to proceed with this relative v for a long period of agreed upon time.

Then O applies a constant acceleration in the opposite direction of the prior acceleration for the shorter time period.

O' applies a constant acceleration in the opposite direction of the prior acceleration for the longer time period such that they are forced back to the original inertial frame but at a much greater distance.

They can apply then Einstein's clock synchronization procedure since they are now in the same inertial frame.

What will the clocks reveal?

Of course, SR claims they will both be dilated by the same value.
I'm not sure yet if they would or not, but it sounds like they might. If they both accelerated in the same way in opposite directions and back, they would definitely read the same times when they meet. What you're saying will probably have to be worked out to be sure.

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Originally Posted by grav
I'm not sure yet if they would or not, but it sounds like they might. If they both accelerated in the same way in opposite directions and back, they would definitely read the same times when they meet. What you're saying will probably have to be worked out to be sure.

They would not meet.

The relative velocity for the long time period creates a long distance between them.

4. Oh okay, so according to an observer C that remains stationary where both accelerating observers A & B started, A accelerates for a time of t1 and decelerates for a time of t2 until she is stationary to C again but at some distance, while B accelerates in the opposite direction for a time of t2 and decelerates for a time of t1 until he is stationary to C also but at some distance in the other direction, right?

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Originally Posted by grav
Oh okay, so according to an observer C that remains stationary where both accelerating observers A & B started, A accelerates for a time of t1 and decelerates for a time of t2 until she is stationary to C again but at some distance, while B accelerates in the opposite direction for a time of t2 and decelerates for a time of t1 until he is stationary to C also but at some distance in the other direction, right?
OK, you can introduce a C, that seems to make the proposal clearer.

But all the accelerations start at agreed upon proper times.

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Actually grav, that C makes the problem even more interesting.

Since they are all inertial at the end of the "burns", ie accelerations, they can apply the clock sync procedure between all of them.

7. Originally Posted by abcdefg
OK, you can introduce a C, that seems to make the proposal clearer.
That C should be the midpoint of the two, when the whole thing is done.

But all the accelerations start at agreed upon proper times.
Then yes, the time that C observes elapsing for A while A accelerates will be the same as the time that C observes elapsing for B while B decelerates over the same time on C's clock, and the time for A's deceleration will be the same as the time for B's acceleration according to C also. The distances travelled in each case will also be the same, so when both come back to the stationary frame, their times will read the same and their distances from C will be the same in opposite directions. The clocks of A and B will read the same, but differently than C's.

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Originally Posted by grav
Then yes, the time that C observes elapsing for A while A accelerates will be the same as the time that C observes elapsing for B while B decelerates over the same time on C's clock, and the time for A's deceleration will be the same as the time for B's acceleration. The distances travelled in each case will also be the same, so when both come back to the stationary frame, their times will read the same and their distances will be the same in opposite directions. The clocks of A and B will read the same, but differently than C's.

I said C is a midpoint and I was wrong.

It is not.

Anyway, O and O' have inertial motion v for some time and very short acceleration times.

Since the acceleration of O is for a longer time initially than that of O', then C will view a relative velocity vO for O and vO' for O' and they will be different and O and O' will have relative motion v.

Then the acceleration occurs at different amounts in the opposite directions for O and O' to force them back to the original inertial frame.

Therefore, O and O' had relative motion for an extended time.

9. Ok I'm going to try to understand your scenario here.

Originally Posted by abcdefg
...
I propose the following interpretation.
O and O' are in the same inertial frame.
OK, got that

Originally Posted by abcdefg
O applies a constant acceleration and so does O' in opposite linear directions but for different time periods such that their relative motion is v.
So their vector is in opposite directions but they are not accelerating for the same amount of time? For the time both were accelerating was the acceleration rate the same?

Originally Posted by abcdefg
O apples the acceleration for the longer period.
OK

Originally Posted by abcdefg
They agree to proceed with this relative v for a long period of agreed upon time.
You've lost me here. If O accelerates for the longer period then doesn't this contradict the statement directly above?

Originally Posted by abcdefg
Then O applies a constant acceleration in the opposite direction of the prior acceleration for the shorter time period.
So O then reverses their vector at a given rate of acceleration.

Originally Posted by abcdefg
O' applies a constant acceleration in the opposite direction of the prior acceleration for the longer time period such that they are forced back to the original inertial frame but at a much greater distance.
You've really lost me here.

Originally Posted by abcdefg
They can then apply Einstein's clock synchronization procedure since they are now in the same inertial frame.

What will the clocks reveal?

Of course, SR claims they will both be dilated by the same value.
They would not be dilated at all if I understand what you are trying to say.

Is this diagram accurate to what you mean?

at the start and end points there would be no dilation between O & O'. Their clocks would match. When they meet they would not be at the same frame as when they started.

Since they are travelling in opposite directions they would only be in the same reference frame for an instant and would not converge again unless one or both changed their vectors.

10. Originally Posted by abcdefg
I said C is a midpoint and I was wrong.

It is not.

Anyway, O and O' have inertial motion v for some time and very short acceleration times.

Since the acceleration of O is for a longer time initially than that of O', then C will view a relative velocity vO for O and vO' for O' and they will be different and O and O' will have relative motion v.

Then the acceleration occurs at different amounts in the opposite directions for O and O' to force them back to the original inertial frame.

Therefore, O and O' had relative motion for an extended time.
Oops, I accidentally said A and B instead of O and O', but can we say A and B for this or O and P so I can just refer to O's clock without having to specify whether it is the clock of O or the clock of O'?

C would be the midpoint for what you had before. I'm not sure what you are saying now, though. Can you specify what each are doing in terms of the inertial speeds and accelerations?

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Originally Posted by WayneFrancis
Ok I'm going to try to understand your scenario here.

OK, got that

So their vector is in opposite directions but they are not accelerating for the same amount of time? For the time both were accelerating was the acceleration rate the same?

OK

You've lost me here. If O accelerates for the longer period then doesn't this contradict the statement directly above?

So O then reverses their vector at a given rate of acceleration.

You've really lost me here.

They would not be dilated at all if I understand what you are trying to say.

Is this diagram accurate to what you mean?

at the start and end points there would be no dilation between O & O'. Their clocks would match. When they meet they would not be at the same frame as when they started.

Since they are travelling in opposite directions they would only be in the same reference frame for an instant and would not converge again unless one or both changed their vectors.

Could you please tell me how you multi quote so I can answer your questions in one post?

12. Originally Posted by grav
Oh okay, so according to an observer C that remains stationary where both accelerating observers A & B started, A accelerates for a time of t1 and decelerates for a time of t2 until she is stationary to C again but at some distance, while B accelerates in the opposite direction for a time of t2 and decelerates for a time of t1 until he is stationary to C also but at some distance in the other direction, right?
oh this is different.

Now they must be accelerating and decelerating at different rates

Take O'
Say O' accelerates at a1m/s for t1 seconds then accelerates in the opposite direction at a2 t2

The thing to remember is that with dilation it isn't really the amount of time an object spends accelerating that causes the dilation but the amount they accelerated.

So someone that takes 1 year to accelerate to .5c will have the same dilation as someone that takes 2 years to accelerate to .5c

Now that doesn't mean the 2 clocks will agree. Just that the time dilation when both are at .5c relative to the 3rd external observer will be the same.

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Originally Posted by grav
Oops, I accidentally said A and B instead of O and O', but can we say A and B for this or O and P so I can just refer to O's clock without having to specify whether it is the clock of O or the clock of O'?

C would be the midpoint for what you had before. I'm not sure what you are saying now, though. Can you specify what each are doing in terms of the inertial speeds and accelerations?
No, C will not be the midpoint, I said that but it is false.

O accelerates away from C for a longer time period initially than O' and thus, O will have a greater relative velocity than O' relative to C. I saw the mistake I made shortly after posting but you quoted before I could edit it.

14. Originally Posted by abcdefg
Could you please tell me how you multi quote so I can answer your questions in one post?
I just copy the put in the extra begin and ending QUOTE tags where I want the break

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Originally Posted by WayneFrancis
oh this is different.

Now they must be accelerating and decelerating at different rates

Take O'
Say O' accelerates at a1m/s for t1 seconds then accelerates in the opposite direction at a2 t2

The thing to remember is that with dilation it isn't really the amount of time an object spends accelerating that causes the dilation but the amount they accelerated.

So someone that takes 1 year to accelerate to .5c will have the same dilation as someone that takes 2 years to accelerate to .5c

Now that doesn't mean the 2 clocks will agree. Just that the time dilation when both are at .5c relative to the 3rd external observer will be the same.
Now they must be accelerating and decelerating at different rates
Well it is a constant acceleration where one applies it for t and the other t + n, then to reverse it later, it is a deceleration for t + n and t respectively.

This creates a relative v for a period of time but a a total acceleration of 0 for each if I thought this through correctly.

16. Originally Posted by WayneFrancis
oh this is different.

Now they must be accelerating and decelerating at different rates

Take O'
Say O' accelerates at a1m/s for t1 seconds then accelerates in the opposite direction at a2 t2

The thing to remember is that with dilation it isn't really the amount of time an object spends accelerating that causes the dilation but the amount they accelerated.

So someone that takes 1 year to accelerate to .5c will have the same dilation as someone that takes 2 years to accelerate to .5c

Now that doesn't mean the 2 clocks will agree. Just that the time dilation when both are at .5c relative to the 3rd external observer will be the same.
For the first scenario, we are just taking, for example, the resulting dilation for the first person that accelerates to .5 c over 1 year, which will be the same as the resulting dilation for the second person decelerating from .5c over 1 year, and the resulting dilation for the acceleration of the second person accelerating to .5c over 2 years will be the same as the resulting dilation for the first person decelerating over 2 years also.

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Originally Posted by WayneFrancis

So their vector is in opposite directions but they are not accelerating for the same amount of time? For the time both were accelerating was the acceleration rate the same?
Yes, they are the same with different times.

Originally Posted by WayneFrancis

You've lost me here. If O accelerates for the longer period then doesn't this contradict the statement directly above?

O accelerates for a long period of time otherwise, they would both accelerate the same way and then remain inertial after the burn.

Originally Posted by WayneFrancis

So O then reverses their vector at a given rate of acceleration.
Yes, they reverse vectors but at the other's time period used in the acceleration.

18. Originally Posted by abcdefg
Now they must be accelerating and decelerating at different rates
Well it is a constant acceleration where one applies it for t and the other t + n, then to reverse it later, it is a deceleration for t + n and t respectively.

This creates a relative v for a period of time but a a total acceleration of 0 for each if I thought this through correctly.
if you accelerate for t then you must accelerate in the opposite direction for t at the same rate or change your rate of acceleration.

IE I can't accelerate at 10m/s for 10 seconds in one direction and then 10m/s for 5 seconds in the other and be standing still relative to my beginning point.

perhaps if you put some actual numbers to O & O' vectors we'll get a better understanding of what your scenario is.

19. Originally Posted by abcdefg
No, C will not be the midpoint, I said that but it is false.

O accelerates away from C for a longer time period initially than O' and thus, O will have a greater relative velocity than O' relative to C. I saw the mistake I made shortly after posting but you quoted before I could edit it.
The relative speed to C will be the same if one accelerates slower for a longer period of time and one accelerates faster for a shorter period. They both have to gain the same relative speed at the point where they begin to decelerate in order for both to end up back in the same original frame by decelerating in the same way the other accelerated, so also over the same total times and distances travelled.

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Originally Posted by WayneFrancis

Is this diagram accurate to what you mean?

Yes, if the vectors are that of acceleration but they are too long, but that is the point. The objective is to create relative motion for the observers and then later remove it.

21. Originally Posted by abcdefg
Now they must be accelerating and decelerating at different rates
Well it is a constant acceleration where one applies it for t and the other t + n, then to reverse it later, it is a deceleration for t + n and t respectively.

This creates a relative v for a period of time but a a total acceleration of 0 for each if I thought this through correctly.
They won't end up back in the original frame that way, but they will be in the same frame as each other, I believe.

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Originally Posted by abcdefg
I am not really sure if this is ATM or not. If not, please close the thread.
This is not an ATM, this is a freshman exercise in applying relativity. So, the thread should be closed.

23. Originally Posted by abcdefg
Yes, they are the same with different times.

O accelerates for a long period of time otherwise, they would both accelerate the same way and then remain inertial after the burn.

Yes, they reverse vectors but at the other's time period used in the acceleration.
I think I get you now.

O & O' both accelerate at a rate of nm/s for t1s
O travels for a further t2 amount of time before accelerating in the opposite direction at a rate of nm/s for t1s
O' travel for a t3 amount of time before accelerating in the opposite direction at a rate of nm/s for t1s
t2 > t3

Is this correct?

If so O & O' will, at the very end of the experiment, be the same reference frame, but will have different clock times. This is due to O remaining at the accelerated frame for a longer period of time.

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Originally Posted by WayneFrancis
if you accelerate for t then you must accelerate in the opposite direction for t at the same rate or change your rate of acceleration.

IE I can't accelerate at 10m/s for 10 seconds in one direction and then 10m/s for 5 seconds in the other and be standing still relative to my beginning point.

perhaps if you put some actual numbers to O & O' vectors we'll get a better understanding of what your scenario is.
I did not put that well I agree.

I will come back later with pencil to paper and diagrams.

I think that is necessary at this point.

25. Originally Posted by macaw
This is not an ATM, this is a freshman exercise in applying relativity. So, the thread should be closed.
Actually it should just be moved to the Q&A forum.

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Originally Posted by abcdefg
I am not really sure if this is ATM or not. If not, please close the thread.
This is not an ATM , it is a freshman exercise in the application of hyperbolic motion. Any student that has passed an introductory class in relativity should have no problem in solving this exercise. You are right, the thread should be closed. Moderators, please close it.

27. Originally Posted by abcdefg
Yes, they are the same with different times.
Okay, right, that won't work then. They must have different accelerations also.

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Originally Posted by WayneFrancis
Actually it should just be moved to the Q&A forum.
Yes.

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Originally Posted by macaw
This is not an ATM, this is a freshman exercise in applying relativity. So, the thread should be closed.
Obviously, you do not understand the mainstream.

Starting with Paul Langevin in 1911, there have been numerous explanations of this paradox, all based upon there being no contradiction because there is no symmetry — only one twin has undergone acceleration and deceleration, thus differentiating the two cases.

You see, I am creating symmetry by using acceleration/deceleration for both twins. I guess you missed that.
Last edited by abcdefg; 2009-Sep-21 at 04:39 AM.

30. Originally Posted by abcdefg
I did not put that well I agree.

I will come back later with pencil to paper and diagrams.

I think that is necessary at this point.
Yes, that's a very good idea.

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