What happens to twin paradox if there is symmetry?

[md65536]

Suppose both twins A and B leave Earth in opposite directions and each travel for one year (rocket time) with gamma = 0.5 (relative to Earth), then return at the same speed.

 

My understanding of relativity is that they'll both return to find the earth aged 4 years while they've each aged 2. They remain the same age.

 

 

But I can't figure this out in terms of what they'd each observe.

Twin A would see Earth time appear to run slow as she retreated from it, and then appear to run fast as she approached.

Would she not also see the same thing happening to Twin B? Twin B would appear to age slowly (slower than Earth) on the outbound trip, and then appear to age fast (faster than Earth) on the return trip.

 

However, I realize that to Twin A, her own outbound trip would appear to take 1 year, and the return trip would appear to take 1 year...

yet to Twin A, B's outbound trip would appear to take more than 1 year, and the return trip much less. This is basically because when A reaches her farthest point, she is 2*0.866 light years rest distance from B's farthest point, so she won't see B reach that point for some time (as the light from such an observation will take time to reach Twin A).

 

 

Have I just answered my own question? Yes, Twin A will see Twin B appear to be aging at a much faster pace than it will see Earth aging, but only for a fraction of the time?

I trust that it'll work out to A and B aging exactly the same amount, calculated from anyone's perspective.

Edited March 16, 2011 by md65536

[md65536]

gamma = 0.5

I mean gamma = 2... I had the reciprocal.

 

Other than that is my post correct?

Or it doesn't make enough sense???

 

Obviously I don't do the math enough, but would a Minkowski diagram show the symmetry as it is observed by the twins or by Earth? Or does no one do relativity calculations based on what is seen vs what is calculated?

 

 

[timo]

Suppose both twins A and B leave Earth in opposite directions and each travel for one year (rocket time) with gamma = 2 (relative to Earth), then return at the same speed.

 

My understanding of relativity is that they'll both return to find the earth aged 4 years while they've each aged 2. They remain the same age.

Yep.

 

 

Twin A would see Earth time appear to run slow as she retreated from it, and then appear to run fast as she approached.

Nope. Twin A would see Earth time to appear running slow during both periods. The impact of the time dilatation factor gamma does not depend on whether two objects approach each other or increase their distance.

Note that twin A would see earth to jump forward in time during the short period of turning around.

 

Would a Minkowski diagram show the symmetry as it is observed by the twins or by Earth?

Drawing an x-over-t diagram in the frame of earth shows the symmetry. You've defined your symmetric scenario in this frame, after all. Drawing x-over-t diagrams in the frame of either twin is probably out of your reach because of the part of the trip where the twins turn around (which I expect to be rather tricky for that purpose).

[md65536]

Nope. Twin A would see Earth time to appear running slow during both periods. The impact of the time dilatation factor gamma does not depend on whether two objects approach each other or increase their distance.

Note that twin A would see earth to jump forward in time during the short period of turning around.

 

 

Drawing an x-over-t diagram in the frame of earth shows the symmetry. You've defined your symmetric scenario in this frame, after all. Drawing x-over-t diagrams in the frame of either twin is probably out of your reach because of the part of the trip where the twins turn around (which I expect to be rather tricky for that purpose).

No, twin A would not see anybody jump forward in time. I used to think that too... see this thread: http://www.sciencefo...post__p__569630

 

Twin A would see B's and Earth's time appearing to run faster at some point. It would calculate that their times are running slower.

What I mean is, suppose everybody was broadcasting a timing signal every second.

As twin A approaches Earth, she is moving toward each subsequent signal from Earth and "observes" them at a rate greater than one per second.

 

Supposing A could turn around instantly, on the return trip she would see Earth time appearing to run fast (call it an illusion if you must), including the 2 additional years that Earth ages -- this will be seen by A.

A will see B continue on an outbound journey for some time, during which A and B will be traveling in the same direction at the same speed, thus A will receive signals from B at a "normal" rate of 1 per second. -- Is this correct?

Then at some point in A's return journey, she will see B turn around, and will receive signals from B at a rate higher than 1 per second (and also higher than the rate from Earth) -- even though B's clock is ticking slower than Earth's which is slower than A's according to A.

 

 

I'm not interested in the "actual time" at B or at Earth, I'm only interested in the appearance of time passing as observed from different perspectives. The "apparent time" is consistent with "actual time" according to the travel speed of light (or any observation) at c. If I can't describe any situation in terms of what an observer will actually see, then my understanding of relativity is not sufficient for the work I'm doing.

Edited March 23, 2011 by md65536

[md65536]

I think I figured out the answer: The symmetry of this scenario is only observed by Earth (or anyone who remains equidistant to both twins). The symmetry is observed as the twins always remaining the same age and acting in synchronization. Other observers (such as the traveling twins) would not observe the scenario occurring with synchronization, due to "lack of simultaneity", so the twins can get out of sync (different age) with each other, before eventually returning to synchronization.

 

For an observer to synchronize the twins' age, they'd have to become equidistant to each twin, which for the twins themselves can only occur when they're at the same place. (In general they would need to be not just equidistant but also have the same relative velocity to each???, which in this case is provided by the symmetrical motion of the twins.)

Edited March 23, 2011 by md65536