Extended Twin Paradox: Explaining Time Lapse and Symmetry Principle

[JustTryingToLearn]

Suppose an observer (O) sees a traveler (T1) pass by at time t=0, moving a speed 3c/5. Five years later (according to O), T1 returns. If we assume that T1 traveled at 3c/5 for half the journey and instantaneously reversed direction, returning at the same speed, we can calculate that T1 aged only 4 years by using the standard Minkowski transformation for time.

If, on the other hand, observer O sees two travelers (T1 and T2) moving in opposite directions, then each should return to O having aged the same amount (symmetry principle). The problem I can't seem to figure out is how this is transferred to the points of view of T1 and T2. Since each is moving at 3c/5 relative to O, their speeds relative to each other should always be 15c/17 (relativistic velocity addition). Thus, it seems that each traveler sees the other traveler aging by 17*4/8=17*4/8=8.5 years, while they each measure their own time lapse to be 4 years.

Can anyone explain where I am not correctly translating the problem?

Thanks.

 

[Ibix]

The relativity of simultaneity, as always. If you naively stitch together two frames at the turnaround, you change your definition of "at the same time as the turnaround", and that change of definition is where the "missing" time is.

Have a look at the Minkowski diagrams I posted here: https://www.physicsforums.com/threads/time-dilation-and-einsteins-theorem.963606/page-2#post-6118229 - see if that helps. It's only for one traveller, but the explanation is the same.

 

[PeroK]

Suppose an observer (O) sees a traveler (T1) pass by at time t=0, moving a speed 3c/5. Five years later (according to O), T1 returns. If we assume that T1 traveled at 3c/5 for half the journey and instantaneously reversed direction, returning at the same speed, we can calculate that T1 aged only 4 years by using the standard Minkowski transformation for time.

If, on the other hand, observer O sees two travelers (T1 and T2) moving in opposite directions, then each should return to O having aged the same amount (symmetry principle). The problem I can't seem to figure out is how this is transferred to the points of view of T1 and T2. Since each is moving at 3c/5 relative to O, their speeds relative to each other should always be 15c/17 (relativistic velocity addition). Thus, it seems that each traveler sees the other traveler aging by 17*4/8=17*4/8=8.5 years, while they each measure their own time lapse to be 4 years.

Can anyone explain where I am not correctly translating the problem?

Thanks.

Fundamentally, neither T1 nor T2 is an inertial observer. When they change direction, they change (whether instantaneously or not) their inertial reference frame. You cannot, therefore, simply apply time dilation for these observers but must take the relativity of simulateity into account at the turnaround point; either directly, or, using the Lorentz Transformation.

One thing you could also do that might be enlightening is to study the whole scenario from another reference frame (not just the rest frame of O). For example, traveller T1 sets off at ##3c/5## in one direction. You could calculate everything in this reference frame:

T1 is at rest until the turnaround point, then moves at ##15c/17## back towards O
O moves at ##3/5c## in this frame throughout
T2 moves at ##15c/17## to begin with in this frame, then stops and remains at rest (until O and T1 coincidentally are reunited with him).

You can then calculate how much time has elapsed for all three: T1, O and T2 when they are all reunited.

 

[Nugatory]

Suppose an observer (O) sees a traveler (T1) pass by at time t=0, moving a speed 3c/5. Five years later (according to O), T1 returns. If we assume that T1 traveled at 3c/5 for half the journey and instantaneously reversed direction, returning at the same speed, we can calculate that T1 aged only 4 years by using the standard Minkowski transformation for time.

That calculation gives us the right answer, but that's mostly just luck - the time dilation formula doesn't apply in this situation, which is why you find yourself with the followon problem:

I can't seem to figure out is how this is transferred to the points of view of T1 and T2.

You don't even need to introduce T2 to see the problem. Even in the first case O is moving at .6c relative to T1 during T1's entire four-year journey, so has the slower clock and ought to have aged less than four years by the same time dilation argument. So clearly something is wrong with that argument.

The problem here is that the time dilation formula is based on relativity of simultaneity - it's about what a distant clock reads "at the same time" that my clock reads T. In the twin paradox however we aren't working with distant clocks - they're colocated at the start of the journey and they're colocated at the end of the journey when we compare them. Thus, we have a different problem: The two clocks traveled along two different spacetime paths between the same two events (separation and reunion). The two paths have different lengths, and just as the odometer of a car measures the length of a path space, the length of a path through spacetime is measured with a clock. Different paths, different lengths, different amounts of time elapsed on the paths... and this has nothing to do with time dilation.