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Suppose body A, at rest in its reference frame, is passed by body B which is travelling at a constant 0.6 c, heading toward Z, a point 6 light years away by A's reckoning, and as bodies will, they synchronise their watches.

A decade later by A's watch (and 8 years later by B's), body B arrives at Z. This will not be observable from A for another 6 years.

Body C, also travelling at 0.6 c, immediately then passes body A, synchronising watches in accordance with ancient spacefaring custom. Body D, making the same speed, passes body B and observes the same politeness. Bodies C and D both head for a point equidistant between bodies A and B to enjoy a good gossip and to compare watches.

Do the watches of C and D show a two year offset? Or are they identical?

If they are offset, how does this square with body B's perspective in which it is body A's watch which is running slow? If they are identical, how is this achieved?

With special relativity based discussions of the twin paradox, the change in reference frames affecting one twin and not the other seems to often be invoked to explain the discrepancy in aging. I would therefore be curious to hear an examination in which both twins changed reference frames at the turnaround point and headed to a simultaneous meeting.

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  • $\begingroup$ In which frame is D traveling at 0.6 c? It's also a little unclear which directions these bodies are traveling. Maybe specify velocities more clearly like such "at 0.6 c in the +x direction measured in A's frame of reference." Also, do you presumably mean the point that was equidistant between A and B at the time that C and D passed A and B? $\endgroup$
    – ApproximatelyTrue
    Commented May 23, 2015 at 5:40
  • $\begingroup$ @ApproximatelyTrue: Does it make a difference precisely which directions C and D are travelling, giving they're heading to meet at an equidistant point and must therefore have symmetric velocities? And yes, I mean a point equidistant between A and Z. $\endgroup$
    – Quirk
    Commented May 23, 2015 at 10:26
  • $\begingroup$ The way you've stated the problem is still problematic. You say that simultaneously with B arriving at Z, A passes by C. However, in relativity, simultaneity is famously frame dependent - if two events do not occur at the same point in space, they can be simultaneous in one frame, but not simultaneous in the other. Thus, D passing B at the same time as B reaches Z is unambiguous (since they occur at the same point in space), but A passing C is not, and thus the offset measured depends on in which frame A passing C is simultaneous with B reaching Z. $\endgroup$
    – ApproximatelyTrue
    Commented May 23, 2015 at 13:10
  • $\begingroup$ D passing B, and B reaching Z, are observed from body A at t0 + 16 years. At t0 + 10 years, body C passes A toward the meeting point equidistant between A and Z. The distance between A and Z remains 6 light years for the duration of the question. $\endgroup$
    – Quirk
    Commented May 23, 2015 at 16:32
  • $\begingroup$ To put this perhaps a little more broadly: if we can say "at the conclusion of B's journey, 8 years have passed for B and 10 for A", we are indicating a pair of singular points along the world lines of each. The case I am interested in is that of two voyagers passing A and B when each has reached those respective points, symmetrically travelling towards each other and meeting. $\endgroup$
    – Quirk
    Commented May 23, 2015 at 16:47

1 Answer 1

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Assuming tentatively that all velocities and notions of simultaneity are in the frame of reference of A, i.e., the events C/D passes A/B occur simultaneously with the event B reaches Z in the frame of reference of A, and C/D are moving past A/B at speeds 0.6c as measured in the frame of reference of A. Then the time will be offset by 2 years. If A and B initially synchronized their watches to t=0 years, then when C/D passes A/B, they will set their watches to t=10/t=8 years respectively. Then, by symmetry, when they meet, their watches will still be offset by 2 years.

Now, presumably you titled the post twin paradox because it would seem that the situation is symmetric in the frame of reference of B. However, there are various problems with this. Firstly, the events where the C/D pass A/B are no longer simultaneous in the frame of reference of B. Secondly, the speeds of the C/D are no longer the same in the frame of reference of B - C is at rest and D is moving towards it and 0.96c. If you work everything out in detail (using the relativistic velocity addition formula, and sticking in gammas in various places), you will find the same result in the frame of reference of B.

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  • $\begingroup$ Thank you for the answer. The title is because this is a modification of en.wikipedia.org/wiki/Twin_paradox. Could you expand with a little more detail as to how the asymmetry manifests? $\endgroup$
    – Quirk
    Commented May 24, 2015 at 19:56
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    $\begingroup$ The way I understand your question, the bodies A/B play the role of the twins in the canonical formulation of the twin paradox. Since we cannot unambiguously say whether A is at rest and B is moving at 0.6c or whether B is at rest and A is moving at 0.6c, we expect time dilation to be symmetric. This is true of measurements by clocks at rest in A/B's frames; however, we are considering clocks at rest in C/D's frames, and as explained above the speeds of C/D in A's frame are totally different in B's frame and so we shouldn't expect symmetrical results. $\endgroup$
    – ApproximatelyTrue
    Commented May 24, 2015 at 20:14

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