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If relativity is symmetrical (triplets moving away and returning to the center) then what happens to the Doppler effect as seen by the moving triplets?

A B C

A should see C moving (Doppler) away/to faster than B C should see A moving (Doppler) away/to faster than B B should see A and C moving (Doppler) away/to at the same rate.

Note that the clocks (long term) in each case should match the Doppler (short term).

Qmechanic
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RLH
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    who is stationary (inertial) during this experiment? Who isn't, and which way are they moving? – JEB May 01 '23 at 22:55
  • The central one B can be considered the stationary one. A and C are moving away and then back to B. – RLH May 02 '23 at 00:18
  • It's traditional to have the 1st symbol be stationary, if you want to be less confusing. Which way are the others going. – JEB May 02 '23 at 01:11
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 02 '23 at 02:25
  • You can rename the symbols to be A at the center and B and C the moving ones if you wish. Does not alter that paradox. – RLH May 03 '23 at 09:12

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If I correctly understood your setup,
this spacetime diagram (on rotated graph paper to help visualize tickmarks along worldlines) encodes some observations, which can be used analogously to determine all the other observations.

robphy-RRGP-triplet

You can interpret the broadcast signals as birthday announcements. Then you can see how those announcements are received by various observers. You can see the Doppler effect for various piecewise sections of the worldlines.

(The diagram is taken from my answer to Triplets paradox and What if two twins flew off in opposite directions and were reunited in a perfectly symmetric way, would they have aged same? .)

robphy
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  • So what Doppler signals are collected by each triplet as they are travelling? A sees that C is travelling faster than B at all times. Same goes for vice versa. C see A and B at the same speed (different directions). There are 3 viewpoints you have to consider. One symmetrical, 2 asymmetrical. – RLH May 03 '23 at 09:08
  • Let me add a related question/observation. The Hafele Keating Experiment is notoriously asymmetrical. https://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment so let's make it fully symmetrical. 4 atomic clocks are required. One in the UK, one in South Africa, two travelling between the end points. Synchronize the 4 when on the ground. Which clock will show the least age and why – RLH May 03 '23 at 12:41
  • @RLH I am going to comment only on the Doppler Effect, which relate the period between receptions and between emissions. Since some of the travelers involved are non-inertial, one has to consider various phases in their relative-motions. This can be read off the diagram to make various plots according to each receiver, which may or may not be easy to interpret or be used to reconstruct the diagram. To me, the spacetime diagram is the primary description of what is happening. This diagram shows how each observer's light-clock tick off, given the implied relative velocities (3/5)c and (15/17)c. – robphy May 03 '23 at 22:02
  • All situations can be considered to be inertial as just information can be exchanged (think pictures) between inertial frames as they pass at each outer limit, so no acceleration is involved. At all. – RLH May 04 '23 at 23:08
  • @RLH In Minkowski spacetime, “anyone (e.g any of the triplets) who separates-then-reunites-along-a-worldline-that-is-not-inertial” has a non-inertial worldline. You can have triplets separate, but they can’t all be inertial to reunite. If a triplet requires an information-handoff to reunite, then the worldline traced out by the relay-pair is noninertial. – robphy May 04 '23 at 23:46