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If the twins embark with equal and opposite velocities and return with equal and opposite velocities, which one is older when they reunite? (My contention is that they age by the same amount, but both will be younger than an observer who remains at their centre of mass. More generally, the reference frame with the maximal lapse of proper time is that of the centre of momentum.)

user140847
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    Indeed, if you remove all other matter in the universe, one twin will still age more. The “symmetry breaker” is that one twin accelerates, and the other does not. – d_b Apr 21 '19 at 22:58
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    There is no frame-invariant "center of mass". – WillO Apr 22 '19 at 00:26
  • Indeed there is and you can prove it with GR because an observer in the centre of mass reference frame is the only one that truly follows a geodesic through spacetime. The bungeed twins have equal and opposite velocities so neither one can claim to be inertial. – user140847 Apr 22 '19 at 02:13
  • They also have equal and opposite accelerations. The Earth imparts an overwhelmingly large bias in the traditional scenario so the earth-bound twin is effectively inertial, but not exactly. – user140847 Apr 22 '19 at 02:20
  • Also, acceleration is a red herring. See for example: https://www1.phys.vt.edu/~jhs/faq/twins.html – user140847 Apr 22 '19 at 02:48
  • And https://physics.stackexchange.com/questions/123106/twin-paradox-without-acceleration?rq=1 – user140847 Apr 22 '19 at 03:05
  • And https://youtu.be/GgvajuvSpF4 (Fermilab physicist) – user140847 Apr 22 '19 at 03:08
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    @MikeGale I take objection to that statement; anyone who says the acceleration is irrelevant does not understand the twin paradox. It is no red herring, it is the core to resolving the paradox: the entire paradox is "But Bob thinks that he is the younger twin than Alice" and to resolve it, you have to know that Bob, when accelerating back towards Alice, sees Alice's clock ticking wildly fast because he is accelerating towards her and she is very far ahead of him. This effect is in some sense the only new effect in relativity, and it is crucial to resolving the paradox. – CR Drost Apr 22 '19 at 03:29
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    @CRDrost : This is a mattter of taste, I suppose, but I find it very unhelpful to describe Bob as seeing Alice's clock ticking "wildly fast". At every point in his journey, Bob is moving relative to Alice and so sees Alice's clock ticking slow. But while he's accelerating, he keeps changing his mind about what Alice's clock shows "right now". You can describe those changes-of-mind as "seeing Alice's clock tick fast", but I've found this often confuses students. I want them to know that in each of Bob''s instantaneous frames, Alice's clock ticks slow, but that his frame changes. – WillO Apr 22 '19 at 03:47
  • PS to @CRDrost : When I say that this way of seeing things is better for students, I am thinking, of course, only of those who are new at this. – WillO Apr 22 '19 at 03:49
  • As long as your students understand that this is not Bob’s voluntary choice I think I am fine with that phrasing... I prefer “seeing Alice’s clock tick fast” pedagogically because it allows us to start talking about general relativity in informal terms much earlier, and emphasizes the first-order effect over the second-order, and I guess also because I am a very concrete thinker and so I want to know what people literally observe... but I will be 100% transparent that honestly I do not teach this much to others and so your experience is probably much stronger than mine. – CR Drost Apr 22 '19 at 04:51

1 Answers1

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There is no frame-independent center of mass.

In the diagram below, I've taken the earth to be massless. If you don't like that assumption, feel free to redraw the diagram giving the earth some positive mass; it won't change the conclusion.

enter image description here

The black frame is the frame of the earthbound twin, the dark blue line is the worldline of the outbound twin, and the gold line is, according to the earthbound twin, the worldline of the center of mass. (That is, along any fixed line $t=t_0$, it is equidistant from the worldlines of the two twins.)

Now let $t'$ be the time coordinate of the outbound twin. Along any line $t'=t'_0$ (such as the dashed line in the drawing), it's clear that the distances to the gold line and the black vertical axis are not equal. Therefore the outbound traveler will not agree that the gold line represents the center of mass.

WillO
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  • Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Apr 22 '19 at 10:12
  • For the record, I am not satisfied with this answer because it does not address the original question. I also stand by my assertion that acceleration is a red herring because the problem can and has posed without it, as evidenced by 2 of the 3 the citations I provided. (On closer inspection, that Stack exchange thread I was thinking of raised the question, but did not answer it by any stretch of the imagination.) – user140847 Apr 23 '19 at 00:35