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If I properly understand relativity, time ticks faster for an object sitting still than for an object passing by.

So, in a universe with only two objects which have the same "age", object A is sitting still and object B is quickly approaching it, then we can assume that when object B stops, object B will be "younger" than object A. That works if you choose object A as your point of reference, but what if you choose object B as your point of reference? In that case, object A is the one moving and therefore will end up being "younger" than object B.

How is this possible? How does relativity account for this paradox?

Qmechanic
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  • @ACuriousMind I could more easily follow the other answer's paper at http://home.earthlink.net/~owl232/twinparadox.pdf But his numbers can only work if he treats the trip away from the earth differently to the trip back. That's not OK in my opinion. Or did I just misunderstand it? –  Apr 09 '15 at 01:11
  • What do you mean by "treat the trip away from the Earth differently to the trip back"? Are you talking about Figure 3? Note that this is drawn from the perspective of a frame where the Earth twin was in motion, while the traveling twin was at rest during the trip away but not the trip back--Figure 2 shows that the two trips are symmetrical in the frame where the Earth twin was at rest. BTW, you might also find this page helpful. – Hypnosifl Apr 09 '15 at 02:25
  • @Hypnosifl My problem is with the second section of the page you referenced. It says "For Stella, each Leg takes about a year. Terence maintains that Stella's turnaround takes place at year 7" Why? If Terence's visual of Stella's clock relates to Stella's actual experience of time, then the same can be said about Stella's visual of Terence's clock relation to Terence's experience of time. Why would their experienced times be different? –  Apr 09 '15 at 06:20
  • The times are different because of the asymmetry introduced by Stella accelerating. Acceleration is not relative, it creates G-forces on Stella's ship that can be measured with an accelerometer. It is true that each sees the other's clock running slower than their own when they're moving apart inertially, and both see the other's clock running fast when they're moving towards each other, but Terence doesn't see the light from Stella accelerating until more than halfway between when she left and when she returns, so to him her clock spends more – Hypnosifl Apr 09 '15 at 12:28
  • (cont) time running slow visually than running fast, whereas Stella experiences the acceleration exactly halfway between the moment she departs and the moment she returns, so she sees Terence's clock running slower than hers for half the trip and faster than hers for half the trip. – Hypnosifl Apr 09 '15 at 12:30
  • @Hypnosifl The paper explicitly excludes any effects acceleration (G-Forces) could have, and explains why under the "wrong answers" section, so acceleration does not add any asymmetry to the experiment. As far as Terence experiencing Stella's turn around more than half way through the experiment, the same goes for Stella. If Terence was to signal Stella with a thumbs up at year 7 (when Stella turns around, but the light of the event hasn't reached Terence yet), Stella would not see the thumbs up until more than half way through the experiment either. –  Apr 09 '15 at 17:04

2 Answers2

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Notice that one of the observers has to stop and turn back in order to compare watches and hence feel an acceleration. Thus they are not symmetrical in this sense. That's why the observer, who experiences the acceleration would be younger. As Feynman puts this fact in his famous lectures:

This [twin paradox] is called a “paradox” only by the people who believe that the principle of relativity means that all motion is relative; they say, “Heh, heh, heh, from the point of view of Paul [the stationary observer], can’t we say that Peter[the moving observer] was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet.” But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship—the rockets went off, things jammed up against one wall, and so on—while Peter felt nothing. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an “absolute” sense, and it is certainly correct.

Gonenc
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  • All papers that explain a solution to the paradox explicitly exclude any effects by acceleration and explain why. An example of these papers is this one: http://home.earthlink.net/~owl232/twinparadox.pdf –  Apr 09 '15 at 16:44
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Special relativity is define has either points A or B being motionless to a third point, even if neither observer at point A or B can define that third point. Only after comparing stopwatches will the observers know which of them was at relative rest and which of them was in relative motion to the third point. Can't define absolute rest...but the classic twin paradox tries to disproof relativity by not defining relative rest to a third point.

ebg
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