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I have seen the classical twin paradox before. It uses a twin stationary on Earth and the other traveling away and back. I have seen many contradictory solutions for it, some use general relativity, others use special relativity, either way, I have never been satisfied with what I've read. They always try to break the symmetry through the traveling twin's acceleration and deceleration, but never quite succeed.

So, let's do away with the classical twin paradox and let's explain a much simpler, perfectly symmetrical version of it where both twins are moving towards each other.

So imagine we have Twin A in a spaceship, and Twin B in another. They are both traveling at the same speed towards each other.

If I understand relativity properly:

  • From Twin A's frame of reference, he's stationary and Twin B is moving at a constant speed towards him, therefore, Twin B is aging slower.
  • From Twin B's frame of reference, he's stationary and Twin A is moving at a constant speed towards him, therefore, Twin A is aging slower.

When they both finally meet, they both think that the other is younger. Which one of them is right?

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marked as duplicate by ACuriousMind, user36790, John Rennie special-relativity Feb 15 '16 at 12:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Depends on how old they actually are. $\endgroup$ – user106422 Feb 14 '16 at 21:47
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    $\begingroup$ This is the well-known twin paradox. Look at How is the classical twin paradox resolved? to see its resolution. $\endgroup$ – ACuriousMind Feb 14 '16 at 21:49
  • $\begingroup$ The usual assumption is that they observe each others' ages accurately, so kid's comment is the whole story. If you believe their observations can be inaccurate (as your final question suggests) then of course this question has nothing to do with relativity and everything to do with whatever assumptions you care to make about the competence of the observers. $\endgroup$ – WillO Feb 14 '16 at 22:55
  • $\begingroup$ @ACuriousMind I have seen that paradox before. It uses a twin stationary on Earth and the other traveling away and back. I have seen many contradictory explanations for it, some use general relativity, others use special relativity, either way, I have never been satisfied with what I've read. They always try to break the symmetry through the traveling twins acceleration and deceleration, but never quite succeed. Do away with the classical twin paradox and explain this much simpler, perfectly symmetrical version I just gave where they both move towards each other. Can someone give it a shot? $\endgroup$ – AxiomaticNexus Feb 16 '16 at 3:09
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They are both right. Person A (call her Alice) and Person B (call him Bob) will both measure their counterpart to be aging more slowly. Alice will see Bob moving and aging more slowly than normal, and Bob will see Alice moving and aging more slowly. This is because Alice and Bob are traveling at different velocities through space. Even if the spaceships pass right next to each other, as long as their velocities are different, Alice will see Bob aging slowly and Bob will see Alice aging slowly.

It is impossible to synchronize clocks that have different motions through space since they will move through time differently. Only clocks that are not moving with respect to each other will tick at the same rate. Similarly, Alice and Bob will only age at the same rate if they are not moving with respect to each other. In fact, if Alice and Bob were the same age at the start of their journeys, then if Alice turns on her engines and accelerates to catch up to Bob, she will end up younger than Bob. If Bob accelerates to catch up to Alice, he will end up younger.

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  • $\begingroup$ So, when they both stop who will actually be younger? They both can't be younger than the other. $\endgroup$ – AxiomaticNexus Feb 16 '16 at 17:45
  • $\begingroup$ If they both stop in the same way (with equal accelerations), they they will be the same age. at the end. Otherwise, their clocks and ages won't match up. I believe whoever brakes harder will end up being younger once they are at rest next to each other. $\endgroup$ – Mark H Feb 16 '16 at 19:21
  • $\begingroup$ How can they be the same age when they both stop if one was aging slower relative to the other throughout the entire experiment? $\endgroup$ – AxiomaticNexus Feb 16 '16 at 21:14
  • $\begingroup$ While the two are in motion, Alice sees Bob aging slower AND Bob sees Alice aging slower. It depends on which clock you are looking at. Neither Alice nor Bob age slower according to their own clocks. Both Alice and Bob age slower according the the other's clock. This will be the case until one of them breaks the symmetry of the experiment by changing speeds to match the other. $\endgroup$ – Mark H Feb 16 '16 at 22:44
  • $\begingroup$ But none of them breaks the symmetry at any point. They both slow down at the same rate to a stop in the end. $\endgroup$ – AxiomaticNexus Feb 17 '16 at 18:41

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