What if I fly by the Earth in a relativistic spaceship (no acceleration involved), sync the clocks with my immortal twin who is staying home, and then take a rather long trip around the closed universe? When I circle all the way to the Earth and sync the clocks again (still no acceleration), whose clock will be more vintage, mine or my twin's?
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the universe is expanding too quickly for that to be possible. But the same principles of relativity apply to this cause, the symmetry is broken by you being the one that travels through curved space to get back to where you started. – Señor O Aug 22 '17 at 19:11
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1I'm not sure if your first point is correct for any cosmological model with any parameters. Note that the question is not tied for any particular cosmology. What if there is so much matter in the universe that it is contracting or staying steady for a while? Not a valid argument against the paradox. Then, replace the Earth with another spaceship flying in the opposite direction and your second argument fails as well. – safesphere Aug 22 '17 at 19:25
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If you specify a model of spacetime, you can compute the proper times for both observers and see which one is longer. – WillO Aug 22 '17 at 19:32
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@WillO: Replace the Earth with the second spaceship flying in the opposite direction. Then the case is symmetric in any symmetric model of the spacetime. Therefore both proper times are equal still contradicting the observed time dialation. – safesphere Aug 22 '17 at 19:39
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Possible duplicates: https://physics.stackexchange.com/q/361/2451 and links therein. – Qmechanic Aug 22 '17 at 20:04
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safesphere: I'm not sure exactly what you mean by "symmetric" in this context. Do you mean that the global symmetry group acts transitively? If so, @HenryDeith 's answer appears to be a counterexample to your claim. – WillO Aug 22 '17 at 21:01
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@WillO: Symmetric means that the proper time is the same for both twins. – safesphere Aug 23 '17 at 20:49
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@safesphere: So the content of your earlier comment is that if we assume that the proper time is the same for both twins, then we can conclude that the proper time is the same for both twins? – WillO Aug 23 '17 at 21:42
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@WillO: No, I just hoped you'd understand without a detailed explanation. So there you have two spaceships flying in the opposite directions with a relativistic speed and there is nothing in the spacetime model that would make the first ship different from the second other than that they are flying in the opposite directions. In other words, the model is mirror reflection symmetric where left can be replaced with right. For example, if there is a preferred frame, then the speeds of the ships are equal in magnitude in this frame. – safesphere Aug 23 '17 at 21:55
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safesphere: In that case (i.e. given that you've once again changed your definition of "symmetric"), I'm back to observing that @HenryDeith's answer provides a counterexample to your claim. – WillO Aug 23 '17 at 22:10
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@WillO: What changed was not my definition, but your understanding. And the answer does not provide a counterexample, because it is not symmetric, quote: "What broke the symmetry between Albert and Betty? Albert's constant time curves are circles that are closed, his is a special frame where his world line has zero winding number. Betty's world line winds around the universe and her constant time curves also wind." – safesphere Aug 23 '17 at 22:24
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This is an excellent question! You are asking: if the universe were a large cylinder (it is sufficient that one direction is cyclic), how does the twin paradox resolve. Unlike the original twin paradox, now you can return to your twin without acceleration.
There is a pedagogical answer by Jeff Weeks in The American Mathematical Monthly (Vol 108 p. 585, 2001), which is available here (pdf). I borrow images from that document, but encourage you to read it yourself.
Background: ordinary twin paradox with spacetime daigrams
The best way to see the solution is to first resolve the regular (infinite-space) twin paradox using spacetime diagrams. Here's a reminder:
The twins are Albert and Betty, where Albert stays put relative to the observer. What's drawn are the lines of constant time at intervals of 5 years. On the left are Albert's constant time slices, while on the right are Betty's constant time slices. These are now tilted because of time dilation. If you're not used to this, I very highly recommend Very Special Relativity by Sander Bias for a readable introduction at any level.
You can see that the twin paradox manifests when Betty changes direction (changes frames): she goes from appearing to by synchronous with an Albert who appears to be 16 when she is 25 to being synchronous with an Albert who is 34. (Betty is traveling at velocity $v=(3/5)c$, but the numbers aren't too important for us.)
Cylinder universe twin paradox with spacetime diagrams
In the closed universe, Betty's constant time slices now form helices around the cylinder. Thus she sees that she is contemporary with many copies of Albert. Her helix of constant time intersects copies of Albert at different ages.
Before going into "which one is older," we should stop here because this is the resolution to the paradox. What broke the symmetry between Albert and Betty? Albert's constant time curves are circles that are closed, his is a special frame where his world line has zero winding number. Betty's world line winds around the universe and her constant time curves also wind.
For the parameters in the example, Jeff Weeks used a cylinder with circumference 30 light years so that if Betty travels with $v=(3/5)c$, she intersects with Albert when she is 40 and he is 50.
Henry Deith
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