Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.

On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?

So the question is: Who will be older? And why?

And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...

How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT?

And if we are back to a situation with a preferred frame: what is the difference to the classical Galilean transform? Don't we get all the problems back that seemed to be solved by RT - e.g. the speed limit of light, because if there was a preferred frame you should be allowed to classically add velocities and therefore also get speeds bigger than c ?!? (I know SRT is only a local theory but I don't understand why the global preferred frame should not 'override' the local one).

Could anyone please enlighten me (please in a not too technical way because otherwise I wouldn't understand!)

EDIT
Because there are still things that are unclear to me I posted a follow-up question: Here

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Hi @Jim. The 'paradox' tag is not allowed, cf. this meta post. –  Qmechanic May 5 '13 at 9:18
@Qmechanic okay, will get rid of it. Thanks –  Jim May 6 '13 at 14:32

The twin paradox in compact spaces
Authors: John D. Barrow, Janna Levin
Phys. Rev. A 63 no. 4, (2001) 044104
arXiv:gr-qc/0101014

Abstract: Twins travelling at constant relative velocity will each see the other's time dilate leading to the apparent paradox that each twin believes the other ages more slowly. In a finite space, the twins can both be on inertial, periodic orbits so that they have the opportunity to compare their ages when their paths cross. As we show, they will agree on their respective ages and avoid the paradox. The resolution relies on the selection of a preferred frame singled out by the topology of the space.

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I feel that that link doesn't really answer this person's question. I read the paper, and I still don't understand how a preferred frame is possible given the symmetry of the topology. Here is another paper discussing the paradox, which I think is better (but it's behind a pay wall), although it still doesn't satisfy me. –  user1247 Feb 28 '12 at 9:07
@user1247: I haven't read the attached paper, but my first instinct is that the point in their paths where the "closest" distance to each other flips from going, say Clockwise, to Counterclockwise will serve the same role that the the acceleration kink would in the standard twin paradox. –  Jerry Schirmer Nov 8 '12 at 23:43
@Jerry Schirmer, that does not explain the asymmetry; such a "kink" would be symmetric from each twin's perspective in this topology. –  user1247 Nov 9 '12 at 7:16
I write this from memory, but I remember the paper concluding that the two twins will not be the same age. The difference (asymmetry) comes from one of them crossing over from one cell to another, while the other twin just stays where he is. The preferred frame is the one at rest wrt. to the cell. –  mtrencseni Nov 9 '12 at 13:02
@user1247:both observers would have the kink, and that of course is how it should be, since they both need to agree on their ages when they meet, that type of kink enables them to say "Ah, I"m not the young one!" –  Jerry Schirmer Nov 9 '12 at 13:44

First, let's imagine the ''standard'' twin paradox--you have one observer coinciding with the $y$-axis. The second observer travels out with an angle greater than $45$ degrees, and then has to have a turnaround time, after which they rejoin the $y$ axis and compare ages. The second observer ages less because they travel a path between the two meeting'' events that takes less proper time. The non-equivalence of the two paths is forced by the kink. In particular, if you draw equal-interval light rays leaving the non-accelerating observer, you will see how the moving observer suddenly sees them age as they accelerate.

Now, let's imagine the situation you describe--to do this, imagine a paper strip that is infinite int he $y$-direction, and has the topology of the asteroids'' game in the $x$-direction. This describes a geometrically flat ring evolving in time (adding more dimensions doesn't change the physical picture for this problem much, but makes the whole thing a lot harder to picture mentally). Our two observers travel out at equal speeds in opposite directions from a common point. They go to the edge of the strip and come out the other end, and meet up in the middle. Who thinks that they're older? No one! The paths obviously have equal proper lengths. How does this match up with their perception of each other? Well, just draw the light rays leaving the two observers on your paper--when they come out the other end, they suddenly see aging in the other observer--the act of going ''around the world'' makes it seem like the apparent age of the other observer is suddenly increasing just enough so that when they sync back up, their ages are equal.

Also, note that you don't really need general relativity here--a cylinder is intrinsically flat (or at least, it can be), and so there is no matter here, and local frames of a size less than the diameter of the universe will be able to map themselves, undistorted, onto ordinary Minkowski space. This is all topology.

EDIT:

In the end, all of this can be worked out, just as I said, by drawing lines on paper, identifying sides with each other, and using the rule $\tau^{2} = t^{2} - x^{2}$, where $x$ is the horizontal 'space' distance traveled, $t$ is the time elapsed in the global frame of the torus (or the preferred frame, if you will), and $\tau$ is the amount of aging of the relevant observer in their frame. All it takes is paper and rulers.

And once you talk about preferred frames, you do violate the spirit of Einstein's postulates. This doesn't mean that you can't talk about things consistently. Special relativity on a torus is still perfectly consistent, it's just not a theory in which all frames are equivalent. Also note that the speed of light limit will still be valid--the underlying local geometry will still be identical to SRT. You just will have differences when observers go 'around the world'. This is the difference between geometry and topology.

EDIT II, electric boogaloo:

The asymmetry comes from the fact that the universe itself has a reference frame, and its size will lorentz contract. This is measurable by the people themselves--all that needs to happen is to send out a light ray and wait for the light ray to go around the world. The 'diameter of the universe' will be (light orbit time)/c. This time will be observed to be smaller the faster the observer is travelling. So all observers will agree that there is a global, absolute notion of motion, and this will pick out who ages when.

If you're worried about the translational invariance, note that you're gluing $x=0$ to $x=L$, but a Lorentz transformation is going to mix space and time coordinates, so to a moving observer, this will be gluing $\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}} = 0$ to the same expression equal to $L$, and will involve a mixing of space and time coordinates--the boundaries will appear to move.

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Both the referenced paper and this one too (ajp.aapt.org/resource/1/ajpias/v51/i9/p791_s1?isAuthorized=no) for example, arrive at a different conclusion, ie that the aging is indeed asymmetric and there is a preferred frame. I believe this understanding is canonical. My issue is that I have a hard time reconciling this asymmetry with the perfect symmetry of the problem. –  user1247 Nov 10 '12 at 9:05
@user1247: in your construction, both observers are moving with respect to the preferred frame at the same relative velocity, and the problem is removed. –  Jerry Schirmer Nov 11 '12 at 22:56
what do you mean "in my construction?" In the construction of the OP (to which you responded with an answer that makes it clear you did not read the paper), there is a preferred frame and asymmetric aging. The interesting thing that I am pointing out is that this is in spite of perfect geometric symmetry of the problem. Neither twin accelerates asymmetrically, and the space is compact and symmetric about the direction of motion. This, to me, represents a paradox that should be explained. –  user1247 Nov 12 '12 at 8:30
@user1247: the preferred frame is the one that is at rest with respect to the boundaries, in which ''the universe is at its maixmum diameter.'' If you have one observer at rest with respect to this frame, than the other one will be the one that ages more slowly. You have BOTH observers moving with respect to this frame, one in one direction and the other in the other direction. Thus, no asymmetric aging, by the argument above. (and in my description). The asymmetric aging happens when you have one stationary observer, and the other one doing a ''lap around the universe.'' –  Jerry Schirmer Nov 12 '12 at 14:30
@user1247: (cot'd) these paths are topologically inequivalent and there is no paradox. In essence, you added symmetry back into the problem, while the papers linked consider the general case of motion in these spaces. –  Jerry Schirmer Nov 12 '12 at 14:31

The traveling astronaut is younger. The situation is not reversible between both astronauts because the traveling astronaut is submitted to an effect which is similar to acceleration because he is following the curvature of space in the fourth dimension.

The solution must consider the geometric/ topological constellation. And topologically, the traveling astronaut makes a roundtrip, even if the universe is flat and even if the movement of the traveling astronaut is completely inert and without any acceleration: because he is not returning the same way he took when he left the other astronaut.

This means - even if there is no acceleration - that this case is comparable with two twins, one making a roundtrip and being younger than the remaining twin when returning to Earth. Another comparison: two Earth satellites, one geostationary satellite and one satellite in orbit. The moving satellite will not be submitted to any acceleration of its 1D- (or 2D-) course, but due to gravity it is permanently accelerated perpendicularly in Earth direction (the 3rd dimension). In the same way, the astronaut is not accelerated in 3D-space, but he is following the curvature of the 4th dimension.

Due to the preponderant impact of the curvature of space in the 4th dimension the current equations regarding the twin paradox are not working and need to be corrected for this special case. There is no acceleration in 3D-space, but the curvature is similar to an acceleration in the fourth dimension (= the dimension in which the 3D-space is curved). The result will be that the traveling astronaut (more exactly: the relative distance between both astronauts) is somehow "accelerated" in the direction of the fourth dimension so that the twin paradox cannot be applied symmetrically, exactly in the same way as for the two satellites.

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