1. Dec 22, 2014

### tom.stoer

I know that this has been discussed several times, but I was not able to find a fully convincing conclusion.

Suppose there are two twins travelling with relative velocity v on a torus.

1st question: is it possible to find one unique inertial frame which can serve as a global rest frame? my answers is "yes" b/c an observer can send out light signals in opposite direction; if and only if she receives these light signals at the same instant of time her rest frame is identical with this global rest frame.

2nd question: does a twin residing in this global rest frame observe time dilation for the second twin orbiting the torus with velocity v? my answer is "yes" b/c the travel of the second twin around the torus will take some time T = L/v where L is the spatial length of the circle (T, L and v measured in the global rest frame); so the proper time for the second twin will be T' < T (calculated via standard methods in the global rest frame).

However, it seems to me that there is no general consensus regarding these two statements.

2. Dec 22, 2014

### WannabeNewton

Could you describe the setup in more detail? Do you have two twins that move in opposite directions along a circle around the torus? If so why not just have a ring instead of a torus? Or do the twins have arbitrary trajectories along the torus?

What frame is the relative velocity $v$ of each observer being measured in? Or do you mean that one of the twins has a velocity $v$ relative to the other? You talk about one of the twins being at rest in a global inertial frame and the other twin moving along some trajectory (a circle?) in this frame. If so then what frame is the statement "Suppose there are two twins travelling with relative velocity v on a torus." being made in and what is its purpose?

3. Dec 22, 2014

### tom.stoer

sorry for the confusion; the spacetime is Tn * R; in the two-dim. scenario this is just S1 * R, so in the simplest case space is a circle;

first I start with two twins having relative velocity v w/o using any special frame; then I identify the global rest frame (1st statement); after that I assume that one twin resides in that global rest frame and the relative velocity of the second one is measured w.r.t. this second twin i.e. w.r.t. the global rest frame (2nd statement);

anyway, the global rest frame cannot be identified by using pure local observations

Last edited: Dec 22, 2014
4. Dec 22, 2014

### Staff: Mentor

5. Dec 23, 2014

### tom.stoer

thanks a lot for the reference; agrees with my expectation T' < T and a breakdown of global Lorentz invariance due to topology

6. Dec 23, 2014

### George Jones

Staff Emeritus
7. Dec 24, 2014

### stevendaryl

Staff Emeritus
What's interesting about the twin paradox on a torus is that for short trips (traveling distances that are small compared with the distance "around" the universe), time-dilation is mutual; each twin can view the other twin as time-dilated. But for longer trips, the twin who travels around the universe will be objectively younger than the twin who stayed still. Here's a way to think about it to realize that this isn't a contradiction:

Suppose we have our two twins, Alice and Bob. Alice stays still (relative to the preferred frame defined by the topology). Bob travels "around" the universe. (For simplicity, just consider one spatial dimension, so space is a circle). Let $(x,t)$ be Alice's coordinates, and $(x', t')$ be Bob's coordinates. Alice views the distance "around" the universe to be $L$. Bob sees the distance around the universe length-contracted to some value $L' < L$. One approach to understanding life on a circle is to transform the problem to an equivalent problem on an infinite universe that has periodic boundary conditions.

From the point of view of Alice, Bob is initially at location $x=0$. But there are exact copies of Bob at $..., x= -2L, x=-L, x=0, x=L, x=2L, ...$. From her point of view, the copy of Bob that is initially at $x=-L$ travels to the point $x=0$. He's time-dilated the whole time, so he ends up younger than Alice. No problem.

From the point of view of Bob, things look different. Bob also sees Alice initially at rest at $x'=0$, and sees copies of Alice at $..., x'= -2L', x'=-L', x'=0, x'=L', x'=2L', ...$. However, these are not exact copies. From Bob's point of view, the Alice at $x'=L'$ is initially older than the Alice at $x'=0$. From his point of view, the old Alice at $x'=L'$ travels to $x'=0$. She is time-dilated the whole time, but since she starts off older than Bob, she is still older when she reaches him.

Translating back from an infinite universe to a circle, in Bob's frame, it seems that Alice initially starts moving away from him (in the negative-x direction). She's time-dilated, so she is aging slower than Bob. But then looking in the positive-x direction, he also sees Alice, but in this direction, she's older than Bob, and moving toward him. Bob (unlike Alice) has no consistent way to determine Alice's age "right now".

8. Dec 25, 2014

### tom.stoer

that doesn't bother me; physically the paradox would show up if and only if they meet in one spacetime point; but exactly in these points there is no paradoxon;

what is also interesting is that the topology affects calculus: suppose you determine the "global rest frame" such that the circle t = const. is a Cauchy surface; on this surface you can setup initial condiditions for differential equations; you can't do that on t' = 0 (defined by a Lorentz boost) b/c c t' = 0 intersects its own forward lightcone!