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Consider two observer in a tree-torus space of size $L$. Observer $A$ is at rest, while observer $B$ moves in the $x$-direction with constant velocity $v$. $A$ and $B$ began at the same event, and while $A$ remains still, $B$ moves once around the universe and comes back to intersect the worldline of $A$ without ever having to accelerate (since the universe is periodic).

I've obtained proper time measured by $A$ (because observer $A$ is at rest, so the trajectory of $A$ is characterized by $v = 0$.):

$$\tau ~=~\int d\tau =\int \sqrt{1-v^{2}} dt~=~t.$$

For particle $B$, it moves at constant velocity $v$:

$$\tau ~=~\int d\tau ~=~\int \sqrt{1-v^{2}} dt~=~t\sqrt{1-v^{2}}.$$

I think this result is inconsistent with my understanding of Lorentz invariance because the times don’t match and no one was accelerated and this is not the usual intuition for Lorentz invariance. Is this really consistent with Lorentz invariance?

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Just write the metric: $ds^2=dt^2-L^2(d\theta_1^2+d\theta_2^2+d\theta_3^2)$ and you'll surely not expect to be able to apply the Lorentz transformations. You can, however measure proper time along worldlines as $\frac{1}{c}\int{ds}$ in the normal way. –  twistor59 Dec 17 '12 at 20:15
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up vote 4 down vote accepted

The three-torus geometry or, more precisely, the $R\times T^3$ spacetime geometry, breaks the Lorentz invariance to nothing. The identification $x\sim x+2\pi R_x$ and similarly for $y,z$ needed to define the torus is an identification that doesn't stay the same when $t,x$ are mixed by the Lorentz transformation, so the spacetime is just not Lorentz-symmetric and this broken Lorentz invariance therefore leaves no constraints on the events in this Universe.

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See also physics.stackexchange.com/questions/361/… –  John Rennie Dec 18 '12 at 7:25
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