Exercise 1.02

Imagine that space (not spacetime) is actually a finite box, or in more sophisticated terms, a three-torus, of size $L$. By this we mean that there is a coordinate system $x^\mu = (t, x, y, z)$ such that every point with coordinates $(t, x, y, z)$ is identified with every point with coordinates $(t, x + L, y, z)$, $(t, x, y + L, z)$, and $(t, x, y, z + L)$. Note that the time coordinate is the same. Now consider two observers; observer $A$ is at rest in this coordinate system (constant spatial coordinates), while observer $B$ moves in the $x$-direction with constant velocity $v$. $A$ and $B$ begin 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). What are the relative proper times experienced in this interval by $A$ and $B$? Is this consistent with your understanding of Lorentz invariance?

Let's call $E_{1}$ and $E_{2}$ the events where $A$ and $B$ meet.

So we have:

$$\Delta\tau_{A}=\sqrt{ (\Delta t)^{2} }=\Delta t\qquad  \qquad \Delta\tau_{B}=\sqrt{ (\Delta t)^{2}-L^{2} }<\Delta\tau_{A}$$

This could make it seem like the proper time of $A$ is smaller in the rest frame of $B$, where it should have been invariant. However, this is not the case if we consider the proper time to be a function of the trajectory. Indeed, if we go in the rest frame of $B$, we obtain something like this:

The proper time of $A$ could not be a function of blue trajectory.