Time Dilation on 3D Torus: Clock Speed & Why It Matters

[appot89]

TL;DR Summary

Does special relativity apply to an object moving in constant velocity parallel to a torus?

Assume that space is a three-dimensional torus ( a 3D donut) . If there is a clock traveling at a CONSTANT speed in a direction parallel to the torus (inside out of the hole) and one clock that is still. Which clock ticks faster and why?

I know that the clock rotating will tick slower, but I cannot understand why it happens to do so if it's speed is constant.
I would greatly appreciate your input!

 

[PeterDonis]

Assume that space is a three-dimensional torus ( a 3D donut) .

For special relativity to apply, it has to be a flat 3-torus. SR only applies if the spacetime geometry is flat. It is possible to have a flat geometry with a 3-torus spatial topology, but it won't be like what you are imagining. See below.

Which clock ticks faster and why?

A simpler case to analyze is the 1+1 spacetime in which the spatial topology is a circle. In other words, the spacetime as a whole has the topology of a cylinder. Everything about this case carries over to the flat 3-torus case.

The nice thing about the 1+1 cylinder case, though, is that you can simply "flatten out" the cylinder without changing its geometry (since an ordinary cylinder already has a flat intrinsic geometry--it only looks curved because of how it is embedded in 3-D Euclidean space, but nothing about that embedding affects how we analyze the 1+1 cylinder spacetime). When you "flatten out" the cylinder, you see that, unlike the ordinary 1+1 Minkowski spacetime (which has an ordinary infinite plane topology), the 1+1 cylinder spacetime has a "preferred frame": the inertial frame whose spatial axis is a closed circle going around the cylinder (and which thus is exactly "horizontal" when the cylinder is flattened out). An observer at rest in this frame will have the fastest ticking clock (more precisely, will age more between meetings with any other observer in relative motion), and this observer's worldline will go "straight up" the cylinder (and will be exactly "vertical" when the cylinder is flattened out).

Any other observer in relative motion will have a worldline that winds around the cylinder, and the "spatial axis" of such an observer's rest frame will not be a closed circle, but a helix (and in fact this poses some technical issues when defining coordinates in such a frame). This should be evident from looking at how such a frame's axes look in the "flattened out" version, and then "rolling up" that flattened out picture into a cylinder again.

 

[Orodruin]

For special relativity to apply, it has to be a flat 3-torus. SR only applies if the spacetime geometry is flat.

I’d even say SR does not apply to a flat torus as the geometry introduces a preferred frame. SR in my mind only applies to Minkowski space.

 

[robphy]

Although the flat torus is flat, I’d argue it’s not quite special relativity unless you have R^4 or restrict to a region of the torus where you don’t see the finite size of the spatial part.