# How does a position based force work on a torus?

Consider a 2 particle system in $\Bbb R^2$. Let's say they have some force acting between them. For the sake of argument let that be $F=\mathbf r/|\mathbf r|^3$. This is a simple inverse square system and the particles happily attract each other.

Now my question is, how can the same be done on a torus $({\Bbb S^1})^2$?

There are two possibilities I can see

• The particles attract each other normally but when one goes over a boundary their positions instantaneously change.

There are several problems with this idea: Firstly this would mean that the acceleration of a particle will no longer have a continuous derivative. Secondly this throws all translational symmetry out of the window, which I suspect is terrible for Noether's theorem meaning there will no longer be conservation of momentum and the likes.

This then likely points to the second possibility:

• The particles are attracted by each other's infinitely many copies.

This seems more reasonable but it would mean that the force of attraction will behave very differently.

Do forces (based on position) acting on a torus behave as I have described them?

Do things like Gauss' law for gravity still have meaning?

How would you find trajectories of particles in simple systems like these? I can think of a naive method but I assume it will give me some sums that cannot be computed directly.

I would like to add that I have been a bit naughty and talked about these forces in $\Bbb R^2$, more sensibly I would talk about $\Bbb R^3$ and it's corresponding 3-torus $(\Bbb S^1)^3$ but this is simpler to think about. When I talk about Gauss' law I am talking about the three dimensional case again.