# Gravity in 1-D torus

1. Jan 20, 2016

### ArielGenesis

I am trying to make understand gravity on a flat 2-D torus. To help myself get my head around it, I simplify the problem into a 1-D torus.

Let's have a 1-D space (or line). Instead of the left part of the line going infinitely to the left and the right part of the line going infinitely to the right, I am going to make them meet, like a torus. So we have circle instead, which radius is R. Limited space, but unbounded.

Let's have a point mass O on the line. The acceleration any particle P will experience towards O (classical mechanic) is a=Gmo/r2 where G is the gravitational constant, m is the mass of O and r is the distance between O and P. (I know there are a lot of issue with using inverse square law in 1-D, but lets just ignore it for the moment). Since this is just for conceptual purposes, I would just assume that G and mo are equal to 1. The formula for acceleration will be reduced to: a=1/r2

However, in a non toroidal space, the gravitational field would touches particle P once and just extend infinitely, never touching P again. Yet, if we say that the particle P is located r distance to the left of O, it will not only experience a=1/r2 going to the right, but also a=1/(R-r)2 going to the left. The resulting acceleration would be a=1/r2 - 1/(R-r)2.

And yet, it doesn't stop there, the gravitational field that to the left, pulling accelerating everything to the right, will meet point P at r, and then circle the whole space and meet point P again at r+R and circle the space again and meet point P again at r+2R and so on. The same also apply with the gravitational field going to the right. The total acceleration would be: sum_(a=1)^infinity(1/(a b+c)^2-1/(a b-c)^2)

Now, Wolfram alpha cannot simplify that sum for me, neither can I myself. http://www.wolframalpha.com/input/?i=sum+1/(ab+c)^2-1/(ab-c)^2,a=1+to+a=infinity

My questions are:
1. Did I made any mistake in my reasoning?
2. Can the sum be simplified?
3. Would the 2-D solution of the same problem be "nice"

2. Jan 20, 2016

### Staff: Mentor

You made up a situation which cannot happen with the physics in our world. You can invent any forces you like.
At least for some special cases I would expect that (e.g. there is a closed formula for the sum over 1/a^2), not sure about the general case.
With 1/r^2 the sum would have a logarithmic divergence. That is certainly not nice.

3. Jan 20, 2016

### zinq

It seems most natural that for a 2D space like the 2-torus, force would attenuate in proportion to the reciprocal of distance.

This suggests that a reasonable potential function could be (proportional to) the logarithm of distance. (It would not work, however, to define potential by integrating force out to infinity, since the integral would diverge.)

4. Jan 24, 2016

### ArielGenesis

Thank you for the replies.

It seems that what I'm trying to is impossible.

My main goal was to create an n-body simulation. But I need limited space, the space cannot go to infinity in all direction. My solution was to wrap it up like a torus. Which then poses the question, how would gravity works in a torus. If the integral doesn't converge, are there any nice trick to do what I'm trying to accomplish?

5. Jan 24, 2016

### Staff: Mentor

I don't know what you try to accomplish. Why do you need to limit space?

If our own universe has a global torus-like structure, this problem would be avoided by the finite age of the universe - things beyond the particle horizon cannot influence us yet, so the sum of gravitational influences is always finite.
A similar problem arises in crystals with the Madelung constant, where you have to sum over all atoms in a crystal with a 1/r potential. The limit depends on the order of atoms to include, and you need some clever way to make the sum properly convergent.

Stopping integration for distances larger than some large threshold could be a workaround.

6. Jan 24, 2016

### zinq

The thing about the Madelung summation is that it takes into account both positive and negative charges. These cancel each other to some extent, which makes it possible for the sum to converge.

But that convergence does not seem to be possible if you try to sum the gravitational force contributions at a point (x,y) in the plane (or even (x,y,z) in space) if you imagine there to be a unit mass at each integer point in the plane or space, respectively (as one might presume in the case of a torus).

7. Jan 25, 2016

### Staff: Mentor

With gravity you have a single type of charge only, but you have the different directions that cancel each other to some extent, if you want to calculate force instead of potential.

8. Jan 25, 2016

### haushofer

What do you mean by "2-dimensional gravity"? It is certainly not GR, and if it is Newtonian it is certainly not 1/r^2.