Gravity on a 1-D Torus: Understanding the Conceptual Simplification

In summary, the conversation discusses the possibility of understanding gravity on a flat 2-D torus. The problem is simplified into a 1-D torus for conceptual purposes, but it is acknowledged that there are issues with using the inverse square law in 1-D. The conversation also explores the possibility of a 2-D solution for the same problem, but it is noted that it may not be possible due to the potential for a logarithmic divergence. The conversation ends with a discussion on the difficulty of summing gravitational force contributions in a 2-D space with a single type of charge.
  • #1
ArielGenesis
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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 let's 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"
 
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  • #2
ArielGenesis said:
1. Did I made any mistake in my reasoning?
You made up a situation which cannot happen with the physics in our world. You can invent any forces you like.
ArielGenesis said:
2. Can the sum be simplified?
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.
ArielGenesis said:
3. Would the 2-D solution of the same problem be "nice"
With 1/r^2 the sum would have a logarithmic divergence. That is certainly not nice.
 
  • #3
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
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
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
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
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
I know there are a lot of issue with using inverse square law in 1-D, but let's just ignore it for the moment.
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.
 

1. What is a 1-D torus?

A 1-D torus, also known as a circle or loop, is a geometric shape that has one dimension (length) and curves back on itself to form a closed loop.

2. How does gravity behave in a 1-D torus?

In a 1-D torus, gravity behaves similarly to how it does in a 2-D or 3-D space. Objects with mass will still be attracted to one another and will follow a curved path towards the center of the torus.

3. What is the equation for calculating gravity in a 1-D torus?

The equation for calculating gravity in a 1-D torus is similar to the equation for calculating gravity in a 2-D or 3-D space. It is given by F = (G*m1*m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

4. Can objects escape the gravitational pull in a 1-D torus?

Yes, objects can escape the gravitational pull in a 1-D torus if they have enough velocity. This is similar to how objects can escape the gravitational pull of a planet if they have enough velocity to overcome it.

5. How does the shape of the 1-D torus affect the strength of gravity?

The shape of the 1-D torus does not affect the strength of gravity. As long as the distance between two objects remains the same, the strength of gravity will remain the same regardless of the shape of the torus.

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