I can give you a Newtonian analysis of the situation, which may or may not be correct. If you want a Newtonian universe, then great. If you want a general relativistic one, you'll need something more complicated.

Example 1: Square domain

As a simple introductory example, let's say your universe is a rectangle with sides of length $l$. In other words, if you travel $l$ units in the $x$- or $y$- directions, you return to where you started.

Consider an object with mass $m_1$ at $p_1=(x_1,y_1)$ and a second object with mass $m_2$ at $p_2=(x_2,y_2)$. For simplicity, I'll set $y_1=y_2$, so the objects are a distance $x=|x_2-x_1|$ units apart. To find the force on $m_1$, we have to create an infinite sum of all the forces on that object from the object on $p_2$. Let's first look at the forces from the $+x$ direction. We have $$\begin{array}{}\text{(1a)}& \sum F_{+x}=\sum _{i=0}^{\mathrm{\infty }}\frac{G{m}_1{m}_2}{(x+il{)}^2}\end{array}$$ Likewise, we can do the same for the forces on $m_1$ in the $-x$ direction: $$\begin{array}{}\text{(1b)}& \sum F_{-x}=\sum _{i=0}^{\mathrm{\infty }}-\frac{G{m}_1{m}_2}{((l-x)+il{)}^2}\end{array}$$ According to Wolfram Alpha, the sum $$\sum _{i=0}^{\mathrm{\infty }}\frac{1}{(x+il{)}^2}$$ converges¹, and since $\text{(1a)}$ is simply this multiplied by $G{m}_1{m}_2$, that must converge. The same logic holds for $\text{(1b)}$, as we've just inserted $(l-x)$ for $x$ and multiplied it by $-1$. If we add two convergent series, the resulting series must also converge. Therefore, on a square domain where the masses are separated by a distance parallel to a side, the force is finite: $$\sum F=\sum F_{+x}+\sum F_{-x}$$

Example 2: $n-1$-dimensional sphere (isotropic universe)

This easily generalizes to a special case. Let's say that we view our universe as the surface of an $n-1$-sphere². In other words, in set-builder notation, $$\text{Universe}=\{p(x_1,x_2,\cdots ,x_n):\sqrt{x_1^2+x_2^2+\cdots +x_n^2}=R,x_1,x_2,\cdots ,x_n\in \mathbb{R}\}$$ where $R$ is the radius of the universe. This is an $n-1$ dimensional universe embedded in $n$-dimensional space. In this $n-1$-dimensional space, there is no preferred direction, i.e. it is isotropic. Therefore, for any two points $p_1$ and $p_2$ in the universe, we can connect them via a one-dimensional geodesic, the shortest distance between them, on the sphere. There are actually going to be two paths, going in opposite directions, just as we had ones with length $x$ and $(l-x)$. One will be the geodesic, and the other will be in the opposite direction. For finite $R$, these paths should have finite length, and so the sums of $\text{(1a)}$ and $\text{(1b)}$ should hold.

Let's be careful, though. In $q$ dimensions, gravity falls of as $$F\propto \frac{1}{r^{q-1}}$$ and so in this universe, we have $$F\propto \frac{1}{r^{n-2}}$$ Setting $p=q-1$, Wolfam Alpha seems to indicate (by repeated trials up to $p=5$) that this type of sum (and thus the total force) converges for all $p\ge 1$, with $p$ an integer. Therefore, for all $n-1$-spheres in two dimensions or more, the force of gravity should converge (it diverges in one dimension, where $p=0$).

A proof for all simply-connected universes

Thanks to some comments by kingledion, we can come up with a rigorous proof for convergence on a number of simply-connected Euclidean spaces. We deal this time with a wraparound $n$-dimensional universe of finite size³: $$\text{Universe}=\{p(x_1,x_2,\cdots ,x_n):p(x_a,x_b,\cdots ,x_i,\cdots ,x_n)=p(x_a,x_b,\cdots ,x_i+l_i,\cdots ,x_n),x_1,x_2,\cdots ,x_n\in \mathbb{R}\}$$ equipped with the standard Euclidean distance metric $$s=\sqrt{x_1^2+x_2^2+\cdots +x_n^2}$$ The $l_i$s may have the same for all $i\in \{1,2,\cdots ,n\}$, as was the case with the $n-1$-spheres, or they may be different, as would be the case if we did Example 1 on a rectangular domain. You can travel along any Cartesian coordinate a certain distance - and you can redefine the coordinate system (i.e. by rotating it), meaning you could travel "diagonally" in Example 1. You'd just need to write out a new coordinate system with new $l_i$s.

We write our net force equation as before: $$\sum F=\sum F_{+x_j}+\sum F_{-x_j}$$ and, as with $\text{(1a)}$, we have $$\sum F_{+x_j}=\sum _{i=0}^{\mathrm{\infty }}\frac{G{m}_1{m}_2}{(x_j+i{l}_j{)}^p}=G{m}_1{m}_2\sum _{i=0}^{\mathrm{\infty }}\frac{1}{(x+i{l}_j{)}^p}=\frac{G{m}_1{m}_2}{x^p}+\sum _{i=1}^{\mathrm{\infty }}\frac{1}{(x+i{l}_j{)}^p}$$ where I've set $p=n-1$. Let $l_j=1$, and let $x>0$⁴. Take a step back and look at the Riemann zeta function, given by $$\zeta (p)=\sum _{i=1}^{\mathrm{\infty }}\frac{1}{i^p}$$ Now, for each positive integer $i$, $$x+i{l}_j>i$$ by our criteria, because $i{l}_j\ge i$ and $x>0$. Therefore, $$\frac{1}{x+i{l}_j}<\frac{1}{i}⟹\frac{1}{(x+i{l}_j{)}^p}<\frac{1}{i^p}$$ It holds, then, that $$\sum _{i=1}^{\mathrm{\infty }}\frac{1}{(x+i{l}_j{)}^p}<\zeta (p)$$ $\zeta (p)$ is finite for all positive integer $p$, and so it holds that the left sum is finite, and thus convergent. The $i=0$ term, which we removed, is also finite for $x>0$, and so the overall sum $\sum _{i=0}^{\mathrm{\infty }}{F}_{+x_j}$ converges, as does $\sum _{i=0}^{\mathrm{\infty }}{F}_{-x_j}$. Therefore, the total force is finite.

There's one important caveat here. Take a point $p_{\ast }\in \text{Universe}$. Move in some direction $x_i$. The idea of a wraparound universe assumes that when you next reach $p_{\ast }$, you've traveling in the same direction as you were when you left. In other words, your velocity vector at the moment you reach $p_{\ast }$ again is parallel to the velocity vector when you left.

What if this wasn't the case? Well, we could imagine a topology where you leave $p_{\ast }$ going in the $x$-direction, and return to it from the $y$ direction. This introduces a totally new direction for the force to act on. While the proof of convergence still holds - because you'd have to travel a distance $l_x$ to finish the full cycle - you also have to consider this weird force in a different direction. So gravity would still be finite, but . . . odd. I'd rather not consider those cases.

More universes

I can't generalize this to all wraparound spaces, although I suspect that gravity may converge for many, if not all, Euclidean surfaces. I can't say much more for many manifolds in general, as the distance metric (for those manifolds equipped with one) will likely not resemble the familiar Euclidean metric I used in the proof.

Indeed, general relativity would be needed for a truly accurate answer here. However, for the simplest sort of universes, using Newtonian gravity, I believe the force of gravity will indeed converge.

Musings

• In their answer, Schwern stated that the speed of light could have an effect, and assuming the Newtonian model (i.e. that gravity travels infinitely fast) would have a different result than modifying it so that gravity travels with some finite speed $c$. I talked this over in chat with kingledion, and I believe this isn't a problem.

  Take two objects at points $p_1$ and $p_2$ with masses $m_1$ and $m_2$, where $p_1,p_2\in \text{Universe}$. This universe can have any classic wraparound topology you want. At time $t=0$, these object are a distance $x_0$ away from each other. However, the first object moves at some non-zero angle from the geodesic connecting the two. Therefore, at any time $t^{\prime }$ the objects will interact as they were some time prior to $t^{\prime }$. The argument was that this would add a weird force vector component not present in the infinitely fast gravitational model. This is because it would take the force of gravity different amounts of time to travel in each geodesic, specifically, a time $t_p=s_p/c$, where $s_p$ is the length of a geodesic.

  Here's why this doesn't work. In normal Newtonian physics problems in our wacky universe, we assume that gravity acts directly between two objects. It instantaneously "knows" the shortest path between them at any time. That's not really the case. Let's say that gravity takes an infinite number of paths between both objects. Each path $C$ has a distance $s_C$, and it thus takes a time $t_C=s_C/c$ for the force to travel that far. However, for that path, we can also create another path $C^{\prime }$ with the same distance, just in the opposite direction, like flipping something along an axis. That axis here is the geodesic between the two points. The only paths that don't cancel are that geodesic and the path opposite it.

  This means, of course, that we don't have to assume that gravity "knows" where to go. That would be absurd. A point object with a mass creates a gravitational potential that is isotropic, and so a mass at any point could feel it. You can define a vector field $\overrightarrow{a}$ at any point giving the acceleration due to gravity. Therefore, in our example, we know that the object is going to be influenced by the gravity of the other no matter whether or not it's moving. We simply have to realize that this time, the paths that don't cancel aren't the original geodesic and its partner, but a new geodesic and its partner, connecting the new position and the position of the other particle.

• In these universe, Bertrand's theorem might not hold. Simply put, it states

  are only two types of central force potentials with the property that all bound orbits are also closed orbits: (1) an inverse-square central force . . . and (2) the radial harmonic oscillator potential.

  Assume our universe is a sphere. Place two particles on opposite sides, such that all of the forces between them cancel out. They're in an unstable equilibrium. Now, give them non-zero velocities such that the velocity vectors are exactly parallel. They'll move around the sphere and come back to where they started. The potential between them wasn't an inverse-square force potential; it was proportional to $r^{-1}$. However, the orbit was still closed! Actually, take a solitary particle on the sphere. Now give it some non-zero velocity. Even though the surrounding potential is zero, the orbit is closed.

  Perhaps this doesn't violate Bertrand's theorem. It's not clear whether these particles are "bound"; there's no force between them, and so gravity isn't really affecting them. However, we could place three particles around the sphere, spaced apart evenly on a circle. Between any two, there is a non-zero force, but they're all at equilibrium because a third particle balances out the interactions. There is a non-zero force between any two particles. Again, this may not a "bound" orbit, as the force isn't influencing the motion at all. So it's not clear whether or not this is a workaround.

  I suspect the above example fails for some reason, at least in a two-spherical topology. However, maybe there are other universes in which there are indeed exceptions to Bertrand's theorem. Exploring the relevant orbital mechanics would be interesting. Anyone up for simulating orbits on an $(n-1)$-sphere?

Notes

¹ Specifically, it converges by the root test, which you could do be hand, if you so desired, and gives us the result $$\sum _{i=0}^{\mathrm{\infty }}\frac{1}{(x+il{)}^2}=\frac{{\psi }^{(1)}\left(\frac{x}{l}\right)}{l^2},l\ne 0$$ Here, ${\psi }^{(n)}(x)$ is the $n$th derivative of the digamma function.
² In this notation (standard in mathematics), the surface of a circle is a one-sphere $S^1$, the surface of a round ball is a two-sphere $S^2$, etc.
³ This set doesn't define the space as being simply connected; it says nothing about its overall topology. However, those assumptions are implicit, and harder to write in set-builder notation. If I can come up with a rigorous proof for all Euclidean universes, I'll add it, but it could be hard - at least, it doesn't seem apparent at the moment. An interesting universe excluded here is the torus, which I could probably figure out if I had the time; it is multiply connected.
⁴ Here, $x$ is really given by $$x=|{\mathbf{x}}_{\mathbf{1}}-{\mathbf{x}}_{\mathbf{2}}|$$ where ${\mathbf{x}}_{\mathbf{1}}$ and ${\mathbf{x}}_{\mathbf{2}}$ are the position vectors of the two masses.

posted about 8 years ago

HDE 226868‭

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