This is a follow up to What would physics be like in a wrap-around universe? PyRulez commented, “Will gravity still make sense? (Namely, will it converge or not)?”

This was bantered about in the comments and one idea was mentioned briefly in one of the answers. But I’d like a more in-depth explanation.

There are several “obvious” answers that disagree. One can draw contour lines representing the potential on a spherical or toroidal map and note that it doesn’t explode or anything; this defines a navigable space useful in a game perhaps. But I think that ad-hoc field doesn’t follow the normal rules of falling off with the square of the distance. If you modeled it as normal space with infinite repeating grid of masses, would the field strength converge, blow up, or become chaotic, or what? Given this, what modification makes sense to produce a sensible wrap-around universe that is still mathematically sensible?

Edit: it appears that much of the “different answers” is due to the chosen topography. In particular, the video-game-screen wrapping which is easier for mental musing is not isotrophic.

I'm less interested in imposing something ahead of time than in learning what does work; conversly, what “interesting” universe parameters might be available for stories/games having more exotic situations.

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    $\begingroup$ I vaguely recall seeing this on astronomy or physics se. The point was that it is hard to prove that our universe isn't looped one, if the loops are big enough. So, how big your universe is? $\endgroup$
    – Mołot
    Mar 10, 2017 at 14:02
  • $\begingroup$ I think any real answer to this question would have to take general relativity into account. Good luck finding one. $\endgroup$
    – praosylen
    Mar 10, 2017 at 15:58
  • $\begingroup$ @Mołot being big enough, the propagation delay makes the problem go away. The edge of the observable universe is all that affects us. Same fix as Olber’s paradox. $\endgroup$
    – JDługosz
    Mar 10, 2017 at 18:54
  • $\begingroup$ The same considerations that apply in a theoretical wrap-around universe apply in a real universe with wormholes. A wormhole can effectively take the place of the edge of the wrap-around universe to create a loop like this. Particularly, Stephen Hawking wrote about this. I don't recall exactly what he said but I think he explained that the effect would stack on top of itself infinitely until enough energy is accumulated to collapse the wormhole (I say effect because it doesn't apply only to gravity but to all other fields as well. Magnetism works just the same). $\endgroup$
    – Annonymus
    Mar 10, 2017 at 21:08
  • $\begingroup$ I think that has to do with wormholes that would provide for cause for causality violations. Wormhole mouths can be at arbitrary space-time coordinates. $\endgroup$
    – JDługosz
    Mar 11, 2017 at 0:16

3 Answers 3


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 $$\sum F_{+x}=\sum_{i=0}^{\infty}\frac{Gm_1m_2}{(x+il)^2}\tag{1a}$$ Likewise, we can do the same for the forces on $m_1$ in the $-x$ direction: $$\sum F_{-x}=\sum_{i=0}^{\infty}-\frac{Gm_1m_2}{((l-x)+il)^2}\tag{1b}$$ According to Wolfram Alpha, the sum $$\sum_{i=0}^{\infty}\frac{1}{(x+il)^2}$$ converges1, and since $\text{(1a)}$ is simply this multiplied by $Gm_1m_2$, 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, $$\text{Universe}=\{p(x_1,x_2,\cdots,x_{n}):\sqrt{x_1^2+x_2^2+\cdots+x_{n}^2}=R,\quad x_1,x_2,\cdots,x_{n}\in \mathbb{R}\}$$ 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 $p\geq1$, 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: $$\text{Universe}=\{p(x_1,x_2,\cdots,x_n):p(x_a,x_b,\cdots,x_i,\cdots,x_n)=p(x_a,x_b,\cdots,x_i+l_i,\cdots,x_n),\quad x_1,x_2,\cdots,x_n\in\mathbb{R}\}$$ 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 $$\sum F_{+x_j}=\sum_{i=0}^{\infty}\frac{Gm_1m_2}{(x_j+il_j)^p}=Gm_1m_2\sum_{i=0}^{\infty}\frac{1}{(x+il_j)^p}=\frac{Gm_1m_2}{x^p}+\sum_{i=1}^{\infty}\frac{1}{(x+il_j)^p}$$ 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 $$\zeta(p)=\sum_{i=1}^{\infty}\frac{1}{i^p}$$ Now, for each positive integer $i$, $$x+il_j>i$$ by our criteria, because $il_j\geq i$ and $x>0$. Therefore, $$\frac{1}{x+il_j}<\frac{1}{i}\implies\frac{1}{(x+il_j)^p}<\frac{1}{i^p}$$ It holds, then, that $$\sum_{i=1}^{\infty}\frac{1}{(x+il_j)^p}<\zeta(p)$$ $\zeta(p)$ 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 $\sum_{i=0}^{\infty}F_{+x_j}$ converges, as does $\sum_{i=0}^{\infty}F_{-x_j}$. Therefore, the total force is finite.

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


  • 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 $t'$ the objects will interact as they were some time prior to $t'$. 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 $C'$ 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 $\vec{a}$ 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?


1 Specifically, it converges by the root test, which you could do be hand, if you so desired, and gives us the result $$\sum_{i=0}^{\infty}\frac{1}{(x+il)^2}=\frac{\psi^{(1)}\left(\frac{x}{l}\right)}{l^2},\quad l\neq0$$ 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 $$x=|\mathbf{x_1}-\mathbf{x_2}|$$ where $\mathbf{x_1}$ and $\mathbf{x_2}$ are the position vectors of the two masses.

  • $\begingroup$ An image of the gravitational potential would be interesting. $\endgroup$ Jan 20, 2019 at 6:45

Using the Gauss formulation of the law of gravitation. The integral of the inward force over a surface is equal to the mass within it. If you have a simply connected spacetime you can expand your volume untill it fills all of space ( even with a non simple spacetime you can fill all of space except a few surfaces and assuming continuous gravitation, no singularities) Then the surface integral becomes 0.

The universe must contain the same amount of positive and negative mass.

If we consider a wraparound square universe containing only a small mass then the gravitational pulls in each direction are infinite, but these cancel to yield a finite gravitational pull. However if the mass is non spherical and rotating, tidal forces fall off inverse cube, mass in a shell increases with the square of distance.Consider two shells of identical thickness but radius differing by a factor of 10. While the mass 10x further gives only 1/1000th of the force per mass, there is 100X more mass and so 1/10 the force. The total force from all the shells is some constant*(1/1+1/2+1/3+1/4...1/n) which diverges to infinity.

These arguments prohibit any wraparound or infinite isotropic universe with locally normal gravity from having existed an infinite time with no damping effects on the propagation of gravity.

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    $\begingroup$ Several problems: The surface integral of gravitational potential (what I assume you mean by inward force) is proportional to but not equal to the mass inside the surface; does our universe contain the same amount of positive and negative mass? that is not self-evident to me; your whole third paragraph needs a more rigorous treatment, since I don't think any of it is correct. $\endgroup$
    – kingledion
    Mar 11, 2017 at 1:24

I'm going to assume your universe isn't infinitely old and follows the same basic rules ours does.

If you modeled it as normal space with infinite repeating grid of masses...

There's problems with that approach. First is gravity propagates at the speed of light. If your universe is large enough, or expanding rapidly enough, gravity won't have had time to "wrap around". If it has, it can only have done so a finite number of times. So no, you wouldn't model it as an infinite grid of masses.

Even if it had, gravity's strength falls off as the square of the distance. Each time it wraps around the universe it's getting exponentially weaker. The effect of subsequent trips around the universe is greatly diminished and will sum to a finite number (I think that's what HDE is proving).

But let's say gravity propagates instantly. There is no additional net force because the force of all those repeats cancels out. Here's why.

Let's assume a 1D universe 10 units wide. Let's say we have two objects, one very massive and one with very little mass. The massive object is at 0, the small one is at 5. So it's feeling the gravity of the massive object from 5 units away.

               *  .
               0  5 10

This universe wraps around, -10 and 10 are the same.

               *  .      
       -10 -5  0  5 10

That means, effectively, there's another massive object out there 15 to the right.

               *  .        *
       -10 -5  0  5 10 -5  0

But wait, there's also one to the left at 25 units.

 *  .          *  .        *
 0  5  -10 -5  0  5 10 -5  0

But wait, there's also another one to the right at 35 units.

 *  .          *  .        *  .          *  .
 0  5  -10 -5  0  5 10 -5  0  5  10  -5  0  5

But wait, there's also another one to the left at 45 units.

But wait, there's also another one to the right at 55 units.

But wait, there's also another one to the left at 65 units.

But wait...

What you wind up producing is a repeating line. There's an infinite number of masses to the right, and an infinite number of masses to the left. The net distance between our object and all the infinite masses is 0. Here's the proof using a generalization of Cesàro summation.

If you sum up the distances between the objects, we get a series.

5 - 15 + 25 - 35 + 45 - 55 + 65 - 75 ...

At first glance that diverges, and it does, but we can get a summation out of it. Does it oscillate around a number? Look at their partial sums.

5, -10, 15, -20, 25, -30, 35, ...

And take their mean values.

5, -5, 5, -5, 5, -5, 5, -5

All the positive values converge at 5. All the negative values converge at -5. So that series is summable to 0. This is the same technique used to show that 1 - 2 + 3 - 4 ... is 1/4.

The net distance is 0, but does that mean net gravity is 0? I don't know how gravity works in 1 dimension, but let's assume it's still Gm/r^2. Set G and m to 1 and we get another series.

d =  5   -   15  +   25  -   35   +   45   - ...
g = 1/25 - 1/225 + 1/625 - 1/1225 + 1/2025 - ...

This is converging around 0.0367. Without a wrap around universe there'd be a pull of 1/25 or 0.04. So that means there's a slight negative pull of about 1/300.

The point of all this is to show that a wrap around universe with infinitely propagating gravity can be resolved. Scaling it up to multiple dimensions is outside my math and physics skills.

  • $\begingroup$ I had not thought of the propagation speed of gravity, but that is highly relevant. If the universe is the size of ours, for example. even if the universe looped around infinitely, each object would be outside the event horizon of itself, given the finite time since the expansion of the universe. This actually contradicts what you said in the second part: you can't have an infinite number of masses to the right and left, only so many of those masses would be within the gravitational event horizon. $\endgroup$
    – kingledion
    Mar 10, 2017 at 18:46
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    $\begingroup$ Are you sure the cancellation isn't just because no matter if you go left or right from one object, the other object is the same distance away? This feels like an effect from just this special case. Plus, no matter what the speed is, the strength of the various contributions will still be different. Maybe I'm just misunderstanding what you're saying. $\endgroup$
    – HDE 226868
    Mar 10, 2017 at 18:50
  • $\begingroup$ That only works on a line along the axis of the tourus (or grid), and it claims the distances are the same in each direction. Generalize it to other positions and different directions, and you’ll see that this is in fact the case that makes me suggest chaotic, with the field varying in direction and strength as you move a little bit. $\endgroup$
    – JDługosz
    Mar 10, 2017 at 18:58
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    $\begingroup$ @kingledion The second part assumes gravity propagates instantly. $\endgroup$
    – Schwern
    Mar 10, 2017 at 19:00
  • $\begingroup$ @HDE226868 I'm not sure about any of this. You're the actual astrophysics student, you tell me! :) $\endgroup$
    – Schwern
    Mar 10, 2017 at 19:01

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