Suppose that $p_1,\ldots,p_n$ are points on the round sphere $S^2$ and that $p_i$ has a mass $m_i$. How would one go about computing the position $\overline p$ of their centre of mass?

I guess one question is, how should we even define the centre of mass? I think it's reasonable to say the notion should correspond the to idea that, if we place the sphere on a surface with the centre of mass being the point of contact with the surface, then the sphere should not roll (i.e. their torques with respect to any great circle through $\overline p$ should cancel). This leads us to note that the antipodal point of the centre of mass will be another centre of mass, though probably only one of them will be the "stable" centre off mass, corresponding to the idea that, if the points are confined to a sufficiently small part of the sphere, there is really only one centre of mass "in the middle of them".

Mike F
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    This is very interesting question, because it is truly 2D, in that treating it as 3D (even for the definition of the centre of mass) leads to, well, more difficulties. In 2D, we are talking about a (curved) surface with (one continuous) periodic boundary. In physics, one could treat the points as either samples of a 2D potential, or fixed nodes adding to a 2D potential, and look for the global extrema of the potential. I could do this numerically, I guess, but my math-fu is insufficient for a pure math, algebraic/geometric solution. – Nominal Animal Aug 29 '16 at 16:15

4 Answers4


Find the center of mass in the $\mathbb R^3$ space where the sphere is embedded (easy), then project it from the center on the sphere surface. Here I'm supposing that your "definition" of center of mass is fine.


Thinking about a natural extension of the center of mass definition which could work on a sphere, one might be tempted to use the same recursive pattern working in $\mathbb R^n$. For two points $A$ and $B$, define their center of mass as the point $G$ along line $AB$ (of course these would be lines in $S^2$, that is circle arcs in $\mathbb R^3$) such that $AG/m_A=BG/m_B$ (this works as long as you don't have two opposite points). For $n+1$ points, the center of mass is computed by applying the preceding rule to one of the points and to the center of mass of the other $n$ points, the latter with a mass equal to the sum of the masses of the $n$ points.

Unfortunately that does not work, because the result depends on which of the $n+1$ points is chosen, while in $\mathbb R^n$ the result is independent of that choice. So I think there is no way to extend to $S^2$ the notion of center of mass.

Intelligenti pauca
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  • What if the $R^3$ center is itself the center of the sphere? – Behnam Esmayli Aug 29 '16 at 14:54
  • Then any point will do. – Intelligenti pauca Aug 29 '16 at 14:54
  • I've just edited my answer to incorporate some thoughts about a possible alternative definition. – Intelligenti pauca Aug 29 '16 at 15:33
  • @Aretino: An additional wrinkle is that on $S^2$, each pair of points have *two* distances from each other -- two distinct $G$ for each pair of points! --, the distances adding up to the circumference of the sphere. So, even the 3D analog itself fails here. Unless, I guess, if the points are limited to one hemisphere? – Nominal Animal Aug 29 '16 at 16:21
  • @NominalAnimal Well, one may choose as distance the shortest path... provided the two points are not opposite. – Intelligenti pauca Aug 29 '16 at 16:28
  • @Aretino: But that is just the same as claiming that of two center of masses, only the closer one matters. Sure, you can absolutely do that, but it is no longer analogous to the original model. The "analogous-ness" breaks. Selection rules like that are (in physics) extremely powerful. Mathematically, we can make or add any rules we want, but if we want to keep the analogous-ness, we do need to have some kind of basis for the added rules -- even if it is just "it works if we do this". Here, it doesn't, and we haven't, so *this* center-of-mass analog breaks (or fails us). – Nominal Animal Aug 29 '16 at 16:49

The notion that the center of mass should correspond to "not rolling" implies that the answer should be consistent with an embedding of $S^2$ in $\mathbb R^3$. (After all, if $S^2$ is the only space that exists, how could the sphere "roll" anywhere?) In fact, we be able to define the center of mass using that embedding, and in particular it seems that this definition must place the center of mass on the line that passes through the center of the sphere and through the $\mathbb R^3$-center of mass of the points (the center of mass of the corresponding collection of points $\mathbb R^3$).

The tricky part is that if we say that a collection of points in $S^2$ with masses somehow acts like a single point with mass $\sum m_i$ at point $\overline p$ in $S^2$, for example for the purpose of finding the new center of mass when this set of points is combined with another set (which works fine for the $\mathbb R^3$-center of mass), we may get inconsistent results; for example, given $n$ points with masses, if we select $n-1$ points, replace them with a point of their combined mass at their center of mass, and then take the center of mass of that point and the $n$th point from the original set, the result may depend on which $n-1$ points we select first. The reason is that projecting the $\mathbb R^3$-center of mass onto the surface of the sphere increases the moment arm of the "center-of-mass" point by some amount that depends on how much the $n-1$ selected points cancel each others' moment arms in the first place. In other words, if center-of-mass calculations are considered as weighted averages, the effect of the projection can "overweight" a subset of the points when the points in that subset are combined together.

So my proposal is this: embed $S^2$ in $\mathbb R^3$ so that the center of $S^2$ is at the origin of $\mathbb R^3$. Compute $\overline p_{\mathbb R^3}$, the $\mathbb R^3$-center of mass of the $n$ points. If $\| \overline p_{\mathbb R^3} \| = 0$, let $\overline p$ be any arbitrary point on $S^2$; otherwise let $\overline p = \frac{\overline p_{\mathbb R^3}} {\| \overline p_{\mathbb R^3} \|}$. Assign the mass $\| \overline p_{\mathbb R^3} \| \sum m_i$ to $\overline p$.

The factor $\| \overline p_{\mathbb R^3} \|$ in the formula for combining the masses of a subset of some points can be considered a correction of the possible "overweighting" of that subset.

In $\mathbb R^3$, the moment of the mass $\| \overline p_{\mathbb R^3} \| \sum m_i$ at $\overline p$ around any axis through the origin is equal to the moment around the same axis of the mass $\sum m_i$ at $\overline p_{\mathbb R^3}$, so when we combine a set of points with masses into a new point on $S^2$ with a new mass in this manner, the resulting point has the same effect on new calculations of the center of mass as the original set of points, just as it does when we compute centers of mass in $\mathbb R^3$.

David K
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This question and its answers help illuminate the difficulty you're having, I think. In brief, you probably would ideally like your "center of mass" functions to have the following properties:

  • They are well-defined (i.e., actually functions).
  • They are continuous.
  • The center of mass of $n$ copies of $p$ is itself $p$.

But there is no function which satisfies all these properties on $S^2$, for any $n \geq 2$. This is true for purely topological reasons. This paper, linked to from the accepted answer to that question, contains a slick proof using homotopy groups. Indeed, it shows that any space which has functions satisfying these properties must be either contractible or pretty weird (not a finite polyhedron).

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  • +1, These obstructions to having globally defined c.o.m. functions are interesting, but don't concern me too much. There are some obvious global problems for what one would like to call the centre of mass (say for antipodal points) on a sphere, but I just wanted something which could be applied locally, i.e. in a small (contractible) patch on the sphere, where the issues you raise are not a problem. I think the suggestion, made by others, to just take the c.o.m. in the ambient space and then project correspond to what I had at mind. – Mike F Aug 30 '16 at 00:08

You may use the definition of the center of mass of a collection of point masses $m_i$ located at positions $\vec r_i$ with total mass $\sum_i m_i = M$. That is, $$ \vec r_\text{cm} = \sum_i \frac{m_i \vec r_i}{M}. $$ Note that the analogue for a body $B$ of volume mass density $\rho(\vec r)$ is the continuum limit of the above: $$ \vec r_\text{cm} = \frac{\int_B dV\, \rho(\vec r) \vec r}{\int_B dV\, \rho(\vec r)}. $$ This definition is common in physics and is effectively a weighted sum of the positions of the masses, each weighted by its mass.

Alex Ortiz
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