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Ordinarily, calculating the center of mass from a mass density (or the center of charge from a charge density, or the center of...) is very straightforward - simply integrate the position vector over the entire domain, weighted by the density function (and then normalize by dividing by the total mass).

But how does this work when calculating the center of mass of a distribution which exists on the surface of a 2-sphere? The normal procedure seems to rely on the ability to meaningfully add vectors together, which does not work on a sphere (or indeed in any other curved space). How do you calculate center of mass in such a case? Is the concept even well-defined?

That was the question - the following is merely context:

The particular case I am faced with is a field of precipitation data, given as a function of latitude and longitude (actually a discrete dataset and not a continuous function, of course). I am trying to calculate the 'center of mass' of such a dataset as required for the calculation of a certain quality measure of weather forecasts. Unfortunately, the paper describing the measure does not go into any details on the subtleties of center of mass on a sphere - they seem to mainly consider small domains and neglect the curvature of the Earth entirely. But I would like to extend the quality measure to larger domains, if possible.

(Please note that I am not talking about the rather more trivial matter of finding the center of mass in ${\rm I\!R^3}$ of the points on the sphere - unless, of course, it should happen that the projection of that point onto the sphere is also equivalent to the center of mass on the sphere. But it is not obvious to me that this should be the case.)

Drubbels
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  • If you actually put some (thin) mass to your surface area of interest, then its physical centre of mass will be the centre of mass in $\Bbb R^3$ (so a point below the surface or even (e.g., with a symmetrical distribution of mass) the centre of the sphere – Hagen von Eitzen Apr 11 '21 at 14:48
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    What traits do you want or need the center of mass to have? You're arguing for a definition after all. The simple projection of the 3D center of mass onto the sphere that you mentioned was the first thing that came to my mind. What requirements are there, and does it satisfy them? – RobertTheTutor Apr 11 '21 at 14:49
  • @HagenvonEitzen If I understand well the question, the OP is considering a sphere as a manifold without any embedding in $\mathbb R^3$. So the center of mass would be based on geodesics or stuff like that. Is the concept even well-defined? is a good question! – mathcounterexamples.net Apr 11 '21 at 14:52
  • @mathcounterexamples.net is correct, I am thinking of the 2-sphere as a two-dimensional surface which simply happens to have curvature properties. There is no embedding in ${\rm I!R^3}$ (unless it should happen to be the case that will help), – Drubbels Apr 11 '21 at 15:50
  • Could not you generalize the definiton of the center of mass of a surface using the definition for a particle system in 3D? Equation for the center of mass R is \sum_{i} m_{i}(r_{i}-R)=0. This equation can be thought as derived from the minimum of the weighted sum of squared distanced \sum_{i} m_{i}(r_{i}-R)^{2} . Can we replace (r_{i}-R)^{2} with squared geodesic distance here and solve for the minimum? – 0kcats Nov 09 '22 at 02:02
  • https://math.stackexchange.com/questions/4345098/a-system-of-particles-distributed-on-the-surface-of-a-ball-what-is-the-center – 0kcats Nov 09 '22 at 02:04
  • https://en.wikipedia.org/wiki/Fr%C3%A9chet_mean – 0kcats Jan 10 '23 at 00:56

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If we follow the idea of the question, a sphere would be considered as a manifold, not embedded in $\mathbb R^3$.

But then, I see a big issue to define the center of mass... Suppose that the complete sphere is having a uniform mass distribution... then there is no reason that the center of mass would be any particular point of the sphere.

Is it sufficient to say that the concept can't be defined?

mathcounterexamples.net
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  • I agree that there is no well-defined center of mass when there is a uniform mass distribution covering the entire sphere. But perhaps that pathological case is the only one where this happens? There does after all exist an obvious candidate for center of mass when, say, a uniform 'disk' covers part of the sphere (i.e. the center of the disk). So it is not obvious to me when the concept can be defined and when it can't. – Drubbels Apr 11 '21 at 16:27
  • You'll have a lot of pathologies. For example, take disks centered on one point of larger and larger radius. The center of mass is fixed. Then when the radius of the disk is large enough, it is becoming the full sphere... and the center of mass is no more defined. So center of mass is not continuous. And so on... – mathcounterexamples.net Apr 11 '21 at 16:37
  • But that last pathology would only appear at radius $\pi$. That's not special to this problem. For example the exponential map (from the tangent plane of a point to the sphere) becomes singular at radius $\pi$. – Lee Mosher Apr 12 '21 at 16:10
  • Regarding when the concept can be defined and when it can't... assuming you want the definition to be symmetric (i.e. commute with symmetries of the sphere), then no, the case of covering the whole sphere isn't the only case where there can't be an answer. There also can't be an answer in any other case where the shape has central symmetry (or in fact any other symmetry which moves every point on the sphere surface). E.g. the union of two same-sized disks, centered at the north and south pole. – Don Hatch Aug 19 '22 at 22:21