Talk:Barycentre

Centre of mass != Centre of gravity in physics

I am not sure about the exact definition (or usage) of either of the terms in geometry (Euklidean or otherwise) but in physics, they describe two slightly but importantly different concepts: The centre of mass is always, as described in the current version of the page,

${\bar {\mathbf {x} }}_{m}=\left(\sum _{i=1}^{n}m_{i}\right){\bar {\mathbf {x} }}=\sum _{i=1}^{n}m_{i}\mathbf {x} _{i}.\,$

Similarly, the centre of gravity can be expressed as an "average" of the forces $F_{i}$ involved:

${\bar {\mathbf {x} }}_{F}=\left(\sum _{i=1}^{n}F_{i}\right){\bar {\mathbf {x} }}=\sum _{i=1}^{n}m_{i}a_{i}\mathbf {x} _{i}.\,$

Hence, ${\bar {\mathbf {x} }}_{m}$ and ${\bar {\mathbf {x} }}_{F}$ are generally only identical if the gravitational field (as expressed in terms of the acceleration $a_{i}$) is constant for all $\mathbf {x} _{i}$, such that $F_{i}=am_{i}$. Naturally, ${\bar {\mathbf {x} }}_{F}$, not ${\bar {\mathbf {x} }}_{m}$, is the point on which forces "may be deemed to act".

However, I am not sure whether these distinctions should be made in the present (geometry-focused) article because I do not remember having seen the use of "barycentre" (or centroid, for that matter) in either of these two physical contexts. --Daniel Mietchen 09:53, 27 November 2008 (UTC)

Centroid is a purely mathematical concept. As for the other 2, the Greeks didn't distinguish weight & mass, so we can't decide the meaning by etymology. Peter Jackson 12:18, 27 November 2008 (UTC)

The point I was trying to make was that for an inverse-square law such as gravitation, the resultant gravitational of a body or system is equal to the gravitational force exerted by a point mass at the barycentre. However I'ld be happy to split off a page of Centre of gravity (physics) and restrict this one to the mathematical concept of centre of mass. Richard Pinch 18:09, 27 November 2008 (UTC)

If barycentre is defined by the formula, then your statement is only exactly true for a spherically symmetric distribution (& even then only in Newtonian gravity, not general relativity). It's approximately true at large distances. Peter Jackson 18:40, 27 November 2008 (UTC)