Where \ vec { r } is the vector from the barycentre to the observer, \ hat { n } is the unit vector from the observer to the object or event, d is the distance from the observer to the observed object or event, and c is the speed of light.
22.
If the two bodies are of equal mass then the barycentre and L1 point will be at the equidistant point on a line joining them, by symmetry ( they are both always on the line joining them and, since there is no way to distinguish the bodies you can't work out which would be closer, so they must be the same distance away ).
23.
Although all the stuff orbiting in the galaxy is chaotic ( it's a colossal n-body problem ) it's not random or arbitrary; for most such things orbiting the galaxy's barycentre, their orbits are of near-mathematical elliptical perfection, over very lengthy periods of time .-- Talk 23 : 43, 25 August 2012 ( UTC)
24.
Planes connecting ( n-1 ) points to the midpoint of each edge should be at the barycentre, which means the simplex created by ( n-1 ) points and this centre point should have the same volume as every other simplex from every other group of ( n-1 ) points and this centre, and finally each of these simplexes is divided exactly in two by the remaining plane, which should make your hypothesis true .-Rainwarrior 06 : 32, 26 June 2007 ( UTC)
25.
1 ) I'm looking at finite subgroups G of the isometries of \ mathbb { R } ^ n, and I've been told to show G fixes some point of \ mathbb { R } ^ n by considering the barycentre ( average ) of the orbit of the origin under G . Now if we let G act on some point x, and note that gG = G ( fairly simple to prove ), then for c = \ frac { \ sum _ { g \ in G } gx } { | G | }, we have gc = c \ forall g \ in G, right?
