Subject: charges on sphere, another

From: elewars <[email-masked]>
Date: Sat, 13 Oct 2001 11:13:38 -0400

2001 Oct 13

Another interesting contribution, from a new angle, to the charges on a
sphere problem:
------------------------

Mitchell Polley <[email-masked]>

I was about to send you this message and I saw Dr Tubert's excellent
response, which made my repsonse somewhat obsolete. However, I've
decided to send it anyway.

Dear Dr. Lewars,

If the problem is reduced to something simpler, it may provide minor
insight into the the more complex picture. While pondering your original

question I simplified it by getting rid of the "charges" on the points
and installing instead fixed radii. Afterall, won't any two point
charges of like polarity repel each other to infinty if no other forces
act? So assuming the distance between two identical point charges will
reach some sort of equilibrium distance (d - measured centre to centre),

and assuming that this is a constant distance, we can rethink the
problem as one of fixed d, similar to a spherical hardshell VDW model of

atoms in close packing circumstances. Obviously, n charges seeking the
equilibrium distance will form a minimum surface area aggregate
(spheroid) and as n increases the packing becomes more complex.
Tetrahedron -> trigonalbipyramid -> octahedron -> pentagonalbipyramid ->

cube/rhombohedron -> etc. After envisioning this growth for a while you
can see that after the tetrahedron we have a shape with three "slices":
two "poles" and an "equator". Clearly the point at which having one of
the "atoms" at the centre of the sphere comes when the radius of the
aggregate, measured from the centroid to the centre of any of the
"atoms", comes when r => d. Under the series envisioned above this wil
be when there are 6 "atoms" in the equatorial slice hexagongally, and 3
"atoms" in each of the polar slices in a triangular arrangement - only
at this point can an extra "atom" fit into the space in the middle, and
the close packing be preserved (thus keeping constant d between ALL
atoms in the aggregate). With this simplified approach the answer (I
believe) is that the 13th "atom" goes in the middle. I'm not sure that
it answers your more complex problem of equidistant charges on a
spherical surface but I beleive it provides at least a different
perspective on how the problem might be conceptualized.

Regards,

Mitchell Polley.

===========================