Re: n - charged particles

In message <3oat03l86bonel6frbqokbqoh5t6ej26js@xxxxxxx>, quasi <quasi@xxxxxxxx> writes
On 31 Mar 2007 15:26:03 +0300, Phil Carmody
<thefatphil_demunged@xxxxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:
On Fri, 30 Mar 2007 11:54:25 +0100, David Hartley <me9@xxxxxxxxxxx>
wrote:
>Looking at http://www.mathpages.com/home/kmath005/kmath005.htm, I can't
>see any symmetry when n=13.

No, there's symmetry there.

Top-right-most point and two bottom-left-most points seem to
be a plane for mirror symmetry. The fact that all edge lengths
appear in pairs apart from 4 of them, seems to support that.

I agree that v[4] (point #4) is asymmetric, which contradicts my
earlier conjecture that for a stable configuration, every point must
be symmetric. However the n=13 configuration does have _some_
symmetry. Just fix v[4] and then rotate space 180 degrees around the
vector v[4].

Rotational symmetry, instead, eh? I can't say it was clear
enough to be sure.

Sorry, I miscalculated. David Hartley is right -- the n=13
configuration has no symmetry of any kind.

I was just about to reply to your earlier post, saying that you and Phil were right! As far as I can make out from the description on the mathpages site, n=13 has both rotational and reflective symmetry: 180 degrees around the axis through vertex 4, reflection in the plane through 4, 9 and 11 and also in the plane through 4 that bisects 9 and 11.

So maybe, as both you and David seem to be suggesting, symmetry is the
exception rather the norm.

My intuition was based on algebra. Since the conditions for stable
equilibrium are algebraically symmetric, my expectation was that any
solutions would naturally tend to exhibit at least some symmetry.

Perhaps instead, the right intuition is that without symmetry and
assuming equilibrium (no net forces on any particle), there's no need
to worry about slipping and sliding.

Ok, but I'm still somewhat biased in favor of symmetry, so before
giving up, I'll retreat to a much weaker conjecture.

With the usual setting -- n equally charged particles on the unit
sphere in R^3, is the following claim true?

If n>1 and n is not a prime or a prime power then any stable
equilibrium must have at least one non-trivial symmetry.

Try n=21. I'm fairly sure it has no rotational symmetry, but haven't ruled out reflections yet.

--
David Hartley
.

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