Re: n - charged particles

On Fri, 30 Mar 2007 11:54:25 +0100, David Hartley <me9@xxxxxxxxxxx>
wrote:

In message <87odmb87nm.fsf@xxxxxxxxxxxxxxxxxxxx>, Phil Carmody
<thefatphil_demunged@xxxxxxxxxxx> writes
quasi <quasi@xxxxxxxx> writes:
Suppose we have one particle at the south pole, two at latitude theta,
longitude 0 and 180 and two at latitude phi, longitude 90 and 270. The
equilibrium pyramid has theta = phi = 14.0755451246 N and potential
energy 6.4836605205167. If we increase theta and decrease phi by 0.001
degrees, the energy becomes 6.4836605205159, (if my calculations are
correct). This doesn't prove it's not a local minimum but it looks very
unlikely.

Putting theta = 0 and phi = 30 N gives an energy of 6.47469...

Ok, I can visualize it now -- I see the resulting motion just as you
described it. That, together with the numerical evidence you provided
makes it clear that the pyramid equilibrium configuration is _not_
stable. To Phil Carmody -- yes, you were hallucinating and worse --
you even lured me into your hallucination!

Goddamn! So many times I wish I still had my old program, that would
have answered the question immediately. Apologies for misremembering.
In retrospect, I do seem to remember square faces being quite rare.
The twist in 8 should have reminded me that it tries to avoid squares.

Ok, so here's a question.

Can a stable equilibrium configuration have an asymmetrical point?

It seems intuitive that the answer is no for all n.

Should I start hallucinating again and report back what I see? :-)

My intuition goes the other way. If reducing the symmetry can lead to a
more stable equilibrium, as with the pyramid and with the cube,

Well for the pyramid and the cube, I'm not sure that reducing symmetry
was responsible for restoring stability so much as switching to a more
interlocking kind of symmetry.

entirely possible that, with more particles, there can be equilibria
with no symmetry.

Perhaps such a configuration exists but all the evidence so far
suggests otherwise.

Looking at http://www.mathpages.com/home/kmath005/kmath005.htm, I can't
see any symmetry when n=13.

No, there's symmetry there.

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].

My geometric intuition was never very strong (I'm working on it), but
even so, I'm fairly certain that all stable configurations must have
some nontrivial symmetry.

quasi
.

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