Re: n - charged particles

On Sat, 24 Mar 2007 16:29:11 -0500, quasi <quasi@xxxxxxxx> wrote:

On Fri, 23 Mar 2007 11:43:03 -0500, quasi <quasi@xxxxxxxx> wrote:

On 23 Mar 2007 12:24:54 +0200, Phil Carmody
<thefatphil_demunged@xxxxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:
On 22 Mar 2007 16:59:11 -0700, "Sunil Sangwal"
<sunil.sangwal@xxxxxxxxx> wrote:

Consider n - positively charged particles (so that they always repel
each other) that are only allowed to move on the boundary of a unit
sphere (and not necessarily lie in a plane).

I assume you meant n positively charged particles of equal charge.

In the equilibrium
position, what is the angle subtended at the center of the sphere by
two nearest particles? (Or equivalently what is the distance between
two particles along the smallest arc joining the two nearest
particles?)

Some obvious solutions for n (based on symmetry) are:

n angle
--- --------
2 180
3 120
4 105.?? (not sure, but this case is tractable)
6 90


In particular, what is the angle when there are 5 particles? Is this
problem analytically tractable?
Thanks.

Sunil Sangwal

Rather than focus on minimal angle, I think the underlying question is
what are all the possible geometric configurations, up to congruence.

It's clear that, at least in some cases, the solution is not unique.
You can always place n points on the equator so that they are the
vertices of a regular n-gon, and so that gives a trivial 2D solution
for any n.

That's not a solution as they haven't repelled each other.

But if that's the initial configuration, they _don't_ repel each
other. Of course, as Michael Jørgensen pointed out to me, the 2D
configurations are unstable (for n>2), so if stability is required,
then I agree they are not solutions. At first glance, it seems that
you are objecting to something else. You say "they haven't repelled
each other". I would interpret that as meaning that an equilibrium
configuration is one that can be achieved (in the limit), with
positive probability, from randomly chosen initial configurations
which repel each other. If so, I think it amounts to the same as
stability since if small perturbations of a claimed equilibrium
configuration return in the limit, then the claimed equilibrium
configuration has a positive probability of being achieved. Is that
right?

For n=4,6,8,12,20 there exists a regular polyhedron with n
vertices so for those values of n you also get a 3D configuration.

Again, not necessarily a solution to the OP's problem.

So some regular polyhedra are unstable in the sense being discussed?

Are there any other configurations for static equilibrium? My
intuition suggests "no", but I'm not sure.

A pyramid is stable for 5, IIRC. What do you think makes it unstable?

I never considered it. But it's not obvious to me.


Phil

Thanks for the feedback. I find the OP's problem quite interesting so
I appreciate being straightened out.

quasi

Ok, so as David Hartley points out, the concept of an equilibrium
configuration simply requires that there is no net charge on any
particle. Equivalently, the sum of all the vector charges on a given
particle must be normal to the surface. This was what I had thought
originally.

So, if we accept the "no net charge" concept of equilibrium
configuration, then I think, by symmetry, it's obvious (as I
originally claimed) that:

(1) a regular n-gon, placed on the equator (or on any great circle)
gives a trivial 2D equilibrium configuration

(2) a regular polyhedron gives a 3D equilibrium configuration.

As David Hartley also points out, there are other configurations, and
I can see now that there are lots of others, based on symmetry.

However for n=5, as far as I can see, a pyramid does not qualify as an
equilibrium configuration.

I am assuming that by a pyramid you mean 4 points forming a square
base and a 5th point directly "above" the center of the square at some
height. By hypothesis, the particles are all on the unit sphere. Let's
assume, without loss of generality that the 5th point is at the south
pole. The other 4 points are located in a square configuration on the
surface of the sphere at some height. But then there will be a net
upwards repulsive charge on the 4 points from the particle at the
south pole, so the configuration is not in equilibrium.

What am I missing?

quasi

Ok, I see what I was missing.

The 4 vertices can't be pushed "up" forever since they will repel each
other strongly if they get too close to the north pole. Hence there is
some height at which equilibrium is achieved.

quasi
.

Relevant Pages

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  • Re: n - charged particles
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  • Re: n - charged particles
    ... I assume you meant n positively charged particles of equal charge. ... vertices so for those values of n you also get a 3D configuration. ... Are there any other configurations for static equilibrium? ... Thus, for any displacement from an equilibrium state, no matter how ...
    (sci.math)
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