Re: n - charged particles
- From: quasi <quasi@xxxxxxxx>
- Date: Thu, 29 Mar 2007 21:49:06 -0500
On Thu, 29 Mar 2007 19:50:34 +0100, David Hartley <me9@xxxxxxxxxxx>
wrote:
In message <2bgk031c54dmdpqglsl8baaresr5vsqkd8@xxxxxxx>, quasi
<quasi@xxxxxxxx> writes
On Wed, 28 Mar 2007 10:03:49 +0100, David Hartley <me9@xxxxxxxxxxx>
wrote:
It can never be a strict minimum as any rotation will give a configuration
with the same energy but we can ignore that. With this definition the
pyramid is not stable for n=5. If two opposite points of the base are
displaced upwards slightly, and the other two downwards the total energy
is slightly lower. The net forces on these particles push them further
apart.
Your argument seems reasonable, but I don't have a clear visualization
of the resulting motion, so I'm not fully convinced. I'll see if I can
actually verify algebraically or numerically that the potential energy
is not a local minimum. That would clinch it.
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!
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.
quasi
.
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