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Re: Electric fields and points of stability



I wrote in part:

The center of the cube is equally unstable along all axes.

That's misleading, although technically true, especially
if you underline the word "axes".

Although there is a conservation law for "lines of force"
(which have to do [with] equilibrium, not stability), there is
no corresponding conservation law for "lines of stability".

I'm even more unhappy with that.

Earnshaw's theorem _does_ give us something rather akin
to a "conservation" law for stability in an electrostatic
field. This has been known for about 160 years.


http://www.phys.ncku.edu.tw/mirrors/physicsfaq/General/Levitation/levitation.html

I suspect Gary's intuition about this played a useful
role in focusing attention on this puzzle.

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

Bernard Cleyet wrote:

"... and will have opposite stability on the axis."

Please help me here.

One (J.D.) has suggested derivative zeroes (higher order zeros). If I read it
correctly, they don't necessarily indicate stability -- "convex" away from the
zero(s). Zeros at the bottom of a concavity would be stable, right?

0) Let's start with the potential.

1) The field is the first derivative of the potential.
It is zero at equilibrium, by definition of equilibrium.

2) The second derivative is what !!usually!! tells us about
stability.

I haven't bothered to calculate it, but my intuition tells
me that for the cube (and not the square) this second
derivative is zero at the center. The argument goes like
this: consider one of the principal axes of the cube, say
the Z axis. By symmetry, the curvature in the X-direction
must equal the curvature in the Y-direction. At the center
of the cube, all three of these must be equal. By an
Earnshaw-like argument, they must all be zero. By a Wigner-
Eckart-like argument, if those three directional second
derivatives are zero, all other second derivatives are zero.

http://www.monmouth.com/~jsd/physics/wig-eck.htm

3) Given that in this case we have the somewhat-atypical
situation where the second derivatives vanish, we must look
to higher-order derivatives to determine stability. The
potential will have high-order-multipole ruffles in it, like
the dimples in the rim of a classical pie-crust. If you
take a third-order step in some directions, the potential will
go up (stability in those modes) while in other directions the
potential will go down (instability in those modes).

Clear enough?

This posting is the position of the writer, not that of Wigner, Eckart,
or Earnshaw.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.