Re: D=2 versus cylinders versus dots on paper

[Bernard Cleyet <anngeorg@PACBELL.NET>]

The equations describing the planes of constant potential for the dipole and
dicylinder are different.   When one slices each with the conductive paper.
Which one does one pick?   In the case of the dipole all the charge is
included, while with the dicylinder only a small portion is.  The xptl data is
confusing.  LW's favours the dipole, while JB's favours the assumption that, at
least, coaxial cylinders may be sliced.   Another experiment would be to use a
larger piece of paper (in the mfging, I suspect they begin with a wide roll and
cut it and ship it) and instead of small dots use  BIG ones.

bc

P.s.Tthe dipole reduces to circles when the diam. of the equipotential is <<
separation of the poles.

"RAUBER, JOEL" wrote:

> John D. wrote:
>
> > 2) For long cylinders in real D=3 world, the (d/dz)^2
> > contribution vanishes by symmetry also.  The argument
> > is actually the same as in case (1), but students
> > often don't recognize it as such.  Students are
> > quick to notice that the cylinders have _translational_
> > symmetry up and down the z axis -- but that doesn't
> > directly serve the purpose.  What we really need is
> > this:  the symmetry group of the cylinder has a
> > subgroup, namely reflection in the xy plane.  (Indeed
> > it has many reflection subgroups, reflection in any
> > plane parallel to the xy plane, but the xy plane itself
> > is the one we really need.)
> >
> > To repeat:  The cylinder (case 2) has the same reflection
> > symmetry as the plane (case 1), plus some other symmetries
> > that are just a distraction.
> >
> > To make this brutally explicit:  The symmetry implies
> > that (d/dz) is equal to (-d/dz) and the only way
> > something can be equal to its negative is if it is zero.
>
> While agree with the above completely in the context of Ludwik's original
> question, namely that all one needs is to note the reflection symmetry about
> the x-y plane. But IMO one can directly use the translational symmetry to
> deduce the d/dz derivatives are zero.
>
> One could argue that more generally the d/dz terms are zero *anywhere* due
> to the translational symmetry, from which one can deduce the myriad
> reflection symmetries that are present.
>
> Of course one could start with the myriad of reflection symmetires to deduce
> the translational symmetry.
>
> IMO, its a bit like the chicken and the egg;  I don't think one is more
> fundamental then the other.