Re: D=2 versus cylinders versus dots on paper

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

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.