Re: "4/3 Problem" Resolution (fwd), comment on

[David Rutherford <drutherford@SOFTCOM.NET>]

On Mon, 14 May 2001 13:35:44 -0500, Jack Uretsky <jlu@HEP.ANL.GOV> wrote:

> Hi all-
>        David's calculation makes no sense to me because it makes use
> of a special gauge for the EM field.  After discussion with David I
> see no way to resolve our disagreement.  I would firmly reject any
> calculation that forswears gauge invariance.  Gauge invariance merely
> says that it is the fields that are physical, not the potentials.
>                                Regards,
>                                        Jack

Div(A) is a gauge-invariant quantity as long as the gauge function /\
satisfies the condition

d^2(/\)/dx^2 +  d^2(/\)/dy^2 +  d^2(/\)/dz^2 = 0       (*)

for the three-dimensional, time independent case I gave in my original
post. Div(A) is physical, since v.E/c^2 (which is the same as -div(A)) is
physical, just as vxE/c^2 (which is the same as curl(A)) is physical.
Therefore div(A) is specified, just as curl(A) is specified. Any gauge
transformation must be chosen so that it leaves curl(A), -grad(phi), _and_
div(A) (for the time independent case) invariant. If one can't be found,
then it's the concept of gauge invariance that must be abandoned, not the
physicality of div(A).

In general, the condition is that /\ need only satisfy
D'Alembertian^2(/\)=0. In the case I gave, the terms d(/\)/dt,
d^2(/\)/dt^2, d(phi)/dt, and d(A)/dt are all zero since, for constant v (as
in this case), there is no time dependence. So the condition
D'Alembertian^2(/\)=0, in this case, reduces to
d^2(/\)/dx^2+d^2(/\)/dy^2+d^2(/\)/dz^2=0, as in (*).

If anyone has any responses to this post, please post them on PHYS-L, and
I'll do my best to answer them, here. I won't answer any more questions by
e-mail.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555