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



On Wed, 16 May 2001 10:19:15 -0500, Jack Uretsky <jlu@HEP.ANL.GOV> wrote:

On Wed, 16 May 2001, David Rutherford wrote:

On Tue, 15 May 2001 23:45:11 -0500, Jack Uretsky <jlu@HEP.ANL.GOV> wrote:

Hi all-
I regard this argument as nonsense, as follows:
On Tue, 15 May 2001, David Rutherford wrote:

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 (*)

The "as long as" says that Div<A> is not a gauge invariant
quantity. Gauge invariance means that an expression does not change
if I add a -grad(/\) term to the vector potential and a d/\/dt term
to the scalar potential. Such an addition
does not change the fields, whic are defined in every gauge, in
terms of the vector and scalar potentials, as
E(vec) = grad(phi) - dA/dt ( A a vector)
and
B(vec) = curl{A}
. It is not permitted to
put any condition on /\ other than it be a scalar function of space
and time. Clearly, div(A) is not a gauge invariant quantity.

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.

Specifying curl(A) is not the same as specifying A. Since the
curl of a gradient is zero, I can add any gradient to A without changing
the value of the curl. But there is an infinity of gradients that will
change the value of div(A). That's part of the principal of gauge
invariance.

Then it's incumbant on you to find a condition under which the value of
div(A) is not changed or you must dump the concept of gauge invariance,
because div(A) is definitely specified. In this three-dimensional, time
invariant case, gauge invariance is only possible if the condition

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

is satisfied. If you don't choose to accept that condition, then you have to
abandon the validity of gauge invariance for this case. Periodisimo.

Regards,
Dave