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On Wed, 16 May 2001, David Rutherford wrote: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>
is
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))
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.