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

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.


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).

Gauge invariance means that div(A) is unspecified,

This whole discussion seems to come down to whether div(A) is specified or
not. How can you admit that vxE/c^2, which is the same as curl(A), is
specified, and then turn around and deny that v.E/c^2, which is the same as
-div(A), is unspecified?


--
Franz Kafka's novels and novella's are so Kafkaesque that one has to
wonder at the enormity of coincidence required to have produced a writer
named Kafka to write them.
Greg Nagan from "The Metamorphosis" in
<The 5-MINUTE ILIAD and Other Classics>