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



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. 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? That makes absolutely no sense. You must either
accept that _both_ are specified or you must accept that _neither_ is
specified. The fact that the combination of the two solves the 4/3 problem,
strongly suggests that both are specified. And if both are specified, then
you need to find a gauge transformation that leaves _both_ unchanged (along
with E), or you are obliged to dump the validity of the concept of gauge
invariance. Specified quantities rule. It's as simple as that. If that fact
escapes you, then you are unreachable.

Regards,
Dave

up to the
transformations already described.

In general, the condition is that /\ need only satisfy
D'Alembertian^2(/\)=0.
That statement is incorrect. Any scalar /\ can do the trick,
as can easily be verified from the Maxwell equations.

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 (*).

All this amounts to a special choice of gauge.

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


Regards,
Jack



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