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

[David Rutherford <drutherford@SOFTCOM.NET>]

On Wed, 16 May 2001 13:39:32 -0400, John S. Denker <jsd@MONMOUTH.COM> wrote:

> I wrote:
> > > 2) It is not generally true that v.E is related to div(A).
>
> Then at 08:35 AM 5/16/01 -0700, David Rutherford wrote:
>
> > In my original post, I showed that it definitely is. Here is my derivation
> >
> > -div(A) = v.(-grad(phi))/c^2 = v.E/c^2
> >
> > I've used the same reasoning that Feynman used to derive curl(A)=vxE/c^2,
> > for constant v and E=-grad(phi) (d(A)/dt=0). If you accept Feynman's
> > derivation you _must_ accept mine.
>
> Well, actually, Mr. Rutherford has not followed Feynman's reasoning.  He
> has only pushed some symbols around in patterns superficially analogous to
> Feynman's calculation.

Here is the calculation for -div(A), in detail, since you seem to have
trouble following the simple vector math in my original post. Taking the
negative divergence of A, you get

-div(A) = -(d(A_x)/dx + d(A_y)/dy + d(A_z)/dz)       (*)

The vector potential A, for an electron in uniform motion with velocity v,
is

A = v(phi)/c^2        (**)

where phi is the scalar electric potential. Using (**), you can write (*) as

-div(A) = -(d(v_x(phi))/dx + d(v_y(phi))/dy + d(v_z(phi))/dz)/c^2   (***)

and since v is constant, (***) becomes

-div(A) = (v_x*(-d(phi)/dx) + v_y*(-d(phi)/dy) + v_z*(-d(phi)/dz))/c^2
(****)

Since E=-grad(phi) for constant v, you can write (****) as

-div(A) = (v_x*E_x + v_y*E_y + v_z*E_z)/c^2

or

-div(A) = v.E/c^2

> Feymnan invoked the Maxwell equations (among others).  The Maxwell
> equations make certain statements about the behavior of curl(A).  There are
> no corresponding equations governing the behavior of div(A).

That naughty old Maxwell! Did he forget to put that in? Well don't worry,
I've fixed it for you.

> You can make up additional equations if you want, but there is no
> theoretical or experimental justification for doing so.

You don't need any justification. All you need is insight into the way
nature really operates. There was no theoretical justification for -dE/dt,
either, but Maxwell put it in anyway. What a joker!

> Arguing that div(A) must behave analogously to curl(A) is like building a
> wooden mockup of a radio, and expecting it to behave like the original.

Conventional EM theory is the "wooden mockup". My theory is the real thing.
Which one shows that mass is totally electromagnetic in origin? No matter
how close you hold your ear to your wooden theory, you'll never hear the
song of truth.

>  http://www.physics.brocku.ca/etc/cargo_cult_science.html
>
>
> > >   3b) It turns out that div(A) is not observable.
> >
> > I would be if anybody chose to try to observe it. Which they haven't.
>
> And how are they supposed to observe it?

>  -- Using the Lorentz force law?   I don't think so!

Nope. Use the force density equations in my theory.

http://www.softcom.net/users/der555/newtransform.pdf

Regards,
Dave