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

[David Rutherford <drutherford@SOFTCOM.NET>]

On Wed, 16 May 2001 22:23:37 -0400, John S. Denker <jsd@MONMOUTH.COM> wrote:

> At 05:49 PM 5/16/01 -0700, David Rutherford wrote:
>
> > Here is the calculation for -div(A), in detail, since you seem to have
> > trouble following the simple vector math in my original post.
>
> The math would be easier to follow if it were correct.
>
> > ....
> > Since E=-grad(phi) for constant v,
>
> No, E is !not! equal to -grad(phi) for constant v (excluding the trivial
> case of v=0).
>
> There is an important contribution from the time derivative
> of A.
>
> See the calculations at the beginning of Feynman section II.26-2
> leading to equation 26.6.

Look at "Lectures on Physics", vol. 2, pg. 26-3, where Feynman derives
B=vxE/c^2 for a charge moving with constant speed v in the x-direction,

"For the z-component [of B],

B_z = d(A_y)/dx - d(A_x)/dy

Since A_y is zero we have just one derivative to get. Notice, however, that
A_x is just v(phi) [he has set c=1], and d/dy of v(phi) is just -v*E_y. So

B_z = v*E_y

Similarly,

B_y = d(A_x)/dz - d(A_z)/dx = +v*d(phi)/dz

and

B_y = -v*E_z

Finally, B_x is zero, since A_y and A_z are both zero. We can write the
magnetic field simply as

B = vxE    [c=1]"

His derivation of B=vxE/c^2, above, is analogous to my derivation of
-div(A)=v.E/c^2, in my last post to you. If you want to argue, argue with
Feynman. If my derivation is wrong, so is his.

> Everything that follows from this false equation is false.
>
>
> At 08:35 AM 5/16/01 -0700, David Rutherford wrote:
> > > > 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.
>
> But then at 05:49 PM 5/16/01 -0700, David Rutherford wrote:
>
> > That naughty old Maxwell! Did he forget to put that in? Well don't worry,
> > I've fixed it for you.
>
> You can't have it both ways.  Either you can argue that your conclusion is
> a consequence of conventional physics in the morning,

I said that it used classical quantities, which it does.

> or you can argue that
> the conventional physics is naughty in the evening, but there is no way to
> make sense of both arguments.
>
> In fact both arguments are nonsense.  Mr. Rutherford's attempt to follow
> Feynman's reasoning violates the rules of algebra as well as the rules of
> physics,

If my reasoning violates the rules of algebra and physics, then so does
Feynman's.

> and the result is not an improvement over the usual Maxwell
> equations.

I guess we'll just have to wait and see, won't we?

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555