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



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.

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