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

[David Rutherford <drutherford@SOFTCOM.NET>]

On Wed, 16 May 2001 06:10:21 -0400, John S. Denker <jsd@MONMOUTH.COM> wrote:

> At 12:35 AM 5/16/01 -0700, David Rutherford wrote:
>
> > "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 specified?"
>
> 1) Let's tone down the rhetoric.  This is not a cross-examination.
>
> 2) It is not generally true that v.E is related to div(A).

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.

> 3) Let's focus on A, and use the standard textbook definition thereof.  And
> let's start by asking which things are _observable_ (which is not quite the
> same as _specified_;  see item (4)).
>
>   3a) It turns out that curl(A) is observable.
>   Specifically:  curl(A) = B, and B is observable.
>
>   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.

> 4) Gauge invariance is a freedom.

It's only free if the transformation leaves curl(A), -grad(phi), and div(A)
unchanged.

>   4a) If you want, you can _specify_ div(A), by exercise of gauge freedom.
>   That's your freedom.

There is _no_ freedom to choose div(A), it's as specified, by v and E, as B
is.

>   4b) If I want, I can specify div(A) differently,  or not at all.
>   That's my freedom.

No it isn't, not any more that you have the freedom to specify curl(A)
differently.

Regards,
Dave