Re: Is charge conserved between frames?

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 11/05/04 

Date: 4 Nov 2004 18:58:01 -0800

dubious@radioactivex.lebesque-al.net (Bilge) wrote in message news:<slrncoim9m.nlv.dubious@radioactivex.lebesque-al.net>...

> Of course not. If it did, we wouldn't need a six-component tensor
> containing E and B to define the field strength tensor. The only
> four-vector in E&M is A_u.

Agreed, we may take a vector A>, and generate components,

  A>.e>_u = A_u (scalar product),

(aka project A> onto basis vectors e>_u).

Let's relate two charges "a" and "b",
we have field F_uv from B>, I'll describe in an new notation,

   f_u on [a] = a*F_uv of [b] *U^v [4-vector of "a"]

  force on "a" = a*[EM field of "b"]*[4-velocity of "a"]

and we agree that is a tensor.

Is this a better description,

   f_u[a] = a*F_uv[b] * U^v[a]

where I use brackets [ ] to refer to the charge
that the tensor is generated from, i.e. in any FoR
what charge is being used to create the components
of the tensor parts in the above equation.

Sorry this is going to get complicated, let's
look at our naked E_u = F_uv U^v, with FoR's
defined above.

  a*E_u[a] = a*F_uv[b] * U^v[a]

  E_u[a] = F_uv[b] * U^v[a]

Outer muliply by U_w[a] and get,

  E_u[a]*U_w[a] = F_uv[b] * U^v[a]*U_w[a]

and resolves to, (using Kronecker Delta),

   E_u*U_w [a] = F_uw [b]

Now watch this, I'll transpose [a] and [b],

   E_u*U_w [b] = F_uw [a]

makes no difference, just notational.

I think Tom Roberts, (and Einstein) has this
problem...allow me to outer multiply with the
metric "g^uv" and watch carefully,

  g^uv[?]*E_u*U_w [b] = g^uv[?]*F_uw [a]

See that? Is g^uv to be evaluated at [a] or [b]?

My research indicates guv is to be evaluated over
a finite length, (in accord with ISU) but that's
a digression....

Sal, when you mix GR and EM get ready for unified
field theory.

Regards
Ken S. Tucker

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