Re: [Phys-l] Current as Vector

[Brian Whatcott <betwys1@sbcglobal.net>]

At 09:32 AM 3/23/2006, you wrote:
> Brian writes:
>
> >   Let us suppose I have a cube of graphite a kilometer on a side.
> ///
> >And suppose I consider a situation near the middle of this mass.
> >Suppose I have a driving voltage field  ///
>  [or]   "electric field."
>
> >Should I say that the current in the middle of this mass is a
> >scalar, now I don't have the invisible directional constraints
> >of conductive wires in air that allow me  ..er.. you to talk as though
> >directionality is a given?

> ...the situation is entirely standard.  Unless the
> graphite is in crystalline form it has an isotropic conductivity and
> the current density (a vector) at any point is proportional to the
> local electric field (another vector.)  If it is in crystalline form
> then we need to use the conductivity tensor instead.
>
> If you really still want to know some "current," then you'll first
> have to specify an open surface and distinguish its two sides so that
> I may calculate the value of the current (a signed scalar) across
> that surface from the flux of the current density.
>
> --
> John "Slo" Mallinckrodt


Ah: mind experiments are SUCH fun, and so appealing to
physics people.
So here I am sitting on an imaginary straight transparent pipe,
which has the splendid property that it will allow just one
 charge carrier to enter at a time, and I can image each
passing charge. It will come as no surprise
that in a conductive material like graphite, these carriers move
rather slowly. There again, they are quite numerous.

I am able to orient the pipe so that it is in the direction of the
local current deep in the block.
   Any other orientation would reduce the current
 that I measure, naturally.   And I measure the current by counting
the charge carriers passing me in unit time.

If I were to attempt an orthodox measure of current using an
area represented by the pipe's entrance, I would be in trouble.
This area is close to infinitesimal, so I would over-estimate a huge
current I imagine.



Brian Whatcott    Altus OK    Eureka!