Re: Capacitor Charged, Right term

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Regarding Joel's question:
> ...
> I intuitively understand what you are saying, regarding the Poisson 
> equation, based on you're wonderful example above that I snipped.  Can you 
> quote any theorems or direct me to any references? ...

How about the Divergence Theorem (applied to a generic orientable curved
manifold)?  The lhs of the Poisson equation is a divergence so the volume
integral of both sides of the equation over the entire closed spatial
manifold (i.e. compact manifold without a boundary) yields the net charge
in the space for the rhs, but the volume integral of the lhs is the net
flux leaving the boundary surface for the region of integration.  If the
space is closed (compact without a boundary) then there is no boundary
surface for the integration region since the entire space is included in
the integration region.  This means the lhs integrates to zero.  Since the
rhs is the net charge on the space this clearly says that the net charge on
the space is zero.  Thus, if the space is *not* charge-neutral so that the
rhs did not integrate to zero we would have a contradiction for any
solution of the Poisson equation whose lhs *must* integrate to zero by the
Divergence Theorem.  Thus, the Poisson equation can have no solution for
such a closed manifold.                                                 QED

Another fancier proof of this is just realizing that the problem is a
special case of Hodge's Theorem applied to 0-forms on a 3-d closed
Riemannian manifold.  For references try almost any textbook on
differential geometry that discusses the integration of differential forms.
The book I happen to have handy in my office is Theodore Frankel's book:
_The_Geometry_of_Physics_an_Introduction_, Cambridge University Press,
1997.  For a particular reference try looking at the discussion following
Hodge's Theorem (14.28) on page 371 in the above book.

David Bowman
dbowman@gtc.georgetown.ky.us