Re: AC electricity

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Ludwik's and leigh's comments:

> > Suppose the load is a high R wire perpendicular to
> > the energy-guiding wires (see below). Near that
> > load the Poyting vector is directed everywhere into
> > the load. The energy of the field "enters the metal"
> > and is dissipated in it. Is this a correct explanation?
>
> That is a "correct" explanation which reifies the "flow
> of energy", renaming it "the Poynting flux". It is, as I
> pointed out, a physically useful way to consider the
> problem, but don't get carried away. In a simple resistor
> carrying a current the Poynting flux which accounts for
> the Joule heating effect comes in radially from infinity.
> Sure, the numbers work out fine, but do you believe that
> God in her infinite omniscience, was also sufficiently
> prescient that she started this energy flowing coherently
> inward toward the resistor at some time in the distant
> past - from all sides!?

Leigh is not quite correct when he says: "In a simple resistor carrying a
current the Poynting flux which accounts for the Joule heating effect
comes in radially from infinity.". It is true that the Poynting flux
converges from the outside (but it's doubtful that the convergence is
typically approximately radial) into the resistor.   But that flux is
certainly *not* "from infinity" (unless maybe you want to consider the
Joule heating induced by the penetration of the Cosmic Microwave
Background).  The Poynting flux tends to be loosely guided by the wires
carrying the currents around the circuit.  If the currents are paired
together into cables whose algebraic sum of the currents carried by the
conductors add to zero and is shielded (such as the case for coaxial
cables) then the Poynting flux is completely internally confined by the
shield and tends to propagate along the inside the cable (mostly) in the
insulation region between the conductors.

In general the "flow" of the Poynting flux is described by Poynting's
Theorem which says that locally the convergence of the Poynting flux,
being the negative of the divergence (- div(S)) of the flux is the local
sum of the rate at which the EM field does work on the charge carriers
(Joule heating the resistor for instance) per unit volume *plus* the
partial time derivative of the local EM energy density in that region.
The equation is:

- div(S) = E(dot)j + du/dt

where div(...) is the differential divergence, E is the electric field,
j is the current density of the charge sources present, u is the EM
energy density, and d(...)/dt is a partial derivative.  One can think of
the E(dot)j term as being a local creation/destruction term that pumps
energy into or out of the EM field from the work done on or by the
moving charged sources, and the other terms are usual continuity equation
terms describing how that EM energy is either accounted for by changes in
concentration (i.e. du/dt) or by flow across the boundary of the
region of interest (i.e. div(S)).

It is true that inside the resistor div(S) is negative (where the
Poynting flux converges) but that convergence does not extend all the way
in from infinity.  Rather, I would expect it to tend to curve around
following the general flow of the current in the wires connected to the
resistor where the direction of the general trend would be determined by
the global electric field configuration.  If the high potential lead of
the resistor was the only high potential object in the vicinity of the
resistor and the resistor's surroundings tended to be at a potential
similar to that of the low potential lead, then I would expect typically
more Poynting flux to enter the resistor preferentially from the
direction of its high potential side than the general direction of its
low potential side (maybe some flux might even leave near the low
potential lead where the current exits), but there would probably also be
significant Poynting flux coming into the resistor from its sides as well.
But if the resistor's surroundings tended to be mostly near the
potential of its high potential lead, then I would expect more of the
flux to enter the resistor from the direction of the low potential end.

David Bowman
David_Bowman@georgetowncollege.edu