> There is a E field component along the x axis within the wire, correct. This provides the force which causes the electrons to accelerate. The field is slight -- remember it's just the voltage from one end of the wire to the other (millivolts) divided by its length. Notably, this component does not impart energy to the load -- it corresponds directly to energy dissipated in the wire itself (i.e. its Poynting vector points toward the center of the wire).

Ok so far that is in accordance with how I understand things to work, assuming a 'good conductor' and low impedance for AC or varying current.

> But along the wire's radius, within the wire, there's no net E field.

Because there are no electrons flowing in that direction, and even if there were there would be others canceling that out (this holds on average for thermal noise as well, which would show up as minor current fluctuations), to get an E field at right angles to the wire you'd have to to see a voltage difference between the center of the wire and the outside of the wire, and barring local effects there will be no such thing. Essentially because there is a dead short between those two points (a very short piece of a section of the wire) Right?

> Each electron (and each proton!) contributes a small field (which does extend immediately from the electron-point-source as you say), but the electrons arrange themselves within the metal such that the net field along these axes in the wire is zero. (If they were not arranged so, the nonzero field would push them toward this arrangement. Since the electrons aren't moving along these axes, the field along these axes is zero.) In the intuitive sense -- uniform static distribution of electrons begets a uniform static field among them.

Except for the limit case of a single electron moving in a conductor, that presumably would result in a net (extremely small) field, but for larger numbers of electrons (any practical measurable current) that would all cancel out within the wire perimeter.

> Outside the wire -- assuming the wire has a net charge (positive or negative) -- we do see a radial electric field, because the electrons are not free to move to/from the wire through free space to neutralize this field.

Because the wire perimeter is defined as the area where the conductor ends (which makes me wonder how that transition would work in a material that consists of an outer core that gradually becomes higher impedance).

> This field is much larger,

Right, because it can't neutralize because it is in a dielectricum.

> particularly if the wires are physically near each other

Yes, but that's not a given and any old wire loop has its wires as far away from each other as you want them to be. Coaxial cable is the way it is to get the fields to cancel out so that the amount of spurious radiation from the cable is kept to a minimum (you can also twist the two wires around each other for some of the same effect).

> since it derives from the voltage difference between the wires, rather than across them.

You mean 'wires on both sides of the load'? Somewhere along the line we switched from discussing a single wire in isolation to two wires.

> (And most importantly, the Poynting vector of this field points toward the load.)

So we're now talking about the electrical field between the two wires on either side of the load?

> You may be right that the region outside the wire where energy flows may be small and near the wire

I don't see how it could be any other way but I'm curious as to the results of your calculations.

> but it's definitely outside the wire for the reasons given above

Ok

> and flows between sources and sinks according to the Poynting field, not along any path that's visually represented in a circuit diagram.

Ok, I think I'm finally seeing what you are getting at, you are talking about the net flow of energy from a 'source' (say a battery or a generator) to a 'load' and that energy clearly flows in one direction even if the electrons themselves make a round trip from one side of the source to the other side of the source (well, technically they don't even do that, they may well reverse direction before they get there), and that the field that carries that energy is the field at right angles to the wire rather than the field that you see across the length of the wire.

edit: thinking about this some more: essentially the two wires you are talking about have some of the properties of a charged capacitor, and the electric field is passing between those two 'plates'. That makes me wonder if the field is mostly concentrated between the 'plane' (it can be quite a complex shape) between the two conductors rather than that most of it is in free space radiating outward. After all an electric field normally does the same thing for a charged capacitor, the bulk of the field sits between the plates with only a small fraction moving outside of the envelope of the plates.