1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3.
This would be true (maybe, idk actually) if the battery were just floating in space with no wires attached. However, that's not the real situation here. One of those wires has a voltage 9V higher than the other, and since E = ∇ V, there an electric field between between the two wires.
If a battery is able to maintain charge separation between its external contacts while in a circuit, then its far E field has to look like that of a dipole from the POV of a circuit that is much bigger than the battery. I can't see how it could be otherwise (though I'm always happy to change my mind on seeing a decent argument against my position).
It is true that there are other E fields in the circuit to consider though - these arise due to charge distributions on the wires caused by the battery - that was the whole point of my post. To get the whole field, we add the dipole field due to battery and local fields due to charge distributions. This is simply the principle of superposition (linear Maxwell's eqns blah blah).
I claim (without having actually analysed the setup carefully) that the contributions of the local E fields in the circuit dominate (in most parts of the circuit) over that of the dipole E field of the battery when calculating S at any point, since a 1/r^3 field will die off too quickly for it to be otherwise.
I cannot stress enough that if you correctly integrate the Poynting vectors over a plane dividing the battery from the load, you will see the correct value for the power being transferred from battery to load. If you disagree, you made a mistake in the calculation
I don't think anyone is disagreeing. I'm certainly not. I suspect that you misunderstood the point that I was making - I have tried to clarify it above.
In short, my rough intuitive analysis tells me that in a circuit as described:
1. there is no E field in the wires as they are ideal conductors so no S vector therein.
2. the E field of the battery is largely inconsequential in calculating S (too small in most places as it goes as 1/r^3)
3. the battery's role is to redistribute charges in the wires to produce a surface charge distribution whose local E field dominates in calculating S (but I'm very hazy on how one would figure out the details of that charge distribution - haven't seen a convincing derivation anyware, and I'm too dim to do it)
4. I'm guessing that the local E field on the wires goes roughly as 1/r from the wires (since it probably looks like a long line of charge on most of the wire)
5. Hence, as B goes as 1/r from the wire, S goes as 1/r^2 from the wire, since S = E x B.
6. So if all that is true, then S is largest near the wires, and dies off inverse-squarely away from them, more-or-less. So the pictures showing lots of lovely S in the middle of square circuits are not incorrect but pretty misleading - most of S will be seen near the wires, and the comment that someone made (here or on another site?) that the wires "guide" S are largely (but not completely) true.
If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.
All IMHO, of course.