Poynting Vector Direction Confusion

[rude man]

I see. That's because the resistance of the wire is negligible compared to the resistor at the end. The original question was about the power loss in a wire due only to the Ohmic resistance of the wire. That's why we've been talking about the power going into the wire.

As I said on several occasions, no one is disputing that the Poynting vector points into the wire's axis from its surface. And that's why I said we're probably just confusing the OP since all he/she was interested in was why the Poynting vector pointed towards the axis rather than from the axis to the surface.

I just contested van's assertion that the Poynting vector is the same outside as in. And I have shown that it isn't, even for the case of a dc current in the (inner) wire. It points in the direction of the dc current. Which is the direction of energy travel.

 

[rude man]

Anyway, after all we agree that the fields in the free space between the inner and the outer conductor are responsible for the energy transport from the source ("battery") at the one end to the other.

Exactly. And that's the way the Poynting vector points. From the battery to the load. Not radially.

 

[rude man]

If you short the far end instead of applying the resistor the numbers wold be less dramatic: for the same 10V battery, i ~ 100A, j ~ 1.9e8, inner E ~ 3.2V/m, outer E same as before = 1.1e4 (near the battery). The outer/inner ratio is still 34000:1.

 

[vanhees71]

Exactly. And that's the way the Poynting vector points. From the battery to the load. Not radially.

Ok, I see it was a bad idea to take as an example the single wire, where the Pointing vector is indeed radial everywhere :-(.

 

[rude man]

Ok, I see it was a bad idea to take as an example the single wire, where the Pointing vector is indeed radial everywhere :-(.

Only in a nonexistent, isolated wire. Any single wire is still part of a circuit, where the Poynting vector points mostly in the direction of current.