Poynting Vector Direction Confusion

[BruceW]

yeah. It is nice to actually show mathematically that the energy flows in the space around the the conductor, and not through the conductor itself.

For the completely general case, we can give a 'logical' or 'reasonable' answer. The charge carriers are moving with some slow drift velocity, and since it is a conductor, this means the electric field inside the conductor must be small and parallel to the direction of the wire. Therefore, the EM energy cannot be carried inside the wire, since the Poynting vector inside the wire cannot point in the direction of the wire. And, the energy must get to different parts of the wire somehow, therefore we can conclude that the energy travels outside the wire.

But it is nice to get a more analytical/specific answer. I think we can agree that for a wire of vanishing thickness, it is possible to calculate the Poynting vector just outside the wire. (i.e. in the limit of being very close to the wire). Since in this case, the electric and magnetic fields just outside the wire will 'look' the same as if we only had one infinitely long wire. I am fairly sure this will be true for any shape of circuit (as long as the wire thickness is small). So we can at least calculate the Poynting vector analytically in the space 'just outside' the wire.

 

[vanhees71]

Exactly! One should, however warn, about the approximation of infinitesimally thin wires, which makes a lot of trouble. Famously even H. Hertz didn't get a meaningful answer when he tried to evaluate the traveling of electromagnetic waves along a single wire, because he couldn't find the correct boundary conditions in this limit. It was solved later by Sommerfeld for both the single wire and the Lecher (double wire) system.

The coax cable is pretty simple. That's why I used this as an example in my FAQ article. You also find it in Sommerfeld's marvelous textbook (Vol. III of "Lectures on Theoretical Physics"), which is available in English (I've only the German edition so that I can't provide a scan of this section) and for sure also in many other textbooks. I'm not aware of any calculation which treats a simple finite closed circuit. So maybe that's not so easy. I'll check for a ring wire (which has to be treated as a torus of finite thickness) as soon as I find some free time, but I guess that's tough. The reason, why infinitely long wires are easy is that you have mostly a plane (2D) problem for the potential equation at hand, and this can be very elegantly solved by using complex functions and conformal-mapping techniques.

Another interesting point is that all of what we have discussed so far is strictly speaking a not fully consistent solution of Maxwell's equations + the flow of the electrons in the wire. It works of course with almost perfect precision, because the drift velocities of the electrons are in the order of mm/s, which is as good as if the electrons are at rest since you have to compare to the speed of light, but a fully relativistic calculation is interesting for principle reasons. I've done this in my FAQ article. What comes out (for the coax or single-wire cable) is that in order to have a current going strictly in direction of the wire, you must take into account the magnetic part of the Lorentz force on an electron moving in the magnetic field caused by all other electrons. So there must build up a radial electric-field component so that the corresponding electric part of the Lorentz force on the electron compensates the magnetic force (this is analogous to the consideration of the Hall effect for a current moving in an external magnetic field, but here it's for self-consistency of the equations of motion for the "electron fluid" and the Maxwell equations for the field). Of course this effect is unmeasurable small for usual house-hold currents and thus can be neglected for any practical purposes, but it's an interesting exercise in relativistic electromagnetics in the medium :-). You find this in my FAQ article (in German):

http://theory.gsi.de/~vanhees/faq/coax/node7.html

 

[rude man]

OK, let's take 1 m of coax, terminated at the far end by 500 ohms, and apply 10V dc to the near end. Sorry, pure phycisists, Système Internationale (SI).

Assume inner conductor #20 AWG Cu for which
A ~ 0.52e-6 m^2.
σ ~ 6e7 S/m
C ~ 100pF/m = 100 pF
ε_r of dielectric (polyethylene) = 2.3
distance from inner conductor to braid ~ 4e-4m

So i = 10/500 = 0.02A,
j = i/A ~ 0.02/0.52e-6 ~ 3.8e4 A/m^2 in inner conductor

Thus E in inner conductor = j/σ ~ 6.3e-4 V/m pointing along current direction

But E outside the inner conductor, i.e. in the dielectric between the inner and outer conductors, is

E ~ (V/4e-4m)/2.3 ~ 1.1e4 V/m pointing radially outward.

So we see that the tangential E outside the inner conductor is practically negligible, that E points almost exactly radially, and that in consequence the Poynting vector points almost exactly in the direction of current travel.

QED

 

[BruceW]

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.

 

[vanhees71]

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. In my FAQ article I made a shortcircuit at the far end instead of a finite resistor. The qualitative result is the same.

 

[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.