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.
> isn't it the electromagnetic field then (that consists of photons) that moves the energy from source to sink because photons are capable of moving energy rather than the electric field between the wires?
Yes, that's my understanding also; I focused on the electric field in my last post because that's what differs inside and outside a conductor; the magnetic field exists and has the same orientation both in and outside a conductor carrying current. But it's the product of the two which defines the Poynting vector, which indicates the flow of energy in the EM field.
What I've calculated so far (for the case of two infinite parallel wires) does indicate that the Poynting vector is concentrated around the wires -- it drops off approximately with the 4th power of the distance from either wire (though there's a decent saddle between them). I plan still to calculate some integrals around the wires because I'm curious actual numbers. For reference the expression I've derived is:
|S| = Pr² / (ln(2r/a) × ((y² - z² - r²)² + 4y²z²))
where:
S = Poynting vector
P = power delivered
2r = distance between wires
a = diameter of wire (this only creeps in when relating voltage between the wires to charge on the wires)
The Poynting field points uniformly from the source to the sink, in the +x direction (same axis as the wires). The y axis points from one wire to the other.
Ah that makes a big difference. Note how all of this started:
> The energy doesn't flow inside of a wire, it flows around the wire.
As though all of the energy flows outside of a wire, which doesn't make any sense at all (and still doesn't!), if it did then the wire itself would never heat up, which given the fact that there is a voltage across the wire (low, but still, it is there) and current flowing through it does not make any sense at all.
And the 'B' component of the fields created by a current carrying wire is in fact the magnetic field (and a magnetic field as I'm sure you are aware doesn't need a 'return wire' as such, as any magnet demonstrates, and there too the field is much stronger at the poles of the magnet), these fieldlines would be best visualized as concentric circles around the wire ( https://www.researchgate.net/figure/The-magnetic-field-lines... ) with the spacing dropping off as you get further away from the wire ( http://physics.bu.edu/~duffy/semester2/c15_inside.html ), and it definitely would be present inside the (single!) wire.
The electrical field by itself doesn't do any work, but has the potential to do so and this potential diminishes the further along the path from 'source' to 'sink' and on the return leg through the wire. The magnetic of the free electrons in a conductor also doesn't do any work. But when you start moving those electrons their magnetic field is doing work. The electrons themselves only move a fraction while they move the magnetic field lines 'along' with them in the direction of their travel and the strength of the electrical field between the two conductors is a measure of the potential energy available to do work. In a pure DC setup this shows up as the voltage across the wires relative to a reference ground (say, a sense wire on the - pole of the battery in the simplest possible circuit, a power source coupled to a single loop of wire without any visible load, the wire is the load). As you move around the circuit you'll see a nice and predictable drop in voltage as the potential of the fields to do work diminishes until it reaches zero at the point of reference. You'd see the same drop in voltage if you picked the + as the reference, only the magnitude would be reversed, but the product of voltage and current across any given section of wire of a particular length would be constant.
In an AC setup some of the radiated energy would 'escape' from the wire and if you engineer your AC setup carefully to maximize such loss you have an antenna. Conversely, that antenna would be able to capture an external magnetic field and turn it into electrical energy again (technically: potential energy, the circuit would have to be closed somehow for current to actually flow). If you carefully engineer this to transfer energy between two points you end up with a closely coupled set of two loops of wire aka a transformer.
I think what is happening here is that OP mistook the Poynting vector (a mathematical construct derived from the cross product of electrical and magnetic fields) for 'reality' in the same way that the map is not the territory, it is a very useful tool to show the flow of energy in a system, essentially it allows you to decompose the 'A' and 'B' fields of a system of wires, for instance a current carrying loop and a load.
But it's wrong to say that the Poynting vector moves the energy outside (or even mostly outside) of the wire, even a single one because you are now talking about the whole system rather than just a single wire and in that system the plane described by the loop of wire is still 'inside' the wire. All the Poynting vector does is to show how strong the cross product of electrical and magnetic fields are at any given point and what the magnitude (power delivered) and direction (where the source is and where the sink is). But that's a simplified view that does not take into account the interior resistance of the source and the interior resistance of the wire and if you don't take into account you end up with a system that is not realistic and which leaves a lot of real world phenomenon unexplained (conductor losses, source losses).
So by switching from one wire to two wires and a load the picture became needlessly complicated, you don't actually need a separate load to demonstrate all of this (the resistance of the wire is a load!). And then all of the rest of the bits fall into place much easier, after all the work done by the magnetic field doesn't care in which direction the current flows and doesn't even care about the direction of the field, it will simply excite the atoms of the material that the carrier is made out of and it will do so proportionally to the resistance of the carrier resulting in heating the material up (which in turn shows up as new photons, IR ones...). This is is self evident: if it didn't you would violate the conservation of energy rules, you can't excite the atoms in the material without energy transfer of some sort, and in this case the energy transfer at the quantum level is mediated by the EM fields of the free electrons interacting ('bumping into', if you want to use a simplified view) with the EM fields of the electrons in the material itself. The more such free electrons (the larger the current) the more such interactions and the higher the voltage the taller the stack of such interactions, hence P = I * V (momentary) and over time that translates into energy delivered.
Too late to edit 'A' should of course have been 'E'. (E -> electrical field, B -> magnetic field).
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).
But along the wire's radius, within the wire, there's no net E field. 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.
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. And if there are two wires of opposite charge (as in the circuit described), the fields sum constructively (as one wire's field is radially outward, the other's radially inward). This field is much larger, particularly if the wires are physically near each other, since it derives from the voltage difference between the wires, rather than across them. (And most importantly, the Poynting vector of this field points toward the load.)
You may be right that the region outside the wire where energy flows may be small and near the wire (working through some math now) but it's definitely outside the wire for the reasons given above, and flows between sources and sinks according to the Poynting field, not along any path that's visually represented in a circuit diagram.