Consider the x component (along the wire's length) and the y and z components (along the wire's radius) separately.
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.
> 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.
Yet another edit, much more thinking later: 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? After all you'd need something moving energy and electromagnetism can transfer energy whereas an electric field (or even a moving charge) does not? (as in: current does not equal 'work', that all depends on the associated voltage). It also would explain the 'single electron' issue outlined above, that wouldn't apply, you get a magnetic fieldline or you don't and if you do that single photon can do useful work.
I think we're mostly on the same page, yes, sorry I wasn't clear, I've been talking about two wires, linking a power source with a load.
> 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:
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.
> I think we're mostly on the same page, yes, sorry I wasn't clear, I've been talking about two wires, linking a power source with a load.
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 single wire, which doesn't make any sense at all (and it 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 shows that there definitely is 'work' being done inside a wire. So there is a non-zero length Poynting vector pointing (heh) to the inside of the wire! (as a result of the resistance of the wire), and that energy is subtracted from the amount of energy available for the load.
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 (it's circulating 'virtual photons' for a bit of physics weirdness), 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. Its presence gives rise to the 'Hall effect', which shows up as a voltage induced in a conductor placed across the field lines.
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 field of the free electrons in a conductor also doesn't do any work. But when you start moving those electrons using an electric field their magnetic field is doing work in the medium that they are moving through. The electrons themselves only move a fraction (on the order of a millimeter per second) 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.
[This also neatly explains why electrical signals propagate at the speed of light: the electrons themselves may move slowly but their magnetic fields propagate at the speed of light, and if you think about it a bit longer it also explains why voltage and current in an AC circuit can be different than the real power drawn by the load because the load itself can temporarily store energy (power factor!). This can even cause zero apparent power while real work is done in the load. This shows up as the angle of phase difference between current and voltage, and is why the electrical company really doesn't like you if you use large inductive loads because they have a very low power factor, all the way down to zero for 'perfect' inductors.]
In a pure DC setup the interactions between the moving electrons and the stationary material 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. The current in such a setup would be mostly limited by the internal resistance of the source (the wire loop is essentially a short). 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.
Adding a load is then a slight modification of that circuit, a localized increase in resistance relative to the resistance of the wires, but work is done in all parts of the circuit (including the source!) proportional to the resistance (and hence the voltage drop) across that fraction of the circuit.
In an AC setup some of the radiated energy would 'escape' from the wire because the changing magnetic fields radiated by the wire would induce currents elsewhere, and if you engineer your AC setup carefully (resonance) to maximize such loss you have an antenna. Conversely, that loop 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), such as in an AM radio (which uses a coil of wire across a ferrite core, essentially an electromagnet).
If you carefully engineer this to transfer energy using magnetic fields between two such circuits with minimal losses you end up with a closely coupled set of two loops of wire aka a transformer.
In each of these cases the part that you try to engineer away shows up as loss, imperfect transfer in a transformer warms up the transformer (eddy currents in the laminations for instance) and imperfect transfer for an antenna system shows up as heat due to reflected power (to the point where a very simple way to blow up a radio transmitter of any serious power is to disconnect the antenna).
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 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 mostly 'inside' the wire (say: a standard powercord where the conductors are close to each other, but they definitely do not have to be, in fact the wires could be of unequal length all the way down to zero). All the Poynting vector does is to show how strong the cross product of electrical and magnetic fields are at any given point of a system 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 physical lay-out of the system, the interior resistance of the source and the interior resistance of the wire and if you don't take these 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, radiated EM field), each of which would require their own miniature Poynting vector to account for all of the imperfections.
So by switching from one wire to two wires and a load the picture became needlessly complicated, and technically incorrect. You don't actually need a separate load to demonstrate any 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 does not depend on in which direction the current flows and doesn't even care about the direction of the field itself, it will simply excite the atoms of the material that the carrier is made out of by virtue of moving the free electrons around and it will do so proportionally to the resistance of the carrier resulting in heating the material up (which in turn shows up as excited atoms with electrons in a higher oribit, and gives rise to new photons when they fall back, 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 moving 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. If the energy really did flow 'mostly outside of the wire' then the wire wouldn't heat up, it heats up because of the fields of all of those electrons interacting with the fields of the captive electrons in the material just like a mirror absorbs light and then spits out brand new photons (the basis for the photo electric effect, think of a solar panel as a 'one way' mirror that turns those absorbed photons into free electrons which are then captured by a diode resulting in a potential difference across that diode). It also explains superconductivity: in a superconductor the current flows unimpeded because the number of interactions between the fields of the electrons moving through the material and the fields of the atoms of the material itself is extremely low, so there is no 'work' done in the conductor itself and hence no potential across the superconductor. If you use a superconductor to short a power source then all of the energy will go into the magnetic field, rather than that it gets converted into heat (which creates all kinds of interesting complications, the fields are so strong that if the electromagnet isn't constructed very carefully it will tear itself apart).
Interesting discussion. Main takeaway: precision matters.
> 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.
Too late to edit 'A' should of course have been 'E'. (E -> electrical field, B -> magnetic field).
I think a big part of the confusion stems from separating the wires from the load: they are all the same thing, wires are load too and there is no 'privileged' reference frame for any particular chunk of wire (or load, or internal resistance in the battery) that deserves special consideration or that needs to be paired with something else to make it work, as long as the circuit is whole current will flow and the moving electromagnetic field is sufficient to explain all of the observed phenomena. The only part where the electric field figures in is as the accelerating force for the free electrons in the circuit, the higher the voltage the more of those electrons that will partake in the movement (which is effectively how you can derive Ohms law).
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.