I Energy flux direction in a conducting wire?

[fluidistic]

On the Internet, I have read that the energy doesn't flow in the wire, for example in a very simple electric circuit made of a battery and a closed loop. When one computes the Poynting vector $\vec S \propto \vec E \times \vec B$, one gets that its direction is towards the center of the wire. The closer to the center of the wire, the smaller the magnitude of $\vec S$, and it vanishes right at the center. So far so good.

Now, from a thermodynamics point of view, there exist a relation between the internal energy $U$ and the electrochemical potential $\overline{\mu}$. This relation implies that the internal energy flux $\vec J_Q=\overline{\mu}\vec J$ where $\vec J$ is the current density (I am ignoring thermoelectric effects for simplicity here). However this means that the energy flux's direction is along the wire, not perpendicular to it, i.e. the direction is perpendicular to that of $\vec S$.

I am thus left confused. What is the direction of the energy flux created by the battery? Why do I get 2 different directions? Are these different energies? That's very confusing. I can derive Joule effect starting from any of the 2 energies mentioned above... so... shouldn't they be the same? If so, why do they have a different direction? What is going on?

 

[Baluncore]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

 

[fluidistic]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

I must say I am a bit confused about the setup. Hmm, magnetic field created by surface currents? I thought the problem assumed a constant and homogeneous $\vec J$ inside the wire. The magnetic field outside the wire is due to the current density of the whole wire, shouldn't it be that way, rather than surface currents?

Also, I do not really see why we need the 2 wires. I can just consider a battery attached to a loop of a single material, right? I mean, I understand that the Poynting vector exists outside the wire, and that it does penetrate the wire too. What I don't understand is why isn't the internal energy flux inside the wire pointing in the same direction than the Poynting vector.

 

[bob012345]

It cannot be understood by considering only one wire. The analysis must consider the two wires that make up the complete circuit.

The electric field is guided by the two conductors from the generator to the load. Between those two conductors, the electric field is proportional to voltage, while the magnetic field is the sum of the fields generated by the currents in the two conductors, (or rather the currents are proportional to the magnetic field being guided by the surface currents on the wires). The cross product of the E and M fields is then the Poynting vector, the energy directed from the generator to the load.

The voltages and currents measured on conductors, are proxies for the electric and magnetic fields that direct the energy through the space between the wires.

Are you referring to a sort of wave guide?

 

[Baluncore]

What I don't understand is why isn't the internal energy flux inside the wire pointing in the same direction than the Poynting vector.

The Poynting vector component into the wire represents the resistive I²R losses in the wire.

The surface of a conductor makes a very good reflector that keeps most of the energy outside the conductor. Energy that enters the wire will be lost as heat. Better conductivity makes a better mirror surface.

Are you referring to a sort of wave guide?

That is one way to consider the model.
The magnetic fields of the two conductors sum between the wires, but cancel away from the wires, so the circuit is really a two wire transmission line.

 

[bob012345]

The Poynting vector component into the wire represents the resistive I2R losses in the wire.

The surface of a conductor makes a very good reflector that keeps most of the energy outside the conductor. Energy that enters the wire will be lost as heat. Better conductivity makes a better mirror surface.That is one way to consider the model.
The magnetic fields of the two conductors sum between the wires, but cancel away from the wires, so the circuit is really a two wire transmission line.

Are arguing the waveguide is along each wire as an inherent property of the wire but you need to complete the circuit or are you arguing the fields go between two wires?

Also, what happens if you use wires that are shielded so there can be no field outside them?

 

[Dale]

Now, from a thermodynamics point of view, there exist a relation between the internal energy U and the electrochemical potential μ―. This relation implies that the internal energy flux J→Q=μ―J→ where J→ is the current density (I am ignoring thermoelectric effects for simplicity here). However this means that the energy flux's direction is along the wire, not perpendicular to it, i.e. the direction is perpendicular to that of S→.

Do you have a source for this? I am skeptical of it.

As you say the electromagnetic energy flux with the Poynting vector is clear and is directed radially inward inside the wire and roughly parallel to the wire outside the wire. This energy flux is derived directly from Maxwell’s equations, so it applies any time Maxwell’s equations apply.

I am not sure where this other flux comes from nor what assumptions are made in its derivation.

 

[Dale]

Are you referring to a sort of wave guide?

This is how a simple DC circuit behaves. No need to consider waves.

 

[bob012345]

This is how a simple DC circuit behaves. No need to consider waves.

Are these static fields then?

 

[Baluncore]

Are arguing the waveguide is along each wire as an inherent property of the wire but you need to complete the circuit or are you arguing the fields go between two wires?

The waveguide (or transmission line) is between the two wires carrying the equal and opposite (surface) currents.

The fields that propagate the energy are mostly through the insulation and space between the conductors. Only for DC or very low frequencies does the skin effect allow the time needed for the current and magnetic field to diffuse into, and flow deep inside the conductors.

Also, what happens if you use wires that are shielded so there can be no field outside them?

The fields remain inside the shield, between the conductor and the shield which completes the circuit.

 

[fluidistic]

Do you have a source for this? I am skeptical of it.

As you say the electromagnetic energy flux with the Poynting vector is clear and is directed radially inward inside the wire and roughly parallel to the wire outside the wire. This energy flux is derived directly from Maxwell’s equations, so it applies any time Maxwell’s equations apply.

I am not sure where this other flux comes from nor what assumptions are made in its derivation.

Yes I do have a reference. Non equilibrium thermodynamics by de Groot and Mazur. Essentially it's just the usual expression of dU in terms of mu dn that becomes a heat flux in terms of a particle flux, which is nothing but the usual current density.

 

[bob012345]

The waveguide (or transmission line) is between the two wires carrying the equal and opposite (surface) currents.

The fields that propagate the energy are mostly through the insulation and space between the conductors. Only for DC or very low frequencies does the skin effect allow the time needed for the current and magnetic field to diffuse into, and flow deep inside the conductors.The fields remain inside the shield, between the conductor and the shield which completes the circuit.

Thanks. I don't mean to sound obtuse but I don't have a clear picture in my mind of what you are saying. I would love to see a picture or diagram of how the energy flows around a simple DC circuit with a battery and a resistor.

Also confused how if there was a shield around the wire that that would complete the circuit?

 

[Dale]

Essentially it's just the usual expression of dU in terms of mu dn that becomes a heat flux in terms of a particle flux, which is nothing but the usual current density.

That doesn’t sound like an electromagnetic energy flux.

 

[Dale]

Are these static fields then?

They are any EM fields, static or not

 

[fluidistic]

That doesn’t sound like an electromagnetic energy flux.

Sorry, I didn't mean a heat flux. I meant an energy flux, of course.

As I wrote in my OP, I can derive Joule heat from it, just like with the poynting vector theorem. That does sound like an electeomagnetic flux, therefore?

 

[Dale]

Sorry, I didn't mean a heat flux. I meant an energy flux, of course.

Yes, but I don’t think it is an electromagnetic energy flux. Different forms of energy need not have parallel fluxes

As I wrote in my OP, I can derive Joule heat from it, just like with the poynting vector theorem. That does sound like an electeomagnetic flux, therefore?

No, why do you think it does? I don’t have that reference, so you will have to evaluate it. But it doesn’t sound like it was derived from Maxwell’s equations.

 

[Baluncore]

Are these static fields then?

The static fields of DC obey exactly the same mathematics as high frequency fields. DC is really just very low frequency AC, you build it and turn it on, then someone disconnects it, or the battery is exhausted later.

Also confused how if there was a shield around the wire that that would complete the circuit?

The inside of the shield is conductive, so it is a mirror. An equal and opposite current will be seen in the mirror, which completes the virtual circuit.
Why is a conductor a mirror? Because the incident magnetic field induces a perpendicular current in the conductive surface, which in turn generates an almost equal and opposite magnetic field that cancels into the mirror but is reflected back. In effect, turning left twice sends you back the way you came; i² = -1.

 

[bob012345]

The static fields of DC obey exactly the same mathematics as high frequency fields. DC is really just very low frequency AC, you build it and turn it on, then someone disconnects it, or the battery is exhausted later.The inside of the shield is conductive, so it is a mirror. An equal and opposite current will be seen in the mirror, which completes the virtual circuit.
Why is a conductor a mirror? Because the incident magnetic field induces a perpendicular current in the conductive surface, which in turn generates an almost equal and opposite magnetic field that cancels into the mirror but is reflected back. In effect, turning left twice sends you back the way you came; i² = -1.

Practically, in a simple DC circuit with say a 1 mm diameter wire, the energy that flows along the wire but outside it, what kind of additional radius essentially contain the fields?

 

[Baluncore]

... what kind of additional radius essentially contain the fields?

The radius is virtually infinite, that is how wireless signals radiate.

Both the E and M fields away from the two wires tend to cancel, so are very small. The magnetic fields between the wires reinforce, so the vast majority of energy is propagated between the wires. That is why two wire lines make such good transmission lines, and such poor antennas.

Any conductive material nearby will restrict the effective radius by shielding, so reflecting the small radiating fields back. Placing a two wire cable in a conductive conduit will minimise the radiation radius.

 

[hutchphd]

I am thus left confused. What is the direction of the energy flux created by the battery? Why do I get 2 different directions? Are these different energies? That's very confusing. I can derive Joule effect starting from any of the 2 energies mentioned above... so... shouldn't they be the same? If so, why do they have a different direction? What is going on?

Consider a battery, a length of twin conductor (flat untwisted) cable, and a resistor (>> battery resistance) at the far end. Estimate the fields between the wires. Calculate the Poynting Vector. Energy flows in the fields.
Questions?

 

[Dale]

what kind of additional radius essentially contain the fields?

The radius is virtually infinite, that is how wireless signals radiate.

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits. I have seen this calculation before, but don’t recall the exact result. I will try to find it.

 

[Baluncore]

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits.

I agree. There is almost no energy out there, but the radius is still theoretically infinite.

 

[fluidistic]

Consider a battery, a length of twin conductor (flat untwisted) cable, and a resistor (>> battery resistance) at the far end. Estimate the fields between the wires. Calculate the Poynting Vector. Energy flows in the fields.
Questions?

Yes. First, it's not as simple as this. In the original problem the current density is uniform in the wire. This means that a charge density is built up on the surface of the wire, and in principle should be computed so that it makes J constant throughout the wire. If you do the math, you should find that the charge builds up linearly from a terminal of the battery to the other. Then I hope the wire is circular, else it might become a nightmare to compute the field you ask. But that's irrelevant, I mean, I already buy that the Poynting vector points inward, in the wire. This is not.the problem.

The problem is that from a thermodynamics point of view, to a non vanishing particle number passing through a cross surface is associated an energy flux. And that this flux has the direction of the current, whereas the Poynting vector, which should be the energy flux, points in a direction perpendicular to this. That's the problem.

 

[Baluncore]

There is little electric field inside the conductor, so there is little energy propagating along the inside of the conductor.

Some of the energy does not reach the resistive load. The resistance of the conductor results in a small voltage drop along the conductor. Some of the energy therefore turns into the conductor and is converted to heat by I²R.

 

[vanhees71]

There is a very nice discussion of these issues for a DC in a coaxial cable in Sommerfeld, Lectures on Theoretical Physics, vol. 3.

 

[fluidistic]

There is little electric field inside the conductor, so there is little energy propagating along the inside of the conductor.

Some of the energy does not reach the resistive load. The resistance of the conductor results in a small voltage drop along the conductor. Some of the energy therefore turns into the conductor and is converted to heat by I²R.

I know this, this doesn't change the problem.
J equals sigma E (sorry on the phone, hard to write latex). Where E equals minus grad phi. The electrochemical potential is worth the chemical potencial plus e grad phi. Imposing the condition that the div of the energy flux vanishes in the steady state, one finds the heat equatiob in the wire, and we see that therr is indeed a joule term.

The problem that the direction of the energy flux does not match the direction of the poynting vector rrmains.

 

[Baluncore]

There is a very nice discussion of these issues for a DC in a coaxial cable in Sommerfeld, Lectures on Theoretical Physics, vol. 3.

Perhaps you can give the article number or the relevant figure number.

 

[vanhees71]

That's a textbook. You can of course find it online (though not legally ;-)).

You find a writeup about it also here (but in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/coax.pdf

For the relativistic part of the infinitely long cylindrical wire but without the discussion of the Poynting vector, see (in English):

https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

 

[Baluncore]

That's a textbook.

I have a copy here of the 1952 English translation, but could not find the relevant section. I would have thought the section numbers and figure numbering would stay the same.

 

[fluidistic]

I am making progress. I thibk I figured out that JU and S are different. Because in steady state, div JU equals 0, and this lead to the heat equation of the wire, it contains the conduxtion term as well as a Joule term. Whereas in steady state, div S equals Joule heat. So the Poynting vector is a different energy flux than the interbal ebergy flux from thermodynamics. And so, maybe, all the youtubers claimjng that energy doesn't flow in the wires are wrong.

I wiol come back to you later, but it's as if S equals mu J, except that.the direction do not match.

 

[vanhees71]

I have a copy here of the 1952 English translation, but could not find the relevant section. I would have thought the section numbers and figure numbering would stay the same.

In this edition it's Paragraph 17 (p. 125ff). A nice diagram is Fig. 23 on p. 129 showing the electric field lines and the those of the Poynting vector.

 

[Dale]

The problem is that from a thermodynamics point of view, to a non vanishing particle number passing through a cross surface is associated an energy flux. And that this flux has the direction of the current, whereas the Poynting vector, which should be the energy flux, points in a direction perpendicular to this. That's the problem.

Why is that a problem? There is no reason to expect that the thermodynamic flux should be the same direction as the EM energy flux.

 

[Dale]

There may be some energy far away from the wire, but the majority of the energy is quite close to the wire for DC circuits. I have seen this calculation before, but don’t recall the exact result. I will try to find it.

I found the calculation for a coaxial cable in section 3.5.3 of this text courtesy of @vanhees71

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

It looks like the key point is that the energy density falls off as $1/R^2$

 

[fluidistic]

Why is that a problem? There is no reason to expect that the thermodynamic flux should be the same direction as the EM energy flux.

I expected that the internal energy flux would be equal to the EM ebergy flux in the isothermal case where grad T vanished.

I mathematically reached that div S equals div (mu times J). However I cannot conclude that S equals mu times J, whixh is pretty clear in that they do not share the same direction.

I just want to understabd what's going on regarding the energy flux inside the wire in such conditions, and in particular whether it is true or not that the energy doesn't travel inside the wire, as is claimed by famous youtubers.

 

[Dale]

I just want to understabd what's going on regarding the energy flux inside the wire in such conditions, and in particular whether it is true or not that the energy doesn't travel inside the wire, as is claimed by famous youtubers.

I don’t think that the YouTube discussion you reference was talking about any thermal flux. They were specifically describing the electromagnetic flux.

 

[fluidistic]

I don’t think that the YouTube discussion you reference was talking about any thermal flux. They were specifically describing the electromagnetic flux.

I know. I am also talking about an energy flux, the internal enerfy flux, whixh I expected to equal the poynting flux. The energy flux can be decomposed into a thermal flux (that I can neglect in isotgermal conditions), and an energy flux due to the current, worth mu times J. By neglecting.the thermal flux, only the em flux should remain, and this should match the poynting flux, though I don't get this mathematically. I get that their div must match, even though their directions are different. Both are able to yield.the joulean dissipation in the material.

 

[fluidistic]

Ok, I think I fibally figured the mystery out! Skipping all mathematical steps, insode the material, S is equal to kappa grad T (note that the direction makes sense, and the fact that they vanish at the center too). There is a non zero poynting vector inside only if the conductivity is finite. If thr thermal gradient ie neglected then it implies that.the electrical conductivity is infinite. Everythibg makes srnse. And claiming thst the ebergy doesn't flow in wires cannot make sense as soon as the electric conductivity is finite. And when it is infinite I am not sure yet, but there is no poynting vector inside the material for sure, no thermal gradient. I am not sure if there is an ebergy associatee to the current, i suppose yes but not a hindrede percebt sure.

Edit 2: when i take the limit of large sigma, i get that grad T and therefore S become small inside.the wire. However, mu times J becomws big, and so does.the energy flux in the current's direction. Taking the limit rho the resistivity equals 0 shows that the energy's direction is pureley along the wire, and non null. The videos are wrong in their claim, i believe.

 

[vanhees71]

Sure, microscopically the resistance/electrical conductivity is due to friction and indeed the energy is converted to heat. The total flux of energy per unit time indeed is $R I^2$, as demonstrated in Sect. 3.5.3 (with the bad clash of notation using $R$ all of a sudden also for the resistance, which I'll correct in a moment).

 

[fluidistic]

I will write up my answer when i get the time in front of a computer instead of my phone. I have everything figured out. The interesring questions that remain are: 1)what would happen in the case of a superconduxtor, how does it differ from a conduxtor with rho equals 0? 2) what fraction of the total energy of tge power source ends up in the wire, as rho is increased from 0 up to. a finite value? So that we'd know if the conduxtivity impacts on the quantity of the total energy being stored in the em fields outide the wire vs inside of it. I already know that for an ideal conduxtor, there is an energy flux flowing along the wire inside every point in the wire, contradixting the claim that energy doesn't flow in the wires.

Sure, there is a joule heat when rho isn't zero, there was no problem with that part. Note that it's not a flux, it's a volume heat generated, it doesn't have a direction if you prefer.

 

[fluidistic]

The long awaited promise has been fullfilled. I join a pdf explaining what is going on. There are still a few things to figure out, but what is crystal clear is that.there is an energy flux along the direction of the wire, regardless of the value of.the resistivity. This goes against.the claim of many youtubers, some of which are experts in electronics, or scientific popularizers.
Have fun!

 

[Baluncore]

My argument is that electrical energy does not propagate from the generator to the load along the inside of the conductor. You seem to be saying that the joule heating is propagated into the wire, which I agree with. But I don't believe joule heating propagates useful electrical energy from the generator to the load.

I have trouble knowing when the energy you are considering is thermal waste energy, or useful electrical energy. It is important to distinguish between the thermal energy wasted and the transmitted electrical energy.

How do you say electrical energy is propagating from the generator to the load along the inside of the conductor, without it being wasted as heat.

 

[fluidistic]

My argument is that electrical energy does not propagate from the generator to the load along the inside of the conductor. You seem to be saying that the joule heating is propagated into the wire, which I agree with. But I don't believe joule heating propagates useful electrical energy from the generator to the load.

I have trouble knowing when the energy you are considering is thermal waste energy, or useful electrical energy. It is important to distinguish between the thermal energy wasted and the transmitted electrical energy.

How do you say electrical energy is propagating from the generator to the load along the inside of the conductor, without it being wasted as heat.

I should wait to go home before replying hastily, but here it goes.
I say that joule heat is produced everywhere in the wire, at every point if you prefer, it is homogrneously produced when no material properties depend on temperature, which I assumed it holds to keep things simple. In steady state, the joule hest generated in any volume element must be conducted out of that element. In particular the heat flux is radial in this homogeneous case (stuff can get much messier if the thermal conductivity is anisotropic and thermoelectric effets are taken into account).

I say that.there is an energy flux that is proportional to the electric current density, i.e. it has the same direction. If we take a superconductor where no power is lost by dissipation, I guess I am saying that the fact that there's a non vanishing electric current implies that there is an energy flux (even though Poynting vector vanishes inside the SC wire) going through the wire. If I integrate this energy flux with respect to position along the wire, I should get a non zero energy. I think this energy is due to the fact that the electrochemical potential does not vanish, and neither does the electric current. (Is this true also in the case of Cooper pairs? I guess so.)

I have just found 2 refs of papers using the same terms albeit with a different notation, regarding the total energy flux. Paper by Callen called The application of Onsager's reciprocal relations to thermoelectric, thermomagnetic, and galvanomagnetic effets''. Paper by Domenicali ''Irreversible thermodynamics of thermoelectricity''. I think they address your question regarding ''wasted'' energy. Callen says entropy increases because of heat conduction and because of the degradation of electrochemical potential (essentially this means that.there's a potential drop). Both vanish when the resistivity vanishes, so in that case no entropy increase, no wasted heat.

 

[vanhees71]

The thermodynamical approach to electrodynamics is rarely treated in the literature afaik. A nice EJP paper is here:

https://iopscience.iop.org/article/10.1088/1361-6404/aa9caf

 

[fluidistic]

Some more thoughts. Maybe the "extra" energy flux term $\overline{\mu}\vec{J_e}$ (note that there's a slight units mistake, the charge of the electrons should hang around, but this doesn't change the discussion) accounts for an energy the power source needs to "use" initially, but needs not to maintain, in the steady state for example.
The thing is, an ideal wire (zero resistivity/resistance) with a current has more energy than the same wire with zero current. This also means that when the current goes away, this energy has to go somewhere (it cannot go away by Joule heat if the resistivity is null), maybe it is radiated away, I do not know. But I do know there is an energy due to the current (comes from integrating $\overline{\mu}\vec{J_e}$ over the volume of the wire).

What I could do soon, is to make a sketch and draw $\vec{J_U}$, $\kappa \nabla T$ (or the Poynting vector if you prefer), and $\mu \vec J_e$, to get the full mental picture of how the energy is flowing inside the wire. When the resistivity is non zero, $\mu$ grows/decrease (I'd have to check the signs) when going along the direction of the current density $\vec J_e$ is constant throughout the volume, so this means that this energy term the youtubers have missed either grows or decrease along the wire, i.e. it's not constant. It is constant radially, for a given position along the wire. The Poynting vector follows what the termal gradient does, i.e. vanishes at the center of the wire and increases as we move away from the center radially. It does not depend on the position along the wire, only depends on the radial coordinate. The total energy flux (which correspond to the internal energy flux) is the sum of these 2 energy fluxes.

In the case of zero resistivity, the first energy flux doesn't increase/decrease anymore along the wire ($\mu$ is constant), and it is non zero.

Therefore, in every single possible cases imaginable, there is an energy flux that goes along the wire, and energy "flows" inside the wire, no matter what happens to the Poynting vector (could vanish, or not, doesn't matter).

 

[fluidistic]

The thermodynamical approach to electrodynamics is rarely treated in the literature afaik. A nice EJP paper is here:

https://iopscience.iop.org/article/10.1088/1361-6404/aa9caf

Thanks for the ref. Lots of insights in there. Eq. 21 is pretty similar to what I wrote in my doc (similar notation too). From a quick glance at the paper, he refers to "ionization energy" for what I have as $\overline{\mu}\vec{J_e}$ but there is more to it, as he splits an electronic contribution to a lattice one...

 

[bob012345]

It seems confusing to me to talk about energy flowing through or around a wire instead of some form of potential energy. Consider a simple circuit consisting of a battery and a resistor and wires completing the circuit in a loop. If energy flows out of the battery and then is virtually all consumed in the resistor as heat, what flows out of the resistor and through or around the wire back to the battery? The current still flows the same in both paths into and out of the resistor. Aren't the Poynting vectors the same? I'm confused. No, I'm really confused.

 

[vanhees71]

I think @fluidistic 's manuscript is pretty clear though I've not checked the details carefully yet. It's clear that here you have a current in a conductor due to electrons, which are driven by the electric field and subject to friction through scatterings with the ion lattice of the metal, which leads to dissipation and thus the production of heat. The manuscript treats this as an energy-transport process in the usual way of hydrodynamical transport equations. The Poynting vector is of course the em. part of the energy flux needed in the energy balance, including in addition the heat flow.

 

[fluidistic]

Thanks Vanhees71. I promise a new version soon (I'm losing my job tomorrow, I may have more free time soon). I have reorganized the doc. a little bit, and there is more juice to squeeze in order to obtain an accurate sketch of the energy flux. For.example, the fact that the potential drop is linear w.r.t. z (because the current density is uniform), etc.

 

[fluidistic]

Here's the new doc.

 

[Baluncore]

Here's the new doc.

"I investigate the claim that energy doesn’t flow in wires, but in the space around them. The claim comes from focusing on the Poynting vector and by assuming that the conducting wire of a circuit has no resistance. However, it turns out that there is energy ”flowing” in the wire in all cases, including when the resistivity vanishes."

You are making a different claim. The web claim is that electrical energy from a generator, that reaches the load, does not travel inside the conductor(s). The web claim does NOT assume the conducting wires of the circuit have no resistance. We know real wires get hot, but that thermal energy does not reach the load as useful electrical energy.