Energy flux direction in a conducting wire?

[fluidistic]

I do not understand this straw man. One cannot build a wire using classical elctromagnetic theory alone. Of course the energy supplied by the influx indicated by the Poynting vector will end up as heat (there are alternate degrees of freedom afforded by QM). This heat will diffuse essentially isotropically.
A wire is complicated. Why is this interesting?

.

Sorry hutch for being ignorant, what you have in mind is likely over my head. I was making allusion to Allan's mention of "the real energy flow" when he mentions Poynting vector. I understood it as "the total energy flux". I am just saying that Poynting's vector is not the whole energy flux inside the wire.

 

[coquelicot]

It's a good analogy. Indeed, coordinates have no direct physical meaning either. Particularly in GR, what's observable are not the coordinates but only the (local) coordinate-independent quantities described by tensors.

So simple when you have grasped the point, isn't it? and that may well open many doors for quantum mechanics etc.

 

[hutchphd]

Sorry hutch for being ignorant, what you have in mind is likely over my head.

Perhaps, but I doubt it. I was trying to point out that there are a panoply of electrectromagnetic interactions happening in the solid and proper treatment thereof would involve very detailed considerations on an atomic scale including the very local and complicated Poynting vector inside. Not useful to calculate. When you treat the wire as "a conductor" you have in some sense decided to only do thermodynamics inside. It is not surprising that the result looks like thermodynamics.

 

[vanhees71]

Of course, the first step is to consider thermal equilibrium and linear-response theory. With this you get the standard constitutive equations explained in all electrodynamics textbooks.

 

[Philip Koeck]

I have a conceptual problem with energy flowing in the wire.
If energy flows from an object A to an object B then I would expect the energy stored in object A to decrease and the energy in B to increase. A good example is a warm object connected to a cold object by a thermal conductor. In that case the energy flow is simply the heat current.
Now if A is the negatively charged plate of a capacitor and B is the positively charged plate, in what sense does the energy stored in A decrease and the energy in B increase when the two plates are connected by a wire? Why is it not the other way round. (Or how can it be either of the two ways?)

 

[renormalize]

Now, let me propose to you a completely observable quantity S′ (according to this definition), different from the Poynting vector and still leading to exactly the same energy transfer results:
A(M1,t)=14πϵ0c2∫j(M2,t′)||M1−M2||dV,
with t′ the retarded time, and
Φ(M1,t)=14πϵ0c2∫ρ(M2,t′)||M1−M2||dV
with the following bound conditions for the integrals: A(M1,0)=0 and Φ(M1,0)=0.
Then define S′=(Φ∇−∇∂A∂t)×B and you are done.

So you have introduced explicit expressions for the scalar potential $\Phi$ and the magnetic vector potential $\boldsymbol{A}$ (which together makeup the relativistic 4-vector potential $A_{\mu}$) and from them defined a modified Poynting vector $$\boldsymbol{S'=}\left(\Phi\nabla-\nabla\boldsymbol{\dot{A}}\right)\times\boldsymbol{B=}\Phi\nabla\times\boldsymbol{B}-\nabla\boldsymbol{\dot{A}}\times\boldsymbol{B}$$ But your operator $\nabla\boldsymbol{\dot{A}}$ as written has two 3-vector indices and it's unclear how you mean to contract and/or cross it with $\boldsymbol{B}$. And does $\nabla$ differentiate both $\boldsymbol{\dot{A}}$ and $\boldsymbol{B}$ or just one of them? Could you please clarify?

This is again exactly analogous to the fact that the energy flux in electromagnetics cannot depend on the four-vector potential, but only on the electric and magnetic fields derived from that potential.

The magnetic and electric fields depend themselves from the potentials, so your last sentence is formally a nonsense.

OK, its my turn to clarify my "nonsense"! The energy flux in electromagnetics depends only on the differentiated 4-vector potential, and only in the specific combinations $\boldsymbol{E}=-\nabla\Phi-\frac{1}{c}\boldsymbol{\dot{A}}$ and $\boldsymbol{B}=\nabla\times\boldsymbol{A}$. What is nonsense is to claim that the undifferentiated 4-potential can appear in the energy flux.

...in the old-new theory "I" propose, the notion of EM flux through an open surface is not intrinsically defined, and need not actually.

I assume you mean to say "...need not actually exist". Given that statement, if EM flux through an open surface might not actually exist in your proposed theory, can you please state clearly what you believe the (open) front face of an illuminated solar panel is receiving from the sun and converting to usable electrical power? After all, the utility of solar panels is an empirical fact independent of any proposed theory.

 

[coquelicot]

So you have introduced explicit expressions for the scalar potential $\Phi$ and the magnetic vector potential $\boldsymbol{A}$ (which together makeup the relativistic 4-vector potential $A_{\mu}$) and from them defined a modified Poynting vector $$\boldsymbol{S'=}\left(\Phi\nabla-\nabla\boldsymbol{\dot{A}}\right)\times\boldsymbol{B=}\Phi\nabla\times\boldsymbol{B}-\nabla\boldsymbol{\dot{A}}\times\boldsymbol{B}$$ But your operator $\nabla\boldsymbol{\dot{A}}$ as written has two 3-vector indices and it's unclear how you mean to contract and/or cross it with $\boldsymbol{B}$. And does $\nabla$ differentiate both $\boldsymbol{\dot{A}}$ and $\boldsymbol{B}$ or just one of them? Could you please clarify?

My bad! the formula was correctly written in the last version of the article I posted. It is
$$\boldsymbol{S'=}\epsilon_0 c^2 \left(\Phi\nabla-\boldsymbol{\dot{A}}\right)\times\boldsymbol{B=}\epsilon_0 c^2(\Phi\nabla\times\boldsymbol{B}-\boldsymbol{\dot{A}}\times\boldsymbol{B}).$$

OK, its my turn to clarify my "nonsense"! The energy flux in electromagnetics depends only on the differentiated 4-vector potential, and only in the specific combinations $\boldsymbol{E}=-\nabla\Phi-\frac{1}{c}\boldsymbol{\dot{A}}$ and $\boldsymbol{B}=\nabla\times\boldsymbol{A}$. What is nonsense is to claim that the undifferentiated 4-potential can appear in the energy flux.

Well, that's the same discussion again and again. I think I have explained the point as far as I could in post #207. Please, take the time to read it carefully; if after that you disagree with my proposition, I think it's a matter of choice and I can do nothing more.

I assume you mean to say "...need not actually exist". Given that statement, if EM flux through an open surface might not actually exist in your proposed theory, can you please state clearly what you believe the (open) front face of an illuminated solar panel is receiving from the sun and converting to usable electrical power? After all, the utility of solar panels is an empirical fact independent of any proposed theory.

My bad again! I meant "the notion of EM flux through an open surface is gauge dependent", which is allowed by "my" theory. In my view, that's the whole body that receives the energy from the sun. But choosing the gauge $\Phi = 0$, you have a useful theorem that says "the energy received by the body is equal to the integral of the flux (equal to the Poynting vector in this gauge) on the exposed surface". That's just because the Poynting vector is directed along the propagation direction of the light.

 

[coquelicot]

I have a conceptual problem with energy flowing in the wire.
If energy flows from an object A to an object B then I would expect the energy stored in object A to decrease and the energy in B to increase. A good example is a warm object connected to a cold object by a thermal conductor. In that case the energy flow is simply the heat current.
Now if A is the negatively charged plate of a capacitor and B is the positively charged plate, in what sense does the energy stored in A decrease and the energy in B increase when the two plates are connected by a wire? Why is it not the other way round. (Or how can it be either of the two ways?)

So, you have also a problem with the density of current $\bf j$. Why is the current flowing in one direction and not the other?
Or, trying to understand the question more in depth, you have also a problem with the circulation of water inside a circular pipe, with a small fan inside the pipe to maintain a constant water stream?

 

[Philip Koeck]

So, you have also a problem with the density of current $\bf j$. Why is the current flowing in one direction and not the other?
Or, trying to understand the question more in depth, you have also a problem with the circulation of water inside a circular pipe, with a small fan inside the pipe to maintain a constant water stream?

A current of charges is no problem. One plate is positively charged to start with and one is negative. The current reduces this unbalance.
I just don't see that the same is true for an energy flow. I can't see that one plate has more energy than the other to start with.

 

[coquelicot]

A current of charges is no problem. One plate is positively charged to start with and one is negative. The current reduces this unbalance.
I just don't see that the same is true for an energy flow. I can't see that one plate has more energy than the other to start with.

Let take a perfect analogy with water. Take a circular pipe full of water. At some point of the pipe, there is a small piston tied to a spring. The spring is tied to the pipe wall. By some mean, you make the piston move from its natural rest position, bending the spring. Then you relax the piston. The piston push the water in the pipe, and you have a current of water, associated with an energy flow. If you have a doubt about that, here is something from Feynman lectures, that will show you that an energy flow is usually associated with the current of particle:

There is an important theorem in mechanics which is this: whenever there is a flow of energy in any circumstance at all (field energy or any other kind of energy), the energy flowing through a unit area per unit time, when multiplied by $1/c^2$, is equal to the momentum per unit volume in the space. In the special case of electrodynamics, this theorem gives the result that g is $1/c^2$ times the Poynting vector:
$$g={1\over c^2}S.$$
So the Poynting vector gives not only energy flow but, if you divide by $c^2$, also the momentum density. The same result would come out of the other analysis we suggested, but it is more interesting to notice this more general result. We will now give a number of interesting examples and arguments to convince you that the general theorem is true.
First example: Suppose that we have a lot of particles in a box—let’s say $N$ per cubic meter—and that they are moving along with some velocity $v$. Now let’s consider an imaginary plane surface perpendicular to $v$. The energy flow through a unit area of this surface per second is equal to Nv, the number which flow through the surface per second, times the energy carried by each one. The energy in each particle is $$m_0c^2\over \sqrt{1−v^2/c^2}.$$ So the energy flow per second is
$$Nvm_0c^2\over \sqrt{1−v^2/c^2}.$$
But the momentum of each particle is $$m_0v\over \sqrt{1−v^2/c^2},$$ so the density of momentum is
$$Nm_0v \over \sqrt{1−v^2/c^2},$$
which is just $1/c^2$ times the energy flow—as the theorem says. So the theorem is true for a bunch of particles.

 

[Philip Koeck]

Let take a perfect analogy with water. Take a circular pipe full of water. At some point of the pipe, there is a small piston tied to a spring. The spring is tied to the pipe wall. By some mean, you make the piston move from its natural rest position, bending the spring. Then you relax the piston. The piston push the water in the pipe, and you have a current of water, associated with an energy flow. If you have a doubt about that, here is something from Feynman lectures, that will show you that an energy flow is usually associated with the current of particle:

I agree that there is an energy flow associated with a particle flow simply because every particle has a momentum and kinetic energy, but I'm not sure that this is what we are discussing.
You get an energy flow with magnitude V I in your theory (using one of the gauges).
I wonder if the flow of kinetic energy of the electrons can be that large.

I also wonder about the direction. If you replace the wire by some device with mobile positive charges and stationary negative charges then the electric current still goes in the same direction, but the particle flow goes in the other direction.

Your result that the power equals V I is obviously right. Maybe the problem just lies in the interpretation. Couldn't it just be that this simply cannot be regarded as an energy flow,
whereas in the other gage (with Φ = 0) you do get an energy flow in form of the Poynting vector?

 

[coquelicot]

I agree that there is an energy flow associated with a particle flow simply because every particle has a momentum and kinetic energy, but I'm not sure that this is what we are discussing.
You get an energy flow with magnitude V I in your theory (using one of the gauges).
I wonder if the flow of kinetic energy of the electrons can be that large.

That's not the flow of kinetic energy (which is negligible), but the flow of electric potential energy, carried by the charges, and equal to $\Phi j$.

I also wonder about the direction. If you replace the wire by some device with mobile positive charges and stationary negative charges then the electric current still goes in the same direction, but the particle flow goes in the other direction.

You are right to say that that's not really the flow of particle (I was aware of that from the beginning, but I didn't want to introduce another inessential problem). That's the algebraic flow of charges that import here. OK, not really a particle flow, but still, sufficiently strongly related. In the same way, the current density is not really the density associated to the current of positive charges, but that's the algebraic flow of the charges (I mean, $J = J_+ - J_-$).

Your result that the power equals V I is obviously right. Maybe the problem just lies in the interpretation. Couldn't it just be that this simply cannot be regarded as an energy flow,
whereas in the other gage (with Φ = 0) you do get an energy flow in form of the Poynting vector?

If you take the example of Feynman in my previous post, you'll see that in addition to the kinetic energy of the particles, Feynman includes their rest energy $m_0c^2$, which is a "kind of" potential energy as far as I understand. So, I see no reason to ban the flow electric potential energy of the charges as well.

 

[Philip Koeck]

That's not the flow of kinetic energy (which is negligible), but the flow of electric potential energy, carried by the charges, and equal to $\Phi j$.
You are right to say that that's not really the flow of particle (I was aware of that from the beginning, but I didn't want to introduce another inessential problem). That's the algebraic flow of charges that import here. OK, not really a particle flow, but still, sufficiently strongly related. In the same way, the current density is not really the density associated to the current of positive charges, but that's the algebraic flow of the charges (I mean, $J = J_+ - J_-$).
If you take the example of Feynman in my previous post, you'll see that in addition to the kinetic energy of the particles, Feynman includes their rest energy $m_0c^2$, which is a "kind of" potential energy as far as I understand. So, I see no reason to ban the flow electric potential energy of the charges as well.

Then I still have the same problems.
I can't see how potential energy can flow to start with.
The other problem is that any flow of energy would mean that there is more energy at one end than the other to start with and the flow evens out this imbalance. I can't see that either.

 

[coquelicot]

Then I still have the same problems.

Ask Feynman

I can't see how potential energy can flow to start with.

Take again the example of the water circuit above, and assume that the pipe is full with a solution containing some chemical energy. Can't you imagine the flow of chemical energy within the pipe? that's just what is said in chemical thermodynamics, or even in the formula that was the subject of this thread.

The other problem is that any flow of energy would mean that there is more energy at one end than the other to start with and the flow evens out this imbalance. I can't see that either.

No, again, you have a circuital flow. In your cap, the potential energy is not stored at the plates, but inside the E-field of the cap (mostly located inside the plates of the cap, but also, in a less extent, outside). To prevent a further question, observe that the discharge of a cap does not produce a steady current, hence truly steady potentials are impossible here. Near the cap, there is an energy flow outside the wires of course. Far from the cap, the potentials are almost steady (with any non foolish gauge), and the energy flow described with respect to this gauge is observed inside the wires.
Of course, you can choose the foolish (in this circumstance) gauge $\Phi = 0$, which leads to a vector potential depending upon the time; then the generalized Poynting vector now reduces to the usual Poynting vector with respect to this gauge, and you are happy. I guess most electrical engineers and thermodynamists will be happy with the steady potentials and the energy power flowing inside the wires.

 

[Philip Koeck]

Ask Feynman Take again the example of the water circuit above, and assume that the pipe is full with a solution containing some chemical energy. Can't you imagine the flow of chemical energy within the pipe? that's just what is said in chemical thermodynamics, or even in the formula that was the subject of this thread.

No, again, you have a circuital flow. In your cap, the potential energy is not stored at the plates, but inside the E-field of the cap (mostly located inside the plates of the cap, but also, in a less extent, outside). To prevent a further question, observe that the discharge of a cap does not produce a steady current, hence truly steady potentials are impossible here. Near the cap, there is an energy flow outside the wires of course. Far from the cap, the potentials are almost steady (with any non foolish gauge), and the energy flow described with respect to this gauge is observed inside the wires.
Of course, you can choose the foolish (in this circumstance) gauge $\Phi = 0$, which leads to a vector potential depending upon the time; then the generalized Poynting vector now reduces to the usual Poynting vector with respect to this gauge, and you are happy. I guess most electrical engineers and thermodynamists will be happy with the steady potentials and the energy power flowing inside the wires.

Energy that flows in a circle is even worse to my way of thinking. Why would it do that?

I'd like to hear what others have to say, though.

The picture that makes sense to me is that heat leaves the wire and this energy loss is balanced by energy that flows from the battery or capacitor via the surrounding space into the wire.
Obviously the total power of this process is V I, but that doesn't indicate an alternative path for the energy flow as far as I'm concerned.

 

[coquelicot]

Energy that flows in a circle is even worse to my way of thinking. Why would it do that?

I have to clarify myself: indeed the energy is flowing in a circle, but it is not constant along the circle. Actually, if the wire as some resistance per unit of length, the energy flow, at the minus of the battery, is null. It is maximal at the + terminal of the battery, and dissipates into heat all along the wire (hence its decreasing). Notice that the situation is worse and much less intuitive with the Poynting vector flow: there the energy flows symmetrically from the two terminals of the battery "near" the wire, decreasing more and more until it becomes null near the middle of the wire (see the article of Harbola I posted somewhere in this thread). That's a bit weird isn't it?

The picture that makes sense to me is that heat leaves the wire and this energy loss is balanced by energy that flows from the battery or capacitor via the surrounding space into the wire.
Obviously the total power of this process is V I, but that doesn't indicate an alternative path for the energy flow as far as I'm concerned.

That's your choice and I respect it, as I use to say.

EDIT: By choosing the gauge $\Phi := -\Phi$, the energy will flow in the opposite direction. So, you have perhaps an answer to your intuitive problem.

 

[Philip Koeck]

Notice that the situation is worse and much less intuitive with the Poynting vector flow: there the energy flows symmetrically from the two terminals of the battery "near" the wire, decreasing more and more until it becomes null near the middle of the wire (see the article of Harbola I posted somewhere in this thread). That's a bit weird isn't it?

That does sound strange, I agree.
My feeling would be that the EM energy flow should balance the heat flow everywhere in steady state.
So if the wire is the same everywhere I would expect the same T, the same heat flow out from the wire and the same energy flow into the wire everywhere along the length of the wire.

 

[coquelicot]

That does sound strange, I agree.
My feeling would be that the EM energy flow should balance the heat flow everywhere in steady state.
So if the wire is the same everywhere I would expect the same T, the same heat flow out from the wire and the same energy flow into the wire everywhere along the length of the wire.

Regarding "the same T", I believe this is almost the case because we have to take into account the heat conduction in the wire. But that's a question of thermodynamics with heat fluxes etc. and I think Fluidistic is better than me to answer it (in fact, I think he has already answered to it, and there are also articles about that). The point is that thermodynamists use the formula that is the subject of the question of the OP, that cannot be justified with the Poynting vector, as far as I can conclude from this thread. This was the starting point of my thoughts, as you know.

 

[vanhees71]

A current of charges is no problem. One plate is positively charged to start with and one is negative. The current reduces this unbalance.
I just don't see that the same is true for an energy flow. I can't see that one plate has more energy than the other to start with.

Of course not, and for a statically charged capacitor the energy flow is 0.

A bit more puzzling is the explanation of the situation, where you have an electrostatic field within a capacitor superimposed by a magnetostatic field. Then the energy flow (Poynting vector), $\vec{E} \times \vec{B} \neq 0$. How to explain this flow in a purely static situation. Hint: Google for "hidden momentum". It's nicely treated in Griffiths's E&M textbook. A very nice collection about these apparent "paradoxes" in E&M is by McDonald:

https://physics.princeton.edu/~mcdonald/examples/

For the question here:

https://physics.princeton.edu/~mcdonald/examples/current.pdf

The answer is that E&M is a relativistic theory and also energy-momentum balance has to be treated relativistically. In fact there is no "hidden momentum" but just the correct definition of the energy-momentum-stress tensor of the em. field + the charged particles.

 

[coquelicot]

A bit more puzzling is the explanation of the situation, where you have an electrostatic field within a capacitor superimposed by a magnetostatic field. Then the energy flow (Poynting vector), $\vec{E} \times \vec{B} \neq 0$. How to explain this flow in a purely static situation. Hint: Google for "hidden momentum". It's nicely treated in Griffiths's E&M textbook. A very nice collection about these apparent "paradoxes" in E&M is by McDonald:

Note: with the theory of relativity of gauges I propose, this is no more a problem. Under any steady gauge, there is no flow of energy in a static state. Another a weird thing that disappears!

 

[vanhees71]

This doesn't make sense. The Poynting vector is the electromagnetic energy-flow density. It's compensated by the corresponding "hidden momentum" of the charges making the current to produce the magnetic field moving in the electric field of the capacitor. All this is of course entirely gauge-independent as it must be for observable phenomena.

 

[coquelicot]

This doesn't make sense. The Poynting vector is the electromagnetic energy-flow density. It's compensated by the corresponding "hidden momentum" of the charges making the current to produce the magnetic field moving in the electric field of the capacitor. All this is of course entirely gauge-independent as it must be for observable phenomena.

In a purely static situation (which I thought was the context), where the magnetic field is created by a magnet and the charge is static, my assertion makes sense because there is no hidden momentum (no charges movement).
In a situation where the magnetic field is created by charges in a solenoid, then we are in the context of magnetostatic, and there is, of course, a field momentum (in my proposed theory). So, no nonsense as well. I will drop your assertion "The Poynting vector is the electromagnetic energy-flow density", since this is a loop inside the same debate, of which I have provided licit arguments.

 

[vanhees71]

There are moving charges (currents). Otherwise there'd be no magnetic field. Of course, to understand permanent magnets you need quantum mechanics to correctly describe it, which is outside of the realm of classical electromagnetism, but also there you have a "current" and "hidden momentum".

 

[coquelicot]

There are moving charges (currents). Otherwise there'd be no magnetic field. Of course, to understand permanent magnets you need quantum mechanics to correctly describe it, which is outside of the realm of classical electromagnetism, but also there you have a "current" and "hidden momentum".

Do you have a reference for you last assertion? Also, does not this hidden momentum statistically cancel?

 

[fluidistic]

In steady state, it's the divergence of the energy flux that must vanish in any volume considered, not the flux itself. Imposing this condition yields the steady state heat equation in the material.

 

[renormalize]

My bad! the formula was correctly written in the last version of the article I posted. It is $$\boldsymbol{S'=}\epsilon_0 c^2 \left(\Phi\nabla-\boldsymbol{\dot{A}}\right)\times\boldsymbol{B=}\epsilon_0 c^2(\Phi\nabla\times\boldsymbol{B}-\boldsymbol{\dot{A}}\times\boldsymbol{B}).$$

Thanks for posting this corrected equation from your article of April 28 in post #176. I have read it and now better understand your motivation for introducing your gauge-dependent energy flux $\boldsymbol{S'}$.
Another form you write for this flux is $$\boldsymbol{S}^{'}=\boldsymbol{S}+\mathrm{\frac{1}{\mu_{0}}}\nabla\times\left(\Phi\boldsymbol{B}\right)$$ where $\boldsymbol{S}=\mathrm{\mathrm{\frac{1}{\mu_{0}}}}\boldsymbol{E}\times\boldsymbol{B}$ is the usual gauge-invariant Poynting vector. This makes it clear that both $\boldsymbol{S}$ and $\boldsymbol{S'}$ have the same divergence and that they are identical whenever the scalar potential $\Phi$ vanishes. Indeed, you deem $\Phi=0$ to be the appropriate gauge for considering a plane wave since the energy flux of EM radiation is well described by the usual Poynting vector. So far so good.
You then introduce another equivalent expression for your energy flux $$\boldsymbol{S}^{'}=\Phi\boldsymbol{J}+\varepsilon_{0}\Phi\boldsymbol{\dot{E}}-\mathrm{\mathrm{\frac{1}{\mu_{0}}}}\boldsymbol{\dot{A}}\times\boldsymbol{B}$$ and consider the steady-state case (like a DC current flowing in a wire) by dropping the time derivatives, resulting in $$\boldsymbol{S}^{'}=\Phi\boldsymbol{J}$$ I note that this flux does indeed have (by construction) the expected divergence $$\nabla\cdot\boldsymbol{S}^{'}=\left(\nabla\Phi\right)\cdot\boldsymbol{J}+\Phi\nabla\cdot\boldsymbol{J}=-\boldsymbol{E}\cdot\boldsymbol{J}$$ (because $\nabla\Phi=-E-\boldsymbol{\dot{A}}$ and both $\boldsymbol{\dot{A}}$ and $\nabla\cdot\boldsymbol{J}$ vanish for a steady state), i.e., this divergence is the negative of the flux of work done by the EM field on the current.
Even so, I remain (as yet) unconvinced when you conclude regarding $\boldsymbol{S}^{'}=\Phi\boldsymbol{J}$ that
The meaning of this equation is that for steady currents, there is no flow of power wherever the space is free of charges: the power is carried by the charges only.
To make this convincing, you need to explicitly solve the relevant Maxwell equations for the scalar potential $\Phi$ and then demonstrate that $\Phi\boldsymbol{J}$ gives a physically reasonable energy flux in the wire.
For example, consider the simple case of a wire of circular cross-section carrying a steady uniform current-density and choose cylindrical-coordinates {$r,\phi,z$} that align the z-axis with the axial-direction of the wire. Since your flux contains just $\Phi$, the only relevant Maxwell equation is Gauss's Law, $\frac{\rho}{\varepsilon_{0}}=\nabla\cdot\boldsymbol{E}=-\nabla^{2}\Phi$. So with no $\phi$-dependence, you have to solve Poisson's equation $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial\Phi}{\partial r}\right)+\frac{\partial^{2}\Phi}{\partial z^{2}}=-\frac{\rho}{\varepsilon_{0}}$$ Because the electric field external to a current-carrying wire is strictly radial as $r\rightarrow\infty$ and since there is nothing in this simple wire problem that appears to depend on $z$, it's tempting to assume that $\Phi$ depends exclusively on $r$. But of course that can't be right for your theory since it yields an electric field $\boldsymbol{E}=-\frac{\partial\Phi\left(r\right)}{\partial r}$ that's strictly radial even inside the conductor, and hence the field can do no work on a current flowing in the axial direction. (In this scenario, some non-electromotive force, like gravity, must exist to propel the current flow against the resistance.)
Instead, I think you're going to have to find a solution $\Phi(r,z)$ of Poisson's equation that yields an electric field $\boldsymbol{E}(r,z)$ which points radially at infinity. As it gets closer to the wire, the field must bend in the axial-direction such that the component $E_{z}$ reaches the just the right value inside the wire to yield the proper energy flux. I can't claim that this is impossible, but I do wish you luck finding such a scalar potential.

 

[Dale]

Temporarily closed for moderation

Edit: after some internal discussion this thread will remain closed. Participants are reminded that personal speculation is prohibited at PF and all posts must be consistent with the professional scientific literature, not merely PF posts that have been put into a .pdf file