Energy flux direction in a conducting wire?

[coquelicot]

If I think of my example with the two slabs connected by a wire (like an H when viewed from the side), shouldn't E be parallel to the wire everywhere inside the H and B would go in circles around the wire.
Then the Poynting vector should point towards the wire everywhere inside the H.
Is that correct?

The Poynting vector should exactly balance the heat radiation from the wire if the whole thing is in vacuum.
(I'm just guessing. Not completely my field.)

Well, my bad. Actually your objection is good even in the the more simple configuration I spoke about above (that is, just a wire connecting the two plates of a capacitor). After having a look at the article of Harbola, it turns out that the EM energy that is flowing parallel to the wire flows from both sides of the battery, and becomes smaller and smaller farther along the wires (until the two opposite vectors meet and then the Poynting vector is null). Here is a picture from the article of Arbola:

So, your objection is a good objection to the theory of fluidistic: why does his theory break the symmetry?
good catch!

 

[Philip Koeck]

Ok, in that case i will spend the time to answer you, hopefukly in a few hours.

You made me thibk about something, it is possible that mu0 is zero (i am not sure). This would mean that there is indeed no energy flowing along the wire in the case of zero resistivity (but only then). I am not sure mu0 should be zero though. If jt isn't zero then the conclusion holds.!(i am guessing it isn't strictly zrro for the simple fact that electeons do carry mass, as small as it may be, but this would be negligible).

There is one thing in the document that caught my eye.
In the text between equations 2 and 3 you write for the entropy flux: J_S=J_Q/T
This should only be true for reversible processes, or at least there should be a reversible process with the same starting and end point as the process you are studying.
I'm not sure if this is possible in this case.
A wire gets hot due to an electric current and gives off heat to the surroundings.
I don't see how to reverse this.

Could this be a problem for your derivation?

 

[fluidistic]

There is one thing in the document that caught my eye.
In the text between equations 2 and 3 you write for the entropy flux: J_S=J_Q/T
This should only be true for reversible processes, or at least there should be a reversible process with the same starting and end point as the process you are studying.
I'm not sure if this is possible in this case.
A wire gets hot due to an electric current and gives off heat to the surroundings.
I don't see how to reverse this.

Could this be a problem for your derivation?

Very good question. The Js=Jq/T is found everywhere in the litterature. I myself do not fully understand it (and I have seen people asking whether it is correct even in physics stack exchange website). From what I could gather, it holds, when the entropy flux expression takes into consideration not only a transfer of entropy, but also the entropy generated in the volume, so the "S" would be a sort of total S. From this relation, one can derive the heat equation(s), I believe this shouldn't be a problem.

 

[Philip Koeck]

Very good question. The Js=Jq/T is found everywhere in the litterature. I myself do not fully understand it (and I have seen people asking whether it is correct even in physics stack exchange website). From what I could gather, it holds, when the entropy flux expression takes into consideration not only a transfer of entropy, but also the entropy generated in the volume, so the "S" would be a sort of total S. From this relation, one can derive the heat equation(s), I believe this shouldn't be a problem.

Well dS = dQ / T only holds for a reversible process, so it could lead to a wrong result if you use it for an irreversible process.

 

[fluidistic]

So, I will do most of the work for you.
I wrote all your equations here:

(1) ##dU = T dS + \bar \mu dN##, with ##U## total energy, ##T## temperature, ##S## entropy, ##\bar \mu## electrochemical potential (battery potential??), N not defined (what is it?)

(2) ##\vec J_U = T\vec J_S + \bar\mu \vec J_e## with ##\vec J_U## total energy flux, ##\vec J_S## entropy flux, ##\vec J_e## density of current;

(3) ##\nabla \cdot(-\kappa \nabla T) - \rho \vec J_e^2 = 0##;

(4) ## \nabla \cdot(-\kappa \nabla T) + \nabla\cdot \vec S = 0## with ##\vec S## poynting vector;

(5) ##\vec S = \kappa \nabla T ##;

(6) ##\nabla T = -\rho {\vec J_e^2 r \over 2\kappa}\hat r##;

(7) ##\bar \mu = \bar \mu_0 + \mu_1 z##;

(8) ##\nabla \bar \mu = \mu_1 \hat z = -\rho |\vec J_e| \hat z##.

Now, I don't ask you to explain me the seven relations above, but how you deduce from them the following equation:
$$\vec J_U = \rho {\vec J_e^2 r\over 2} \hat r + (\bar \mu_0 - \rho |\vec J_e| z )|\vec J_e| \hat z.$$
(Admittedly, I may be missing something and this is probably a stupid question, but I don't see).

EDIT: I'm more or less OK with the right most member, after the "+", assuming you use (7) and (8) to inject inside the right most member of (2).

For me, you contradict yourself: you said that you need some energy to create a current inside a circuit, even if the wire has no resistance, and that your energy flow expresses this fact. But in such a circuit, the only energy needed to create a current is the magnetic energy stored inside the circuit, that is ##{1\over 2} LI^2## (##I## intensity of the current). If you omit the inductance of the circuit, you need absolutely no energy to create a current, since there is no resistance. But because every circuit is a loop and has an inductance ##L##, in a circuit with 0 resistance, you have ##V = LdI/dt## with ## V## electromotive force, hence the current rises linearly if V is supposed constant, until it reaches the desired value. During this process, you have provided an energy equal to ##{1\over 2} LI^2##, and that's all (neglecting very small other effects).

Alrighty, I have some minutes (but no pen nor paper!).
So, we start with the standard eq. ##dU=TdS+\overline{\mu}dN## which describes the change in internal energy in the wire (the wire as a thermodynamics system out of equilibrium, but not too far off either), we ignore the change in volume. Now we imagine the wire as a cylinder, or torus if you prefer, the thing is, it has a cross section and a direction along which current can flow. The eq. becomes eq. 2, i.e. changes in thermodynamics variables become fluxes. The dN part becomes a particle flux. For the wire, the particles are charged, and so the usual electric current density ##\vec {J_e}## appears.

Now, we impose the condition of a steady state (if you don't, you'll get time derivatives, which just complicate things), i.e. nothing depends on time anymore. In that case, there can be no energy accumulation, nor charge accumulation in every single part of the wire. Mathematically, this means ##\nabla \cdot \vec{J_U}=\nabla \cdot \vec{J_e}=0## (sorry to bring those relations again, but you actually need to use them to derive the heat eq., which is itself required to derive eq. 4).

You also know from thermodynamics that ##T\vec{J_S}=\vec{J_Q}## (closing your eyes on the usual inequality, as Philip Koeck pointed out). The eq. 2 becomes ##\vec{J_U}=\vec{J_Q}+\overline{\mu}\vec{J_e}##. Note that we could already stop here, since we already see that there is an energy flux pointing in the direction of the current, something which is denied in Veritasium's video (and many others), because neither ##\mu## nor ##\vec{J_e}## vanishes. But let's continue.

Fourier's law says ##\vec{J_Q}=-\kappa \nabla T##, we can plug it back into our last equation, call this eq. 100. Then we mathematically evaluate the condition of the divergence of the energy flux must equal 0. By doing so, after a few mathematical steps (chain rule for the gradient), we find the heat eq. that the temperature satisfies inside the wire: ##\nabla \cdot (-\kappa \nabla T)-\rho |\vec J_e|^2=0##. Note here that the first term comes from the div of J_Q, whereas the second term comes from the div of mu J_e. (that's relevant, I believe).

So, I solved this heat equation using cylindrical coordinates, with Dirichlet boundary conditions, which gave me T(r), and so ##\nabla T##, too. I also pointed out that the equation tells us that whenever there is a Joule heat in the wire, there must be a thermal gradient too. It is impossible to keep the whole wire at uniform temperature in that case (I found that interesting on its own).

After this, I plugged back the expression of ##\nabla T## into the expression I had (eq. 100). So we have the first part of eq. 4.

For the second part, as I wrote above, I assumed that ##\vec{J_e}## was constant throughout the wire, which implies a linear potential drop (or a constant electrochemical potential gradient). This condition yields ##\mu = \mu_0+\mu_1 z## (just an integration). But looking at Ohm's law ##\vec{J_e}=-\sigma \nabla \overline{\mu}##, I could identify that ##\mu_1 = -\rho|\vec{J_e}|##. This complete the puzzle to reach eq. 4.Now, I have a comment. Note that I didn't bring the Poynting vector at all in the picture. There was no need for it, it is already subtetly included in ##\vec{J_U}##. But out of curiosity, when I actually computed what it was worth, I saw it was equal to ##\kappa \nabla T##, in other words, it is worth (minus) the thermal energy flux. It points in the same direction than it. It is the term that conducts away the heat generated by the Joule effect. It is not the term that produces Joule heat in this thermodynamics derivation (!).

Two seemingly completely different approaches to show that the Poynting vector was there in the internal energy flux. And it is evident that Poynting vector ##\vec P## is not the whole energy flux (sorry Veritasium). In doubt, just compute ##\nabla \cdot \vec P## and you'll see it isn't worth 0, therefore it cannot be the whole energy flux.

I hope this is clearer.

 

[fluidistic]

Well dS = dQ / T only holds for a reversible process, so it could lead to a wrong result if you use it for an irreversible process.

I used J_S=J_Q/T, might be a subtle but important difference. I do not have access to it, but apparently this chaper's book: https://www.sciencedirect.com/sdfe/pdf/download/eid/3-s2.0-B0123694019007129/first-page-pdf contains a formal derivation of the equality (regarding the fluxes), for irreversible processes.

It's quite likely proven elsewhere too. Like I said, it is used everywhere (I have never seen the inequality when it comes to fluxes). So, I trusted what I had seen. The interesting question is then, why does the equality hold for irreversible processes, when it comes to fluxes, rather than "warning, since dQ/T < dS for irreversible processes, then Js = dQ/dT is probably wrong".

Edit: Look at the derivation of eq. 2-38 in https://digitalcommons.lsu.edu/cgi/viewcontent.cgi?article=2476&context=gradschool_disstheses.

 

[bob012345]

Alrighty, I have some minutes (but no pen nor paper!).
So, we start with the standard eq. ##dU=TdS+\overline{\mu}dN## which describes the change in internal energy in the wire (the wire as a thermodynamics system out of equilibrium, but not too far off either), we ignore the change in volume. Now we imagine the wire as a cylinder, or torus if you prefer, the thing is, it has a cross section and a direction along which current can flow. The eq. becomes eq. 2, i.e. changes in thermodynamics variables become fluxes. The dN part becomes a particle flux. For the wire, the particles are charged, and so the usual electric current density ##\vec {J_e}## appears.

What exactly is this internal energy change in the wire? Can you supply a simple example with numbers as to what you say are the energy flows as conventionally understood vs. your calculations?

Also, your document suggests the current is uniform in the wire when it is really more of a skin effect. Does that change anything?

 

[fluidistic]

What exactly is this internal energy change in the wire? Can you supply a simple example with numbers as to what you say are the energy flows as conventionally understood vs. your calculations?

Also, your document suggests the current is uniform in the wire when it is really more of a skin effect. Does that change anything?

Internal energy change in the wire is just the well known thermodynamics relation, as in https://en.wikipedia.org/wiki/Funda...n#The_first_and_second_laws_of_thermodynamics, neglecting the volume change. This "U" contains all the energy inside the wire, including the energy due to Poynting vector.

I am not sure what you mean by "conventionnally understood vs your calculations". It's not like I have done something new, I just employed well known thermodynamics to the wire, with a few assumptions. Tell me more about the conventionally understood part and I might give a better, more detailed answer.

No idea why you bring the skin effect. The current density for a DC in a wire is pretty much uniform throughout the whole volume of the wire. There is no AC involved here. What do you have in mind exactly?

The charge density on the surface of the wire is linear along the direction of the wire, be it for a straight line or a torus (see e.g. eq. 12 of https://arxiv.org/pdf/1207.2173.pdf). I had found a nice website about this, here it is: http://www1.astrophysik.uni-kiel.de/~hhaertel/CLOC/Circuit/html/2-surface-charges.htm. That's another reason why Veritasium's video is misleading, since it displays an evenly distributed + and - charges along the wires, while in reality this is inaccurate.

 

[coquelicot]

Now, we impose the condition of a steady state (if you don't, you'll get time derivatives, which just complicate things), i.e. nothing depends on time anymore. In that case, there can be no energy accumulation, nor charge accumulation in every single part of the wire. Mathematically, this means ##\nabla \cdot \vec{J_U}=\nabla \cdot \vec{J_e}=0## (sorry to bring those relations again, but you actually need to use them to derive the heat eq., which is itself required to derive eq. 4).

You also know from thermodynamics that ##T\vec{J_S}=\vec{J_Q}## (closing your eyes on the usual inequality, as Philip Koeck pointed out). The eq. 2 becomes ##\vec{J_U}=\vec{J_Q}+\overline{\mu}\vec{J_e}##. Note that we could already stop here, since we already see that there is an energy flux pointing in the direction of the current, something which is denied in Veritasium's video (and many others), because neither ##\mu## nor ##\vec{J_e}## vanishes. But let's continue.

I still don't see this is true at this point, since ##\vec J_Q## is not known, and might kill ##\vec{J_e}## in the expression ##\vec{J_Q}+\overline{\mu}\vec{J_e}##. But Ok since you show later that ##\vec J_Q = \vec S##, and the poynting vector is radial.

The only thing that is not entirely clear for me now is how ##\nabla\cdot \bar\mu J_e = \rho \vec J_e^2##, despite I believe this is Ohm law somehow.

 

[bob012345]

Internal energy change in the wire is just the well known thermodynamics relation, as in https://en.wikipedia.org/wiki/Funda...n#The_first_and_second_laws_of_thermodynamics, neglecting the volume change. This "U" contains all the energy inside the wire, including the energy due to Poynting vector.

I am not sure what you mean by "conventionally understood vs your calculations".

Aren't you are saying the Poynting vector approach is basically wrong? In my understanding that is not the generally accepted view. If I understand you, you are not just saying you have an equivalent method to get the same answer, you are saying the energy flow is not mainly due to the Poynting vector outside the wire but is mostly inside the wire. Is that correct? It was already established that there is a small component of the Poynting vector pointing inside the wire which accounts for Joule heating. The "U" would include that part. That doesn't include the energy flow that ends up in the load. I think you're saying it does?

It's not like I have done something new, I just employed well known thermodynamics to the wire, with a few assumptions. Tell me more about the conventionally understood part and I might give a better, more detailed answer.

Well, I think those assumptions need to be vetted. I'd like to see your approach in a peer reviewed journal.

No idea why you bring the skin effect. The current density for a DC in a wire is pretty much uniform throughout the whole volume of the wire. There is no AC involved here. What do you have in mind exactly?

Sorry, I was confused by the surface charges being around the circuit as the papers I referenced earlier showed.

The charge density on the surface of the wire is linear along the direction of the wire, be it for a straight line or a torus (see e.g. eq. 12 of https://arxiv.org/pdf/1207.2173.pdf). I had found a nice website about this, here it is: http://www1.astrophysik.uni-kiel.de/~hhaertel/CLOC/Circuit/html/2-surface-charges.htm. That's another reason why Veritasium's video is misleading, since it displays an evenly distributed + and - charges along the wires, while in reality this is inaccurate.

See the Jackson paper regarding Charge density. I do agree the video is a bit misleading but not because the theoretical concept is wrong, but I think the graphics are misleading as to the scale of the fields and energy flow.

 

[coquelicot]

I used J_S=J_Q/T, might be a subtle but important difference. I do not have access to it, but apparently this chaper's book: https://www.sciencedirect.com/sdfe/pdf/download/eid/3-s2.0-B0123694019007129/first-page-pdf contains a formal derivation of the equality (regarding the fluxes), for irreversible processes.

It's quite likely proven elsewhere too. Like I said, it is used everywhere (I have never seen the inequality when it comes to fluxes). So, I trusted what I had seen. The interesting question is then, why does the equality hold for irreversible processes, when it comes to fluxes, rather than "warning, since dQ/T < dS for irreversible processes, then Js = dQ/dT is probably wrong".

Edit: Look at the derivation of eq. 2-38 in https://digitalcommons.lsu.edu/cgi/viewcontent.cgi?article=2476&context=gradschool_disstheses.

I am probably missing something, but the author just after eq. 2-38 says that the reduced law (2.39) you used is valid only in the absence of mass and charge flows.

 

[Philip Koeck]

I used J_S=J_Q/T, might be a subtle but important difference. I do not have access to it, but apparently this chaper's book: https://www.sciencedirect.com/sdfe/pdf/download/eid/3-s2.0-B0123694019007129/first-page-pdf contains a formal derivation of the equality (regarding the fluxes), for irreversible processes.

It's quite likely proven elsewhere too. Like I said, it is used everywhere (I have never seen the inequality when it comes to fluxes). So, I trusted what I had seen. The interesting question is then, why does the equality hold for irreversible processes, when it comes to fluxes, rather than "warning, since dQ/T < dS for irreversible processes, then Js = dQ/dT is probably wrong".

Edit: Look at the derivation of eq. 2-38 in https://digitalcommons.lsu.edu/cgi/viewcontent.cgi?article=2476&context=gradschool_disstheses.

Wouldn't the term with J_e simply cancel if you use (2-38) from the thesis you quote above instead of using (2-39)?

 

[fluidistic]

I still don't see this is true at this point, since ##\vec J_Q## is not known, and might kill ##\vec{J_e}## in the expression ##\vec{J_Q}+\overline{\mu}\vec{J_e}##. But Ok since you show later that ##\vec J_Q = \vec S##, and the poynting vector is radial.

The only thing that is not entirely clear for me now is how ##\nabla\cdot \bar\mu J_e = \rho \vec J_e^2##, despite I believe this is Ohm law somehow.

I do not understand what you are missing. ##\vec J_Q## can be expressed in terms of the thermal conductivity and the thermal gradient through Fourier's law, like I have done.
And yes, you are correct, ##\vec J_Q## and ##\vec J_e## point in different directions, they can't cancel each other out (unless you set the current to zero, in this case the wire would be isothermal, so ##\vec J_Q## would be zero too).

For the last part, the key is the chain rule coupled with the condition of steady state:
##\nabla \cdot (\overline{\mu} \vec{J_e})=\nabla \overline{\mu} \cdot \vec{J_e} + \overline{\mu}\underbrace{\nabla \cdot \vec{J_e}}_{=0}##. Then use Ohm's law ##-\rho \vec{J_e}=\nabla \overline{\mu}##, plug ##\nabla \overline{\mu}## into the expression just obtained, and you get the relation you quoted.

 

[fluidistic]

Aren't you are saying the Poynting vector approach is basically wrong? In my understanding that is not the generally accepted view. If I understand you, you are not just saying you have an equivalent method to get the same answer, you are saying the energy flow is not mainly due to the Poynting vector outside the wire but is mostly inside the wire. Is that correct? It was already established that there is a small component of the Poynting vector pointing inside the wire which accounts for Joule heating. The "U" would include that part. That doesn't include the energy flow that ends up in the load. I think you're saying it does?

Well, I think those assumptions need to be vetted. I'd like to see your approach in a peer reviewed journal.

Sorry, I was confused by the surface charges being around the circuit as the papers I referenced earlier showed.

See the Jackson paper regarding Charge density. I do agree the video is a bit misleading but not because the theoretical concept is wrong, but I think the graphics are misleading as to the scale of the fields and energy flow.

Not at all. Poynting's approach does what it's supposed to do, it is not wrong at all, my point is that it doesn't take into account the full energy flux in the wire. This whole thread is just about pointing out that Poynting's vector does not account for the whole energy flux inside the wire. There is an energy flux that goes along the wire. This is nothing new, this can be found in hundreds or thousands of papers. But apparently Veritasium (and many other youtubers) missed this, and have done their analysis thinking and claiming that the energy doesn't flow along the current. I have found nothing new.

Thanks for the Jackson's reference, I might take a look.

 

[coquelicot]

I do not understand what you are missing. ##\vec J_Q## can be expressed in terms of the thermal conductivity and the thermal gradient through Fourier's law, like I have done.
And yes, you are correct, ##\vec J_Q## and ##\vec J_e## point in different directions, they can't cancel each other out (unless you set the current to zero, in this case the wire would be isothermal, so ##\vec J_Q## would be zero too).

I'm not missing anything (except perhaps something that is trivial for you according to your knowledge of this domain), I just say that you cannot stop here since you have still not shown that ##\vec J_Q## and ##\vec J_e## point in different directions up to now. You do that later in eq. (4) (even without the Poynting vector). Thanks for you explanations.

EDIT: Ah yes, regarding what I am missing, I now understand that you refer to my other post about formula (2.38) and (2.39) of the thesis you provided. That's just that the author says that the formula you used is valid only in the absence of charge and mass flows. But I think that here, there is a charge flow. Well, I'm not very experimented in this domain, so, feel free not to answer me if that's too bad for you.

 

[Philip Koeck]

EDIT: Ah yes, regarding what I am missing, I now understand that you refer to my other post about formula (2.38) and (2.39) of the thesis you provided. That's just that the author says that the formula you used is valid only in the absence of charge and mass flows. But I think that here, there is a charge flow. Well, I'm not very experimented in this domain, so, feel free not to answer me if that's too bad for you.

I'm also wondering about (2-38) versus (2-39).

 

[bob012345]

Not at all. Poynting's approach does what it's supposed to do, it is not wrong at all, my point is that it doesn't take into account the full energy flux in the wire. This whole thread is just about pointing out that Poynting's vector does not account for the whole energy flux inside the wire. There is an energy flux that goes along the wire. This is nothing new, this can be found in hundreds or thousands of papers. But apparently Veritasium (and many other youtubers) missed this, and have done their analysis thinking and claiming that the energy doesn't flow along the current. I have found nothing new.

Thanks for the Jackson's reference, I might take a look.

Ok, but I simply want to know what percentage of power flows through the wire vs. along the wire from outside the wire according to your calculations?

 

[fluidistic]

Ok, but I simply want to know what percentage of power flows through the wire vs. along the wire from outside the wire according to your calculations?

What I can answer is what fraction of the total internal energy is flowing radially as opposed to along the wire. The answer is "it depends where in the wire".

Fixing my wrong units, take eq. 4 and replace the electrochemical potential by the voltage (and throw out ##\mu_0## (or ##V_0##) if you want. You'll end up with ##\frac{|\vec{J_Q}|}{|V\vec{J_e}|}=\frac{\rho |\vec{J_e}|r}{2V_1z}##, a quantity that depends on both ##r## and ##z##. Here I assumed that ##V=V_1z##, i.e. that the voltage goes from 0 up to a certain value, from an electrode to the other. Then it's just a matter of plugging and chugging with literature's data, i.e. take reasonable values, something like copper for the resistivity, a 0.5 m long wire of radius 5 mm and you're all set having fun evaluating this fraction anywhere you want. You can also see that no matter where along the wire, the heat flux vanishes at the center of the wire, so along that line, all the energy flux is directed in the wire's direction.

We can also integrate these fluxes in the whole volume, to get an answer to your question, I suppose... Try it and let me know :)Edit: Not sure if this answers your question, but I get that (and please read me well!) all the power that enters at the surface of the base of the cylinder/wire (i.e. where the current is injected) in the direction along the wire is equal to the power removed at the open surface of the cylinder (i.e. whole surface of the wire, except for the 2 caps). I guess that's not really surprising. This also means that at the other electrode (or top of the cylinder/wire, where the current "escapes the system"), there's no more energy flowing in the direction of the wire. Note that this is an approximation, where I neglected the ##\mu_0## term, while in reality it is non zero (e.g. because the electrons possesses mass), therefore, technically and being extremely strict, there would still be some energy flux left escaping the wire in the direction of the wire, but we can ignore it in this particular case, I believe.

 

[coquelicot]

While I have no more questions regarding the math, there are still unsolved big problems with this theory:

1. Fluidistic has not addressed the question of Philip Koeck regarding eq.(2.38) and (2.39) of the thesis mentioned by him (posts #116 and #117).

2. Also, no answer regarding the lack of symmetry of this theory (Philip Koeck, and also see my add about that, posts #104 to #106): why should the energy flowing in one direction and not the other? It has been pointed out that not so regarding the Poynting vector: the EM energy outside the wire is flowing from both electrodes, approximately parallel to the wires, and both sides meet somewhere in the middle.
Energy is a quadratic quantity, that has nothing to do with signs of charges and direction of vectors. I can hardly imagine that energy could break the symmetry of a system.

3. Most importantly, there is a big question of principle that I have not succeeded to understood until now: Fluidistic himself says (and shows) that the EM energy flux, carried by the Poynting vector, is "equal" (or more precisely opposite) to the heat flux that dissipate the heat outside the wire, maintaining the whole system in a steady state. These fluxes are radial to the wire, and everyone agree with that. But then, what does the new axial energy flux of Fluidistic along the wire do? this energy is not transformed in heat since that's the role of the EM energy flux. So, where is this energy going?
Or maybe he pretends that it's the contrary: the axial energy flux is producing the heat, which escapes the wire radially. If so, what does the EM energy do and where is it going?
There is a redundancy of energy flux somewhere, apparently. I would really like to understand this point.

4. What is the nature of the energy carried by the axial flux?
We already know that the EM energy flux is radial, and so is the heat energy flux that balances it. So, the energy is not EM, and (see no 3 above, apparently), not heat too. What remains? well, the rest energy mc^2 of the electrons, and their kinetic energy 1/2 mv^2 (or more concisely, the total relativistic energy of the electrons). Both of them, I guess, are minuscule. Most importantly, I think, they are irrelevant to electrical effect. For example, the electrical power dissipated by the wire is a quadratic function of the current: ##P = RI^2##. But the total mass energy of the electrons flowing through a cross section of the wire is simply proportional to the current, since their speed remains sensibly constant (as far as I know). So, the mystery is open.

EDIT: let me compare the situation with a circular pipe of water, arranged like a circuit. At some point of the pipe, there is a small turbine that moves the water at constant speed inside the pipe. We assume steady state. So, the kinetic energy flux of the water in a cross section of the pipe is equal to ##{1\over 2} m v^2 \hat z##, where m is the mass of water that flows through the cross section. In other words, it is ##{1\over 2} \rho I v^2 \hat z##, with ##v## constant. At any time, the energy provided by the turbine is converted to heat by friction with the pipe, and escapes radially from the pipe. But their remains the kinetic energy of the water, directly related to the axial flux.

In an electrical wire, the analog of the kinetic energy of the water is the magnetic energy of the circuit ##{1\over 2} LI^2##, where ##L## is the inductance of the circuit. If the theory of Fluidistic is true, that's the only relevant energy I see which could be a candidate for his axial flux.
So, assuming an elementary section of the wire ##d\ell##, and introducing the notion of "inductance per unit length" of the wire, say ##\lambda##, it should be possible to show that the axial flux flowing through ##d\ell## is equal to something like ##{1\over 2}\lambda I^2 \hat z## (neglecting minuscule other forms of energies).

 

[fluidistic]

While I have no more questions regarding the math, there are still unsolved big problems with this theory:

1. Fluidistic has not addressed the question of Philip Koeck regarding eq.(2.38) and (2.39) of the thesis mentioned by him (posts #116 and #117).

2. Also, no answer regarding the lack of symmetry of this theory (Philip Koeck, and also see my add about that, posts #104 to #106): why should the energy flowing in one direction and not the other? It has been pointed out that not so regarding the Poynting vector: the EM energy outside the wire is flowing from both electrodes, approximately parallel to the wires, and both sides meet somewhere in the middle.
Energy is a quadratic quantity, that has nothing to do with signs of charges and direction of vectors. I can hardly imagine that energy could break the symmetry of a system.

3. Most importantly, there is a big question of principle that I have not succeeded to understood until now: Fluidistic himself says (and shows) that the EM energy flux, carried by the Poynting vector, is "equal" (or more precisely opposite) to the heat flux that dissipate the heat outside the wire, maintaining the whole system in a steady state. These fluxes are radial to the wire, and everyone agree with that. But then, what does the new axial energy flux of Fluidistic along the wire do? this energy is not transformed in heat since that's the role of the EM energy flux. So, where is this energy going?
Or maybe he pretends that it's the contrary: the axial energy flux is producing the heat, which escapes the wire radially. If so, what does the EM energy do and where is it going?
There is a redundancy of energy flux somewhere, apparently. I would really like to understand this point.

4. What is the nature of the energy carried by the axial flux?
We already know that the EM energy flux is radial, and so is the heat energy flux that balances it. So, the energy is not EM, and (see no 3 above, apparently), not heat too. What remains? well, the rest energy mc^2 of the electrons, and their kinetic energy 1/2 mv^2 (or more concisely, the total relativistic energy of the electrons). Both of them, I guess, are minuscule. Most importantly, I think, they are irrelevant to electrical effect. For example, the electrical power dissipated by the wire is a quadratic function of the current: ##P = RI^2##. But the total mass energy of the electrons flowing through a cross section of the wire is simply proportional to the current, since their speed remains sensibly constant (as far as I know). So, the mystery is open.

EDIT: let me compare the situation with a circular pipe of water, arranged like a circuit. At some point of the pipe, there is a small turbine that moves the water at constant speed inside the pipe. We assume steady state. So, the kinetic energy flux of the water in a cross section of the pipe is equal to ##{1\over 2} m v^2 \hat z##, where m is the mass of water that flows through the cross section. In other words, it is ##{1\over 2} \rho I v^2 \hat z##, with ##v## constant. At any time, the energy provided by the turbine is converted to heat by friction with the pipe, and escapes radially from the pipe. But their remains the kinetic energy of the water, directly related to the axial flux.

In an electrical wire, the analog of the kinetic energy of the water is the magnetic energy of the circuit ##{1\over 2} LI^2##, where ##L## is the inductance of the circuit. If the theory of Fluidistic is true, that's the only relevant energy I see which could be a candidate for his axial flux.
So, assuming an elementary section of the wire ##d\ell##, and introducing the notion of "inductance per unit length" of the wire, say ##\lambda##, it should be possible to show that the axial flux flowing through ##d\ell## is equal to something like ##{1\over 2}\lambda I^2 \hat z## (neglecting minuscule other forms of energies).

I agree, these are good questions. I didn't take the time to fully understand Phillip's questions.

Regarding the symmetey breaking, I thibk i can answer. Take a look at the mu dN sign, or more precisely, the mu Je sign. Mu can be decomposed into.the chemical potential plus eV. This is what one should focus on to figure out this ebergy flux's direction.

 

[hutchphd]

May I suggest to all an hour with Prof. Feynman (paricularly section 27-2 through 27-4)

https://www.feynmanlectures.caltech.edu/II_27.htmlThe last section of 27-4 says it all

"You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong."

/

 

[fluidistic]

May I suggest to all an hour with Prof. Feynman (paricularly section 27-2 through 27-4)

https://www.feynmanlectures.caltech.edu/II_27.htmlThe last section of 27-4 says it all

"You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong."

/

I think Feynman is saying that the energy from the EM field goes inside the wire radially, which would come from the charge density that has built up on the surface of the wire. In other words, the charges on the surface of the conductor provide the energy that creates the uniform current density in the wire. However inside the wire, the thermodynamics approach is the one that gets the full picture of the energy flux, since the energy density of the EM field is not conserved (as Feynman says himself), whereas the thermodynamics internal energy is.

Therefore, even though the power source doesn't provide the energy from within the direction that goes along the wire, the fact that there's an electric current going along the wire still makes up for an energy flux that does go along the wire. This energy flux is different than the Poynting vector.

 

[coquelicot]

May I suggest to all an hour with Prof. Feynman (paricularly section 27-2 through 27-4)

https://www.feynmanlectures.caltech.edu/II_27.htmlThe last section of 27-4 says it all

"You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong."

/

Yeah I knew this part of Feynan Lectures. Personally, I have no problem with the Poynting vector intuition, especially after it can be shown that most of the EM energy is flowing very close to the wires.
But Fluidistic has a nice theory that extends the classic Poynting vector theory with thermodynamics, and he has also the maths. But there remain questions to be elucidated. I definitely think that his axial flow has something to do with the magnetic energy stored in the whole circuit (see my post #124). This may even be not too difficult to prove. If this is right, I would suggest to call this flux the "magnetic inertial flux", or simply "the inertial flux". This would not contradict the current theory with the Poynting vector: the EM flux would still be the flux which carries the energy to the various elements of the circuit, which is eventually transformed in heat. But there would exist another flux, which could be exhibited by involving thermodynamics, and which could be interpreted as an "inertial magnetic flux".

 

[fluidistic]

Yeah I knew this part of Feynan Lectures. Personally, I have no problem with the Poynting vector intuition, especially after it can be shown that most of the EM energy is flowing very close to the wires.
But Fluidistic has a nice theory that extends the classic Poynting vector theory with thermodynamics, and he has also the maths. But there remain questions to be elucidated. I definitely think that his axial flow has something to do with the magnetic energy stored in the whole circuit (see my post #124). This may even be not too difficult to prove. If this is right, I would suggest to call this flux the "magnetic inertial flux", or simply "the inertial flux". This would not contradict the current theory with the Poynting vector: the EM flux would still be the flux which carries the energy to the various elements of the circuit, which is eventually transformed in heat. But there would exist another flux, which could be exhibited by involving thermodynamics, and which could be interpreted as an "inertial magnetic flux".

I do not know whether this has to do with the magnetic energy stored in the whole circuit.
This energy flux along the wire would still exist even if the moving particles were not charged. We could imagine a neutron flux, or something like that, although I am not quite sure what their electrochemical potential would look like.

 

[Philip Koeck]

I do not know whether this has to do with the magnetic energy stored in the whole circuit.
This energy flux along the wire would still exist even if the moving particles were not charged. We could imagine a neutron flux, or something like that, although I am not quite sure what their electrochemical potential would look like.

I'd like to discuss a few things in your text.
Is there an updated version maybe? Maybe even including the intermediate steps.

Anyway, here are a few questions on what I've read so far:

Question 1:
What is the system you are applying the thermodynamics to? Is it the whole wire or a short section of the wire or maybe a small volume inside the wire?

Question 2:
Why do use Fourier's law to replace the heat current dQ/dt by a temperature gradient?
This is only valid for heat conduction. What about the heat radiated away by the wire?

Question 3:
Still the same: dS = dQ / T for reversible processes. If you divide by dt you get J_S = J_Q/T, so that should also be restricted to reversible processes. Maybe you should give equation (2-38) from the thesis a try.

 

[coquelicot]

I do not know whether this has to do with the magnetic energy stored in the whole circuit.
This energy flux along the wire would still exist even if the moving particles were not charged. We could imagine a neutron flux, or something like that, although I am not quite sure what their electrochemical potential would look like.

Your derivation involves Ohm law, so, it has to do with charged particles.
In fact, I have come to the conclusion that something here is not well defined: namely, an ill notion of flux seems to be used here. Let me explain.

Flux is defined in the following circumstances:

• You have a vector field ##F## already defined. Then the flux of F through a surface is the integral of ##F.d\vec S## on the surface; That is, you decide that ##F## is a flux, which is licit.
• You have something that can be described mathematically like a fluid in movement. In other words, you have the density of some object as a function of ##x##, say ##\rho(x)## together with a field of velocities ##\vec v (x, t)##. Then you can define the flux vector of the fluid by ##\rho(x) \vec v##. Now, you are in the first case.

Examples:

• heat flux: the flux is defined a priori by Fourier Law ##\phi = - k\nabla T##; So, it uses a known object, the temperature.
• Radiative flux is the amount of power radiated through a given area, in the form of photons or other elementary particles, typically measured in W/m2. So, you know about what particles you are speaking, and their velocities.
• Sound energy flux: the product of the sound pressure, and the particle velocity;
• Any form of energy flux: each form is defined in a particular way involving some velocity, or involving a priori some already known vector field.
• Entropy flux: equal to the heat flux divided by the temperature, an a priori definition again.

In any case, you have to know about what you are speaking, that is, either you already know your vector field, or at least you have the density of something, and an associated velocity. But in your case, what is the "total energy flux". You have to precise a priori about what you are speaking. For example, you have to consider such or such particles with THEIR MOVEMENT, and such or such notion attached to the particles (the DENSITY OF SOMETHING). Then, and only then, you can speak about the flux of this "something". But "total energy" has no movement associated a priori with it. Of course, you have written ##J_U = TJ_S + \bar\mu J_e##, and this can be seen as a definition a priori of your energy flux. But then, this is only what it is, the definition of a vector field from two already known vector fields, and you cannot claim that ##J_U## is the flux of "all possible forms of energies". That's simply not well defined.

That's why I think one of the main problem is to know how to interpret your ##J_U##.

 

[fluidistic]

Your derivation involves Ohm law, so, it has to do with charged particles.
In fact, I have come to the conclusion that something here is not well defined: namely, an ill notion of flux seems to be used here. Let me explain.

Flux is defined in the following circumstances:

• You have a vector field ##F## already defined. Then the flux of F through a surface is the integral of ##F.d\vec S## on the surface; That is, you decide that ##F## is a flux, which is licit.
• You have something that can be described mathematically like a fluid in movement. In other words, you have the density of some object as a function of ##x##, say ##\rho(x)## together with a field of velocities ##\vec v (x, t)##. Then you can define the flux vector of the fluid by ##\rho(x) \vec v##. Now, you are in the first case.

Examples:

• heat flux: the flux is defined a priori by Fourier Law ##\phi = - k\nabla T##; So, it uses a known object, the temperature.
• Radiative flux is the amount of power radiated through a given area, in the form of photons or other elementary particles, typically measured in W/m2. So, you know about what particles you are speaking, and their velocities.
• Sound energy flux: the product of the sound pressure, and the particle velocity;
• Any form of energy flux: each form is defined in a particular way involving some velocity, or involving a priori some already known vector field.
• Entropy flux: equal to the heat flux divided by the temperature, an a priori definition again.

In any case, you have to know about what you are speaking, that is, either you already know your vector field, or at least you have the density of something, and an associated velocity. But in your case, what is the "total energy flux". You have to precise a priori about what you are speaking. For example, you have to consider such or such particles with THEIR MOVEMENT, and such or such notion attached to the particles (the DENSITY OF SOMETHING). Then, and only then, you can speak about the flux of this "something". But "total energy" has no movement associated a priori with it. Of course, you have written ##J_U = TJ_S + \bar\mu J_e##, and this can be seen as a definition a priori of your energy flux. But then, this is only what it is, the definition of a vector field from two already known vector fields, and you cannot claim that ##J_U## is the flux of "all possible forms of energies". That's simply not well defined.

That's why I think one of the main problem is to know how to interpret your ##J_U##.

In my mind, ##\vec{J_U}## is well defined, U being the internal energy we see in thermodynamics. The fact that div J_U = 0 yields the correct heat equation in the wire means that it catches all the energy fluxes. If some energy was missing, it would have an impact on the temperature distribution, but, and I repeat, you can find in literature plenty of papers describing the heat equation in a wire. It's just Fourier's conduction term plus Joule heat equals 0 in its simplest form.

In reality, if we do not neglect thermoelectricity, then we have to use a generalized Ohm and Fourier's law. This creates a new term in the heat equation, another heat source besides Joule heat, called the Thomson effect (which can be a heat sink, not necessarily a heat source).

If we allow the volume to change (i.e. we do not neglect thermal expansion) and consider thermoelectricity, it's almost a big bang regarding the heat equation (I haven't found the eq. in papers, but having worked through it myself for fun). In all cases, everything starts from the thermodynamics expression of the internal energy.

 

[fluidistic]

I'd like to discuss a few things in your text.
Is there an updated version maybe? Maybe even including the intermediate steps.

Anyway, here are a few questions on what I've read so far:

Question 1:
What is the system you are applying the thermodynamics to? Is it the whole wire or a short section of the wire or maybe a small volume inside the wire?

Question 2:
Why do use Fourier's law to replace the heat current dQ/dt by a temperature gradient?
This is only valid for heat conduction. What about the heat radiated away by the wire?

Question 3:
Still the same: dS = dQ / T for reversible processes. If you divide by dt you get J_S = J_Q/T, so that should also be restricted to reversible processes. Maybe you should give equation (2-38) from the thesis a try.

1) In my particular case, a resistive wire through which a uniform current is passing due to a gradient in the electrochemical potential (or simply a non zero voltage along it).

2) Because that's a correct expression to use in "bulk" materials, where this law holds. It doesn't hold as is for example in microscopic/mesoscopic system where a ballistic heat transport takes place, for example. Or when thermoelectricity isn't neglected. The law transforms as ##\vec{J_Q}=-\kappa \nabla T + ST\vec{J_e}##, in other words, the heat flux at every point in the material is the sum of the standard Fourier's conduction term plus the Peltier heat (when the Peltier heat changes along the material, heat is released or absorbed, causing what people call the Peltier effect. However Peltier heat is present everywhere in the material, and not just at junctions of different materials, but let's not open this can of worm. Thermoelectricity needs a modern rewrite of a textbook that doesn't follow historical discoveries. That field is plagued with misconceptions due to this.)

The radiation only occurs at the surface of the material. In my case, if I wanted not to neglect it, then I should include it as a boundary condition. This only complicates the problem, in my opinion. The problem I solved (heat equation of the wire), assumes the wire's surface is kept at a fixed temperature, so it's valid if I had say a PID (and possibly equivalent to any steady state problems, too, including with radiative effects since the temperature will end up being uniform on the surface of the wire, and time independent in all cases).

3) If you do that, you end up with "Joule heat equals 0", or said differently "The electric current vanishes". There is no thermal gradient anymore, the problem is completely uninteresting, there is nothing to study in that case.
I have found another justification of TJs=Jq, and where both eqs. 2-38 and 2-39 appear. The reference is nothing less than Callen's Thermodynamics textbook, chapter 14.

 

[coquelicot]

In my mind, ##\vec{J_U}## is well defined, U being the internal energy we see in thermodynamics.

Well, the internal energy U in thermodynamics is defined up to a constant, because it is impossible to know all the energy forms. But let forget that and say that you are right: the internal energy is well defined in thermodynamics.
On the other hand, I have googled hours and haven't found something like this ##\vec {J_U}##. Even the thesis of Craig you provided defines everything rigorously in (2-13), but not a general "internal energy flux". I can conceive you have top books that go further, so could you provide a source to where someone else has used ##J_U## just like that?

but, and I repeat, you can find in literature plenty of papers describing the heat equation in a wire. It's just Fourier's conduction term plus Joule heat equals 0 in its simplest form.

Of course, but heat equation in a wire is not the problem here. The problem for me is your "free" use of fluxes. I am waiting for your references, if possible, hoping my eyes will be opened.

 

[Philip Koeck]

1) In my particular case, a resistive wire through which a uniform current is passing due to a gradient in the electrochemical potential (or simply a non zero voltage along it).

2) Because that's a correct expression to use in "bulk" materials, where this law holds. It doesn't hold as is for example in microscopic/mesoscopic system where a ballistic heat transport takes place, for example. Or when thermoelectricity isn't neglected. The law transforms as ##\vec{J_Q}=-\kappa \nabla T + ST\vec{J_e}##, in other words, the heat flux at every point in the material is the sum of the standard Fourier's conduction term plus the Peltier heat (when the Peltier heat changes along the material, heat is released or absorbed, causing what people call the Peltier effect. However Peltier heat is present everywhere in the material, and not just at junctions of different materials, but let's not open this can of worm. Thermoelectricity needs a modern rewrite of a textbook that doesn't follow historical discoveries. That field is plagued with misconceptions due to this.)

The radiation only occurs at the surface of the material. In my case, if I wanted not to neglect it, then I should include it as a boundary condition. This only complicates the problem, in my opinion. The problem I solved (heat equation of the wire), assumes the wire's surface is kept at a fixed temperature, so it's valid if I had say a PID (and possibly equivalent to any steady state problems, too, including with radiative effects since the temperature will end up being uniform on the surface of the wire, and time independent in all cases).

3) If you do that, you end up with "Joule heat equals 0", or said differently "The electric current vanishes". There is no thermal gradient anymore, the problem is completely uninteresting, there is nothing to study in that case.
I have found another justification of TJs=Jq, and where both eqs. 2-38 and 2-39 appear. The reference is nothing less than Callen's Thermodynamics textbook, chapter 14.

1) If the wire as a whole is the system I'm not sure whether it makes sense to calculate the gradient of the various quantities in the fundamental equation. I'm not sure about switching to currents within the wire either. I would understand currents going into and out of the system.

2) Isn't the radiation the main thermal process here?

3) Are you making some connection between electric current and thermal gradient or am I misunderstanding?

 

[fluidistic]

1) If the wire as a whole is the system I'm not sure whether it makes sense to calculate the gradient of the various quantities in the fundamental equation. I'm not sure about switching to currents within the wire either. I would understand currents going into and out of the system.

2) Isn't the radiation the main thermal process here?

3) Are you making some connection between electric current and thermal gradient or am I misunderstanding?

1) I just focused on the wire itself, since I was interested in the claim that energy doesn't flow in the wires. But you can imagine a bigger system if you want, which includes the power generator/battery.

2) Again, it doesn't matter an iota, and not necessarily. What radiation and/or convection and any other boundary condition that you can imagine will only shift the temperature of the whole system. The thermal gradient will remain unchanged, everywhere in the wire. The result will remain completely unchanged. That's why I focused on Dirichlet's boundary conditions, which are the easiest to deal with, and it is without any loss of generality. If not convinced, do the math yourself with a few cases, i.e. solve the heat equation with different boundary conditions, and then compute the thermal gradient.

3) Not only. My main point is that TJ_S = J_Q is fully justified, it isn't an approximation. Again, if I take the eq. you told me to try, I get that it's equivalent to a trivial system where no current is flowing, no Joule heat, nothing but a wire standing still.

 

[fluidistic]

Well, the internal energy U in thermodynamics is defined up to a constant, because it is impossible to know all the energy forms. But let forget that and say that you are right: the internal energy is well defined in thermodynamics.
On the other hand, I have googled hours and haven't found something like this ##\vec {J_U}##. Even the thesis of Craig you provided defines everything rigorously in (2-13), but not a general "internal energy flux". I can conceive you have top books that go further, so could you provide a source to where someone else has used ##J_U## just like that?

Of course, but heat equation in a wire is not the problem here. The problem for me is your "free" use of fluxes. I am waiting for your references, if possible, hoping my eyes will be opened.

Well, maybe start with the Callen ref. I posted earlier (he uses ##\vec W## while I use the notation ##\vec{J_U}##), or the one I posted around 1 hour before your post. Google gives me a djvu version of his Thermodynamics textbook (maybe the 4th or 5th google result), I said to check chapter 14.

Or eq. 72 of Domenicali's "Irreversible thermodynamics of thermoelectricity". Or page 241 of that same article, where it is written

Following Callen, we introduce a heat current ##\vec Q## as the difference between the total energy current ##\vec W## and the electrochemical potential energy current ##\overline{\mu}\vec{J_e}## : ##\vec Q = \vec W - \overline{\mu}\vec J_e##. If we associate our entropy current ##\vec{J_s}## with Callen's heat current ##\vec Q## in the manner ##\vec Q=T\vec{J_s}## (...)

. Or the page just before that (240), he fully justifies in words and arguments that ##\vec Q =T\vec S##, please read that part if you really want to dig this up.

 

[Philip Koeck]

1) I just focused on the wire itself, since I was interested in the claim that energy doesn't flow in the wires. But you can imagine a bigger system if you want, which includes the power generator/battery.

2) Again, it doesn't matter an iota, and not necessarily. What radiation and/or convection and any other boundary condition that you can imagine will only shift the temperature of the whole system. The thermal gradient will remain unchanged, everywhere in the wire. The result will remain completely unchanged. That's why I focused on Dirichlet's boundary conditions, which are the easiest to deal with, and it is without any loss of generality. If not convinced, do the math yourself with a few cases, i.e. solve the heat equation with different boundary conditions, and then compute the thermal gradient.

3) Not only. My main point is that TJ_S = J_Q is fully justified, it isn't an approximation. Again, if I take the eq. you told me to try, I get that it's equivalent to a trivial system where no current is flowing, no Joule heat, nothing but a wire standing still.

1) I was rather thinking of a smaller system such as a small volume within the wire, not a larger one.
In the fundamental relationship dU, dS, dV and dN stand for changes of these quantities in the system. If you introduce dQ, for example, this must therefore be an amount of heat that leaves the system or enters the system. dQ is not within a system.
My question is really: What is the fundamental equation referring to in your model?

2) In which direction is there a temperature gradient in the wire?

3) Are you saying that there is no current in the wire if there is no temperature gradient along the wire?

 

[fluidistic]

1) I was rather thinking of a smaller system such as a small volume within the wire, not a larger one.
In the fundamental relationship dU, dS, dV and dN stand for changes of these quantities in the system. If you introduce dQ, for example, this must therefore be an amount of heat that leaves the system or enters the system. dQ is not within a system.
My question is really: What is the fundamental equation referring to in your model?

2) In which direction is there a temperature gradient in the wire?

3) Are you saying that there is no current in the wire if there is no temperature gradient along the wire?

1) Right. But you need not to restrain yourself to very small elements. For example, you can compute the total flux passing through a given cross section, or surface. I have done that for the whole wire (open cylinder without caps), and the 2 caps of the wire (it's not in my document). Those quantities make sense. I am not sure about what you are asking, i.e. which fundamental equation you mention. I only used thermodynamics relations.

2) Look up my document. The thermal gradient is radial (has the direction of ##-\hat r##). The temperature profile inside the wire is a parabola, whose maximum is reached right at the center of the wire. If you change the boundary conditions, this parabola will be shifted up/down, but it won't change any more than that.

3) Almost. The thermal gradient is not along the wire, it is radial. If you modify your sentence to "Are you saying that there is no current in the wire if there is no temperature gradient in the wire?" then the answer is yes. I was saying it the other way around, but yes, that's a consequence of the math. If there is a non zero current, it is impossible for the wire to be at uniform temperature if the resistivity is not 0, and the thermal conductivity is not infinite. You can see it in my doc.

 

[coquelicot]

Well, maybe start with the Callen ref. I posted earlier (he uses ##\vec W## while I use the notation ##\vec{J_U}##), or the one I posted around 1 hour before your post. Google gives me a djvu version of his Thermodynamics textbook (maybe the 4th or 5th google result), I said to check chapter 14.

Assuming you refer to Callen Chap 14 eq. 14.33 and 14.34: there is absolutely no derivation for this formula, and even not a hint about what "energy" he speaks about. This formula is simply parachuted here. Not to speak about (14.33) he says "can be derived like..." which is not obvious at all. If you refer to another section or equation of this chapter, please let me know.

Or eq. 72 of Domenicali's "Irreversible thermodynamics of thermoelectricity". Or page 241 of that same article, where it is written

. Or the page just before that (240), he fully justifies in words and arguments that ##\vec Q =T\vec S##, please read that part if you really want to dig this up.

Again and again and again, I have no problem with heat and entropy fluxes, nor with heat equations. These are well defined notions. I have a problem with your alleged "internal energy flux" which "takes into account all possible forms of energies". I don't see that it is defined, and it is even not obvious that the "local energy flux" of Callen is the same as your energy flux.
I don't have access to the article of Domenicali, so, if you can post a snapshot of p 72 to show how he defines your "internal energy flux" (unless that's not that but licit heat equations, involving entropy and heat fluxes, whose I have no problem, again and again and again).