I Energy flux direction in a conducting wire?

[Philip Koeck]

There's an energy flux which is a consequence of the thermodynamics relation $dU=TdS+\overline{\mu}dN$. A particle flux (associated to dN) is associated to an energy flux. The direction will depend on the sign of $\overline{\mu}$ and that of the particle's flux itself.

In the salt water filled tube there is a particle flux in both directions.
Does the energy also flow in both directions then?

 

[fluidistic]

In the salt water filled tube there is a particle flux in both directions.
Does the energy also flow in both directions then?

Yes. In that case there might be a net energy flux from the particle's motion. It depends on the value of $\overline{\mu_1}\vec J_1+\overline{\mu_2}\vec J_2$. I didn't check the specifics of your case, but if that quantity vanishes, then the energy flux coming from the particle's motion cancels out. If the quantity does not vanish, then there will be a net energy flux coming from the particle's motion.

The point is that in any case, there will be non vanishing energy fluxes associated to the particle's motions. These energy fluxes might cancel each other out, but they still exist.

 

[jartsa]

The problem is that from a thermodynamics point of view, to a non vanishing particle number passing through a cross surface is associated an energy flux. And that this flux has the direction of the current,

Like in hydraulic?

Well, I guess inside an electric wire pressure of particles (electrons) is zero, which then results in zero energy flux inside the wire.

(Except for the radial energy flux caused by the electrons inside the wire using the energy outside the wire to move along the resistive wire)

 

[coquelicot]

Here's the new doc.

I have some remarks:

1. In your doc, I don't understand how you derive eq. (4) from eq (2); could you elaborate?

2. At the beginning of your doc, you say

from this expression, it is evident that the Poynting vector does not catch the whole energy involved in the system, because in steady state, the divergence of the (total) energy flux must vanish

But the Poynting vector had never been said to catch the whole energy involved in the system: it is everywhere said that the electromagnetic energy (stemming from the Poynting vector flux) entering in the wire is equal to the energy dissipated in heat by the wire. In other words, obviously, the integral of the flux of the thermal energy + integral of the the electromagnetic energy flux is null in steady state.
Thence, it is natural to suspect that not only the integrals, but in fact "the flux of the thermal energy + the electromagnetic energy flux is null in steady state". If I'm not wrong, that is what you have demonstrated in your Appendix A, and that's perfectly fine. So, what contradiction remains?

Note: You are apparently considering another form of energy flux flowing along the wire (despite I don't understand how you derive it). But this form of energy seems to play no role in the electrical process. In the same way, taking into account the mass $m$ of the electrons, and the fact that $E = mc^2$, we could introduce another form of energy flux by considering the electron mass entering or exiting an element of volume. Would it be relevant?

 

[fluidistic]

I have some remarks:

1. In your doc, I don't understand how you derive eq. (4) from eq (2); could you elaborate?

2. At the beginning of your doc, you say

But the Poynting vector had never been said to catch the whole energy involved in the system: it is everywhere said that the electromagnetic energy (stemming from the Poynting vector flux) entering in the wire is equal to the energy dissipated in heat by the wire. In other words, obviously, the integral of the flux of the thermal energy + integral of the the electromagnetic energy flux is null in steady state.
Thence, it is natural to suspect that not only the integrals, but in fact "the flux of the thermal energy + the electromagnetic energy flux is null in steady state". If I'm not wrong, that is what you have demonstrated in your Appendix A, and that's perfectly fine. So, what contradiction remains?

Note: You are apparently considering another form of energy flux flowing along the wire (despite I don't understand how you derive it). But this form of energy seems to play no role in the electrical process. In the same way, taking into account the mass $m$ of the electrons, and the fact that $E = mc^2$, we could introduce another form of energy flux by considering the electron mass entering or exiting an element of volume. Would it be relevant?

1. I computed $\nabla T$ by solving the heat equation of the wire with Joule heat taken into account. This allowed me to have an expression for $\nabla \cdot (\kappa \nabla T)$. Recalling that $T\vec{J_S}=\vec{J_Q}=\nabla \cdot (\kappa \nabla T)$, I reached the first term of the right hand side in eq. 4. For the second term of eq. 4, I assumed that the electric current was along $\hat z$, and that the potential drop was linear with distance (which is consistant with a homogeneous electric current density). Let me know if you wish to have more details.

2. Right, Poynting vector should not catch the whole energy of the system. However if you listen to Veritasium's video, near the beginning he talks about the energy of the system, conservation of energy and then straight jumps towards the Poynting vector, as if this term was the whole deal. It isn't, as you point out. However this is not clear for Veritasium (and apparently many people), although it appears trivial to you. He confuses the "total energy flux" (or the "internal energy flux" to use common thermodynamics naming convention) with the Poynting vector. That's one the reasons he misses that there's an energy flux that goes along the wire, inside the wire, regardless of the resistivity value (could be 0 or not). Note that I've showed that the Poynting vector is mathematically equal to the heat flux in the wire, that's why it's radial and vanishes if there's no Joule heat (resistivity equals 0). From eq. 2 it is clear that it isn't the whole energy flux.

Response to your note: It is relevant. The degradation of the electrochemical potential gives rise to the Joule heat. The divergence of that energy flux yields the Joule heat. If the resistivity is zero, then this energy flux is constant throughout the wire. When integrated over the whole volume, it yields a non zero energy. So the power source had to provide that energy, even though in steady state it doesn't need to "inject" energy anymore to sustain this energy flux. When the current stops, this energy has to go somewhere, probably radiated away since there is zero resistivity in the wire, and therefore cannot be dissipated there. I claim that Veritasium (and many others), completely dismissed this energy flux, and this energy, too.

Think about comparing the situation of having an idealized zero resistivity wire, and no current, with the same wire but with a non zero current. I claim that the wire having an electric current has an extra energy flux inside of it, which amounts for an extra energy (when the flux is integrated over the length of the wire), even though there is no dissipation of energy anywhere. The power source had to provide the energy to create the current, even though the resistivity vanishes.

 

[coquelicot]

1. I computed $\nabla T$ by solving the heat equation of the wire with Joule heat taken into account. This allowed me to have an expression for $\nabla \cdot (\kappa \nabla T)$. Recalling that $T\vec{J_S}=\vec{J_Q}=\nabla \cdot (\kappa \nabla T)$, I reached the first term of the right hand side in eq. 4. For the second term of eq. 4, I assumed that the electric current was along $\hat z$, and that the potential drop was linear with distance (which is consistant with a homogeneous electric current density). Let me know if you wish to have more details.

Yes, I'll probably need more details. I understand the intermediate steps (eq. 2 and 3), but my question is how you reach from these steps eq. (4) (I would be happy if you write down the derivation).

Regarding Veritasium, I would say: don't believe him to much, just use good books and articles.

Finally, I have a problem regarding your zero resistance thought experiment: first, it is impossible to have a difference of potential in this case, as the current would go to infinity.
You say that an energy should be provided to make a current flow in a superconductor. That's probably true taking into account Faradays laws of induction, when the current is created, until it has reached its steady state. But that seems related to AC and not to DC: it would be surprising that your energy flow is linked somehow to induction laws.

EDIT: Building upon my argument above, is it possible that your energy flow is related somehow to the magnetic energy ${1\over 2} L I^2$ due to the inevitable inductance $L$ of the circuit? In this case, this would be a sort of "potential energy flow".(I have to go, so, I may answer later to further posts).

 

[fluidistic]

Yes, I'll probably need more details. I understand the intermediate steps (eq. 2 and 3), but my question is how you reach from these steps eq. (4) (I would be happy if you write down the derivation).

Regarding Veritasium, I would say: don't believe him to much, just use good books and articles.

Finally, I have a problem regarding your zero resistance thought experiment: first, it is impossible to have a difference of potential in this case, as the current would go to infinity.
You say that an energy should be provided to make a current flow in a superconductor. That's probably true taking into account Faradays laws of induction, when the current is created, until it has reached its steady state. But that seems related to AC and not to DC: it would be surprising that your energy flow is linked somehow to induction laws.

EDIT: Building upon my argument above, is it possible that your energy flow is related somehow to the magnetic energy ${1\over 2} L I^2$ due to the inevitable inductance $L$ of the circuit? In this case, this would be a sort of "potential energy flow".(I have to go, so, I may answer later to further posts).

I do not have the time right now to write down the fully detailed calculations. However if you've reached eq. 2, and the expression for $\nabla T$ I obtained by solving the heat eq. of the wire, then there is only the last $\overline{\mu}\vec{J_e}$ part that remains.

Ok about Veritasium, he's eye opener in general, but here I see he's wrong regarding that part (people focussed on other things in that video).

Regarding the zero resistivity, here is the correct approach. We have the relation $\vec{J_e}=-\sigma \nabla \overline{\mu}$, this is Ohm's law (to have the correct units, $\overline{\mu}$ should be divided by the charge of the electron, but this doesn't change anything). We can rewrite it as $\nabla \overline{\mu}=-\rho \vec{J_e}$. You fix the current, and so the current density is fixed. Taking the limit $\rho \to 0$ shows that you need to apply a smaller and smaller voltage (or electrochemical potential gradient) across the wire, to produce the current we want. When the limit is reached, the electrochemical potential gradient required to establish the given current is 0. There is nothing wrong with this. And it's not related to AC with superconductors, here I only considered a normal material whose resistivity gets smaller and smaller, eventually reaching 0 (but without the other superconducting properties).

And so we have that $\overline{\mu}=\mu_0 + \mu_1 z$ (warning, the mu_i's don't have the same units), $z$ being the cylindrical coordinate that increases along the wire's direction. As I wrote in the doc., we can go a little further and reach $\mu_1 =-\rho|\vec{J_e}|$. We can therefore rewrite $\mu$ as $\mu_0-\rho|\vec{J_e}|$. That's how I reached eq. 4.

 

[coquelicot]

I do not have the time right now to write down the fully detailed calculations. However if you've reached eq. 2, and the expression for $\nabla T$ I obtained by solving the heat eq. of the wire, then there is only the last $\overline{\mu}\vec{J_e}$ part that remains.

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).

And it's not related to AC with superconductors, here I only considered a normal material whose resistivity gets smaller and smaller, eventually reaching 0 (but without the other superconducting properties).

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).

 

[Philip Koeck]

Yes. In that case there might be a net energy flux from the particle's motion. It depends on the value of $\overline{\mu_1}\vec J_1+\overline{\mu_2}\vec J_2$. I didn't check the specifics of your case, but if that quantity vanishes, then the energy flux coming from the particle's motion cancels out. If the quantity does not vanish, then there will be a net energy flux coming from the particle's motion.

The point is that in any case, there will be non vanishing energy fluxes associated to the particle's motions. These energy fluxes might cancel each other out, but they still exist.

Let's go back to the wire: The only particles that can move in the wire are electrons, so I assume the energy flux should be in the same direction as the flow of electrons.
This would mean that there is some sort of excess energy on the negative slab of the capacitor in the beginning. Then, as the current flows through the wire this excess energy is transported to the positive slab until the potential difference is zero.
What exactly was this excess energy in the negative slab and why didn't the positive slab have the same amount of excess energy?

 

[fluidistic]

I am not home anymore, can't write latex easily on phone. You don't want to deduce these eqs starting from the final expression of Ju, it's the other way around, you want to deduce that expression starting from the thermodynamics relation for the internal energy. For that, you use the fact that the divergence of the internal enerfy flux vanishes (I am starting to get bored to explain tbhis again). N is thr number of particle in the system, this is standard thermodynamics notation.

Regarding the rest, I don't have time now to reply, but i stand my ground. I feel my doc hasn't been understood, too bad but got no time. More important tghings to solve for me. Not enough time left to spend on this.

 

[coquelicot]

I am not home anymore, can't write latex easily on phone. You don't want to deduce these eqs starting from the final expression of Ju, it's the other way around, you want to deduce that expression starting from the thermodynamics relation for the internal energy. For that, you use the fact that the divergence of the internal enerfy flux vanishes (I am starting to get bored to explain tbhis again). N is thr number of particle in the system, this is standard thermodynamics notation.

Regarding the rest, I don't have time now to reply, but i stand my ground. I feel my doc hasn't been understood, too bad but got no time. More important tghings to solve for me. Not enough time left to spend on this.

No, your doc has almost been understood (by me), but you are unable to provide a decent answer to my question. I have numbered all the equations for you, and all you have to do is to tell me something like: "take eq. 6, inject into eq. 4, then take the divergence and use eq. 2". That's not that complicated, and you even don't need Latex. But I need precise instructions, and the fact that the div of the internal energy flux vanishes don't help as well (Note: as I said, I think I have understood the right side of the right part of the equation, assuming you are trying to use eq. (2). So, you have only to justify the left part of the right side).
Also, I don't understand why you disdain my argument regarding the magnetic energy, and also the excellent argument of Philip Koek. Regarding phonons, admittedly... :-)

 

[coquelicot]

Let's go back to the wire: The only particles that can move in the wire are electrons, so I assume the energy flux should be in the same direction as the flow of electrons.
This would mean that there is some sort of excess energy on the negative slab of the capacitor in the beginning. Then, as the current flows through the wire this excess energy is transported to the positive slab until the potential difference is zero.
What exactly was this excess energy in the negative slab and why didn't the positive slab have the same amount of excess energy?

I think I can try to answer. Why don't you ask the same question for the Poynting vector? it is well known that the energy flows almost parallel to the wires outside, and only in the vicinity of the wire does the Poynting vector become radial. So, you can ask: why does the electromagnetic flux flows from one side and not the other one outside the wires. That's exactly the same problem. I have not a certain answer for that, but I think this is related somehow to the orientation and the sign conventions. So, this is perhaps not a real objection to the theory of Fluidistic.

 

[fluidistic]

No, your doc has almost been understood (by me), but you are unable to provide a decent answer to my question. I have numbered all the equations for you, and all you have to do is to tell me something like: "take eq. 6, inject into eq. 4, then take the divergence and use eq. 2". That's not that complicated, and you even don't need Latex. But I need precise instructions, and the fact that the div of the internal energy flux vanishes don't help as well (Note: as I said, I think I have understood the right side of the right part of the equation, assuming you are trying to use eq. (2). So, you have only to justify the left part of the right side).
Also, I don't understand why you disdain my argument regarding the magnetic energy, and also the excellent argument of Philip Koek. Regarding phonons, admittedly... :-)

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).

 

[Philip Koeck]

I think I can try to answer. Why don't you ask the same question for the Poynting vector? it is well known that the energy flows almost parallel to the wires outside, and only in the vicinity of the wire does the Poynting vector become radial. So, you can ask: why does the electromagnetic flux flows from one side and not the other one outside the wires. That's exactly the same problem. I have not a certain answer for that, but I think this is related somehow to the orientation and the sign conventions. So, this is perhaps not a real objection to the theory of Fluidistic.

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.)

 

[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.)

I was not aware of your H configuration (smart, indeed). Regarding this configuration, I'm not sure. The wire, as a conductor (even if it has some resistance), may well modify the field between the two slabs. In the electrostatic case, the E field is always normal to the surface of the conductor. Here, some current flows so this does not hold, but still, there are surface charges etc. and I'm not so sure that the E-field remains intact as if there were no wire.
Have you think about a more simple configuration (as I thought yours was); I mean, just a cap whose two plates are wired in a loop. Why is the poynting vector outside the wire directed toward such or such direction? After all, this breaks the symmetry too.

Regarding the Poynting vector - heat balance, yes, that's what I learned too (and not only the heat radiation, but the heat in general). I'm not sure the theory of Fluidistic contradicts this fact, and that's a point to make clear, after the math has been checked.

 

[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.