Energy flux direction in a conducting wire?

[bob012345]

Yes perhaps I can point out why.

The Poynting Vector shows the energy flow in the fields. There is no "return" path except by our naming convention

Their is no return path! I think you need to look at the Veritasium video again carefully. The Poynting vector always comes out of the battery. Everywhere. On one "leg" it is parallel to the current and on the other it is antiparallel. At the resistor it goes into the resistor. The pertinant image is at 8:00:


I hope that helps

Thanks. I am troubled by a couple of statements such as the energy is going out through the sides of the battery (7:47) and the energy is coming in from all around the bulb (8:13). I am not arguing the concept is wrong I just have questions.

Suppose the battery and or the light bulb were completely shielded (not to mention the wires)? One might accept the idea of a conventional filament light bulb glowing with stray EM fields impinging on it from all angles like a Nikola Tesla demonstration but a modern LED bulb? Or a precision motor where stray fields would disrupt the operation? The Poynting vector model is highly geometry dependent yet the result is completely geometry independent?

 

[hutchphd]

Honestly I don't know the detailed answers, because the physics makes perfect sense where I do know the answers. I would ask you to describe in detail how you would create the shielded ssituations you describe. I don't think you can create those situations. If so we will address them.

 

[bob012345]

Honestly I don't know the detailed answers, because the physics makes perfect sense where I do know the answers. I would ask you to describe in detail how you would create the shielded situations you describe. I don't think you can create those situations. If so we will address them.

I would simply build a Faraday cage around the battery and another Faraday cage around the actual bulb allowing only grounded coax leads through the plane of the cage to the bulb. But we don't need to get into a big debate on the subject. All I am really saying is some situations are not always as idealized as the image in the video. It is a very complex subject and I do not think everything is known as to the details.

 

[hutchphd]

The issue is how to feed the power wires into the Faraday cage. I do not desire a debate nor do I care about your unsupported opinion. I just want to analyze an actual situation. It is not really complex, but it may be subtle. You need to discuss specifics.

 

[vanhees71]

I would simply build a Faraday cage around the battery and another Faraday cage around the actual bulb allowing only grounded coax leads through the plane of the cage to the bulb. But we don't need to get into a big debate on the subject. All I am really saying is some situations are not always as idealized as the image in the video. It is a very complex subject and I do not think everything is known as to the details.

Yes, that's why the "very long coax cable" is a nice setup, because it's on the one hand simple due to its high symmetry and thus can be easily solved analytical and on the other hand it's complete, because it describes a complete closed circuit.

 

[bob012345]

FYI, here is a J.D. Jackson paper on the distribution of charge and the roles that plays in circuits. He analyzes a simple case which turns out not to be so simple.

https://pdfcoffee.com/surface-charges-on-circuit-wires-and-resistors-play-three-roles--pdf-free.html

Here is a bit lower level discussion of the topic;

https://matterandinteractions.org/wp-content/uploads/2016/07/circuit.pdf

Here is Jefimenko's early demonstration of surface charges around a circuit;

http://sharif.edu/~aborji/25733/files/Demonstration of the Electric Fields of Current-Carrying Conductors.pdf

 

[hutchphd]

FYI, here is a J.D. Jackson paper on the distribution of charge and the roles that plays in circuits. He analyzes a simple case which turns out not to be so simple.

https://pdfcoffee.com/surface-charges-on-circuit-wires-and-resistors-play-three-roles--pdf-free.html

Thanks for the article it is pretty interesting. Perhaps you misunderstand my point here. I am not saying the charge and currents can be ignored. The electrodynamic system consists of charges, currents, and fields. (If you want thermodynamics in materials you can couple to other degrees of freedom too!). One should not treat the current like water in a pipe. Wherever there are charges and currents there will be E and H and so they need to be included.
The Poynting Vector is often the simplest picture of the energetics and it is certainly not independent of the system spatial parameters and all the other degrees of freedom.

 

[bob012345]

Thanks for the article it is pretty interesting. Perhaps you misunderstand my point here. I am not saying the charge and currents can be ignored. The electrodynamic system consists of charges, currents, and fields. (If you want thermodynamics in materials you can couple to other degrees of freedom too!). One should not treat the current like water in a pipe. Wherever there are charges and currents there will be E and H and so they need to be included.
The Poynting Vector is often the simplest picture of the energetics and it is certainly not independent of the system spatial parameters and all the other degrees of freedom.

And I was not trying to say anything is wrong with this picture although I was perhaps expressing a little incredulity at the way the video graphically showed it by my questions.

 

[Philip Koeck]

I would also think that no energy (thermal or otherwise) is actually flowing along the wire.

I can offer the following thought experiment:
Let's say you have a very large charged capacitor, essentially a negatively charged slab of metal somewhere and parallel to that some distance away a positively charged slab.
The whole setup can be in vacuum.
Now you connect the two slabs with a wire.
The wire gets hot and gives of thermal energy in form of radiation so clearly there is some flow of energy.
If one wants to think that energy is flowing along the wire the first question would be: In which direction? I don't think there is a sensible answer to this question.

Assuming one believes that energy flows from the negative to the positive charge, that would mean that the amount of energy in the negative slab decreases during this process.
The problem with this picture is that there wasn't any excess energy localized in the negative slab to start with. The available energy was in the form of potential energy involving the potential difference between these slabs, so in a sense the available potential energy was in both slabs.
So it's really hard to maintain that the energy should flow in a particular direction along the wire when you connect the two slabs.

Does my thinking make sense?

 

[fluidistic]

So your point can be summed up as :
"Of course the energy flows in wires, they get hot"

And your paper explains that there is heating that is quadratic in J. I don't see why this disputes the Veritasium picture at all.

That's not how to sum up my point.

My main point can be summed up as $\vec{J_U}=J_r \hat r + \textcolor{red}{J_z \hat z}$, where the red part is what the youtubers claim is 0, but I show it isn't zero regardless of whether the resistivity vanishes or not.

I used cylindrical coordinates, z being along the wire, r being radial to the wire.

 

[hutchphd]

where the red part is what the youtubers claim is 0

Can you reference this? How can there be no J_z in any useful system ? I am lost as to your intent here...

 

[fluidistic]

Can you reference this? How can there be no J_z in any useful system ? I am lost as to your intent here...

Sure, e.g. Veritasium video's thumbnail is an example. And several of his statements (already alluded to in this thread).

 

[hutchphd]

Sure, e.g. Veritasium video's thumbnail is an example. And several of his statements (already alluded to in this thread).

They claim that there is zero net current along the wire? Definite reference please.

 

[fluidistic]

They claim that there is zero net current along the wire? Definite reference please.

Sorry for.the confusing notation, it's supposed to be energy flux. So their claim would be 0 energy flux along the direction of the wire, or along the direction of the current.

 

[hutchphd]

That's why I want the reference....I have lost track of what you are trying to say. This is why references are necessary and useful (and hearsay is not allowed in court).. Specific references please.

 

[fluidistic]

That's why I want the reference....I have lost track of what you are trying to say. This is why references are necessary and useful (and hearsay is not allowed in court).. Specific references please.

Ok, here are a few "Energy doesn't flow in wires" (thumbnail of veritasium).

Around minute 8:56 in Veritasium's video, the guy says that energy flux goes one way from the battery to the bulb (while in reality part of it goes from the bulb to the battery, but this is not showed nor mentioned anywhere in the video. They only focus on the Poynting vector when it comes to energy flux.)

Around minute 9:49 they claim the electrons don't carry the energy (dang, they again missed an energy flux term that is not part of Poynting's vector).

Then it continues like this, i.e. we can focus only on the Poynting vector to check the energy flux's direction. But this is wrong.

I have spent enough time on this for now, I'm done with this topic, I have already written a document for the curious. Too bad if it's not understood, but the thoughts are there.

 

[hutchphd]

the guy says that energy flux goes one way from the battery to the bulb (while in reality part of it goes from the bulb to the battery, but this is not showed nor mentioned anywhere in the video.

What are you talking about? The net energy flux is from battery to bulb. The electrical part is given by the Poynting Vector. Their are some small adjustments from nonideal wire.

 

[fluidistic]

What are you talking about? The net energy flux is from battery to bulb. The electrical part is given by the Poynting Vector. Their are some small adjustments from nonideal wire.

I am talking about the energy flux inside the wire, specifically. If the wire is ideal, Poynting vector vanishes there, yet there is still an energy flux going along the wire in the current's direction. That's the term in red I mentioned above. The term the youtubers mention its non existence.

 

[Philip Koeck]

I am talking about the energy flux inside the wire, specifically. If the wire is ideal, Poynting vector vanishes there, yet there is still an energy flux going along the wire in the current's direction. That's the term in red I mentioned above. The term the youtubers mention its non existence.

I have conceptual problems with this.
In my previous post I describe a setup with a capacitor and a wire something like this:

+H-

The vertical lines of the H are charged slabs of metal (with the charges indicated by + and -) and the horizontal line is a wire, which is the only load in this circuit.

Why would energy flow inside the wire in the direction of the electric current (from - to + ?)?
Why not in the opposite direction?
If I use a tube filled with salt water in stead of the wire, in which direction does the energy flow then?

 

[fluidistic]

I have conceptual problems with this.
In my previous post I describe a setup with a capacitor and a wire something like this:

+H-

The vertical lines of the H are charged slabs of metal (with the charges indicated by + and -) and the horizontal line is a wire, which is the only load in this circuit.

Why would energy flow inside the wire in the direction of the electric current (from - to + ?)?
Why not in the opposite direction?
If I use a tube filled with salt water in stead of the wire, in which direction does the energy flow then?

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.

 

[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.