coquelicot said:
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.
