fluidistic said:
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).
