# Direction of energy flux in conducting wire confusion

From what I can read on the Internet, in a very simple electrical circuit consisting of a battery connected to a wire with finite resistance in a loop, the energy flux is given by the Poynting vector $$\vec S \propto \vec E \times \vec B$$ and its direction is perpendicukar to both the $$\vec E$$ and $$\vec B$$ fields, and points radially inward the wire, not along the wire. So far so good.

Inside the wire, which has a conductivity sigma, $$|\vec S|$$ gets attenuated the closer to the center one gets, because $$|\vec B|$$ diminishes as less current is enclosed in the cross section that gets smaller and smaller as one gets to the center of the wire. Ok.

However, from thermodynamics, we know there is a relationship between the internal energy and the the number of particles of the system: $$U = (...) + \overline{\mu}N$$ This translates to a link between the internal energy flux and the electric current (as well as with a thermal gradient in case there would be one): $$\vec J_U =-\kappa \nabla T + \overline{\mu}\vec J$$. Now, for an isothermal conducting wire, this implies that the internal energy flux is proportional to the electric current density, and that it has the same direction. However this direction is along the wire, unlike the Poynting vector! I do not undertand where the discrepancy comes from. I am confused. It's as if there was an energy flux represented by the Poynting vector, and another one related to a different (?) energy, the internal energy. So.... would that mean that the total energy would be the sum of both energies? It would mean that the direction of the energy flux created by the battery is neither along nor orthogonal to the wire... What is wrong with my thoughts?

## 1 Answer

If a current carrying wire has reached thermal equilibrium, the energy flux radially inward (associated with the Poynting vector) must be matched by the thermal flux which flows radially outward (not along the wire).

• I do not really understand why this would be the case, could you precise? As far as I understand, Joule heat is generated homogeneously in the wire (because $\vec J$ is constant throughout). Thermal equilibrium does not imply thermodynamics equilibrium, and in fact, since there's a current the situation is not at equilibrium. (although everything is kept at the same temperature). Mar 22, 2022 at 18:09