Question

Can Polarization Current Density be a tensor quantity?

Answer 1

To avoid overcomplicating the situation, think about just one dipole, with a positive charge +q+q and a negative charge −q−q. Let the vector pointing from −q−q to +q+q be l⃗ l→, so the dipole moment is p⃗ =ql⃗ p→=ql→. When the dipole's center of mass moves, so long as the vector l⃗ l→ remains constant, the motion will contribute no net flow of charge. It's a dipole. So any current has to come from the change of l⃗ l→, i.e. relative motion between the opposite charges ±q±q.

There are two ways the vector l⃗ l→ can change: it can change its magnitude and orientation. When the relative motion l⃗ ˙∥l⃗ l→˙∥l→ is along the direction of l⃗ l→, the magnitude |l⃗ ||l→| changes. And when the relative motion l⃗ ˙⊥l⃗ l→˙⊥l→ is perpendicular to l⃗ l→, the orientation l^l^ changes. Either motion would cause l⃗ l→ to change and thus change the dipole moment p⃗ =ql⃗ p→=ql→. So in writing down j⃗ =dp⃗ /dtj→=dp→/dt, we are not assuming l⃗ ˙∥l⃗ l→˙∥l→. We have considered all possible currents.

Since p⃗ p→ is a vector, so is j⃗ =dp⃗ /dtj→=dp→/dt. The moment contributed by all dipoles per unit volume gives the current density. The current due to dipoles is always a vector. Quadruples may contribute more complicated currents.

Answer 2

the current density expression is fully descriptive when defining the charge current from moving dipoles. Thank you, that makes sense. \Looking at my units, I suppose what I define as G has Am and H has Am/m^3. I suppose what I'm still wondering is if people ever define a moment current instead of a charge current.