Can Polarization Current Density be a tensor quantity? I've seen a definition of Polarization Current Density (usually given when explaining displacement current) given by:
$$\vec{J}_P=\frac{d\vec{P}}{dt}$$
But this seems to not contain all the information I may want. If we picture charge current as a flow of charges, then Polarization current would be viewed as a flow of dipoles. The above definition assumes that the dipole moments are flowing in the direction of the polarization. But why can't the be pointed other ways? If we are trying to be truly analogous to current and charge, where the continuity equations are defined:
$$ \nabla\cdot\vec{J}=-\frac{d\rho}{dt}$$
$$I =\int{ \vec J }\cdot d\vec a$$
Then I think it would be more general to define a Polarization Current Density tensor (I'll call it H for no particular reason) and Polarization Current vector (I'll call it G):
$$ \nabla\cdot\textbf{H}=-\frac{d\vec{P}}{dt}$$
$$\vec G =\int{ \textbf H }\cdot d\vec a$$
This would allow one to account for a current of dipoles oriented in any direction. I don't think this would particularly be useful for describing displacement current, but I've encountered a modeling problem where this could be a useful analysis. Is anyone aware of this tensor formulation? Or is there a good reason why it is not needed or useful? Does continuity still make sense in this case?
 A: To avoid overcomplicating the situation, think about just one dipole, with a positive charge $+q$ and a negative charge $-q$. Let the vector pointing from $-q$ to $+q$ be $\vec{l}$, so the dipole moment is $\vec{p}=q\vec{l}$. When the dipole's center of mass moves, so long as the vector $\vec{l}$ 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 $\vec{l}$, i.e. relative motion between the opposite charges $\pm q$.
There are two ways the vector $\vec{l}$ can change: it can change its magnitude and orientation. When the relative motion $\dot{\vec{l}}\parallel\vec{l}$ is along the direction of $\vec{l}$, the magnitude $|\vec{l}|$ changes. And when the relative motion $\dot{\vec{l}}\perp\vec{l}$ is perpendicular to $\vec{l}$, the orientation $\hat{l}$ changes. Either motion would cause $\vec{l}$ to change and thus change the dipole moment $\vec{p}=q\vec{l}$. So in writing down $\vec{j}=d\vec{p}/dt$, we are not assuming $\dot{\vec{l}}\parallel\vec{l}$. We have considered all possible currents.
Since $\vec{p}$ is a vector, so is $\vec{j}=d\vec{p}/dt$. 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.
