Continuity equation for charge density

Question

Let $\rho$ be the charge density and $M_i$ the momentum density. The article I am reading states that the continuity equations for this system are given by, \begin{equation} \frac{\partial \rho}{\partial t} + \nabla \cdot j=0 \end{equation} and \begin{equation} \frac{\partial M_i}{\partial t} + \nabla_i \cdot \tau_{ij} =0 \end{equation}

The second equation makes sense to me since flux is defined as the rate at which the quantity flows divided by the area which the quantity flows through. Thus, \begin{equation} \phi_M = \frac{\partial(mv)}{\partial t}A^{-1} = \frac{ma}{A}=\frac{F}{A} \end{equation} which gives stress so that makes sense. However for the first equation, I do not understand how one obtains current for the flux. It would seem to me that $j$ should be the current density instead. Since, \begin{equation} \phi_{\rho} = \frac{\partial q}{\partial t}A^{-1} = \frac{j}{A} \end{equation} Which corresponds to current density.

Answer 1

The continuity equation in EM is analogous to the hydrodynamical continuity equation:

$$ \partial_{t} \rho + \nabla \cdot(\rho {\bf u}) = 0 $$ where the quantity $ \rho \mathbf{u}$ represents a kind of "flux" or "flux density", this is exactly the same as the form of the current density $\mathbf{j}$, which is $\mathbf{j} = \rho \mathbf{u}$, where $\rho$ is the charge density and $\mathbf{u}$ the particle drift velocity. Typically, we just call $\mathbf{j}$ current for simplicity.