# Continuity equation for charge density

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, $$$$\frac{\partial \rho}{\partial t} + \nabla \cdot j=0$$$$ and $$$$\frac{\partial M_i}{\partial t} + \nabla_i \cdot \tau_{ij} =0$$$$

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, $$$$\phi_M = \frac{\partial(mv)}{\partial t}A^{-1} = \frac{ma}{A}=\frac{F}{A}$$$$ which gives stress so that makes sense. However for the first equation, I do not understand how one obtains current for the flux. It would seem to me that $$j$$ should be the current density instead. Since, $$$$\phi_{\rho} = \frac{\partial q}{\partial t}A^{-1} = \frac{j}{A}$$$$ Which corresponds to current density.

• It’s a current density in the sense that you need to integrate it to get an actual current:$$I=\iint\vec jd\vec S$$ Also, I think that you have a terminology mix up. Charge flux (current) is the totality over the while surface, it has unit $charge/time$
– LPZ
Commented Aug 10, 2022 at 22:55
• So if the paper just calls $j$ current, can I assume that it is referring to what you describe? Commented Aug 10, 2022 at 23:04
• Yes, the notation is standard and can be inferred from the continuity equation
– LPZ
Commented Aug 10, 2022 at 23:39

$$\partial_{t} \rho + \nabla \cdot(\rho {\bf u}) = 0$$ 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.