# Relation between electric charge and electric current distributions via velocity

Assume we have a charge $$\rho(\vec{r},t)$$ and current $$\vec{\jmath}(\vec{r},t)$$ distributions given in some region of space. From Maxwell's equations we know that the only relation between these two distributions is the continuity equation $$\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{\jmath} = 0.$$ Now, in standard fluid dynamics it is possible to define a velocity field $$\vec{v}=\vec{\jmath}/\rho$$ which corresponds to the Eulerian velocity of a mass density $$\rho$$. However, in electrodynamics, we can easily have $$\rho = 0$$ and $$\vec{\jmath} \neq 0$$ and, therefore, cannot infer the velocity $$\vec{v}$$ of charge density. Is this reasoning correct? If so, what did we additionally assume in fluid mechanics so that we could define $$\vec{v}$$ as above?

• How can you have $\rho=0$ but $j\neq 0$? – Deep Mar 14 '19 at 6:22
• @Deep: imagine a neutral current carrying wire. – Fizikus Mar 14 '19 at 13:27
• You mean positive and negative charges cancelling out? In contrast mass comes only in one variety. – Deep Mar 15 '19 at 5:35
• @Deep: exactly! – Fizikus Mar 15 '19 at 10:38