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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.