# Source terms in Maxwell's Equations and retarded positions

Smacal1072
Hi All,

Thanks again to all the great mentors and contributors to this forum.

I wanted to ask a question about the Gauss's law/Ampere's law equations in Maxwell's Equations:

$\nabla \bullet \textbf{E} = \frac{\rho}{\epsilon_0} \\ \\ \nabla \times \textbf{B} = \mu \left( \textbf{J} + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)$

For charge distributions or currents that are accelerating, I was taught that you need to consider the retarded potentials in order to derive the fields. Why are Maxwell's Equations not written like this:

$\nabla \bullet \textbf{E} = \frac{\rho_r}{\epsilon_0} \\ \\ \nabla \times \textbf{B} = \mu \left( \textbf{J}_r + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)$

Where $\textbf{J}_r \mbox{ and } \rho_r$ are the retarded charge and retarded current density?

Edit: In retrospect, I should have used the integral version of the equations, in particular:

$\oint \textbf{B} \bullet dl = \mu_0 \iint \left( \textbf{J} + \epsilon_0 \frac{\partial \textbf{E}}{\partial t} \right) \bullet d\textbf{S}$

For example, if we instantly switch on a current element at the origin at t = 0, then calculate $\oint \textbf{B} \bullet dl$ at a radius of a million miles, we'll get zero, even though at that instant, a current may be flowing. Unless the current density and "displacement current" cancel out, the inequality won't hold...

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## Answers and Replies

mikeph
There are some catches here. The differential forms don't have retarded potentials because they are, by definition, locally defined.

And as soon as you turn the current on, what makes you think dE/dt is zero? increasing J creates an increasing rot(B) which creates an increasing rot(E), and dE/dt should be in the opposite direction of the original increase in current. I'm not sure how these actually manifest in the retarded potential, but my lazy guess is that J and e0*dE/dt will cancel until the resulting EM radiation reaches your loop integral.

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