# Source terms in Maxwell's Equations and retarded positions

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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