# The truest/most general Maxwell's equations in isotropic, linear, inhomogeneous media with sources

Sources use $\mu H=B$ and $\epsilon E= D$, assuming homogeneous media.

Obviously if $\mu$ is space varying, $\nabla . (\mu H)$ need not be equal to $\nabla . B$

What is the most general form for Maxwell's equations in isotropic, linear, inhomogeneous media with sources.

$$\nabla\cdot\vec{D} = \rho$$ $$\nabla\cdot\vec{B} = 0$$ $$\nabla \times \vec{E} = -\partial_t\vec{B}$$ $$\nabla \times \vec{H} = \vec{J} + \partial_t\vec{D}$$
These are the set you use in linear, isotropic inhomogeneous mediums. So, for example, from the Gauss law for magnetism, you get $\mu\,\nabla\cdot\vec{H} + \nabla\mu \cdot \vec{H}=0$. The displacement and induction vectors $\vec{D}$ and $\vec{B}$ are defined so that their fluxes through a closed surface measure the total free electric or magnetic charge inside that surface; Likewise, $\vec{E}$ and $\vec{H}$ are defined so that their fluxes through a surface with boundary is the relevant "motive force" (EMF or MMF). Once we have the two Gauss laws, which vectors enter the Faraday and Ampère laws on the right hand side is clear: since $\nabla\cdot\nabla\times \vec{F} = 0$ for any vector field $\vec{F}$ with continuous second derivatives, we must have, from Faraday's law $-\partial_t\nabla\cdot\vec{B}=0$ (we'd get a contradiction with Gauss's law for magnetism if it were the $\vec{H}$ vector in this equation) and also, from Ampère's law, $\nabla\cdot \vec{J} + \partial_t\nabla\cdot\vec{D} = \nabla\cdot \vec{J} + \partial_t\rho = 0$, which is the charge continuity equation. We wouldn't get this if it were $\vec{E}$ that entered this equation.
• It might add to the utility of this correct answer to indicated explicitly the relationships between $B$ and $H$ and between $D$ and $E$. – dmckee Jul 11 '15 at 17:42