EM-wave equation in conductors with source terms

Question

The traditional modified Maxwell's equations to express em wave inside conductors that I have come across are:

$$ \nabla\cdot\mathbf E = 0 \\\nabla\cdot\mathbf B = 0 \\\nabla\times\mathbf E = -\frac{d\mathbf B}{dt} \\\nabla\times\mathbf B = \mu\sigma\mathbf E+\mu\epsilon\frac{\partial\mathbf E}{\partial t}$$

where use of $\mathbf J=\sigma \mathbf E$ has been made

This makes sense, inside a conductor wherever there is an electric field, there is a corresponding current density due to free charge throughout

However I am not sure of the physical significance of setting the divergence of $\mathbf E$ to be zero?

Why is this done, and for that matter in the general wave equation in free space why does this also occur? As an EM wave needs to be generated by a source (the free space wave equation I'm guessing is to show the field itself behaves like a wave in general).

But for inside conductors, what would adding that the $\nabla\cdot\mathbf E = \rho/\epsilon$ actually physically mean, and what is the difference between the two?

I have had a good go at solving for the potentials $\phi$ and $\mathbf A$ with source terms.

Solving for $\mathbf A$ is pretty straightforward, providing the gauge choice is

$$\nabla\cdot\mathbf A - \mu\sigma\phi - \mu\epsilon\frac{\partial \phi}{\partial t} = 0 $$

and the equation for the magnetic vector potential I get is:

$$ - \nabla^2\mathbf A+\mu\sigma\frac{d\mathbf A}{dt} + \mu\epsilon\frac{d^2\mathbf A}{dt^2}=0 $$ (correct me if I'm wrong)

The standard potential formulation equation for $\phi$ is initially unchanged, however adding the gauge condition mentioned earlier you get a relatively complicated equation.

Another idea to use (most likely useless):

However is it valid to substitute $\rho/\epsilon$ for $\frac{\sigma\mathbf E}{\epsilon\mathbf v}$ where $\mathbf v$ is the velocity field, then exchange $\mathbf E$ for the potentials $\mathbf A$ and $\phi$, (as obviously $\mathbf J = \rho\mathbf v = \sigma\mathbf E$)?

Edit: doesnt setting p to be zero contradict the statement that $$ J = \sigma E $$ as $$ J = \rho * V $$ so must conclude that E =0

Answer 1

I'm not sure if this is the result you're looking for, but if you take the divergence of Ohm's law $$ \nabla·\vec{J}=\sigma\nabla·\vec{E} \to \nabla·\vec{J}=\frac{\sigma\rho}{\epsilon_0} $$ Now using the continuity equation $$ \nabla·\vec{J}=-\frac{d\rho}{dt} $$ And if we combine both results $$ \frac{\sigma\rho}{\epsilon_0}=-\frac{d\rho}{dt} \to \rho=\rho_0e^{-\frac{\sigma}{\epsilon_0}t} $$ What we conclude from this is that, when the wave creates the current in the conductor, some charge distribution will appear as a consequence, but it will decrease exponentially, decaying faster if we're dealing with a good conductor where $\sigma\to\infty$. So assuming a rapid decay of the charge, we can consider that after some relatively short time $\tau$ we'll have $\rho=0$ and hence $\nabla·\vec{E}=0$ , as first stated. You can always wait until this result is valid.