# EM-wave equation in conductors with source terms

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

• Please use MathJax to format equations and symbols.
– noah
Mar 16, 2021 at 14:04
• How? can i do that :/ Mar 16, 2021 at 14:07
• Have a look here math.meta.stackexchange.com/questions/5020/…
– noah
Mar 16, 2021 at 14:10
• You can't divide by a vector field Mar 16, 2021 at 14:52
• Correction then: Charge density = conductivity * abs(E/B) Mar 16, 2021 at 15:05

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.