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

  • 1
    $\begingroup$ Please use MathJax to format equations and symbols. $\endgroup$
    – noah
    Commented Mar 16, 2021 at 14:04
  • $\begingroup$ How? can i do that :/ $\endgroup$ Commented Mar 16, 2021 at 14:07
  • $\begingroup$ Have a look here math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – noah
    Commented Mar 16, 2021 at 14:10
  • $\begingroup$ You can't divide by a vector field $\endgroup$ Commented Mar 16, 2021 at 14:52
  • $\begingroup$ Correction then: Charge density = conductivity * abs(E/B) $\endgroup$ Commented Mar 16, 2021 at 15:05

1 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.

  • $\begingroup$ Thanks for your input , and yes this is helpful as a way of understanding why some people would simplify the problem. However now i have 2 issues, how does this now come into play: the fact that when a conductor has an e field inside, all of the charge is now on the surface. but the equation you have just derived seemingly contradicts this? $\endgroup$ Commented Mar 18, 2021 at 19:07
  • $\begingroup$ my guess( if you can shed some insight on this) is because these equations do not possess a boundary condition that the charge is not free to move infinitely and this equation shows the dispersal of charge if it were an infinite universe of conductor ( much like charge disperses to the surface of a finite conductor . my second thing now to pick at is that the standard wave equation also DECAYS exponentially so shouldnt it also still make a difference if charge density decays exponentially, also if there is a current density there must be atleast some charge density no?? $\endgroup$ Commented Mar 18, 2021 at 19:07
  • $\begingroup$ so in the case of normal E fields inside NON charged conductors how can there be a current density if it has zero charge density ( unless a very complex distribution whose integral is zero). Also one last thing... by setting the divergence of E to be zero... does that mean you are excluding the effects of the charges E fields themselfs $\endgroup$ Commented Mar 18, 2021 at 19:07
  • $\begingroup$ Okay, I am not 100% sure about the things I'm going to say, take that in mind. First, the result I derived shouldn't contradict the fact that in conductors have some superficial charge density, since it's only saying that rho (volume density) is zero. That means indeed that charges cannot occupy a certain volume, but it does not impose any restriction on the surface. $\endgroup$ Commented Mar 18, 2021 at 19:23
  • $\begingroup$ Yeah on second thought. i dont think this is done because it is uncharged. as if you supposedly wait for p to be zero. then the assumption of ohms law itself is UNTRUE as how can there be a current density without charge density? $\endgroup$ Commented Mar 18, 2021 at 19:24

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