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I am questioning the validity of the standard formula for EM waves in conductors

The standard treatment of EM waves in conductors are:

$$\nabla \cdot \vec{E} = 0$$

$$\nabla \cdot \vec{B} = 0$$

$$\nabla × \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

$$\nabla × \vec{B} = \mu_0 [\sigma \vec{E}] + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$$

Here the use of ohms law, $\vec{J} = \sigma \vec{E}$, has been made.

Here, the equations tell us there where ever there is an electric field, there is also a current density.

This seems to imply that the equations are finding the EM field of a plane wave passing through an INFINITE conductor.

What if we want to find the fields of a finite conductor?

Ohms law should only apply to a restricted domain, with the rest of the current density being zero.

With this restriction, we also cannot have $\nabla \cdot \vec{E} = 0$ as the divergence of this current density function is not zero on the boundary[and thus $\frac{\partial \rho}{\partial t} ≠ 0$] so zero is not a valid charge density satisfying the continuity equation. There should be a buildup of charge on either side of the conductor affecting the fields.

I think a valid way to describe this situation is: $$\vec{J} = \sigma \vec{E}\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\int_{z_{1}}^{z_{2}} \delta ( x-x') \delta (y-y') \delta (z-z')dx'dy'dz'$$

And then finding the corresponding $\rho$ from the continuity equation.

So my final question is, am I correct on the assumptions placed by the standard formula for EM waves in conductors, is my "fix" correct. And does it actually make a big difference on the fields?

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2 Answers 2

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Here, the equations tell us there where ever there is an electric field, there is also a current density.

Only if the conductivity $\sigma$ is non-zero. The conductivity of empty space vanishes. The rest of your question seems to be finding a way to impose that $\sigma \mathbf E\rightarrow 0$ in the vacuum, which is already true.

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  • $\begingroup$ Very true!! However in solving we assume $\sigma$ is independant on $\vec{r}$ When trialing a test function to obtain the dispersion relation, So limitations on the standard formula used are still in question. So my append to ohms law can be chosen to group the scalar sigma and delta terms into one term, the "full conductivity" that is dependant on position $\endgroup$ Aug 3, 2022 at 20:22
  • $\begingroup$ @jensenpaull We certainly don't make that assumption if we're talking about the fields inside and outside of a finite conductor, in which case it is manifestly false. $\endgroup$
    – J. Murray
    Aug 3, 2022 at 20:25
  • $\begingroup$ The only reference to waves in conductors I can find is when we treat sigma as constant. Which is where the standard e.g skin depth formula comes from. Do you know of any other resources I can find for a treatment of finite conductors in the context of waves $\endgroup$ Aug 3, 2022 at 20:27
  • $\begingroup$ @jensenpaull If you want to treat a finite conductor, you need to solve for the fields inside the conductor (where $\sigma$ is constant) and outside the conductor (where $\sigma=0$) and then apply the relevant boundary conditions at the surface to match the solutions together. $\endgroup$
    – J. Murray
    Aug 3, 2022 at 20:32
  • $\begingroup$ @jensenpaull See e.g. Griffiths, section 9.4.2 (or basically any other EM text). $\endgroup$
    – J. Murray
    Aug 3, 2022 at 20:37
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The case of a finite conducting medium is no different from the case of a finite dielectric medium, and the approach is the same in both cases: we have to solve the Maxwell equations in the two regions and then connect them correctly at the bounding surface using the boundary conditions derived from the Maxwell equations. This does allow for sheets of charge at the boundary wherever there is a discontinuty in $\mathbf(E_n)$.

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