How to derive the permittivity conductivity relation?

Question

I would like to derive the common (and important - to me atleast) equation:

$\epsilon = \epsilon_a + j\frac{\sigma}{\omega\epsilon}$

I have seen some resources calling this a definition (in which case, it may not have a derivation), but it looks like it definitely should and it should fall out of Maxwell's equations, but I cannot do it!

So, starting with,

$\nabla \times \vec{H} = j\omega\vec{D}+\vec{J}$

and using Ohm's law and a constitutive relation, which for convenience is:

$\vec{J} = \sigma \vec{E}$ and $\vec{D} = \epsilon\vec{E}$

One can arrive at:

$\nabla \times \vec{H} = j\omega\epsilon\vec{E}+\sigma\vec{E}$ (Eq. 1)

Keeping the above in mind for a moment and working on this:

$\nabla \times \vec{E} = -j\omega\mu\vec{H}$

one can take the curl of both sides:

$\nabla \times (\nabla \times \vec{E}) = -j\omega\mu\nabla \times \vec{H}$

Using the vector identity $\nabla \times (\nabla \times \vec{A}) = (\nabla(\nabla\cdot\vec{A})-\nabla^2 \vec{A})$, recalling Eq. 1 and noting that for a region without a source $\nabla\cdot\vec{E} = 0$, the above becomes:

$\nabla^2\vec{E}=-\omega^2\epsilon\mu\vec{E}+j\omega\mu\sigma\vec{E}=0$

$\nabla^2\vec{E}+\omega^2\epsilon\mu\vec{E}(1-j\frac{\sigma}{\omega\epsilon})=0$

This wave equation for a lossy material is as far as I can get...

Answer 1

It is actually just the rhs of one of the Maxwell's equations $$ \nabla\times\mathbf{H}=\mathbf{J}+\frac{\partial \mathbf{D}}{\partial t}. $$ If we perform Fourier transform in time (or simply take $\mathbf{H},\mathbf{J},\mathbf{D},\mathbf{E}\propto e^{j\omega t}$) then $$ \nabla\times\mathbf{H}(\omega)=\mathbf{J}(\omega)-j\omega \mathbf{D}(\omega) = \sigma \mathbf{E}(\omega)-j\omega \epsilon\mathbf{E}(\omega)=\\-j\omega\left(\frac{\sigma}{-j\omega}+\epsilon\right)\mathbf{E}(\omega)=-j\omega\epsilon_{eff}\mathbf{E}(\omega)=-j\omega\mathbf{D}_{eff}(\omega), $$ where the effective permittivity is introduced in analogy with $\mathbf{D}=\epsilon\mathbf{E}$.

It could be also be generalized to frequency dependent conductance and permittivity, taking the correct real-time relation between the quantities: $$ \mathbf{J}(\omega)=\sigma(\omega)\mathbf{E}(\omega)\Leftrightarrow \mathbf{J}(t)=\int_{-\infty}^tdt'\sigma(t-t')\mathbf{E}(t'),\\ \mathbf{D}(\omega)=\epsilon(\omega)\mathbf{E}(\omega)\Leftrightarrow \mathbf{D}(t)=\int_{-\infty}^tdt'\epsilon(t-t')\mathbf{E}(t') $$

Answer 2

Let's define a new wave equation that electric field will be subject to. In phasor notation we have $\frac{\partial}{\partial t}\to j\omega$, which gives us $$\nabla^2\vec{E}+\omega^2\frac{1}{v^2}\vec{E}=0$$

We can compare this to your last equation which read $$\nabla^2\vec{E}+\omega^2\epsilon\mu\vec{E}(1-j\frac{\sigma}{\omega\epsilon})=0$$ $$\nabla^2\vec{E}+\omega^2\epsilon\mu\vec{E}(1-j\frac{\sigma}{\omega\epsilon})=0=\nabla^2\vec{E}+\omega^2\frac{1}{v^2}\vec{E}$$ $$\implies \omega^2\epsilon\mu\vec{E}(1-j\frac{\sigma}{\omega\epsilon})-\omega^2\frac{1}{v^2}\vec{E}=0$$ $$\implies \omega^2 (\epsilon\mu-j\frac{\sigma\mu}{\omega}-\frac{1}{v^2}) \vec E=0 \implies \epsilon\mu-j\frac{\sigma\mu}{\omega}-\frac{1}{v^2}=0$$

Normally, we would have $v^2=\frac{1}{\epsilon\mu}$. We can still define effective dielectric and magnetic constants to reflect the characteristics of this system. As it is usual, we set $\mu_{eff}=\mu$ since magnetic effects are usually ignored in these settings, $$\epsilon\mu-j\frac{\sigma\mu}{\omega}=\epsilon_{eff}\mu_{eff}=\epsilon_{eff}\mu$$ $$\implies \boxed{\epsilon_{eff} = \epsilon-j\frac{\sigma}{\omega}}$$

First, this doesn't match exactly your formula in the beginning, however the Wikipedia page actually has this result.

I actually don't know if in free space $\sigma\ne0$ results in any other behavior than the "lossy" EM wave propagation. But if that's all, this assumption accurately models the system (since the wave equations are of correct shape).