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...