Dielectric constant and conductivity in Maxwell's macroscopic equation

Question

In my experimental physics lecture we are looking at the Maxwell equations in matter (macroscopic Maxwell equations) and there is a point where we jump from $$ \nabla \times \vec B = \mu_0 \biggl( \vec j + \frac{\partial \vec D_{bound}}{\partial t}\biggr) $$ to $$ \nabla \times \hat{\vec{B}} = -i \frac{\omega}{c^2} \biggl( \hat{\epsilon}_{bound} + i\frac{\hat{\sigma}}{\epsilon_0 \omega} \biggr)\hat{\vec{E}} $$

We were told that the step is trivial and we should look at it in our free time. My question would be, how do I get from the first equation to the second?

If I'm not mistaken a metal can be considered as a material with a dielectric constant $\hat{\epsilon}_{bound}$ due to bound electrons and a conductivity $\hat{\sigma}$ due to free electrons. Both should be depending on the frequency $\hat{\epsilon}_{bound}(\omega),\hat{\sigma}(\omega) $. Looking at the first equation $\vec D_{bound}$ is due to bound electrons and $\vec j$ is due to the free electrons and in complex notation, $\hat{\vec{E}}=\vec{E}_0\cdot e^{-i\omega t}$ is the electric Field of a plane wave with frequency $\omega$. I don't know if it is imporntant but we looked at a metal as a dielectric with a dielectric constant $\hat{\epsilon}(ω) = \hat{\epsilon}_{bound}(ω) + \hat{\epsilon}_{free}(ω)$, where $\hat{\epsilon}_{free}(ω)$ is the contribution from free electrons. That's not part of the question but I would also be interested how the dielectric constant and the conductivity are related.

Answer 1

Current density and the electric field are related by $\vec{J} = \sigma \vec{E}$.

Your second term is equivalent to $$\mu_0 \vec{J} = \mu_0 \sigma \vec{E} = \frac{\sigma}{\epsilon_0 c^2} \vec{E}\ ,$$ since $\mu_0 \epsilon_0 = c^{-2}$.

The first term is
$$\mu_0 \frac{\partial \vec{D}}{\partial t} = \mu_0 \epsilon \frac{\partial \vec{E}}{\partial t} $$ and if $E = E_0 \exp(-i\omega t)$, then $$\mu_0 \frac{\partial \vec{D}}{\partial t} = -i\omega \mu_0 \epsilon \vec{E} = -i\frac{\omega}{c^2} \epsilon_r \vec{E}\ , $$ where $\epsilon = \epsilon_r \epsilon_0$ and $\epsilon_r$, the relative permittivity, would correspond to your $\epsilon_{\rm bound}$.