Does Gauss's law (Maxwell 1st eq) have to change when the conductivity of the material is non-zero?

Question

Studying Lorentz-Drude model on any book (e.g. Optical proerties of solids by Wooten, Fisica II by Mencucini-Silvestrini et c.) one finds the following equation relating the relative dielectric function $\epsilon_r$ and the average polarizability $\alpha$: $$\epsilon_r=1+4\pi\alpha N$$The polarizability becomes complex in a dissipative medium, hence $\epsilon_r$ becomes complex as well. I am quite convinced of the fact that it is convenient to define a complex $\epsilon_r$ in the presence of an harmonic electric field. Indeed if $\vec{E}$ has to verify $\nabla^2\vec{E}=\frac{\mu}{c^2}\left(\epsilon_r\frac{d^2\vec{E}}{dt^2}+4\pi\sigma\frac{d\vec{E}}{dt}\right)$, taken a harmonic $\vec{E}=\vec{E_0}e^{i(\vec{q}\cdot\vec{r}-\omega t)}$ it is evident that $\nabla^2\vec{E}=\frac{\mu}{c^2}\left(\epsilon_r+i\frac{4\pi\sigma}{\omega}\right)\frac{d^2\vec{E}}{dt^2}$, and all wil go as if the $\sigma$ term was embodied in a new complex dielectric function $\hat{\epsilon_r}=\epsilon_r+i\frac{4\pi\sigma}{\omega}$. This $\hat{\epsilon_r}$ must be the same of the equality containing the polarizability shown at the beginning. But that equality comes from $\vec{D}=\hat{\epsilon_r}\vec{E}=\vec{E}+4\pi\vec{P}$. Therefore, the definition of $\vec{D}$ changes to include the complex term due to $\sigma$. Now the question: if Gauss equation states that $$\vec{\nabla}\cdot\vec{D}=4\pi\rho,$$ how do I justify the advent of an extra imaginary term in the definition of $\vec{D}$? Can I do sme reasonin similar to the one i did manipulating the 3rd and 4th Maxwell's equation? If so, where would the extra charge density distribution allowing me to redifine $\vec{D}$ in Gauss law come from?

Answer 1

No, Maxwell's equations already took that into account. I will work in SI units and the microscopic equations, and leave it to you to translate it to what the book has. $$\vec\nabla\times\vec E=-\partial_t\vec B\\ \vec\nabla\times(\vec\nabla\times\vec E)=-\partial_t\vec\nabla\times\vec B\\ \vec\nabla(\vec\nabla\cdot\vec E)-(\vec\nabla\cdot\vec\nabla)\vec E=-\partial_t(\mu_0\vec j+\mu_0\epsilon_0\partial_t\vec E)\\ \frac1{\epsilon_0}\vec\nabla\rho-(\vec\nabla\cdot\vec\nabla)\vec E=-\mu_0\partial_t(\sigma\vec E)-\frac1{c^2}\partial_t^2\vec E\\ \vec\nabla\rho-\epsilon_0\nabla^2\vec E=\frac{i\omega\sigma+\omega^2\epsilon_0}{c^2}\vec E$$ where the RHS clearly has the desired form, if you just factorise out an $\omega^2$

i.e. you just needed to appropriately remember to insert Ohm's Law $\vec j=\sigma\vec E$ into standard Maxwell's equations.