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

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?

• The definition of $\vec{D}$ relevant here seems to be $\vec E + 4\pi \vec{P}$. This definition remains, even if you introduce complex $\hat{\epsilon}_r$ to describe dissipative processes. Commented May 12, 2023 at 15:17
• You've introduced $\hat{\epsilon}_r$, which, as opposed to $\epsilon_r$, describes also dissipative behaviour of the medium. This does not influence definition of $\vec{D}$ or validity of the Gauss law. Commented May 12, 2023 at 15:20

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.