What is the value of $\vec{\nabla}\cdot\vec{E}$ inside a conductor?

Question

I thought that since a conductor as a whole, an electrically neutral medium, $\vec{\nabla}\cdot\vec{E}=0$ inside a conductor. But while reading Ashcroft and Mermin's Solid state physics, I found out that at equation $1.31$ they assumed $\vec{\nabla}\cdot\vec{E}=0$ but at $1.43$, they assumed $\vec{\nabla}\cdot\vec{E}\neq 0$. I cannot understand this. Please help.

First, they derive the expression for complex frequency-dependent effective permittivity $\epsilon(\omega)=1-\frac{\omega_p^2}{\omega^2}$ assuming $\vec{\nabla}\cdot\vec{E}=0$ (page $17$-$18$). Then on page $19$, they assumed an equation $\vec{\nabla}\cdot\vec{E}(\omega)=4\pi\rho(\omega)$ to show the onset of plasma oscillations.

Answer 1

They assume electrical neutrality in 1.31 as an approximation. An applied AC electric field will cause positive charges in some places and negative charges in others, but this effect is usually pretty small. The charge density wave (plasmon) they describe later (Eqs. 1.42-45) are an exception where the earlier approximation fails. However, these plasmons only occur at specific frequencies. If you apply a field at any other frequency, you don't get a plasmon and the approximation of zero charge density will probably work.

Sometimes you do need more accuracy than the approximation can provide. In that case, you need a more advanced theory, such as the one that produces Eq. 17.60. However, this type of thing gets complicated very quickly, and as Ashcroft and Mermin note, the more advanced theory reduces to the simple theory in the right limit. So, don't make your life complicated if you don't have to.