I was just learning about EM fields in matter, about displacement, dielectric polarization... You have the following equations: $\textbf D = \epsilon_0\textbf E + \textbf P$, where $\textbf D$ denotes the displacement, $\textbf P$ the polarization and $\textbf E$ the electric field. I know that $\textbf P = \frac{d\textbf p}{dV}$, where $\textbf p$ denotes the dipole moment...
My question 1: Can I just say that $\textbf P$ is basically the electric field produced by the bound charge, divided by $\epsilon_0$ and the same for $\textbf D$ with free charge, since $ \nabla.\textbf D=\rho_{free}$? So the resulting electric field $\textbf E$ is the sum of $\textbf D/\epsilon_0\space+\textbf P/\epsilon_0$, the dielectric dissorts the "original" electric field with it's polarization: $\textbf E=\textbf E_{free}+\textbf E_{bound}$ ?
My question 2: Why is the relationship $\textbf D=\epsilon_0 \epsilon \textbf E$ only valid for linear dielectrics, can't $\epsilon$ be a tensor and the relationship be valid in general?