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?
Best Answer
It's because the divergence alone does not determine a vector field. You must know both the divergence and the curl to specify the field. (This is just math, no physics.)
Now the physics:
In general, curl($ \mathbf D $) = curl($\mathbf P$), so you can't simply think of it as an electric field due to the free charge.
In linear media, curl($ \mathbf{D}$)=0, so you in that case you can think of $\mathbf D$ as just the field due to free charges.
Basically, you are thinking if you can use intuition from the $\mathbf E$ field for the $\mathbf D$ field. But for the intuition part, you are most likely assuming electrostatic intuition, so that requires div($\mathbf E$)=$ \frac{\rho}{\epsilon_{0}} $ AND curl($\mathbf{E} $)=0.
Hence, for the $\mathbf D$ field you already know that div($\mathbf D$) = $\rho_{free}$ and curl($\mathbf D$) = curl($\mathbf P$). These equations will look "exactly" like the electrostatic div($\mathbf E$) =$\frac{\rho}{\epsilon_{0}}$, curl($\mathbf E$)=0 only if curl($\mathbf D$) vanishes. General speaking, it doesn't. But it does in linear media.