# Polarization, Displacement explanation

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?

• There is no reason for the response to be linear. If you put stronger and stronger electric fields, eventually you can even ionise the material - so the linearity has to break down. Commented Jan 18, 2017 at 16:08
• For question (2), you can see en.m.wikipedia.org/wiki/Dielectric and en.m.wikipedia.org/wiki/Polarization_density the relationship between electric permittivity and susceptibility and the fact that this one can be a tensor, as it is used for example in non-linear optics Commented Jan 18, 2017 at 16:10
• For your resulting electric field in question (1), I think there is a wrong sign... Shouldn't it be E=1/ε_0 (D-P)? Commented Jan 18, 2017 at 16:15

## 1 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.