# Change in electric displacement field = change in electric field?

(If any of the following steps are wrong, please correct me)

The well known relationship, generalized for anisotropic materials, relate the electric displacement field with the electric field and the polarization field like so:

$$D_i = \epsilon_0 E_i + P_i$$ which, by using the dielectric permittivity $$\epsilon$$ will be written (at first order):

$$D_i = \epsilon_{ij} E_j$$

Conversely, the impermittivity tensor sometimes written $$\eta$$ relates the electric field to the electric displacement field:

$$E_i = \eta_{ij} D_j \tag{1} (1)$$

Now, if suddenly a change in the dielectric permittivity happens, this also affects the dielectric displacement field i.e.

$$\Delta D_i = \Delta \epsilon_{ij} E_j$$ Note however that this, a priori, did not modify the electric field. Indeed, if some matter-related phenomenon happens, then my understanding is that the polarization field $$P_i$$ would be the culprit here for the change in displacement field: \begin{align} \Delta D_i =& \Delta P_i \\ \Delta E_i =& 0 \end{align}

I can now get to my essential point: I find the usual notation quite ambiguous. Is the obtained $$E_i$$ in Eq. (1) in fact only the electric field (which I expect not to change upon entering a dielectric medium), or does it actually includes the polarization field too?

This is of importance when considering electric wave equations that undergo a change in displacement field, so as to know if the inclusion of $$\Delta E$$ is in order or not (again, I expect this quantity to be 0). The wave equation for $$E$$ does not concern $$P$$ (where I expect $$\Delta P \neq 0$$).

Furthermore, this means that $$E$$ could be written, indinstinctively, as \begin{align} E_i =& (\eta + \Delta \eta) (D + \Delta D) \\ =& \eta D \end{align} so that if I follow this reasoning, I could write the wave disturbed wave equation for $$E$$ as

$$\nabla \times(\nabla \times \overrightarrow{\mathbf{E}})-\frac{1}{c^{2}} \frac{\partial^{2} \epsilon \overrightarrow{\mathbf{E}}}{\partial t^{2}}=\frac{1}{\epsilon_{0} c^{2}} \frac{\partial^{2} \Delta \epsilon \overrightarrow{\mathbf{E}}}{\partial t^{2}}$$

and getting the equation for the displacement field $$D$$ :

$$\nabla \times(\nabla \times (\eta\overrightarrow{\mathbf{D}}))-\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{\mathbf{D}}}{\partial t^{2}}=\frac{1}{\epsilon_{0} c^{2}} \frac{\partial^{2} (\epsilon \Delta \eta \overrightarrow{\mathbf{D}})}{\partial t^{2}} \tag{2}$$ (where in the right-hand part I have used the fact that $$\Delta \epsilon_{ij}\eta_{jk}= - \epsilon_{ij}\Delta \eta_{jk}$$)

Does this last Eq. (2) seem correct?

$$D_i=\epsilon_{ij}E_j$$
Not all materials are linear dielectrics, and there are indeed situations where $$P$$ is less dependent on $$E$$ (see, for example, ferroelectric materials and electrets). In those cases, the equation above is not valid.
• @elkevn It's unusual to see the permittivity tensor $\epsilon$ change as a function of applied field. The permittivity of a material at a given frequency is usually only a function of the material itself, not the applied field. Jun 23 '20 at 19:50
• To be transparent, my actual problem has to do with electrostriction. In which case it is customary to write that $\Delta \eta_{ij} = p_{ijkl}S_{kl}$ with $p$ the elasto-optic tensor, $S$ the strain tensor; in the case of electrostriction, specifically, the stress induced is then proportional to $p_{ijkl} D_i D_j$. However, setting elecrostriction aside, suppose the change in $\epsilon$ comes from an acoustic wave (whic is true with electrostriction too). From this, what do you think of my assumption that $E$ is constant when $D$ changes? Jun 24 '20 at 7:10