I wonder why the displacement field $\mathbf{D}$ at a normal mode are parallel to the major and minor axis of the ellipse which is the intersection of index ellipsoid and plane orthogonal to the $\mathbf{k}$.

The explaination in my textbook is below:

The Maxwell equation in anisotropic medium is $$ \nabla \times \mathbf{E}=-\mu_0\frac{\partial \mathbf{H}}{\partial t} $$ $$\nabla \times \mathbf{H}= \frac{\partial \mathbf{D}}{\partial t} $$ Therefore, $$\nabla \times \nabla \times \frac{\epsilon_r ^{-1}}{\epsilon_0}\mathbf{D}=-\mu_0\frac{\partial^2\mathbf{D}}{\partial t^2} $$ In the form of plane wave, we can write above equation as below: $$-\mathbf{k}\times\mathbf{k}\times\epsilon_r ^{-1}\mathbf{D}=\mu_0\epsilon_0\omega^2\mathbf{D}=k_0^2 \mathbf{D} $$ where $\mathbf{k}=k\mathbf{\hat{u}}$ and $k=k_0n$. So we get $$-\mathbf{\hat{u}}\times\mathbf{\hat{u}}\times \epsilon_r^{-1}\mathbf{D}=\frac{1}{n^2} \mathbf{D} $$ Here the operator $(-\mathbf{\hat{u}}\times\mathbf{\hat{u}}\times)$ means the projection to the normal plane of $\mathbf{\hat{u}}$, denoted by $P_\mathbf{u}$. So this is an eigenvalue problem: $$P_\mathbf{u}\epsilon_r^{-1} \mathbf{D}=\frac{1}{n^2}\mathbf{D}$$ where $\mathbf{D}$ is an eigenvector with an eigenvalue $1/n^2$ of the linear map $P_\mathbf{u}\epsilon_r^{-1}$

However, I cannot connect this eigenvalue problem and the geometrical problem about an index ellipsoid quantitatively. Why should the eigenvector be parallel to the axis of ellipse?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.