normal modes of a plasma

Question

I am trying to learn plasma physics and it says that the equation: $$ \nabla^2 \mathbf{E}-\nabla(\nabla \cdot \mathbf{E})+\frac{\omega^2}{c^2}\mathbf{\epsilon}\mathbf{E}=0 $$

where $\mathbf{E}$ is the electric field, $\omega$ is the angular frequency of the plasma and $\mathbf{\epsilon}$ is the dielectric tensor for the plasma. It now says if we assume solutions of the form $\exp(i\mathbf{k\cdot r}-\omega t)$ the equation becomes:

$$ \mathbf{M\cdot E}=0 $$ where $$ \mathbf{M}=\frac{\omega^2}{c^2}\mathbf{\epsilon} +\mathbf{kk}-k^2 \mathbf{I} $$

Firstly, what does $\mathbf{kk}$ mean, I assume it is the tensor product but I am not sure?
Secondly how do we get from the first equation to the last? Thirdly, it says the determinant of M are the normal modes of the system, (I assume that means of the field) again, why is this the case?

Answer 1

1) Yes, your assumption is correct. $\mathbf{k}\mathbf{k}$ is a tensor.

2) If you set $\mathbf{E}=\mathbf{\tilde{E}}\exp(i(\mathbf{k\cdot r}-\omega t))$, where $\mathbf{\tilde E}$ is a constant vector, you will get $\mathbf{M\cdot\tilde E}=0$. The operator $\nabla \nabla \cdot$ becomes $-\mathbf{kk}$ and $\nabla^2$ becomes $-k^2$. This is called a linearization of your wave equation.

3) The equation $\mathbf{M\cdot\tilde E}=0$ constitutes an eigenvalue problem, with eigenvalues $k^2$. The corresponding eigenvectors are called eigenmodes, and any solution of the linearized wave equation can be expressed by a linear superposition of these eigenmodes.