- #1
Karl86
- 40
- 3
- Homework Statement
- Let ##\mathbf{E}## be an electric field that behaves like ##e^{i\mathbf{k}\cdot\mathbf{r} -\omega t}## and consider the equation relating ##\mathbf{E}## to the displacement field ##\mathbf{D}##: $$\mathbf{k}^2 \mathbf{E}=\frac{\omega^2}{c^2}\mathbf{D}$$ ##\mathbf{D}## is related to ##\mathbf{E}## by a permittivity matrix ##\epsilon_{ij}## such that ##D_i=\sum_j\epsilon_{ij} E_j##, but this should not be relevant. If I know that the solutions ##\mathbf{E}## to the equation that are transverse to ##\mathbf{k}##, are given by the ##(E_1,E_2)## such that
$$ \begin{pmatrix} \omega^2 - c^2 k^2 -\frac{\omega_p^2 \omega^2}{\omega^2 - \omega_c^2} & i\frac{\omega_p^2 \omega_c \omega^2}{\omega^2-\omega_c^2} \\ -i\frac{\omega_p^2 \omega_c \omega^2}{\omega^2-\omega_c^2} & \omega^2 - c^2 k^2 -\frac{\omega_p^2 \omega^2}{\omega^2 - \omega_c^2} \end{pmatrix} \begin{pmatrix} E_1 \\ E_2 \end{pmatrix} = 0 $$ It is claimed that the dispersion relation of such a wave solution is $$c^2 k^2 = \omega^2 \left(1 - \frac{\omega_p^2}{\omega(\omega \pm \omega_c)}\right) $$ I don't understand this.
- Relevant Equations
- ##\omega^2=\Omega^2(k)## ??
I have no idea how this dispersion relation was deduced, and also what's the meaning of including plus and minus in the formula.