How to calculate this dispersion relation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Karl86
Messages
40
Reaction score
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.
 
Physics news on Phys.org
The non-trivial solutions have the determinant of the 2x2 matrix on the left equal to zero. Perhaps that helps. Otherwise, for a complete treatment, see Ichimaru's Basic Principles of Plasma Physics A Statistical Approach Section 3.1.
 
  • Like
Likes   Reactions: Karl86
Charles Link said:
Somewhere, there seems to be an error or two in that 2x2 matrix. The (1,2) and (2,1) elements are not dimensionally correct.
There is one extra ##\omega## factor in both but I can no longer edit the post. It all worked out anyway, except that I thought there was a general strategy to obtain dispersion relations for generic waves, instead it looks like there are ad hoc methods, so to speak.
 
And for this one, you need to do two cases separately: ## \omega > \omega_c ##, and ## \omega < \omega_c ##. I worked the first case and got the minus sign in the denominator. The second one is trickier, but it no doubt gives the plus in the denominator.
 
  • Like
Likes   Reactions: Karl86