How to calculate this dispersion relation

  • Thread starter Karl86
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Problem 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.
 

Charles Link

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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.
 

Charles Link

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Somewhere, there seems to be an error or two in that 2x2 matrix. The (1,2) and (2,1) elements are not dimensionally correct.
 
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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.
 

Charles Link

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Yes, I see it now also. It should be first power of ## \omega ## in the numerator of the (2,1) and (1,2) terms.
 

Charles Link

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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.
 

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