Sorry for the awkwardly phrased title, I wasn't sure how to properly word it.

I want to do the following:

I have a vector $\vec J$ and a vector $\vec E$ with the following relation (with the parallel index referring to a unit-vector $\vec n = \vec e_3$: $$J_\| = i\epsilon_0\sum_s \frac{\omega_{ps}^2}{\omega}E_\|$$ $$\vec J_\bot = \epsilon_0\sum_s \omega_{ps}^2\frac{i\omega \vec E_\bot+\Omega_s \vec n \times \vec E_\bot}{\omega^2-\Omega_s^2}$$

I now want to find the tensor/matrix $\vec{\vec\sigma}$ so that:

$$\vec J = \vec{\vec{\sigma}}\vec E$$

The $\sigma_{33}$ component is simple and straightforward, but I have no idea how to get the other components of the tensor. I've tried writing both sides out component wise and relating that to my original equations but that didn't get me anywhere.

My textbook says that $\sigma_{11}$ = $\sigma_{22}$ and $\sigma_{12}$ = $-\sigma_{21}$ together with their values, but that doesn't make it click for me either.

  • $\begingroup$ Clearly the $(i,3)$ and $(3,i)$ components are zero for $i \in \{1,2\}$. The summation over $s$ is a red herring, you can simply write $J_\perp = \alpha E_\perp + \beta (e_3 \times E_\perp)$. Now recall that due to the cross product's convention, $e_3 \times e_2 = \pm e_1$, and $e_3 \times e_1 = \mp e_2$ i.e., precisely one of those two products will have a negative sign. Thus, $J_1 = \alpha E_1 \pm \beta E_2$, and $J_2 = \alpha E_2 \mp \beta E_1$. Presumably your textbook's statement now makes sense. $\endgroup$ Jun 17, 2016 at 20:23

1 Answer 1


$$J_1 = \epsilon_0\sum_s \omega_{ps}^2\frac{i\omega E_1-\Omega_s E_2}{\omega^2-\Omega_s^2}=\sigma_{11}E_1+\sigma_{12}E_2.$$

Compare and find $\sigma_{11}$ and $\sigma_{12}$. Do the same for $J_2$.


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