Warning: jsMath requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

2005 II 4

From QED

< PlasmaWiki | Generals(Link to this page as [[PlasmaWiki/Generals/2005 II 4]])

Jump to: navigation, search

The wave equation is:

$\mathbf{n}\times\left(\mathbf{n}\times\mathbf{E}\right)+\epsilon\cdot\mathbf{E}=0$

The two transverse waves have:

$\left(\chi+\left(1-n^{2}\right)\mathbf{1}\right)\cdot\mathbf{E}=0$

We can take the first two equations:

$\left(1+\chi_{\perp}-n^{2}\right)E_{x}+i\chi_{X}E_{y}=0$ $-i\chi_{X}E_{x}+\left(1+\chi_{\perp}-n^{2}\right)E_{y}=0$

And combine them:

$-\frac{\chi_{X}^{2}}{\left(1+\chi_{\perp}-n^{2}\right)}+\left(1+\chi_{\perp}-n^{2}\right)=0$

Expanding:

$-\chi_{X}^{2}+\left(1+\chi_{\perp}\right)^{2}-2n^{2}\left(1+\chi_{\perp}\right)+n^{4}=0$

Which can be factored to:

$\left(1+\chi_{\perp}+\chi_{X}-n^{2}\right)\left(1+\chi_{\perp}-\chi_{X}-n^{2}\right)=0$

Giving dispersion relations:

$1+\chi_{\perp}+\chi_{X}-n^{2}=0$ $1+\chi_{\perp}-\chi_{X}-n^{2}=0$

The last equation is longitudinal, so:

ε z z = 0

Or:

1 + χ z z = 0

The term that goes to infinity at $ω = Ω H$ is $Z\left(\zeta_{1}\right)$ . This term appears in the dispersion relation:

$1+\chi_{\perp}-\chi_{X}-n^{2}=1+\sum_{s}\frac{\omega_{p}^{2}}{\omega^{2}}\frac{\omega}{k_{\|}v_{t}}Z\left(\zeta_{1}\right)-n^{2}=0$

Using the dispersion relation:

$n^{2}=1+\sum_{s}\frac{\omega_{p}^{2}}{\omega^{2}}\frac{\omega}{k_{\|}v_{t}}Z\left(\zeta_{1}\right)$

For $ω$ real and very close to $Ω H$ , $\zeta_{1}^{\left(e\right)}\gg1$ [note: this assumption comes from Stix, we assume it is based on $kv Ti ˜Ω H$ . This makes sense if it is reasonable to keep the ion term as $Z(ζ)$ , but is not necessarily true], so we take the first term in the electron expansion:

$n^{2}=1-\frac{\omega_{pe}^{2}}{\omega^{2}}\frac{\omega}{k_{\|}v_{te}}\left(\frac{k_{\|}v_{te}}{\omega-\Omega_{e}}\right)+\frac{\omega_{pi}^{2}}{\omega^{2}}\frac{\omega}{k_{\|}v_{ti}}Z\left(\zeta_{1}^{\left(i\right)}\right)$

Since $\omega\ll\Omega_{e}$ :

$n^{2}=1+\frac{\omega_{pe}^{2}}{\omega\Omega_{e}}+\frac{\omega_{pi}^{2}}{\omega^{2}}\frac{\omega}{k_{\|}v_{ti}}Z\left(\zeta_{1}^{\left(i\right)}\right)$

And by quasineutrality, $\omega_{pe}^{2}/\Omega_{e}=-\omega_{pi}^{2}/\Omega_{i}$ , so we will find: