2004 I 4

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The electrostatic dispersion relation is:

k^{2}\left(1+\sum_{s}\chi_{s}\right)=0

Plugging in for a two species plasma:

1+\frac{2\omega_{pi}^{2}}{k^{2}v_{Ti}^{2}}\left\{ 1+\frac{\omega-kv_{D}}{kv_{Ti}}Z\left(\zeta_{i}\right)\right\} +\frac{2\omega_{pe}^{2}}{k^{2}v_{Te}^{2}}\left\{ 1+\frac{\omega-kv_{D}}{kv_{Te}}Z\left(\zeta_{e}\right)\right\} =0

Expanding the ion term in \zeta\ll1:

\begin{array}{rcl} 1+\frac{2\omega_{pi}^{2}}{k^{2}v_{Ti}^{2}}\left\{ 1+\frac{\omega-kv_{D}}{kv_{Ti}}\left(-\left(\frac{kv_{Ti}}{\omega-kv_{D}}\right)-\frac{1}{2}\left(\frac{kv_{Ti}}{\omega-kv_{D}}\right)^{3}\right)\right\} \\ +\frac{2\omega_{pe}^{2}}{k^{2}v_{Te}^{2}}\left\{ 1+\frac{\omega-kv_{D}}{kv_{Te}}\left(i\sqrt{\pi}e^{-\zeta^{2}}-2\frac{\omega-kv_{D}}{kv_{Te}}\right)\right\} = 0\end{array}

This becomes for \omega\ll kv_{Te}:

1-\frac{\omega_{pi}^{2}}{k^{2}v_{Ti}^{2}}\left(\frac{kv_{Ti}}{\omega-kv_{D}}\right)^{2}+\frac{2\omega_{pe}^{2}}{k^{2}v_{Te}^{2}}=0

Taking vD = 0, and since \omega_{ps}^{2}/\omega^{2}\gg1:

\frac{2\omega_{pe}^{2}}{k^{2}v_{Te}^{2}}=\frac{\omega_{pi}^{2}}{\omega^{2}}

Then:

\frac{2m_{i}m_{e}}{m_{e}T_{e}}=\frac{k^{2}}{\omega^{2}}

Or:

\omega^{2}=k^{2}\frac{T_{e}}{2m_{i}}

This page was recovered in October 2009 from the Plasmagicians page on Generals_2004_I_4 dated 22:22, 9 May 2007.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Generals/2004_I_4"
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