Generals 2002 II 7

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[edit] Part a

We have the wave equation:

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


For transverse waves, this becomes:

\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


Plugging in:

\chi_{\perp}=\sum_{s}\frac{1}{2}\frac{\omega_{p}^{2}}{\omega^{2}}\zeta_{0}\left[Z\left(\zeta_{1}\right)+Z\left(\zeta_{-1}\right)\right]


\chi_{X}=\sum_{s}\frac{1}{2}\frac{\omega_{p}^{2}}{\omega^{2}}\zeta_{0}\left[Z\left(\zeta_{1}\right)-Z\left(\zeta_{-1}\right)\right]


We get:

n^{2}=1+\sum_{s}\frac{\omega_{p}^{2}}{\omega^{2}}\zeta_{0}Z\left(\zeta_{1}\right)


n^{2}=1+\sum_{s}\frac{\omega_{p}^{2}}{\omega^{2}}\zeta_{0}Z\left(\zeta_{-1}\right)


[edit] Part b

The wave electric field will be unpolarized, since Ex,Ey = 0. We get the dispersion relation for longitudinal waves:

εzz = 0


This gives:

1 + χzz = 0


[edit] Part c

The dispersion relation was:

1 + χzz = 0


Plugging in:

-1=\sum_{s}\frac{2\omega_{ps}^{2}}{k_{\|}^{2}v_{ts}^{2}}\left[1+\frac{\omega-k_{\|}v_{Ds}}{k_{\|}v_{ts}}Z\left(\zeta_{0}^{\left(s\right)}\right)\right]


For ions, \omega/k\gg v_{ti}, so that \zeta_{0}^{i}=\omega/k_{\|}v_{ti}\gg1. We can take the expansion for the ion term then:

-1=\frac{2\omega_{pe}^{2}}{k_{\|}^{2}v_{te}^{2}}\left[1+\frac{\omega-k_{\|}v_{D}}{k_{\|}v_{te}}Z\left(\zeta_{0}^{\left(e\right)}\right)\right]+\frac{2\omega_{pi}^{2}}{k_{\|}^{2}v_{ti}^{2}}\left[1+\frac{\omega}{k_{\|}v_{ti}}\left(i\sigma\sqrt{\pi}e^{-\zeta_{i}^{2}}-\frac{k_{\|}v_{ti}}{\omega}\left(1+\frac{k_{\|}^{2}v_{ti}^{2}}{2\omega^{2}}\right)\right)\right]


Which becomes, discarding the exponential which is small:

-1=\frac{2\omega_{pe}^{2}}{k_{\|}^{2}v_{te}^{2}}\left[1+\frac{\omega-k_{\|}v_{D}}{k_{\|}v_{te}}Z\left(\zeta_{0}^{\left(e\right)}\right)\right]+\frac{\omega_{pi}^{2}}{\omega^{2}}


Multiplying through by \omega^{2}/\omega_{pe}^{2}, the last term is small (since it goes like the mass ratio me / mi), and we find:

\frac{\omega^{2}}{\omega_{pe}^{2}}=\frac{2}{k_{\|}^{2}v_{te}^{2}}\left[1+\frac{\omega-k_{\|}v_{D}}{k_{\|}v_{te}}Z\left(\zeta_{0}^{\left(e\right)}\right)\right]


Writing ω = ωr + iγ:

\frac{\omega_{r}^{2}+2i\gamma\omega_{r}-\gamma^{2}}{\omega_{pe}^{2}}=\frac{2}{k_{\|}^{2}v_{te}^{2}}\left[1+\frac{\omega_{r}+i\gamma-k_{\|}v_{D}}{k_{\|}v_{te}}Z\left(\zeta_{0}^{\left(e\right)}\right)\right]


Then, setting the imaginary part of both sides equal:

\frac{2i\gamma\omega_{r}}{\omega_{pe}^{2}}=\frac{2}{k_{\|}^{2}v_{te}^{2}}\frac{\omega_{r}}{k_{\|}v_{te}}\mathrm{Im}\left[Z\left(\zeta_{0}^{\left(e\right)}\right)\right]


Or:

\gamma=\frac{\omega_{pe}^{2}}{\left(k_{\|}v_{te}\right)^{3}}\mathrm{Im}\left[Z\left(\zeta_{0}^{\left(e\right)}\right)\right]
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