2001 II 5

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The electrostatic approximation assumes that \mathbf{E}=-\nabla\phi=-i\mathbf{k}\phi. This is valid for short wavelengths. The condition is that n^{2}\gg\left|\epsilon_{ij}\right| for all elements of the dielectric tensor \mathbf{\epsilon}.

The perturbed distribution function is:

f_{1}\left(v_{\perp},v_{z},\phi,t\right)=-qe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\int_{0}^{\infty}d\tau\, e^{i\beta}\left(\begin{array}{c} U\cos\left(\phi+\Omega\tau\right)\\ U\sin\left(\phi+\Omega\tau\right)\\ \frac{\partial f_{0}}{\partial v_{z}}-V\cos\left(\phi+\Omega\tau\right)\end{array}\right)\cdot\mathbf{E}

With

U=\frac{\partial f_{0}}{\partial v_{\perp}}+\frac{k_{z}}{\omega}\left(v_{\perp}\frac{\partial f_{0}}{\partial v_{z}}-v_{z}\frac{\partial f_{0}}{\partial v_{\perp}}\right)
V=\frac{k_{x}}{\omega}\left(v_{\perp}\frac{\partial f_{0}}{\partial v_{z}}-v_{z}\frac{\partial f_{0}}{\partial v_{\perp}}\right)
\beta=-\frac{k_{x}v_{\perp}}{\omega}\left[\sin\left(\phi+\Omega\tau\right)-\sin\phi\right]+\left(\omega-k_{z}v_{z}\right)\tau
f_{0}=f_{0}\left(v_{\perp},\, v_{z}\right)

The electrostatic approximation gives \mathbf{E}=-\nabla\phi=-i\mathbf{k}\Phi. Then:

E_{x}=-ik_{x}\Phi;\qquad E_{z}=-ik_{z}\Phi

Rewriting the distribution function:

\begin{array}{rcl} f_{1}\left(v_{\perp},v_{z},\phi,t\right)  =  -qe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\int_{0}^{\infty}d\tau\, e^{-ik_{x}v_{\perp}\left[\sin\left(\phi+\Omega\tau\right)-\sin\phi\right]/\omega+i\left(\omega-k_{z}v_{z}\right)\tau}\cdot\\     \left[-ik_{x}\Phi U\cos\left(\phi+\Omega\tau\right)-ik_{z}\Phi\frac{\partial f_{0}}{\partial v_{z}}+ik_{z}\Phi V\cos\left(\phi+\Omega\tau\right)\right]\end{array}

Integrating over angles φ:

\begin{array}{rcl} f_{1}\left(v_{\perp},v_{z},t\right)  =  2\pi iqe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\sum_{n=-\infty}^{\infty}\int_{0}^{\infty}d\tau\, e^{i\left(\omega-k_{z}v_{z}\right)\tau}e^{-in\Omega\tau}\cdot\\   J_{n}^{2}\left(\frac{k_{x}v_{\perp}}{\omega}\right)\left[k_{x}\Phi U\frac{n\omega}{k_{x}v_{\perp}}+k_{z}\Phi\frac{\partial f_{0}}{\partial v_{z}}-k_{z}\Phi V\frac{n\omega}{k_{x}v_{\perp}}\right]\end{array}

If \omega-k_{z}v_{z}\neq n\Omega, this integral will average out to 0. Otherwise the integral is infinity. This gives:

\begin{array}{rcl} f_{1}\left(v_{\perp},v_{z},t\right)  =  2\pi iqe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\sum_{n=-\infty}^{\infty}J_{n}^{2}\left(\frac{k_{x}v_{\perp}}{\omega}\right)\cdot\\   \left[k_{x}\Phi U\frac{n\omega}{k_{x}v_{\perp}}+k_{z}\Phi\frac{\partial f_{0}}{\partial v_{z}}-k_{z}\Phi V\frac{n\omega}{k_{x}v_{\perp}}\right]\end{array}

Integrating over velocity space:

\begin{array}{rcl} n_{1}\left(t\right) = \int v_{\perp}dv_{\perp}\int dv_{z}\,2\pi iqe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\sum_{n=-\infty}^{\infty}\delta\left(\omega-k_{z}v_{z}-n\Omega\right)\cdot\\   J_{n}^{2}\left(\frac{k_{x}v_{\perp}}{\omega}\right)\left[k_{x}\Phi U\frac{n\omega}{k_{x}v_{\perp}}+k_{z}\Phi\frac{\partial f_{0}}{\partial v_{z}}-k_{z}\Phi V\frac{n\omega}{k_{x}v_{\perp}}\right]\end{array}

One integral is over a delta function:

\begin{array}{rcl} n_{1}\left(t\right) = \int dv_{\perp}\, v_{\perp}2\pi iqe^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\sum_{n=-\infty}^{\infty}J_{n}^{2}\left(\frac{k_{x}v_{\perp}}{\omega}\right)\cdot\\   \left[k_{x}\Phi U\frac{n\omega}{k_{x}v_{\perp}}+k_{z}\Phi\frac{\partial f_{0}}{\partial v_{z}}-k_{z}\Phi V\frac{n\omega}{k_{x}v_{\perp}}\right]_{v_{z}=\left(\omega-n\Omega\right)/k_{z}}\end{array}

Using Poisson's equation:

-\left(k_{x}^{2}+k_{z}^{2}\right)\Phi=-4\pi qn_{1}

So:

\begin{array}{rcl} \left(k_{x}^{2}+k_{z}^{2}\right) = 8\pi^{2}iq^{2}e^{i\mathbf{k}\cdot\mathbf{r}-i\omega t}\int dv_{\perp}\, v_{\perp}\sum_{n=-\infty}^{\infty}J_{n}^{2}\left(\frac{k_{x}v_{\perp}}{\omega}\right)\cdot\\ \left[k_{x}U\frac{n\omega}{k_{x}v_{\perp}}+k_{z}\frac{\partial f_{0}}{\partial v_{z}}-k_{z}V\frac{n\omega}{k_{x}v_{\perp}}\right]_{v_{z}=\left(\omega-n\Omega\right)/k_{z}}\end{array}

This page was recovered in October 2009 from the Plasmagicians page on Generals_2001_II_5 dated 19:45, 25 April 2007.

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