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Next: Collisional damping Up: Waves in cold plasmas Previous: Resonances

The resonant layer

Consider the situation under investigation in the preceding section, in which a plane wave, polarized in the $y$-direction, is launched along the $z$-axis, from an antenna located at large positive $z$, and absorbed at a resonance located at $z=0$. In the vicinity of the resonant point, the electric component of the wave satisfies
\begin{displaymath}
\frac{d^2 E_y}{dz^2} + \frac{k_0^2\,b}{z+{\rm i}\,\epsilon} E_y = 0,
\end{displaymath} (583)

where $b>0$ and $\epsilon<0$.

The time-averaged Poynting flux in the $z$-direction is written

\begin{displaymath}
P_z = - \frac{(E_y\,B_x^{~\ast} + E_y^{~\ast}\,B_x)}{4\,\mu_0}.
\end{displaymath} (584)

Now, the Faraday-Maxwell equation yields
\begin{displaymath}
{\rm i}\,\omega\,B_x = -\frac{d E_y}{dz}.
\end{displaymath} (585)

Thus, we have
\begin{displaymath}
P_z = -\frac{{\rm i}}{4\,\mu_0\,\omega} \left(\frac{d E_y}{dz}\, E_y^{~\ast}
- \frac{d E_y^{~\ast}}{dz} \,E_y\right).
\end{displaymath} (586)

Let us ascribe any variation of $P_z$ with $z$ to the wave energy emitted by the plasma. We then have

\begin{displaymath}
\frac{d P_z}{dz} = W,
\end{displaymath} (587)

where $W$ is the power emitted by the plasma per unit volume. It follows that
\begin{displaymath}
W = -\frac{{\rm i}}{4\,\mu_0\,\omega}\left(\frac{d^2 E_y}{dz^2}\,E_y^{~\ast}
- \frac{d^2 E_y^{~\ast}}{dz^2}\,E_y\right).
\end{displaymath} (588)

Equations (583) and (588) yield
\begin{displaymath}
W = \frac{2\,k_0^{~2}\,b}{4\,\mu_0\,\omega} \frac{\epsilon}{z^2 + \epsilon^2}\, \vert E_y\vert^2.
\end{displaymath} (589)

Note that $W<0$, since $\epsilon<0$, so wave energy is absorbed by the plasma. It is clear from the above formula that the absorption takes place in a narrow layer, of thickness $\vert\epsilon\vert$, centred on the resonance point, $z=0$.


next up previous
Next: Collisional damping Up: Waves in cold plasmas Previous: Resonances
Richard Fitzpatrick 2006-02-16