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Next: Resonant Layers Up: Waves in Cold Plasmas Previous: Cutoffs

Resonances

Suppose, now, that a resonance is located at $z=0$, so that
\begin{displaymath}
n^2 = \frac{b}{z + {\rm i}\,\epsilon} +O(1)
\end{displaymath} (581)

in the immediate vicinity of this point, where $b>0$. Here, $\epsilon$ is a small real constant. We introduce $\epsilon$ at this point principally as a mathematical artifice to ensure that $E_y$ remains single-valued and finite. However, as will become clear later on, $\epsilon$ has a physical significance in terms of damping or spontaneous excitation.

In the immediate vicinity of the resonance point, $z=0$, Eqs. (555) and (581) yield

\begin{displaymath}
\frac{d^2 E_y}{d\hat{z}^2} + \frac{E_y}{\hat{z}+{\rm i}\,\hat{\epsilon}} = 0,
\end{displaymath} (582)

where
\begin{displaymath}
\hat{z} = (k_0^{~2}\,b)\,z,
\end{displaymath} (583)

and $\hat{\epsilon} = (k_0^{~2}\,b)\,\epsilon$. This equation is singular at the point $\hat{z} = -{\rm i}\,\hat{\epsilon}$. Thus, it is necessary to introduce a branch-cut into the complex-$\hat{z}$ plane in order to ensure that $E_y(\hat{z})$ is single-valued. If $\epsilon>0$ then the branch-cut lies in the lower half-plane, whereas if $\epsilon<0$ then the branch-cut lies in the upper half-plane--see Fig. 15. Suppose that the argument of $\hat{z}$ is $0$ on the positive real $\hat{z}$-axis. It follows that the argument of $\hat{z}$ on the negative real $\hat{z}$-axis is $+\pi$ when $\epsilon>0$ and $-\pi$ when $\epsilon<0$.

Figure 15: Branch-cuts in the $z$-plane close to a wave resonance.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter04/sing.eps}}
\end{figure}

Let

$\displaystyle y$ $\textstyle =$ $\displaystyle 2\,\sqrt{\hat{z}},$ (584)
$\displaystyle E_y$ $\textstyle =$ $\displaystyle y\,\psi(y).$ (585)

In the limit $\epsilon\rightarrow 0$, Eq. (582) transforms into
\begin{displaymath}
\frac{d^2\psi}{dy^2} + \frac{1}{y}\frac{d\psi}{dy} + \left(1-\frac{1}{y^2}\right)\!
\psi = 0.
\end{displaymath} (586)

This is a standard equation, known as Bessel's equation of order one,[*]and possesses two independent solutions, denoted $J_1(y)$ and $Y_1(y)$, respectively. Thus, on the positive real $\hat{z}$-axis we can write the most general solution to Eq. (582) in the form
\begin{displaymath}
E_y(\hat{z}) = A\,\sqrt{\hat{z}}\,J_1(2\,\sqrt{\hat{z}}) +
B\,\sqrt{\hat{z}}\,Y_1(2\,\sqrt{\hat{z}}),
\end{displaymath} (587)

where $A$ and $B$ are two arbitrary constants.

Let

$\displaystyle y$ $\textstyle =$ $\displaystyle 2\,\sqrt{a\hat{z}},$ (588)
$\displaystyle E_y$ $\textstyle =$ $\displaystyle y\,\psi(y),$ (589)

where
\begin{displaymath}
a = {\rm e}^{-{\rm i}\,\pi\,{\rm sgn}(\epsilon)}.
\end{displaymath} (590)

Note that the argument of $a\hat{z}$ is zero on the negative real $\hat{z}$-axis. In the limit $\epsilon\rightarrow 0$, Eq. (582) transforms into
\begin{displaymath}
\frac{d^2\psi}{dy^2} + \frac{1}{y}\frac{d\psi}{dy} - \left(1+\frac{1}{y^2}\right)\!
\psi = 0.
\end{displaymath} (591)

This is a standard equation, known as Bessel's modified equation of order one,[*]and possesses two independent solutions, denoted $I_1(y)$ and $K_1(y)$, respectively. Thus, on the negative real $\hat{z}$-axis we can write the most general solution to Eq. (582) in the form
\begin{displaymath}
E_y(\hat{z}) = C\,\sqrt{a\hat{z}}\,I_1(2\,\sqrt{a\hat{z}}) +
D\,\sqrt{a\hat{z}}\,K_1(2\,\sqrt{a\hat{z}}),
\end{displaymath} (592)

where $C$ and $D$ are two arbitrary constants.

Now, the Bessel functions $J_1$, $Y_1$, $I_1$, and $K_1$ are all perfectly well-defined for complex arguments, so the two expressions (587) and (592) must, in fact, be identical. In particular, the constants $C$ and $D$ must somehow be related to the constants $A$ and $B$. In order to establish this relationship, it is convenient to investigate the behaviour of the expressions (587) and (592) in the limit of small $\hat{z}$: i.e., $\vert\hat{z}\vert\ll 1$. In this limit,

$\displaystyle \sqrt{\hat{z}}\,J_1(2\,\sqrt{\hat{z}})$ $\textstyle =$ $\displaystyle \hat{z} + O(\hat{z}^2),$ (593)
$\displaystyle \sqrt{a\hat{z}} \,I_1(2\,\sqrt{a\hat{z}})$ $\textstyle =$ $\displaystyle - \hat{z} + O(\hat{z}^2),$ (594)
$\displaystyle \sqrt{\hat{z}}\,Y_1(2\,\sqrt{\hat{z}})$ $\textstyle =$ $\displaystyle -\frac{1}{\pi}\left[
1 -\left\{ \ln\vert\hat{z}\vert + 2\,\gamma-1\right\}\hat{z}\,\right]$  
    $\displaystyle + O(\hat{z}^2),$ (595)
$\displaystyle \sqrt{a\hat{z}}\,K_1(2\,\sqrt{a\hat{z}})$ $\textstyle =$ $\displaystyle \frac{1}{2}\left[
1 -\left\{ \ln\vert\hat{z}\vert + 2\,\gamma-1\right\}\hat{z} - {\rm i}\,{\rm arg}(a)\,
\hat{z}\,\right]$  
    $\displaystyle + O(\hat{z}^2),$ (596)

where $\gamma$ is Euler's constant, and $\hat{z}$ is assumed to lie on the positive real $\hat{z}$-axis. It follows, by a comparison of Eqs. (587), (592), and (593)-(596), that the choice
$\displaystyle C$ $\textstyle =$ $\displaystyle -A +{\rm i}\,\frac{\pi}{2}\,{\rm sgn}(\epsilon)\,D = -A - {\rm i}\,
{\rm sgn}(\epsilon)\,B,$ (597)
$\displaystyle D$ $\textstyle =$ $\displaystyle - \frac{2}{\pi}\,B,$ (598)

ensures that the expressions (587) and (592) are indeed identical.

Now, in the limit $\vert\hat{z}\vert\gg 1$,

$\displaystyle \sqrt{a\hat{z}} \,I_1(2\,\sqrt{a\hat{z}})$ $\textstyle \sim$ $\displaystyle \frac{\vert\hat{z}\vert^{1/4}}{2\sqrt{\pi}}\,
{\rm e}^{+2\sqrt{\vert\hat{z}\vert}},$ (599)
$\displaystyle \sqrt{a\hat{z}}\,K_1(2\,\sqrt{a\hat{z}})$ $\textstyle \sim$ $\displaystyle \frac{\sqrt{\pi}\,\vert\hat{z}\vert^{1/4}}{2}\,
{\rm e}^{-2\sqrt{\vert\hat{z}\vert}},$ (600)

where $\hat{z}$ is assumed to lie on the negative real $\hat{z}$-axis. It is clear that the $I_1$ solution is unphysical, since it blows up in the evanescent region $(\hat{z}<0)$. Thus, the coefficient $C$ in expression (592) must be set to zero in order to prevent $E_y(\hat{z})$ from blowing up as $\hat{z}\rightarrow-\infty$. According to Eq. (597), this constraint implies that
\begin{displaymath}
A = -{\rm i}\,{\rm sgn}(\epsilon)\,B.
\end{displaymath} (601)

In the limit $\vert\hat{z}\vert\gg 1$,

$\displaystyle \sqrt{\hat{z}}\,J_1(2\,\sqrt{\hat{z}})$ $\textstyle \sim$ $\displaystyle \frac{\hat{z}^{1/4}}{\sqrt{\pi}}\cos\!\left(2\,
\sqrt{z}
-\frac{3}{4}\,\pi\right),$ (602)
$\displaystyle \sqrt{\hat{z}}\,Y_1(2\,\sqrt{\hat{z}})$ $\textstyle \sim$ $\displaystyle \frac{\hat{z}^{1/4}}{\sqrt{\pi}}\sin\!\left(2\,
\sqrt{z}
-\frac{3}{4}\,\pi\right),$ (603)

where $\hat{z}$ is assumed to lie on the positive real $\hat{z}$-axis. It follows from Eqs. (587), (601), and (602)-(603) that in the non-evanescent region ($\hat{z}> 0$) the most general physical solution takes the form
$\displaystyle E_y(\hat{z})$ $\textstyle =$ $\displaystyle A'\,\left[{\rm sgn}(\epsilon) + 1\right]\,\hat{z}^{1/4}\,
\exp\!\left[
+{\rm i}\,2\sqrt{\hat{z}} - \frac{3}{4}\,\pi\right]$  
    $\displaystyle +A'\,
\left[{\rm sgn}(\epsilon) - 1\right]\,\hat{z}^{1/4}\,\exp\!\left[
-{\rm i}\,2\sqrt{\hat{z}} + \frac{3}{4}\,\pi\right],$ (604)

where $A'$ is an arbitrary constant.

Suppose that a plane electromagnetic wave, polarized in the $y$-direction, is launched from an antenna, located at large positive $z$, towards the resonance point at $z=0$. It is assumed that $n=1$ at the launch point. In the non-evanescent region, $z>0$, the wave can be represented as a linear combination of propagating WKB solutions:

\begin{displaymath}
E_y(z) = E\, n^{-1/2}\,\exp\left(- {\rm i}\, k_0 \!\int_0^z ...
...,n^{-1/2}\,\exp\left(+{\rm i}\, k_0 \!\int_0^z \!n\,dz\right).
\end{displaymath} (605)

The first term on the right-hand side of the above equation represents the incident wave, whereas the second term represents the reflected wave. Here, $E$ is the amplitude of the incident wave, and $F$ is the amplitude of the reflected wave. In the vicinity of the resonance point (i.e., $z$ small and positive, or $\hat{z}$ large and positive) the above expression reduces to
\begin{displaymath}
E_y(\hat{z}) \simeq (k_0 b)^{-1/2}\left[
E\,\hat{z}^{1/4}\ex...
...t{z}^{1/4}\exp\left(+{\rm i}\,2\,\sqrt{\hat{z}}\right)\right].
\end{displaymath} (606)

A comparison of Eqs. (604) and (606) shows that if $\epsilon>0$ then $E=0$. In other words, there is a reflected wave, but no incident wave. This corresponds to the spontaneous excitation of waves in the vicinity of the resonance. On the other hand, if $\epsilon<0$ then $F=0$. In other words, there is an incident wave, but no reflected wave. This corresponds to the total absorption of incident waves in the vicinity of the resonance. It is clear that if $\epsilon>0$ then $\epsilon$ represents some sort of spontaneous wave excitation mechanism, whereas if $\epsilon<0$ then $\epsilon$ represents a wave absorption, or damping, mechanism. We would normally expect plasmas to absorb incident wave energy, rather than spontaneously emit waves, so we conclude that, under most circumstances, $\epsilon<0$, and resonances absorb incident waves without reflection.


next up previous
Next: Resonant Layers Up: Waves in Cold Plasmas Previous: Cutoffs
Richard Fitzpatrick 2011-03-31