Resonances

Suppose, now, that a resonance is located at $z=0$ , so that

$$n^2 = \frac{b}{z + {\rm i}\,\epsilon} +{\cal O}(1)$$ (6.29)

in the immediate vicinity of this point, where $b>0$ . Here, $\epsilon$ is a small real constant. We introduce $\epsilon$ in our analysis 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 the damping or the spontaneous excitation of waves.

In the immediate vicinity of the resonance point, $z=0$ , Equations (6.3) and (6.29) yield

$$\frac{d^2 E_y}{d\hat{z}^{\,2}} + \frac{E_y}{\hat{z}+{\rm i}\,\skew{3}\hat{\epsilon}} = 0,$$ (6.30)

where

$$\hat{z} = (k_0^{\,2}\,b)\,z,$$ (6.31)

and $\skew{3}\hat{\epsilon} = (k_0^{\,2}\,b)\,\epsilon$ . This equation is singular at the point $\hat{z} = -{\rm i}\,\skew{3}\hat{\epsilon}$ . Thus, it is necessary to introduce a branch-cut into the complex-$\hat{z}$ plane, so as 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 Figure 6.1.) 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: Branch-cuts in the $z$ -plane close to a wave resonance.

Let

$$y = 2\sqrt{\hat{z}},$$ (6.32)

$$E_y(y) = y\,\psi(y).$$ (6.33)

In the limit $\epsilon\rightarrow 0$ , Equation (6.30) transforms into

$$\frac{d^2\psi}{dy^2} + \frac{1}{y}\frac{d\psi}{dy} + \left(1-\frac{1}{y^2}\right) \psi = 0.$$ (6.34)

This is a standard equation, known as Bessel's equation of order one (Abramowitz and Stegun 1965c), 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 Equation (6.30) in the form

$$E_y(\hat{z}) = A\sqrt{\hat{z}}\,J_1\!\left(2\sqrt{\hat{z}}\right) + B\sqrt{\hat{z}}\,Y_1\!\left(2\sqrt{\hat{z}}\right),$$ (6.35)

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

Let

$$y =2\sqrt{a\,\hat{z}},$$ (6.36)

$$E_y(y) = y\,\psi(y),$$ (6.37)

where

$$a = \exp\left[-{\rm i}\,\pi\,{\rm sgn}(\epsilon)\right].$$ (6.38)

Note that the argument of $a\,\hat{z}$ is zero on the negative real $\hat{z}$ -axis. In the limit $\epsilon\rightarrow 0$ , Equation (6.30) transforms into

$$\frac{d^2\psi}{dy^2} + \frac{1}{y}\frac{d\psi}{dy} - \left(1+\frac{1}{y^2}\right) \psi = 0.$$ (6.39)

This is a standard equation, known as Bessel's modified equation of order one (Abramowitz and Stegun 1965c), 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 Equation (6.30) in the form

$$E_y(\hat{z}) = C\sqrt{a\,\hat{z}}\,I_1\!\left(2\sqrt{a\,\hat{z}}\right) + D\sqrt{a\,\hat{z}}\,K_1\!\left(2\sqrt{a\,\hat{z}}\right),$$ (6.40)

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

The Bessel functions $J_1(z)$ , $Y_1(z)$ , $I_1(z)$ , and $K_1(z)$ are all perfectly well-defined (i.e., analytic) for complex arguments, so the two expressions (6.35) and (6.40) 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 behavior of the expressions (6.35) and (6.40) in the limit of small $\hat{z}$ : that is, $\vert\hat{z}\vert\ll 1$ . In this limit,

$$\sqrt{\hat{z}}\,J_1\!\left(2\sqrt{\hat{z}}\right) = \hat{z} + {\cal O}\left(\hat{z}^{\,2}\right),$$ (6.41)

$$\sqrt{a\,\hat{z}} \,I_1\!\left(2\sqrt{a\,\hat{z}}\right) = - \hat{z} + {\cal O}\left(\hat{z}^{\,2}\right),$$ (6.42)

$$\sqrt{\hat{z}}\,Y_1\!\left(2\sqrt{\hat{z}}\right) = -\frac{1}{\pi}\left[ 1 -\left(\ln\vert\hat{z}\vert + 2\,\gamma-1\right)\hat{z}\,\right]+ {\cal O}\left(\hat{z}^{\,2}\right),$$ (6.43)

$$\sqrt{a\,\hat{z}}\,K_1\!\left(2\sqrt{a\,\hat{z}}\right) = \frac{1}{2}\left[ 1 -\left( \ln\vert\hat{z}\vert + 2\,\gamma-1\... ... {\rm i}\,{\rm arg}(a)\, \hat{z}\,\right] + {\cal O}\left(\hat{z}^{\,2}\right),$$ (6.44)

where $\gamma$ is Euler's constant (Abramowitz and Stegun 1965a), and $\hat{z}$ is assumed to lie on the positive real $\hat{z}$ -axis. It follows, by a comparison of Equations (6.35), (6.40), and (6.41)-(6.44), that the choice

$$C = -A +{\rm i}\,\frac{\pi}{2}\,{\rm sgn}(\epsilon)\,D = -A - {\rm i}\, {\rm sgn}(\epsilon)\,B,$$ (6.45)

$$D = - \frac{2}{\pi}\,B,$$ (6.46)

ensures that the expressions (6.35) and (6.40) are indeed identical.

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

$$\sqrt{a\,\hat{z}}\,\,I_1\!\left(2\sqrt{a\,\hat{z}}\right) \simeq \frac{1}{2\sqrt{\pi}}\,\vert\hat{z}\vert^{\,1/4}\,\exp\left(+2\sqrt{\vert\hat{z}\vert}\right),$$ (6.47)

$$\sqrt{a\,\hat{z}}\,\,K_1\!\left(2\sqrt{a\,\hat{z}}\right) \simeq \frac{\sqrt{\pi}}{2}\,\vert\hat{z}\vert^{\,1/4}\,\exp\left(-2\sqrt{\vert\hat{z}\vert}\right),$$ (6.48)

where $\hat{z}$ is assumed to lie on the negative real $\hat{z}$ -axis. It is clear that the $I_1$ solution is unphysical, because it blows up in the evanescent region $(\hat{z}<0)$ . Thus, the coefficient $C$ in expression (6.40) must be set to zero in order to prevent $E_y(\hat{z})$ from blowing up as $\hat{z}\rightarrow-\infty$ . According to Equation (6.45), this constraint implies that

$$A = -{\rm i}\,{\rm sgn}(\epsilon)\,B.$$ (6.49)

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

$$\sqrt{\hat{z}}\,\,J_1\!\left(2\sqrt{\hat{z}}\right) \simeq \frac{1}{\sqrt{\pi}}\,\hat{z}^{\,1/4}\,\cos\left(2\sqrt{z}-\frac{3}{4}\,\pi\right),$$ (6.50)

$$\sqrt{\hat{z}}\,\,Y_1\!\left(2\sqrt{\hat{z}}\right) \simeq \frac{1}{\sqrt{\pi}}\,\hat{z}^{\,1/4}\,\sin\left(2\sqrt{z} -\frac{3}{4}\,\pi\right),$$ (6.51)

where $\hat{z}$ is assumed to lie on the positive real $\hat{z}$ -axis. It follows from Equations (6.35), (6.49), (6.50), and (6.51) that, in the non-evanescent region ( $\hat{z}> 0$ ), the most general physical solution takes the form

$$E_y(\hat{z}) = A'\,\left[{\rm sgn}(\epsilon) + 1\right]\,\hat{z}^{\,1/4}\, \exp\left[ +{\rm i}\left(2\sqrt{\hat{z}} - \frac{3}{4}\,\pi\right)\right]$$

$$\phantom{=}+A'\, \left[{\rm sgn}(\epsilon) - 1\right]\,\hat{z}^{\,1/4}\,\exp\left[ -{\rm i}\left(2\sqrt{\hat{z}} + \frac{3}{4}\,\pi\right)\right],$$ (6.52)

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 $z$ , toward 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:

$$E_y(z) = E\, n^{\,-1/2}\,\exp\left(- {\rm i}\, k_0 \!\int_0^z \!n\,dz'\right) + F\,n^{\,-1/2}\,\exp\left(+{\rm i}\, k_0 \!\int_0^z \!n\,dz'\right).$$ (6.53)

The first term on the right-hand side of the previous 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, which corresponds to $\hat{z}$ large and positive), the previous expression reduces to

$$E_y(\hat{z}) \simeq (k_0 \,b)^{-1/2}\left[ E\,\hat{z}^{\,1/4}\exp... ...}\right) + F\,\hat{z}^{\,1/4}\exp\left(+{\rm i}\,2\sqrt{\hat{z}}\right)\right].$$ (6.54)

A comparison of Equations (6.52) and (6.54) 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.