WKB Reflection Coefficient

Let us write $\hat{z} = x + {\rm i}\,y$ , where $x$ and $y$ are real variables. Consider the solution of the differential equation

$$w'' + h^{\,2} \,q(x)\, w = 0,$$ (1196)

where $q(x)$ is a real function, $h$ is a large number, $q>0$ for $x<0$ , and $q<0$ for $x>0$ . It is clear that $\hat{z}=0$ represents a simple zero of $q(\hat{z})$ . Here, we assume, as seems eminently reasonable, that we can find a well-behaved function of the complex variable $q(\hat{z})$ such that $q(\hat{z})=q(x)$ along the real axis. The arrangement of Stokes and anti-Stokes lines in the immediate vicinity of the point $\hat{z}=0$ is sketched in Figure 24. The argument of $q(\hat{z})$ on the positive $x$ -axis is chosen to be $-\pi$ . Thus, the argument of $q(\hat{z})$ on the negative $x$ -axis is 0 .

Figure: The arrangement of Stokes lines (dashed) and anti-Stokes lines (solid) in the complex $\hat{z}$ plane. Also shown is the branch cut (wavy line).

On $OB$ , the two WKB solutions (1176)-(1177) can be written

$$(0, x) = q^{-1/4}(x) \,\exp\left(\,{\rm i}\,h\int_0^{x} q^{\,1/2}(x')\, dx'\right),$$ (1197)

$$(x, 0) = q^{-1/4}(x)\,\exp\left(-{\rm i}\,h\int_0^{x} q^{\,1/2}(x')\, dx'\right).$$ (1198)

Here, we can interpret $(0,x)$ as a wave propagating to the right along the $x$ -axis, and $(x,0)$ as a wave propagating to the left. On $OA$ , the WKB solutions take the form

$$(0,x)_d = {\rm e}^{\,{\rm i}\,\pi/4} \,\vert q(x)\vert^{-1/4} \,\exp\left( +h\int_0^x \vert q(x')\vert^{\,1/2}\,dx'\right),$$ (1199)

$$(x, 0)_s ={\rm e}^{\,{\rm i}\,\pi/4}\,{\vert q(x)\vert^{-1/4}} \,\exp\left( -h \int_0^x \vert q(x')\vert^{\,1/2}\,dx'\right).$$ (1200)

Clearly, $(x,0)_s$ represents an evanescent wave that decays to the right along the $x$ -axis, whereas $(0,x)_d$ represents an evanescent wave that decays to the left. If we adopt the boundary condition that there is no incident wave from the region $x\rightarrow +\infty$ , the most general asymptotic solution to Equation (1198) on $OA$ is written

$$w(x,h) =A\, (x,0)_s,$$ (1201)

where $A$ is an arbitrary constant.

Let us assume that we can find an analytic solution, $w(\hat{z},h)$ , to the differential equation

$$w'' + h^{\,2}\,q(\hat{z})\,w = 0,$$ (1202)

which satisfies $w(\hat{z}, h) = w(x,h)$ along the real axis, where $w(x,h)$ is the physical solution. From a mathematical point of view, this seems eminently reasonable. In the domains 1 and 2, the solution (1203) becomes

$$w(1) = A\, (\hat{z},0)_s,$$ (1203)

and

$$w(2) = A\,(\hat{z},0)_s,$$ (1204)

respectively. Note that the solution is continuous across the Stokes line $OA$ , because the coefficient of the dominant solution $(0,\hat{z})$ is zero: thus, the jump in the coefficient of the subdominant solution is zero times the Stokes constant, ${\rm i}$ . In other words, it is zero. Let us move into domain 3. In doing so, we cross an anti-Stokes line, so the solution becomes

$$w(3) = A\,(\hat{z},0)_d.$$ (1205)

Let us now move into domain 4. In doing so, we cross a Stokes line. Applying the general rule derived in the preceding section, the solution becomes

$$w(4) = A\, (\hat{z},0)_d + {\rm i}\,A\,(0,\hat{z})_s.$$ (1206)

Finally, on $OB$ the solution becomes

$$w(x,h) = A\,(x,0) + {\rm i}\,A\,(0,x).$$ (1207)

Suppose that there is a point $a$ on the negative $x$ -axis where $q(x)=1$ . It follows from Equations (1201) and (1209) that we can write the asymptotic solution to Equation (1198) as

$$w(x,h) = q^{-1/4}(x)\,\exp\left(\,{\rm i}\,h\int_a^x q^{\,1/2}(x')\,dx'\right)$$ (1208)

$$- {\rm i}\, \exp\left(2\,{\rm i}\,h\int_a^0 q^{1/2}(x')\,dx'\right)\,\, q^{-1/4}(x)\, \exp\left(-{\rm i}\,h\int_a^x q^{\,1/2}(x')\,dx'\right),$$

in the region $x<0$ , and

$$w(x,h) = \exp\left(\,{\rm i}\,h\int_a^0 q^{\,1/2}(x')\,dx'\right)... ...rt q(x)\vert^{-1/4}\, \exp\left(-h\int_0^x \vert q(x')\vert^{\,1/2}\,dx'\right)$$ (1209)

in the region $x>0$ . Here, we have chosen

$$A = -{\rm i}\,\exp\left(\,{\rm i}\,h\int_a^0 q^{\,1/2}(x')\,dx' \right).$$ (1210)

If we interpret $x$ as a normalized altitude in the ionosphere, $q(x)$ as the square of the refractive index in the ionosphere, the point $a$ as ground level, and $w$ as the electric field strength of a radio wave propagating vertically upwards into the ionosphere, then Equation (1210) tells us that a unit amplitude wave fired vertically upwards from ground level into the ionosphere is reflected at the level where the refractive index is zero. The first term in Equation (1210) is the incident wave, and the second term is the reflected wave. The reflection coefficient (i.e., the ratio of the reflected to the incident wave at ground level) is given by

$$R = -{\rm i}\, \exp\left(2\,{\rm i}\,h\int_a^0 q^{\,1/2}(x')\,dx'\right).$$ (1211)

Note that $\vert R\vert=1$ , so the amplitude of the reflected wave equals that of the incident wave. In other words, there is no absorption of the wave at the level of reflection. The phase shift of the reflected wave at ground level, with respect to that of the incident wave, is that associated with the wave propagating from ground level to the reflection level and back to ground level again, plus a $-\pi/2$ phase shift at reflection. According to Equation (1211), the wave attenuates fairly rapidly (in the space of a few wavelengths) above the reflection level. Of course, Equation (1213) is completely equivalent to Equation (1089).

Note that the reflection of the incident wave at the point where the refractive index is zero is directly associated with the Stokes phenomenon. Without the jump in the coefficient of the subdominant solution, as we go from domain 3 to domain 4, there is no reflected wave on the $OB$ axis. Note, also, that the WKB solutions (1210) and (1211) break down in the immediate vicinity of $q=0$ (i.e., at the reflection point). Thus, it is possible to demonstrate that the incident wave is totally reflected at the point $q=0$ , with a $-\pi/2$ phase shift, without having to solve for the wave structure in the immediate vicinity of the reflection point. This demonstrates that the reflection of the incident wave at $q=0$ is an intrinsic property of the WKB solutions, and does not depend on the detailed behavior of the wave in the region where the WKB solutions break down.