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Ionospheric Radio Wave Propagation

To a first approximation, the Earth's ionosphere consists of an unmagnetized, horizontally stratified, partially ionized gas (Budden 1985). The dispersion relation for the electromagnetic plasma wave takes the form [see Equation (5.98)]

$\displaystyle {\cal M} = \omega^2 - k^2 c^2 - {\mit\Pi}_e^{\,2} = 0,$ (6.109)


$\displaystyle {\mit\Pi}_e = \sqrt{\frac{N\,e^2}{\epsilon_0\,m_e}}.$ (6.110)

Here, $ N=N(z)$ is the density of free electrons in the ionosphere, and $ z$ is a coordinate that measures height above the surface of the Earth. (The curvature of the Earth, the Earth's magnetic field, and collisions, are neglected in the following analysis.)


$\displaystyle \frac{\partial{\cal M}}{\partial\omega}$ $\displaystyle = 2\,\omega,$ (6.111)
$\displaystyle \frac{\partial{\cal M}}{\partial {\bf k}}$ $\displaystyle = - 2\,c^2\,{\bf k},$ (6.112)
$\displaystyle \frac{\partial{\cal M}}{\partial{\bf r}}$ $\displaystyle = -\nabla{\mit\Pi}_e^{\,2},$ (6.113)
$\displaystyle \frac{\partial{\cal M}}{\partial t}$ $\displaystyle = 0.$ (6.114)

Thus, the ray equations, (6.106)-(6.108), yield

$\displaystyle \frac{d{\bf r}}{dt}$ $\displaystyle = \frac{c^2}{\omega}\,{\bf k},$ (6.115)
$\displaystyle \frac{d{\bf k}}{dt}$ $\displaystyle = - \frac{\nabla{\mit\Pi}_e^{\,2}}{2\,\omega},$ (6.116)
$\displaystyle \frac{d\omega}{d t}$ $\displaystyle = 0.$ (6.117)

Evidently, the frequency of a radio pulse does not change as it propagates through the ionosphere, provided that $ N(z)$ does not vary in time. Furthermore, it follows from Equations (6.115)-(6.117), and the fact that $ {\mit\Pi}_e={\mit\Pi}_e(z)$ , that a radio pulse that starts off at ground level propagating in the $ x$ -$ z$ plane, say, will continue to propagate in this plane.

For pulse propagation in the $ x$ -$ z$ plane, we have

$\displaystyle \frac{dx}{dt}$ $\displaystyle = \frac{c^2\,k_x}{\omega},$ (6.118)
$\displaystyle \frac{dz}{dt}$ $\displaystyle = \frac{c^2\,k_z}{\omega},$ (6.119)
$\displaystyle \frac{d k_x}{dt}$ $\displaystyle = 0.$ (6.120)

The dispersion relation (6.109) yields

$\displaystyle n^2 = \frac{(k_x^{\,2} + k_z^{\,2})\,c^2}{\omega^2} = 1-\frac{{\mit\Pi}_e^{\,2}}{\omega^2},$ (6.121)

where $ n(z)$ is the refractive index.

Let us assume that $ n=1$ at $ z=0$ , which is equivalent to the reasonable assumption that the atmosphere is non-ionized at ground level. It follows from Equation (6.120) that

$\displaystyle k_x= k_x(z=0) = \frac{\omega}{c}\,S,$ (6.122)

where $ S$ is the sine of the angle of incidence of the pulse, with respect to the vertical axis, at ground level. Equations (6.121) and (6.122) yield

$\displaystyle k_z = \pm \frac{\omega}{c}\sqrt{n^2-S^2}.$ (6.123)

According to Equation (6.119), the plus sign corresponds to the upward trajectory of the pulse, whereas the minus sign corresponds to the downward trajectory. Finally, Equations (6.118), (6.119), (6.122), and (6.123) yield the equations of motion of the pulse:

$\displaystyle \frac{dx}{dt}$ $\displaystyle = c\,S,$ (6.124)
$\displaystyle \frac{dz}{dt}$ $\displaystyle = \pm c\sqrt{n^2-S^2}.$ (6.125)

The pulse attains its maximum altitude, $ z=z_0$ , when

$\displaystyle n(z_0) = \vert S\vert.$ (6.126)

The total distance traveled by the pulse (i.e., the distance from its launch point to the point where it intersects the Earth's surface again) is

$\displaystyle x_0 = 2\,S\int_0^{z_0(S)} \frac{dz}{\sqrt{n^2(z)-S^2}}.$ (6.127)

In the limit in which the radio pulse is launched vertically (i.e., $ S=0$ ) into the ionosphere, the turning point condition (6.126) reduces to that characteristic of a cutoff (i.e., $ n=0$ ). The WKB turning point described in Equation (6.126) is a generalization of the conventional turning point, which occurs when $ k^2$ changes sign. Here, $ k_z^{\,2}$ changes sign, while $ k_x^{\,2}$ and $ k_y^{\,2}$ are constrained by symmetry (i.e., $ k_x$ is constant, and $ k_y$ is zero).

According to Equations (6.115)-(6.117) and (6.121), the equation of motion of the pulse can also be written

$\displaystyle \frac{d^2{\bf r}}{dt^2} = \frac{c^2}{2}\,\nabla n^2.$ (6.128)

It follows that the trajectory of the pulse is the same as that of a particle moving in the gravitational potential $ -c^2\,n^2/2$ . Thus, if $ n^2$ decreases linearly with increasing height above the ground [which is the case if $ N(z)$ increases linearly with $ z$ ] then the trajectory of the pulse is a parabola.

next up previous
Next: Exercises Up: Wave Propagation Through Inhomogeneous Previous: Ray Tracing
Richard Fitzpatrick 2016-01-23