Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The ionosphere

The region of the earth's atmosphere from roughly $60 km$ to $1000 km$ altitude is called the ionosphere. The gas at these heights is so thin that ultraviolet radiation from the sun can detach electrons from neutral atoms and molecules when it is absorbed, and the electrons recombine only slowly. The ionosphere therefore contains free electrons and positive ions. These charged particles are also bound by the earth's magnetic field, and they affect the propagation of radio waves. Their density changes with the sunspot cycle, with the season and with the hour of the day. An ionised gas with equal densities of electrons and positive charge is called a plasma. Figure 6.2.9 shows a typical daytime profile with its layers, D (about $60$ to $90 km$ ), E ( $90$ to $150 km$ ) and F (above $150 km$ ). At night the D layer disappears and the F layer weakens.

$A typical daytime electron density profile of the ionosphere with its D, E and F layers. The dashed lines mark the densities whose plasma frequencies are 0.9 and 9\,\mathrm{MHz}.$

Figure 6.2.9. A typical daytime electron density profile of the ionosphere with its D, E and F layers. The dashed lines mark the densities whose plasma frequencies are

0.9

and

9 MHz

The motion of a free electron

Consider a free electron, of charge $- e$ and mass $m$ , in the electric field of a time-harmonic wave. From Coulomb's law and Newton's second law of motion,

- e E = m \frac{d ^{2} r}{d t ^{2}}, (6.2.32)

where $r$ is its displacement from the rest position. In time-harmonic form, $d / d t \to j ω$ , this becomes $- e E = - m ω^{2} r$ , which yields

r = \frac{e}{m ω ^{2}} E . (6.2.33)

The ions are at least $1800$ times heavier and are taken at rest. Collisions with neutral molecules and the earth's magnetic field are neglected, so the plasma is lossless and isotropic. The constants are $e = 1.602 \times 1 0^{- 19} C$ and $m = 9.11 \times 1 0^{- 31} kg$ . The displacement of the electron from its ion gives rise to an electric dipole moment

p = - e r = - \frac{e ^{2}}{m ω ^{2}} E . (6.2.34)

The force on the electron points against $E$ , yet the displacement of Eq. (6.2.33) points along $E$ : an oscillating mass is always $18 0^{\circ}$ out of phase with the force that drives it (Figure 6.2.10a). The dipole moment therefore points against the field, the opposite of what happens in an ordinary dielectric (Section 3.4).

(b) N electrons per unit volume. — (a) One electron and its ion.

Polarisation, plasma frequency and the equivalent permittivity

If there are $N$ electrons per unit volume, the polarisation vector is

P = N p = - \frac{N e ^{2}}{m ω ^{2}} E, (6.2.35)

and the electric flux density becomes

D = ε_{0} E + P = ε_{0} (1 - \frac{N e ^{2}}{m ε _{0} ω ^{2}}) E . (6.2.36)

Writing this as $D = ε_{0} (1 - ω_{p}^{2} / ω^{2}) E$ defines the plasma angular frequency and the plasma frequency.

Plasma frequency and equivalent permittivity

ω_{p} f_{p} = \frac{N e ^{2}}{m ε _{0}}, = \frac{1}{2 π} \frac{N e ^{2}}{m ε _{0}} \approx 8.9836 N \approx 9 N Hz (N in m^{- 3}), (6.2.37)

ε_{p} k η_{p} = ε_{0} (1 - \frac{ω _{p}^{2}}{ω ^{2}}) = ε_{0} (1 - \frac{f _{p}^{2}}{f ^{2}}), = \frac{ω}{c} 1 - \frac{f _{p}^{2}}{f ^{2}}, = \frac{η _{0}}{1 - f _{p}^{2} / f ^{2}} . (6.2.38)

In plain words

A plasma is a medium with a permittivity smaller than that of vacuum, and the smaller the lower the frequency. Below $f_{p}$ the permittivity is negative and no wave can propagate: $f_{p}$ is the cutoff frequency of the plasma.

The numerical form of $f_{p}$ follows with $e = 1.602 \times 1 0^{- 19} C$ , $m = 9.11 \times 1 0^{- 31} kg$ and $ε_{0} = 1 0^{- 9} /36 π F/m$ . The polarisation of Eq. (6.2.35) points against $E$ , so the susceptibility $χ_{e} = - ω_{p}^{2} / ω^{2}$ is negative: a plasma is "less than vacuum". The plasma frequency is also the natural frequency at which the electron cloud oscillates about the ions when it is displaced and released. At $f = f_{p}$ the permittivity vanishes, and $D = 0$ while $E \neq = 0$ : this is the plasma oscillation.

The wave number in the plasma follows from $k = ω^{2} μ_{0} ε_{p}$ as in Eq. (6.2.38). Two regimes appear (Figure 6.2.11a):

for $f < f_{p}$ , $k$ is purely imaginary, so there is no propagation,
for $f > f_{p}$ , $k$ is purely real, so there is no attenuation.

The electron density of the ionosphere ranges from about $1 0^{10} m^{- 3}$ in the lowest layer to $1 0^{12} m^{- 3}$ in the highest, which gives plasma frequencies from $0.9$ to $9 MHz$ (Figure 6.2.11b). For communication through the ionosphere, with a satellite, the frequency of the wave must therefore be well above $9 MHz$ . Since $f_{p} \propto N$ , a hundredfold denser plasma has only a tenfold higher cutoff. Satellite links (GPS at $1.6 GHz$ , television at $12 GHz$ ) are far above every ionospheric $f_{p}$ , but the plasma around a re-entering spacecraft is not (Problem 6.2.2).

(b) f_p\approx9\sqrt N. — (a) Permittivity and wave number.

Below the plasma frequency: evanescence and total reflection

With $γ = j k$ the two regimes read

f < f_{p} : γ = α = \frac{ω}{c} \frac{f _{p}^{2}}{f ^{2}} - 1 = \frac{2 π}{c} f_{p}^{2} - f^{2}, β = 0, (6.2.39)

f > f_{p} : γ α = j β, β = \frac{2 π}{c} f^{2} - f_{p}^{2}, = 0. (6.2.40)

Below cutoff the field does not oscillate in space at all: it only decays, which is called an evanescent field (Figure 6.2.12). Its impedance is purely imaginary. The root of $1 - f_{p}^{2} / f^{2} < 0$ must be taken as $- j f_{p}^{2} / f^{2} - 1$ , so that $γ = j k$ is positive and the field decays towards $+ z$ , and then

η_{p} ⟹ S_{a v} = \frac{η _{0}}{1 - f _{p}^{2} / f ^{2}} = j \frac{η _{0}}{f _{p}^{2} / f ^{2} - 1} = \frac{j ω μ _{0}}{α} = \frac{1}{2} Re {E \times H^{*}} = 0. (6.2.41)

The last form is Eq. (6.2.13), $η = j ω μ / γ$ , with $γ = α$ . Electric and magnetic fields are in quadrature and no power flows on average. The field decays, but nothing is absorbed, since there are no collisions: the energy must go back. A wave meeting a plasma with $f_{p} > f$ is therefore totally reflected.

Reflection by a plasma layer

A wave of frequency $f$ incident vertically on a layer whose electron density $N (h)$ grows with height turns back at the height where $f_{p} (h) = f$ , that is where

N (h) = (\frac{f}{8.98})^{2} m^{- 3} (f in Hz) . (6.2.42)

The highest frequency returned at vertical incidence is the critical frequency $f_{o} F_{2} = 8.98 N_{m a x}$ of the layer.

In plain words

A plasma acts as a mirror for every frequency below its plasma frequency and as a window, slightly slower in its group velocity, for every frequency above it.

Attenuation and phase constants of a collisionless plasma. Below f_p (shaded) the wave is evanescent, \beta=0, and is reflected without loss. Above f_p it propagates without attenuation, with a \beta that approaches the vacuum line from below. — **Figure 6.2.12.** Attenuation and phase constants of a collisionless plasma. Below $f_{p}$ (shaded) the wave is evanescent, $β = 0$ , and is reflected without loss. Above $f_{p}$ it propagates without attenuation, with a $β$ that approaches the vacuum line from below.

The ionosonde

Sweeping the frequency upwards at vertical incidence and timing the echoes maps the profile $N (h)$ : each frequency turns back at its own height, Eq. (6.2.42), and the highest frequency still returned is the critical frequency $f_{o} F_{2}$ . This is how profiles like Figure 6.2.9 are measured. Strictly, the echo time gives the group delay, the subject of Section 6.2.7.

Quick check

The F layer has $N_{m a x} = 1 0^{12} m^{- 3}$ . Is a $5 MHz$ wave sent straight up returned, and at which electron density?

$f_{o} F_{2} = 8.98 MHz > 5 MHz$ , so the wave is returned, at the height where $N = (5 \times 1 0^{6} /8.98)^{2} = 3.1 \times 1 0^{11} m^{- 3}$ , on the lower side of the layer.

Application 6.2.2 · Ionospheric HF skywave

A short-wave (HF, $3$ to $30 MHz$ ) signal sent obliquely upwards is turned back by the ionosphere and lands hundreds or thousands of kilometres away. One hop via the F $_{2}$ layer covers up to about $4000 km$ , and several hops go round the world. Short-wave broadcasting, amateur radio, aircraft and ship communication and over-the-horizon radar all use this path.

To see why oblique rays return frequencies that pierce the layer overhead, slice the layer into thin horizontal slabs. In each slab the refractive index is

n (h) = \frac{ε _{p} ( h )}{ε _{0}} = 1 - \frac{f _{p}^{2} ( h )}{f ^{2}}, (6.2.43)

which decreases upwards as $N$ grows. Snell's law of refraction (Section 6.4), applied from slab to slab, keeps the product $n sin θ$ constant:

sin θ_{i} = n (h) sin θ (h), (6.2.44)

with $θ_{i}$ the angle of incidence at the bottom of the layer, where $n = 1$ . As $n$ decreases the ray bends away from the vertical, and it turns back where it becomes horizontal, $θ (h) = 9 0^{\circ}$ , that is where $n (h) = sin θ_{i}$ :

1 - \frac{f _{p}^{2} ( h )}{f ^{2}} = sin^{2} θ_{i} ⟹ f = f_{p} (h) sec θ_{i} . (6.2.45)

At the peak of the layer this gives the highest frequency that is still returned.

The secant law

f_{MUF} = f_{o} F_{2} sec θ_{i}, (6.2.46)

the maximum usable frequency for a ray incident on the layer at the angle $θ_{i}$ from the vertical.

Application 6.2.2. An oblique ray is bent back to the earth inside the F layer at the height where f_p(h)=f\cos\theta_i. A steeper ray, with a smaller \theta_i, needs a larger f_p and escapes if the layer does not have it. Around the transmitter lies a skip zone that no sky wave reaches. — **Figure 6.2.13.** Application 6.2.2. An oblique ray is bent back to the earth inside the F layer at the height where $f_{p} (h) = f cos θ_{i}$ . A steeper ray, with a smaller $θ_{i}$ , needs a larger $f_{p}$ and escapes if the layer does not have it. Around the transmitter lies a skip zone that no sky wave reaches.

With a peak density $N_{m a x} = 1 0^{12} m^{- 3}$ the critical frequency is $f_{o} F_{2} = 8.98 MHz$ , and at $θ_{i} = 7 0^{\circ}$

f_{MUF} = \frac{8.98}{cos 7 0 ^{\circ}} = 26.3 MHz . (6.2.47)

Higher frequencies, or steeper rays, escape into space. This is why satellite links use VHF and higher frequencies, and why a skip zone surrounds the transmitter (Figure 6.2.13): near the transmitter the rays are steep, $sec θ_{i}$ is small, and a frequency above $f_{o} F_{2}$ is not returned until $θ_{i}$ is large enough.

Day, night and the D layer

By day the D layer absorbs the low HF frequencies through electron collisions, the loss we neglected (Exercise 6.2.13 shows that a $1 MHz$ sky wave loses about $67 dB$ in it). At night the D layer vanishes, $f_{o} F_{2}$ drops to a few megahertz, and the lower HF bands open for long distances while the higher ones close. Home exercise 6.2.2 works the secant law with numbers.

Change $θ_{i}$ and $f$ below and see which frequencies are returned to earth. With $f_{o} F_{2} = 8.98 MHz$ and $θ_{i} = 7 0^{\circ}$ every frequency below $26.3 MHz$ comes back.

InteractiveLoading simulation…

6.2.6 Plasmas and the ionosphere

The ionosphere

The motion of a free electron

Polarisation, plasma frequency and the equivalent permittivity

Below the plasma frequency: evanescence and total reflection