University Physics — Year 2 · Bachelor Year 2

14Electromagnetic Waves in Plasmas, Conductors and Dielectrics

A shortwave broadcast crosses an ocean by bouncing off the upper atmosphere, which reflects it like a mirror — yet the same layer lets television and satellite signals through. A metal spoon is shiny, a metal mesh in the door of a microwave oven holds in the waves while you watch your food through it, and a sheet of glass is transparent to light but opaque to ultraviolet. In each case a wave enters a material, the material’s charges respond, and their response changes the wave: its speed, its attenuation, even whether it propagates at all. This chapter treats the three standard models of a medium — a gas of free electrons, an ohmic conductor, and bound electrons — with the one method that handles them all: add the induced current to Maxwell’s equations and read off the dispersion relation.

14.1 Waves in a plasma

Definition 14.1 (Plasma; the free-electron model)

A plasma is a globally neutral gas of free electrons (density $n$ , mass $m$ , charge $-e$ ) and positive ions, which, being thousands of times heavier, are taken as fixed. The upper atmosphere (ionosphere, $n \sim 10^{11}$ – $10^{12}$ m $^{-3}$ ), a discharge tube, the solar corona, and — at optical frequencies — the conduction electrons of a metal are plasmas. Collisions are neglected (the wave’s period is short compared with the time between collisions). Its characteristic frequency is the plasma frequency

\omega_p = \sqrt{\frac{ne^2}{\varepsilon_0m}} , \qquad f_p = \frac{\omega_p}{2\pi} \approx 9\,\text{Hz}\times\sqrt{n\,[\text{m}^{-3}]} :

$3\,\mathrm{MHz}$ to $9\,\mathrm{MHz}$ for the ionosphere, $2 \times 10^{15}\,\mathrm{Hz}$ (ultraviolet) for a metal.

Theorem 14.2 (Dispersion relation of a plasma)

For a transverse PPH wave $\underline{\vect E} = \underline{\vect E}_0\eu^{\iu(\omega t - kx)}$ in a plasma, the electrons oscillate with $\underline{\vect v} = -e\underline{\vect E}/\iu m\omega$ and carry the current density $\underline{\vect j} = \underline\gamma\underline{\vect E}$ with $\underline\gamma = ne^2/\iu m\omega$ ; Maxwell’s equations then give

k^2 = \frac{\omega^2 - \omega_p^2}{c^2} .

Above $\omega_p$ the wave propagates, with $v_\varphi = c/\sqrt{1 - \omega_p^2/\omega^2} > c$ and $v_g = c\sqrt{1 - \omega_p^2/\omega^2} < c$ , $v_\varphi v_g = c^2$ ; below $\omega_p$ , $k = -\iu\kappa$ with $\kappa = \sqrt{\omega_p^2 - \omega^2}/c$ : the wave is evanescent, penetrates over $1/\kappa$ and is totally reflected (Chapter 15); the plasma is then a mirror. At $\omega \gg \omega_p$ the electrons cannot follow and the plasma is transparent.

Proof. Equation of motion $m\,\dd\vect v/\dd t = -e\vect E$ (the magnetic force $ev B \sim evE/c$ is negligible for $v \ll c$ , and the ions’ motion is $m/M$ times smaller): $\iu m\omega\underline{\vect v} = -e\underline{\vect E}$ . Then $\underline{\vect j} = -ne\underline{\vect v}$ . The plasma stays neutral for a transverse wave ( $\vect k\cdot\vect E = 0$ gives $\rho = 0$ by Gauss), so Maxwell–Ampère reads $-\iu\vect k\wedge\underline{\vect B} = \mu_0\underline\gamma\underline{\vect E} + \iu\omega\mu_0\varepsilon_0\underline{\vect E}$ and Maxwell–Faraday $\underline{\vect B} = \vect k \wedge\underline{\vect E}/\omega$ ; combining, $k^2\underline{\vect E} = \mu_0(\iu\omega\underline\gamma - \omega^2\varepsilon_0)(-1)\underline{\vect E}$ , i.e. $k^2 = \omega^2\mu_0\varepsilon_0 - \iu\omega\mu_0 \underline\gamma = (\omega^2 - ne^2/\varepsilon_0m)/c^2$ . The velocities follow as in Chapter 8. ∎

Remark 14.3 (Where the energy goes)

The conductivity is imaginary: $\langle\vect j\cdot\vect E\rangle = 0$ , the electrons take energy from the wave during half a cycle and give it back during the next — a lossless medium, in which the wave’s energy is shared between the fields and the electrons’ kinetic energy, and travels at $v_g$ . Collisions add a small real part to $\underline\gamma$ and a small absorption: the ionosphere’s D layer absorbs medium waves by day, which is why distant AM stations come in at night.

Example 14.4 (The ionosphere and radio)

With $n = 1 \times 10^{12}\,\mathrm{m}^{-3}$ in the F layer, $f_p = 9\,\mathrm{MHz}$ : shortwave broadcasts below that are reflected (and can hop around the Earth between the ionosphere and the ground), FM ( $100\,\mathrm{MHz}$ ), television and satellite links pass through. A wave at $3\,\mathrm{MHz}$ penetrates only $1/\kappa = c/\sqrt{\omega_p^2 - \omega^2} = 6\,\mathrm{m}$ into the layer before turning back. A spacecraft re-entering the atmosphere is sheathed in a plasma of $n \sim 10^{18}$ m $^{-3}$ , $f_p \sim 10\,\mathrm{GHz}$ : radio contact is lost for minutes.

Left: the dispersion relation of a plasma — no propagation below _p, a hyperbola approaching the light line above. Right: shortwaves below the plasma frequency are reflected by the ionosphere and hop around the Earth; higher frequencies escape to space. — Left: the dispersion relation of a plasma — no propagation below $\omega_p$ , a hyperbola approaching the light line above. Right: shortwaves below the plasma frequency are reflected by the ionosphere and hop around the Earth; higher frequencies escape to space.

14.2 Waves in an ohmic conductor: the skin effect

Theorem 14.5 (Skin effect)

In an ohmic conductor ( $\vect j = \gamma\vect E$ , $\gamma$ real, at frequencies with $\omega\tau \ll 1$ and $\varepsilon_0\omega \ll \gamma$ , i.e. below $\sim1 \times 10^{16}\,\mathrm{Hz}$ for copper) the displacement current is negligible and the fields obey the diffusion equation $\Delta\vect E = \mu_0\gamma\,\partial_t\vect E$ . A PPH wave entering the conductor at $x = 0$ has

\underline k^2 = -\iu\mu_0\gamma\omega , \qquad \underline k = \frac{1 - \iu}\delta , \qquad \delta = \sqrt{\frac2{\mu_0\gamma\omega}} ,

so that $\underline E = E_0\,\eu^{-x/\delta}\eu^{\iu(\omega t - x/\delta)}$ : it is damped over the skin depth $\delta$ , with a phase that also turns by one radian per $\delta$ — a wavelength $2\pi\delta$ , a phase velocity $\omega\delta \ll c$ . The current $\vect j = \gamma\vect E$ is confined to a layer of thickness $\delta$ under the surface, and $\vect B$ lags $\vect E$ by $45{}^{\circ}$ with $|B| = |E|\sqrt2/\omega\delta$ . For copper, $\delta = 9.2\,\mathrm{mm}$ at $50\,\mathrm{Hz}$ , $0.21\,\mathrm{mm}$ at $100\,\mathrm{kHz}$ , $2.1\,\text{µ}\mathrm{m}$ at $1\,\mathrm{GHz}$ .

Proof. $\operatorname{\vect{curl}}\vect B = \mu_0\gamma\vect E$ (displacement dropped) and $\operatorname{\vect{curl}}\vect E = -\partial_t\vect B$ : taking the curl of the second, $-\Delta\vect E = -\mu_0\gamma\partial_t\vect E$ (the conductor is neutral, $\operatorname{div} \vect E = 0$ ). For $\eu^{\iu(\omega t - kx)}$ : $k^2 = -\iu\mu_0\gamma\omega$ , and $\sqrt{-\iu} = (1 - \iu)/\sqrt2$ gives $\underline k = (1 - \iu)/\delta$ (the root that decays into the conductor). Faraday: $\underline B = \underline k\underline E/\omega = (1 - \iu) \underline E/\omega\delta$ , of modulus $\sqrt2E/\omega\delta$ and phase $-\pi/4$ . ∎

Left: the field inside a conductor — a damped oscillation, dead within a few skin depths. Right: the skin depth of copper and of sea water against frequency: millimetres at mains frequency, micrometres at gigahertz; the sea lets only the longest radio waves in.

Example 14.6 (Consequences of the skin effect)

(i) A wire of radius $a \gg \delta$ carries its current in a ring of thickness $\delta$ : its resistance rises from $1/\gamma\pi a^2$ to about $1/\gamma 2\pi a\delta$ per metre — for a $1\,\mathrm{mm}$ copper wire, four times the DC value at $1\,\mathrm{MHz}$ , a hundred times at $1\,\mathrm{GHz}$ . High-frequency coils use "Litz" wire (many insulated strands) or silver-plated tubes; the power grid’s conductors are stranded with an aluminium sheath. (ii) A metal box attenuates a wave by $\eu^{-d/\delta}$ per thickness $d$ : a $1\,\mathrm{mm}$ aluminium enclosure blocks $1\,\mathrm{MHz}$ by $100\,\mathrm{dB}$ but $50\,\mathrm{Hz}$ hardly at all — low-frequency magnetic fields need iron. (iii) Submarines receive only very-long-wave radio ( $\delta = 2.3\,\mathrm{m}$ at $10\,\mathrm{kHz}$ in sea water); mines are detected by the eddy currents a coil induces in them (Chapter 11, the diffusion time $\mu_0\gamma d^2$ ).

14.3 Waves in a dielectric: the bound electron

Proposition 14.7 (Elastically bound electron; complex index)

In an insulator each electron is bound to its atom: a displacement $\vect r$ from equilibrium calls the restoring force $-m\omega_0^2\vect r$ and a damping $-m\Gamma\,\dd\vect r/\dd t$ (the energy radiated or handed to the lattice). Driven by the wave, $m\ddot{\vect r} = -m\omega_0^2\vect r - m\Gamma\dot{\vect r} - e\vect E$ ; the displaced electrons give the medium the polarization (dipole moment per unit volume) $\vect P = -ne\vect r = \varepsilon_0\underline\chi(\omega) \underline{\vect E}$ with the susceptibility

\underline\chi(\omega) = \frac{\omega_p^2}{\omega_0^2 - \omega^2 + \iu\Gamma\omega} , \qquad \omega_p^2 = \frac{ne^2}{\varepsilon_0m} ,

and the bound current $\vect j = \partial_t\vect P$ . Maxwell’s equations then give

\underline k^2 = \frac{\omega^2}{c^2}\,(1 + \underline\chi) = \frac{\omega^2}{c^2}\,\underline n^2 , \qquad \underline n = n' - \iu n'' ,

with $n'$ the refractive index (phase velocity $c/n'$ ) and $n''$ the extinction (the intensity decays as $\eu^{-2n''\omega x/c}$ ). Far below a resonance ( $\omega \ll \omega_0$ , $\Gamma$ negligible), $n'^2 \approx 1 + \omega_p^2/\omega_0^2 + \omega_p^2\omega^2/\omega_0^4$ : the index rises with frequency — normal dispersion, Cauchy’s law $n \approx A + B/\lambda^2$ — and absorption is negligible; near $\omega_0$ the medium absorbs, and just above it $n'$ falls with $\omega$ (anomalous dispersion).

Proof. Complex amplitudes: $(\omega_0^2 - \omega^2 + \iu\Gamma\omega)\underline{\vect r} = -e\underline{\vect E} /m$ ; $\underline{\vect P} = -ne\underline{\vect r}$ . Then, as for the plasma, $k^2 = \omega^2\mu_0 \varepsilon_0 - \iu\omega\mu_0\underline\gamma$ with $\underline{\vect j} = \iu\omega\underline{\vect P} = \iu\omega\varepsilon_0 \underline\chi\underline{\vect E}$ , i.e. $\underline\gamma = \iu\omega\varepsilon_0\underline\chi$ : $k^2 = (\omega^2/c^2) (1 + \underline\chi)$ . (The plasma is the case $\omega_0 = 0$ , $\Gamma = 0$ .) Expansion: $1/(\omega_0^2 - \omega^2) \approx (1 + \omega^2/\omega_0^2)/\omega_0^2$ . ∎

The bound-electron (Lorentz) model: the real part of the index rises with frequency below the resonance (normal dispersion: violet is slower than red in glass), the imaginary part peaks at the resonance (absorption), and the index falls through the resonance (anomalous dispersion).

Example 14.8 (Glass, water, air)

Glass has its electronic resonances in the ultraviolet ( $\omega_0 \sim 10^{16}$ rad/s): in the visible it is transparent and dispersive, $n = 1.51$ at $700\,\mathrm{nm}$ , $1.53$ at $400\,\mathrm{nm}$ , enough for a prism to spread a rainbow and for a fibre’s pulses to spread (Chapter 8); in the ultraviolet it absorbs — you do not tan behind a window. Water has, besides, a rotational resonance of its polar molecules near $20\,\mathrm{GHz}$ (a Debye relaxation rather than a sharp line): it absorbs microwaves strongly, which heats the food and blinds radar in rain, and has $n'' \approx 0$ in a narrow visible window — the window through which eyes evolved. Air, with $n - 1 = 2.9 \times 10^{-4}$ , still bends the setting Sun by half a degree.

Remark 14.9 (One equation, three media)

Every linear medium enters Maxwell’s equations through its induced current $\underline{\vect j} = \underline\gamma(\omega)\underline{\vect E}$ , and the dispersion relation is always $\underline k^2 = \omega^2/c^2 - \iu\omega\mu_0\underline\gamma(\omega)$ :

medium	$\underline\gamma(\omega)$	$\underline k^2$
plasma	$ne^2/\iu m\omega$ (imaginary)	$(\omega^2 - \omega_p^2)/c^2$ : cut-off
ohmic conductor	$\gamma$ (real)	$-\iu\mu_0\gamma\omega$ : skin effect
dielectric	$\iu\omega\varepsilon_0\underline\chi(\omega)$	$(\omega^2/c^2)(1 + \underline\chi)$ : index

A metal is all three in turn: ohmic below $1 \times 10^{13}\,\mathrm{Hz}$ , a plasma (reflecting) in the visible, transparent in the far ultraviolet.

Method 14.10 (Waves in a medium)

(1) Write the equation of motion of the charges and get $\underline{\vect j} = \underline\gamma\underline{\vect E}$ . (2) Insert into Maxwell–Ampère with the PPH ansatz: $\underline k^2 = \omega^2/c^2 - \iu\omega\mu_0\underline\gamma$ . (3) Take the root with $\operatorname{Im}k \le 0$ (decay in the direction of travel), split into real (propagation, $v_\varphi = \omega/k'$ ) and imaginary (attenuation) parts. (4) Identify the regimes in frequency — propagating, evanescent, absorbing — and the characteristic frequency ( $\omega_p$ , $1/\mu_0\gamma\delta^2$ , $\omega_0$ ). (5) Find $\vect B$ from Faraday and, if needed, the Poynting flux and the dissipated power.

14.4 Exercises

Exercise 14.1 ★

Plasma frequencies for $n = 1 \times 10^{11}\,\mathrm{m}^{-3}$ (ionosphere by night), $1 \times 10^{12}\,\mathrm{m}^{-3}$ (by day), $1 \times 10^{18}\,\mathrm{m}^{-3}$ (re-entry sheath), $8.5 \times 10^{28}\,\mathrm{m}^{-3}$ (copper), $1 \times 10^{5}\,\mathrm{m}^{-3}$ (interstellar space). For each, which of the following passes: a $1\,\mathrm{MHz}$ AM station, a $100\,\mathrm{MHz}$ FM station, a $1.5\,\mathrm{GHz}$ GPS signal, visible light?

Solution

Solution of Exercise 14.1.

$f_p \approx 9\sqrt n$ Hz: $2.8\,\mathrm{MHz}$ , $9\,\mathrm{MHz}$ , $9\,\mathrm{GHz}$ , $2.6 \times 10^{15}\,\mathrm{Hz}$ ( $\lambda_p = 115\,\mathrm{nm}$ ), $2.8\,\mathrm{kHz}$ . AM at $1\,\mathrm{MHz}$ : only through interstellar space. FM: through the ionosphere, not the sheath. GPS: everything but the sheath. Light: everything but copper.

Exercise 14.2 ★

Skin depth of copper ( $\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ ) at $50\,\mathrm{Hz}$ , $1\,\mathrm{kHz}$ , $1\,\mathrm{MHz}$ , $1\,\mathrm{GHz}$ ; of aluminium ( $3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ ) at $1\,\mathrm{MHz}$ ; of sea water ( $5\,\mathrm{S}/\mathrm{m}$ ) at $10\,\mathrm{kHz}$ and $1\,\mathrm{GHz}$ ; of wet soil ( $1 \times 10^{-2}\,\mathrm{S}/\mathrm{m}$ ) at $1\,\mathrm{MHz}$ . Which radio frequencies reach a submarine at $10\,\mathrm{m}$ ?

Solution

Solution of Exercise 14.2.

Copper: $9.2\,\mathrm{mm}$ , $2.1\,\mathrm{mm}$ , $65\,\text{µ}\mathrm{m}$ , $2.1\,\text{µ}\mathrm{m}$ ; aluminium at $1\,\mathrm{MHz}$ : $84\,\text{µ}\mathrm{m}$ ; sea water: $2.3\,\mathrm{m}$ at $10\,\mathrm{kHz}$ , $7\,\mathrm{mm}$ at $1\,\mathrm{GHz}$ ; soil: $5\,\mathrm{m}$ at $1\,\mathrm{MHz}$ . A submarine at $10\,\mathrm{m}$ needs $\delta \gtrsim 10\,\mathrm{m}$ : below about $500\,\mathrm{Hz}$ .

Exercise 14.3 ★

A glass obeys $n = 1.50 + 4.5 \times 10^{3}\,\mathrm{nm}^{2}/\lambda^2$ . Index at $400\,\mathrm{nm}$ , $550\,\mathrm{nm}$ , $700\,\mathrm{nm}$ ; phase velocities; group index $n_g = n - \lambda\,\dd n/\dd\lambda$ at $550\,\mathrm{nm}$ ; angular spread of a white beam refracted at $45{}^{\circ}$ incidence into the glass.

Solution

Solution of Exercise 14.3.

$n = 1.528$ , $1.515$ , $1.509$ ; $v = 1.963$ , $1.980$ , $1.987 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ ; $n_g = n + 2B/\lambda^2 = 1.545$ . Refraction angles $\arcsin(0.707/n)$ : $27.57{}^{\circ}$ and $27.93{}^{\circ}$ : a spread of $0.36{}^{\circ}$ .

Exercise 14.4 ★

A $5\,\mathrm{MHz}$ wave meets a layer with $n = 2 \times 10^{11}\,\mathrm{m}^{-3}$ : is it reflected? Penetration depth; same at $2\,\mathrm{MHz}$ ; minimum frequency that crosses a layer of $n = 1.5 \times 10^{12}\,\mathrm{m}^{-3}$ . Why do shortwave bands "open" and "close" with the hour and the solar cycle?

Solution

Solution of Exercise 14.4.

$f_p = 4.0\,\mathrm{MHz}$ : $5\,\mathrm{MHz}$ passes; $2\,\mathrm{MHz}$ is reflected, penetrating $c/\sqrt{\omega_p^2 - \omega^2} = 14\,\mathrm{m}$ . For $n = 1.5 \times 10^{12}\,\mathrm{m}^{-3}$ : $11\,\mathrm{MHz}$ . The density follows the Sun’s ionizing flux — hour, season, and the eleven-year cycle — so the highest usable frequency moves with them.

Exercise 14.5 ★★

AC resistance. A copper wire of radius $a = 1.0\,\mathrm{mm}$ . (a) DC resistance per metre. (b) At $1\,\mathrm{MHz}$ : $\delta$ ; approximate resistance per metre with the current confined to a ring of thickness $\delta$ ; ratio to DC. (c) At $100\,\mathrm{MHz}$ . (d) Why does a coil of $N$ turns of this wire have a quality factor that stops improving with $N$ at high frequency, and what is Litz wire?

Solution

Solution of Exercise 14.5.

(a) $1/\gamma\pi a^2 = 5.3\,\mathrm{m}\Omega/\mathrm{m}$ . (b) $\delta = 65\,\text{µ}\mathrm{m}$ ; $R \approx 1/\gamma2\pi a\delta = 41\,\mathrm{m}\Omega/\mathrm{m}$ , $7.7$ times DC. (c) $\delta = 6.5\,\text{µ}\mathrm{m}$ , $0.41\,\Omega/\mathrm{m}$ , $77$ times. (d) $R \propto \sqrt f$ grows as fast as $L\omega$ once the skin dominates, and every turn adds resistance; Litz wire bundles many insulated strands thinner than $\delta$ , so the whole copper conducts.

Exercise 14.6 ★★

Shielding. A box of aluminium ( $\gamma = 3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ ) $1.0\,\mathrm{mm}$ thick. Attenuation of the field amplitude crossing the wall, in dB, at $50\,\mathrm{Hz}$ , $10\,\mathrm{kHz}$ , $1\,\mathrm{MHz}$ (the factor $\eu^{-d/\delta}$ , reflection at the surfaces neglected). Why is the mains’ magnetic field shielded with iron (high permeability) rather than copper?

Solution

Solution of Exercise 14.6.

$\delta_{\text{Al}} = 11.9$ , $0.84$ , $0.084\,\mathrm{mm}$ : $d/\delta = 0.08$ , $1.2$ , $12$ : $0.7$ , $10$ , $103\,\mathrm{dB}$ . At $50\,\mathrm{Hz}$ the conductor is transparent; iron of high $\mu_r$ diverts the flux (magnetic shielding) and also shrinks $\delta$ by $\sqrt{\mu_r}$ .

Exercise 14.7 ★★

Energy in a plasma. For the transverse wave of Theorem 14.2: (a) electron velocity amplitude and the kinetic energy density $\tfrac12nm\langle v^2\rangle$ . (b) Mean electric and magnetic energy densities. (c) Show that the total energy density is $\tfrac12\varepsilon_0E_0^2(1 + \dots)$ — compute the bracket — and that $\langle\Pi\rangle$ divided by it equals $v_g$ . (d) What fraction of the energy is in the electrons at $\omega = 2\omega_p$ ?

Solution

Solution of Exercise 14.7.

(a) $v_0 = eE_0/m\omega$ ; $\tfrac12nm\langle v^2\rangle = ne^2E_0^2/4m\omega^2 = \tfrac14\varepsilon_0 E_0^2\,\omega_p^2/\omega^2$ . (b) $\tfrac14\varepsilon_0E_0^2$ and $\tfrac14\varepsilon_0E_0^2(1 - \omega_p^2/ \omega^2)$ . (c) The sum is $\tfrac12\varepsilon_0E_0^2$ (the bracket is $1$ ); $\langle\Pi \rangle = E_0^2k/2\mu_0\omega$ , and the ratio is $c^2k/\omega = v_g$ . (d) $\tfrac14\cdot\tfrac14 /\tfrac12 = 1/8$ .

Exercise 14.8 ★★

Pulsar dispersion. Radio pulses from a pulsar cross a distance $L$ of interstellar plasma ( $n \ll$ anything, $\omega \gg \omega_p$ ). (a) Show that the group delay is $t \approx (L/c)(1 + \omega_p^2/2\omega^2)$ . (b) The delay between the arrivals at $f_1$ and $f_2$ is $\Delta t = \dfrac{e^2}{8\pi^2 \varepsilon_0mc}\,nL\Bigl(\dfrac1{f_1^2} - \dfrac1{f_2^2}\Bigr)$ : compute the constant. (c) A pulsar at $L = 1\,\mathrm{kpc} = 3.1 \times 10^{19}\,\mathrm{m}$ through $n = 3 \times 10^{4}\,\mathrm{m}^{-3}$ : delay between $400\,\mathrm{MHz}$ and $1.4\,\mathrm{GHz}$ . (d) Astronomers measure $\Delta t$ and know $n$ : what do they get? (The dispersion measure, the standard distance gauge for pulsars.)

Solution

Solution of Exercise 14.8.

(a) $1/v_g = (1/c)(1 - \omega_p^2/\omega^2)^{-1/2} \approx (1 + \omega_p^2/2\omega^2)/c$ . (b) $\Delta t = (L\omega_p^2/2c)(1/\omega_1^2 - 1/\omega_2^2)$ with $\omega_p^2 = ne^2/\varepsilon_0m$ and $\omega = 2\pi f$ : constant $e^2/8\pi^2\varepsilon_0mc = 1.34 \times 10^{-7}\,$ (SI). (c) $nL = 9.3 \times 10^{23}\,\mathrm{m}^{-2}$ , $(1/f_1^2 - 1/f_2^2) = 5.7 \times 10^{-18}\,\mathrm{s}^{2}$ : $\Delta t = 0.72\,\mathrm{s}$ . (d) The column $nL$ (the dispersion measure), hence the distance for a model of $n$ .

Exercise 14.9 ★★

A spectral line. In a dilute gas ( $\underline n \approx 1 + \underline\chi/2$ ) the Lorentz model gives $n'' = \tfrac12\operatorname{Im}\underline\chi$ . (a) Show that near $\omega_0$ , $n'' \approx \dfrac{\omega_p^2}{4\omega_0}\,\dfrac{\Gamma/2}{(\omega_0 - \omega)^2 + \Gamma^2/4}$ , a Lorentzian of full width $\Gamma$ . (b) Absorption coefficient $\alpha = 2n''\omega/c$ at the centre. (c) Sodium vapour, $n = 1 \times 10^{17}\,\mathrm{m}^{-3}$ , $\lambda_0 = 589\,\mathrm{nm}$ , $\Gamma = 6 \times 10^{7}\,\mathrm{s}^{-1}$ : $\alpha$ at the centre and the thickness that absorbs $90\%$ (ignore the Doppler broadening, which in fact widens the line a hundredfold). (d) Sketch $n' - 1$ across the line; where is the dispersion anomalous?

Solution

Solution of Exercise 14.9.

(a) $\operatorname{Im}\underline\chi = \omega_p^2\Gamma\omega/[(\omega_0^2 - \omega^2)^2 + \Gamma^2\omega^2]$ with $\omega_0^2 - \omega^2 \approx 2\omega_0(\omega_0 - \omega)$ : $n'' = \tfrac12\operatorname{Im}\underline\chi$ is the stated Lorentzian. (b) $n''(\omega_0) = \omega_p^2/2\omega_0\Gamma$ , $\alpha = 2n''\omega_0/c = \omega_p^2/\Gamma c$ . (c) $\omega_p^2 = 3.2 \times 10^{20}\,\mathrm{s}^{-2}$ : $\alpha = 1.8 \times 10^{4}\,\mathrm{m}^{-1}$ ; $\ln10/\alpha = 0.13\,\mathrm{mm}$ . (d) $n' - 1 \propto (\omega_0 - \omega)/[(\omega_0 - \omega)^2 + \Gamma^2/4]$ : positive below, negative above, decreasing within $\pm\Gamma/2$ of the line — anomalous there.

Exercise 14.10 ★★★

Joule heating in the skin. A wave $E_0\eu^{\iu\omega t}$ (along $y$ ) enters a conductor filling $x > 0$ . (a) Write $\underline E(x, t)$ and deduce $\underline B(x, t)$ (along $z$ ) from Faraday. (b) Mean Poynting vector at the surface: show $\langle\Pi_x(0)\rangle = E_0^2/2\mu_0\omega\delta$ . (c) Mean Joule power per unit area, $\int_0^\infty\tfrac12\gamma|\underline E|^2\dd x$ : show it equals (b). (d) Express the result as $\tfrac12R_sH_0^2$ with $H_0 = B_0/\mu_0$ the surface magnetic field and $R_s = 1/\gamma\delta$ the surface resistance; value for copper at $1\,\mathrm{GHz}$ , and the loss of a microwave cavity whose walls carry $H_0 = 100\,\mathrm{A}/\mathrm{m}$ .

Solution

Solution of Exercise 14.10.

(a) $\underline E = E_0\eu^{-x/\delta}\eu^{\iu(\omega t - x/\delta)}$ , $\underline B = \underline k\underline E/\omega = (1 - \iu)\underline E/\omega\delta$ . (b) $\langle\Pi_x\rangle = \tfrac12\operatorname{Re}(\underline E \underline B^*)/\mu_0 = E_0^2/2\mu_0\omega\delta$ at $x = 0$ . (c) $\int_0^\infty\tfrac12\gamma E_0^2 \eu^{-2x/\delta}\dd x = \gamma E_0^2\delta/4 = E_0^2/2\mu_0\omega\delta$ using $\gamma = 2/\mu_0\omega\delta^2$ . (d) $H_0 = \sqrt2E_0/\mu_0\omega\delta$ , so $\tfrac12R_sH_0^2 = E_0^2/\gamma\mu_0^2\omega^2\delta^3 = E_0^2/2\mu_0\omega\delta$ . Copper at $1\,\mathrm{GHz}$ : $R_s = 1/\gamma\delta = 8\,\mathrm{m}\Omega$ ; $\tfrac12 \times 8 \times 10^{-3} \times 10^4 = 40\,\mathrm{W}/\mathrm{m}^{2}$ .

Exercise 14.11 ★★★

Why metals shine. Treat silver’s conduction electrons ( $n = 5.9 \times 10^{28}\,\mathrm{m}^{-3}$ ) as a collisionless plasma at optical frequencies. (a) $\omega_p$ and the corresponding wavelength. (b) At $500\,\mathrm{nm}$ : $k$ is imaginary — penetration depth; the wave is totally reflected (no absorption in this model): silver is a mirror. (c) At $200\,\mathrm{nm}$ ? Hence the "ultraviolet transparency" of alkali metals. (d) Real silver reflects $95\%$ , gold looks yellow: what does the model miss (think of collisions, and of bound electrons whose resonances lie in the visible for gold)?

Solution

Solution of Exercise 14.11.

(a) $\omega_p = 1.4 \times 10^{16}\,\mathrm{rad}/\mathrm{s}$ , $\lambda_p = 2\pi c/\omega_p = 140\,\mathrm{nm}$ . (b) $\omega = 3.8 \times 10^{15}\,\mathrm{rad}/\mathrm{s} < \omega_p$ : $1/\kappa = c/\sqrt{\omega_p^2 - \omega^2} = 23\,\mathrm{nm}$ , total reflection. (c) Still below $\omega_p$ : reflecting; transparency only beyond $140\,\mathrm{nm}$ (in sodium, $210\,\mathrm{nm}$ ). (d) Collisions give a real part to $\underline\gamma$ and a few percent of absorption; bound-electron resonances in the visible (gold, copper) absorb the blue and colour the metal.

Exercise 14.12 ★★★

Water in the microwave. The orientation of water’s polar molecules gives a susceptibility $\underline\chi = \chi_s/(1 + \iu\omega\tau)$ (Debye relaxation), $\chi_s \approx 80$ , $\tau = 8\,\mathrm{ps}$ , on top of a constant $\chi_\infty \approx 4$ . (a) Real and imaginary parts of $\underline\varepsilon_r = 1 + \chi_\infty + \underline\chi$ at $2.45\,\mathrm{GHz}$ and at $20\,\mathrm{GHz}$ . (b) Complex index and the absorption length $1/(2n''\omega/c)$ of the intensity at $2.45\,\mathrm{GHz}$ : how deep does an oven cook? (c) Power absorbed per unit volume for a field amplitude $E_0$ in the water: $\tfrac12\omega\varepsilon_0\varepsilon''E_0^2$ ; value for $E_0 = 2\,\mathrm{kV}/\mathrm{m}$ . (d) Why is $2.45\,\mathrm{GHz}$ used rather than the $20\,\mathrm{GHz}$ of maximum absorption?

Solution

Solution of Exercise 14.12.

(a) $\omega\tau = 0.12$ : $\underline\varepsilon_r = 84 - 9.7\iu$ ; at $20\,\mathrm{GHz}$ , $\omega\tau = 1$ : $45 - 40\iu$ . (b) $\underline n \approx 9.2 - 0.53\iu$ ; $c/2n''\omega = 1.8\,\mathrm{cm}$ : the oven cooks the outer two centimetres, conduction does the rest. (c) $\tfrac12 \times 1.54 \times 10^{10} \times 8.85 \times 10^{-12} \times 9.7 \times 4 \times 10^6 = 2.6\,\mathrm{MW}/\mathrm{m}^{3}$ . (d) At $20\,\mathrm{GHz}$ the absorption length would be a millimetre: only the skin would cook; $2.45\,\mathrm{GHz}$ penetrates, and is a free band.

A shortwave transmitting station at dusk: its waves, a few megahertz below the plasma frequency of the ionosphere, are about to be reflected back to the ground a thousand kilometres away.

The aurora australis photographed from the International Space Station: particles from the Sun, guided by the Earth’s magnetic field, excite the plasma of the upper atmosphere — the ionosphere whose plasma frequency reflects radio waves (NASA).

14.5 Problem: The ionosphere, the oven door and the prism

Problem 14.1

Weekend problem — three materials seen by a wave: the plasma that bends GPS signals, the metal that keeps microwaves in, and the glass that spreads white light

Part I — The ionosphere and GPS. Electron density in the F layer $n = 1 \times 10^{12}\,\mathrm{m}^{-3}$ ; the total column of electrons along a vertical path is $N_T = \int n\,\dd z = 2 \times 10^{17}\,\mathrm{m}^{-2}$ by day. GPS: $f_1 = 1575\,\mathrm{MHz}$ , $f_2 = 1228\,\mathrm{MHz}$ .

Plasma frequency of the F layer; does GPS pass? Does a $5\,\mathrm{MHz}$ shortwave?
Show that for $\omega \gg \omega_p$ the group velocity is $v_g \approx c(1 - \omega_p^2/2\omega^2)$ and the phase velocity $c(1 + \omega_p^2/2\omega^2)$ .
Extra group delay of the GPS signal crossing the layer: show $\Delta t = \dfrac{e^2}{8\pi^2\varepsilon_0mc}\,\dfrac{N_T}{f^2}$ and compute it for $f_1$ ; the range error $c\,\Delta t$ .
The receiver uses both frequencies: from the two delays it removes the error. Express $N_T$ in terms of $\Delta t_1 - \Delta t_2$ ; what precision on the delay difference gives $N_T$ to $1\%$ ?
How many extra carrier cycles does the delay of question 3 represent at $f_1$ ?
The phase of the carrier is advanced by the same amount as the group is delayed: why (signs of the two corrections)? Which one matters for a receiver that counts carrier cycles?
At night $N_T$ falls tenfold: error for a single-frequency receiver by day and by night.
A solar flare raises $n$ in the D layer ( $80\,\mathrm{km}$ ) to $1 \times 10^{10}\,\mathrm{m}^{-3}$ with many collisions: which radio services are hit, and why does GPS only slightly degrade?

Part II — The oven door. The cavity walls and the door mesh are steel ( $\gamma = 1.0 \times 10^{6}\,\mathrm{S}/\mathrm{m}$ ); the mesh holes are $1.0\,\mathrm{mm}$ wide in a sheet $0.5\,\mathrm{mm}$ thick; $f = 2.45\,\mathrm{GHz}$ , $P = 800\,\mathrm{W}$ .

Skin depth in the steel at $2.45\,\mathrm{GHz}$ ; is the wall "thick"?
Surface resistance $R_s = 1/\gamma\delta$ ; with a surface magnetic field of amplitude $H_0 = 30\,\mathrm{A}/\mathrm{m}$ on the walls ( $1\,\mathrm{m}^{2}$ ), power lost in the walls, $\tfrac12R_sH_0^2$ per unit area; fraction of $P$ .
In a hole of width $a \ll \lambda$ the field cannot propagate (the hole is a waveguide below cut-off, Chapter 16): it decays as $\eu^{-\pi z/a}$ across the thickness $z$ . Attenuation of the amplitude across the $0.5\,\mathrm{mm}$ sheet, in dB.
Why can you see through the mesh (wavelength of light against the hole size) while the microwaves cannot get out?
The glass window is a dielectric with $\underline\varepsilon_r = 4 - 0.02\iu$ : index, and the intensity attenuation length in it; is the glass heated?
A metal fork in the oven: the field at its tips is enhanced; with the skin depth of question 8, estimate the current density at the surface for $H_0 = 300\,\mathrm{A}/\mathrm{m}$ (use $j \approx H_0/\delta$ at the surface) and the local heating $j^2/\gamma$ — why sparks?

Part III — The wire at radio frequency. A copper wire of radius $a = 0.5\,\mathrm{mm}$ ( $\gamma = 6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ ) in a $13.56\,\mathrm{MHz}$ induction heater coil carries $20\,\mathrm{A}$ rms.

Skin depth; effective conducting section; AC resistance per metre against DC.
Power dissipated per metre of coil wire; temperature the wire would reach with convective cooling $h = 20\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K})$ .
The coil induces currents in a steel workpiece ( $\gamma = 1 \times 10^{6}\,\mathrm{S}/\mathrm{m}$ , $\mu_r \approx 1$ when hot): skin depth there; why does induction heating heat only a surface layer, and how is that used for hardening gears?
To reduce the coil’s loss the wire is replaced by a $3\,\mathrm{mm}$ copper tube with water inside: new resistance per metre; does the inside of the tube carry current?
At what frequency would the skin depth in copper equal the wire radius? Below it, what approximation replaces the "thick-wire" formula?

Part IV — The prism. A glass has $n(\lambda) = 1.500 + 5.0 \times 10^{3}\,\mathrm{nm}^{2}/\lambda^2$ , with its resonance at $\lambda_0 = 100\,\mathrm{nm}$ .

Check that the Cauchy form follows from the Lorentz model far below resonance, and that $\omega_p^2/\omega_0^2 = n^2(\infty) - 1$ : value of $\omega_p^2/\omega_0^2$ and of the effective $n$ for this glass.
Index and phase velocity at $400\,\mathrm{nm}$ and $700\,\mathrm{nm}$ ; group index at $550\,\mathrm{nm}$ ; delay between the two colours over $1\,\mathrm{km}$ of such glass in a fibre.
A $60{}^{\circ}$ prism at minimum deviation for $550\,\mathrm{nm}$ : deviation (use $n\sin(A/2) = \sin((A + D)/2)$ ); angular distance between $400\,\mathrm{nm}$ and $700\,\mathrm{nm}$ (differentiate); on a screen $2\,\mathrm{m}$ away, the width of the spectrum.
Why is the glass opaque at $200\,\mathrm{nm}$ , and what is its colour if a trace of iron adds a weak resonance at $700\,\mathrm{nm}$ ?
In the Lorentz picture, why is blue deviated more than red by the prism?
In one table give, for the F layer, the steel, the copper and the glass at their working frequencies, the nature of $\underline\gamma$ and the fate of the wave.

Solution

Solution of Problem 14.1.

1. $f_p = 9\,\mathrm{MHz}$ : GPS passes, the shortwave is reflected.

2. $k = (\omega/c)\sqrt{1 - \omega_p^2/\omega^2}$ : $v_\varphi = \omega/k \approx c(1 + \omega_p^2/ 2\omega^2)$ , $v_g = \dd\omega/\dd k = c^2k/\omega \approx c(1 - \omega_p^2/2\omega^2)$ .

3. $\Delta t = \int(1/v_g - 1/c)\dd z = \int\omega_p^2\dd z/2c\omega^2 = e^2N_T/2\varepsilon_0 mc\omega^2 = (e^2/8\pi^2\varepsilon_0mc)N_T/f^2 = 1.34 \times 10^{-7} \times 2 \times 10^{17}/2.48 \times 10^{18} = 11\,\mathrm{ns}$ : $3.2\,\mathrm{m}$ .

4. $N_T = (\Delta t_1 - \Delta t_2)/K(1/f_1^2 - 1/f_2^2)$ ; the difference is $7\,\mathrm{ns}$ : $1\%$ needs $0.07\,\mathrm{ns}$ .

5. $v_\varphi$ exceeds $c$ by as much as $v_g$ falls short: the carrier phase arrives early while the modulation arrives late. A carrier-phase receiver applies the correction with the opposite sign.

6. $3.2\,\mathrm{m}$ by day, $0.3\,\mathrm{m}$ by night.

7. $f_p = 0.9\,\mathrm{MHz}$ with collisions: medium and short waves are absorbed (a radio blackout); GPS at $1.5\,\mathrm{GHz}$ feels an absorption $\propto 1/\omega^2$ , negligible.

8. $f_1\Delta t = 1.575 \times 10^9 \times 1.1 \times 10^{-8} = 17$ cycles.

9. $\delta = \sqrt{2/\mu_0\gamma\omega} = 10\,\text{µ}\mathrm{m}$ : a millimetre wall is a hundred skin depths.

10. $R_s = 1/\gamma\delta = 0.1\,\Omega$ ; $\tfrac12R_sH_0^2 = 45\,\mathrm{W}/\mathrm{m}^{2}$ : $45\,\mathrm{W}$ , $6\%$ .

11. $\eu^{-\pi \times 0.5} = 0.21$ : $-14\,\mathrm{dB}$ in amplitude, $-27\,\mathrm{dB}$ in power (and the tiny hole couples little to begin with).

12. Light, $\lambda = 0.5\,\text{µ}\mathrm{m}$ , is two thousand times smaller than the holes and passes; the microwave, $\lambda = 12\,\mathrm{cm}$ , is a hundred times larger.

13. $\underline n = 2 - 0.005\iu$ ; $c/2n''\omega = 1.9\,\mathrm{m}$ : the glass is hardly warmed.

14. $j \approx H_0/\delta = 3 \times 10^{7}\,\mathrm{A}/\mathrm{m}^{2}$ , $j^2/\gamma = 9 \times 10^{8}\,\mathrm{W}/\mathrm{m}^{3}$ in a ten-micrometre layer: tips heat in milliseconds, emit electrons, ionize the air — sparks.

15. $\delta = 18\,\text{µ}\mathrm{m}$ ; section $2\pi a\delta = 5.6 \times 10^{-8}\,\mathrm{m}^{2}$ against $\pi a^2 = 7.9 \times 10^{-7}\,\mathrm{m}^{2}$ : $R_{\text{ac}} = 0.30\,\Omega/\mathrm{m}$ , fourteen times DC.

16. $RI^2 = 120\,\mathrm{W}/\mathrm{m}$ ; surface $\pi d = 3.1 \times 10^{-3}\,\mathrm{m}^{2}$ per metre: $\Delta T \approx 1900\,\mathrm{K}$ — it would melt; such coils are water-cooled tubes.

17. $\delta = 0.14\,\mathrm{mm}$ : the induced current, and the heat, stay in a tenth of a millimetre; a short pulse followed by a quench hardens the surface of a gear while the core stays tough.

18. $R = 1/\gamma2\pi a\delta = 0.10\,\Omega/\mathrm{m}$ , three times less; the inside carries nothing — hence a tube, with the coolant where the copper would be wasted.

19. $\delta = a$ at $f = 1/\pi\mu_0\gamma a^2 = 17\,\mathrm{Hz}$ ; below, the current is uniform and the DC formula holds.

20. $n^2 \approx 1 + \omega_p^2/\omega_0^2 + (\omega_p^2/\omega_0^2)(\lambda_0/\lambda)^2$ : Cauchy’s form with $n(\infty)^2 - 1 = \omega_p^2/\omega_0^2 = 1.25$ for $n(\infty) = 1.5$ .

21. $n = 1.531$ and $1.510$ , $v = 1.96$ and $1.99 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ ; $n_g(550) = 1.550$ ; $n_g(400) - n_g(700) = 0.063$ : $210\,\mathrm{ns}$ over a kilometre.

22. $\sin((A + D)/2) = 1.5165 \times 0.5$ : $D = 38.6{}^{\circ}$ ; $\dd D = 2\sin (A/2)\dd n/\cos((A + D)/2) = 1.53\,\dd n$ : $\Delta D = 1.85{}^{\circ}$ , $6.5\,\mathrm{cm}$ at $2\,\mathrm{m}$ .

23. At $200\,\mathrm{nm}$ the frequency approaches the resonance and $n''$ grows: opaque. A weak resonance at $700\,\mathrm{nm}$ absorbs the red: green glass.

24. Blue is closer to the ultraviolet resonance: the bound electrons respond more, $n$ is larger, and the deviation too.

25. F layer at $1.5\,\mathrm{GHz}$ : imaginary $\underline\gamma$ , propagation with a small group delay. Steel at $2.45\,\mathrm{GHz}$ : real $\underline\gamma$ , skin effect, reflection. Copper at $13.56\,\mathrm{MHz}$ : the same, a skin of $18\,\text{µ}\mathrm{m}$ . Glass in the visible: $\iu\omega\varepsilon_0\chi$ , a real index, a transparent dispersive medium.

Terms defined in this chapter

Plasma; the free-electron model

See all 393 terms in the glossary