University Physics — Year 2 · Bachelor Year 2

8Dispersion and Wave Packets

Drop a stone in a pond and watch the ring of ripples: new crests are born at its inner edge, run outward through the ring, and die at its outer edge — the ring as a whole moves slower than the crests in it. Sit on a beach after a distant storm and the long, slow swell arrives a day before the short chop: the sea sorts waves by wavelength. The first transatlantic telegraph cable, in 1858, turned crisp dots and dashes into a smear that took minutes to read. In all three cases the speed of a wave depends on its frequency: the medium is dispersive. This chapter introduces the dispersion relation, the complex wavenumber that carries both propagation and absorption, the group velocity at which a packet — and its energy, and its information — actually travels, and the line along which telegraphy, television and the internet have travelled: the coaxial cable.

8.1 The dispersion relation

Definition 8.1 (Dispersion relation; phase velocity)

A linear, homogeneous, time-invariant medium admits the plane monochromatic waves $\underline s = A\,\eu^{\iu(\omega t - kx)}$ (complex notation; the physical signal is the real part) provided $\omega$ and $k$ satisfy the dispersion relation of the medium, $\mathcal D(\omega, k) = 0$ , solved as $k(\omega)$ or $\omega(k)$ . For real $k$ the wave travels without deforming at the phase velocity

v_\varphi = \frac\omega k .

The medium is non-dispersive when $v_\varphi$ does not depend on $\omega$ (then $\omega = ck$ and every signal propagates undeformed — the d’Alembert equation) and dispersive otherwise.

Proof. Inserting $\eu^{\iu(\omega t - kx)}$ into a linear equation with constant coefficients turns each $\partial_t$ into $\iu\omega$ and each $\partial_x$ into $-\iu k$ , leaving an algebraic relation. The string, the sound wave and the rod gave $\omega^2 = c^2k^2$ . ∎

Example 8.2 (Dispersive and non-dispersive)

(i) A string on an elastic bed (restoring force $-Ky$ per unit length, or a chain of pendulums), $\mu\partial_t^2y = T\partial_x^2y - Ky$ : $\omega^2 = \omega_c^2 + c^2k^2$ with $\omega_c = \sqrt{K/\mu}$ — the Klein–Gordon relation, the same as a plasma’s and a waveguide’s (Chapters 14 and 16): $v_\varphi = c/\sqrt{1 - \omega_c^2/\omega^2} > c$ , and no real $k$ below the cut-off $\omega_c$ . (ii) Deep-water gravity waves: $\omega^2 = gk$ (admitted), $v_\varphi = \sqrt{g/k} = \sqrt{g\lambda/2\pi}$ : long waves are faster — $12.5\,\mathrm{m}/\mathrm{s}$ for $100\,\mathrm{m}$ , $40\,\mathrm{m}/\mathrm{s}$ for $1\,\mathrm{km}$ . (iii) The chain of atoms, $\omega = 2\omega_0|\sin(ka/2)|$ (Chapter 6). (iv) Light in glass, $k = n(\omega)\omega/c$ , which is why a prism spreads a spectrum.

Left: three dispersion relations — a straight line for a non-dispersive medium, the Klein–Gordon hyperbola with its cut-off, the parabola of deep-water waves. Right: for the Klein–Gordon relation the phase velocity exceeds c and the group velocity stays below it, with v_ v_g = c2. — Left: three dispersion relations — a straight line for a non-dispersive medium, the Klein–Gordon hyperbola with its cut-off, the parabola of deep-water waves. Right: for the Klein–Gordon relation the phase velocity exceeds $c$ and the group velocity stays below it, with $v_\varphi v_g = c^2$ .

Definition 8.3 (Complex wavenumber: propagation and attenuation)

When the dispersion relation gives, for real $\omega$ , a complex $\underline k = k' - \iu k''$ , the wave is

\underline s = A\,\eu^{-k''x}\,\eu^{\iu(\omega t - k'x)} :

it propagates at $v_\varphi = \omega/k'$ and its amplitude decays as $\eu^{-x/\delta}$ with the attenuation length $\delta = 1/k''$ (it must decay in its direction of propagation: $k'$ and $k''$ of the same sign). If $\underline k$ is purely imaginary, $\underline k = -\iu\kappa$ , the wave is evanescent: $A\,\eu^{-\kappa x}\eu^{\iu\omega t}$ , a standing oscillation whose amplitude dies over the distance $1/\kappa$ without propagating anything — the case of the Klein–Gordon medium below its cut-off, $\kappa = \sqrt{\omega_c^2 - \omega^2}/c$ , and of a metal at optical frequencies.

Example 8.4 (A damped string)

A string in a viscous fluid, $\mu\partial_t^2y + \alpha\partial_ty = T\partial_x^2y$ : $\underline k^2 = (\omega^2 - \iu\alpha\omega/\mu)/c^2$ . For weak damping ( $\alpha \ll \mu \omega$ ), $\underline k \approx \dfrac\omega c\Bigl(1 - \iu\dfrac{\alpha}{2\mu\omega}\Bigr)$ : the wave keeps its speed and loses amplitude over $\delta = 2\mu c/\alpha = 2Z/\alpha$ , independent of frequency — a dissipative, non-dispersive loss, $8.7\,\mathrm{dB}$ per length $\delta$ .

8.2 Wave packets and group velocity

Proposition 8.5 (Beats; group velocity)

Two waves of equal amplitude and neighbouring frequencies $\omega \pm \delta\omega$ , wavenumbers $k \pm \delta k$ , superpose into

s = 2A\cos(\delta\omega\,t - \delta k\,x)\cos(\omega t - kx) :

a carrier at $(\omega, k)$ moving at $v_\varphi = \omega/k$ , modulated by a slow envelope moving at $\delta\omega/\delta k$ . A wave packet — a superposition of monochromatic waves with wavenumbers in a narrow band around $k_0$ — travels, as a whole, at the group velocity

v_g = \frac{\dd\omega}{\dd k}\Bigr|_{k_0} ,

which is the velocity of its energy and of the information it carries. In a non-dispersive medium $v_g = v_\varphi = c$ ; otherwise the packet deforms as it goes (it spreads), the faster the larger $\dd^2\omega/\dd k^2$ and the narrower its spectrum’s complement, its spatial width.

Proof. The sum-to-product formula gives the beats. For a packet, write $s = \int a(k)\eu^{\iu(\omega(k)t - kx)}\dd k$ (a Fourier superposition — the Fourier integral is studied in the Year 3 mathematics volume; here only its interpretation as a continuous sum of sinusoids is used), with $a(k)$ peaked at $k_0$ . Expand $\omega(k) \approx \omega_0 + v_g(k - k_0)$ : then $s \approx \eu^{\iu(\omega_0t - k_0x)}\int a(k)\eu^{\iu(k - k_0)(v_gt - x)}\dd k$ , a carrier times an envelope that is a function of $x - v_gt$ only — the envelope moves at $v_g$ . The next term, $\tfrac12\omega''(k_0)(k - k_0)^2t$ , dephases the components in time and spreads the envelope. The energy of a packet is located where its envelope is; so is any signal. ∎

Beats: two neighbouring frequencies produce a fast carrier (moving at the phase velocity) under a slow envelope (moving at the group velocity); in a dispersive medium the two speeds differ and the crests slide through the envelope.

Example 8.6 (Deep water: the crests outrun the group)

$\omega = \sqrt{gk}$ : $v_g = \tfrac12\sqrt{g/k} = \tfrac12v_\varphi$ . A group of swell moves at half the speed of its crests: watching a wave train, you see crests appear at the back of the group, travel through it and vanish at the front — exactly the pond’s ring. A storm $3000\,\mathrm{km}$ away sends its $15\,\mathrm{s}$ swell ( $\lambda = gT^2/2\pi = 350\,\mathrm{m}$ , $v_g = gT/4\pi = 11.7\,\mathrm{m}/\mathrm{s}$ ) in three days, its $8\,\mathrm{s}$ waves ( $v_g = 6.2\,\mathrm{m}/\mathrm{s}$ ) in five and a half: from the arrival times of the different periods, oceanographers locate the storm.

The wake of a motorboat on a calm lake: the feathered pattern stays inside a wedge of fixed angle because the energy of each water wave travels at half the speed of its crests — dispersion made visible.

Example 8.7 (Klein–Gordon: faster than light, and not)

For $\omega^2 = \omega_c^2 + c^2k^2$ : $v_\varphi = \omega/k$ and $v_g = c^2k/\omega$ , so $v_\varphi v_g = c^2$ : the phase velocity exceeds $c$ (in the ionosphere, a $10\,\mathrm{MHz}$ wave’s crests move faster than light) but the group velocity, which carries the signal, stays below it. A phase velocity transports no energy and no information — the crests are like the spot of a lighthouse beam sweeping a distant cloud.

Remark 8.8 (Spreading)

A packet of spatial width $\Delta x$ contains wavenumbers over $\Delta k \sim 1/\Delta x$ (the Fourier reciprocity, admitted); its components’ group velocities differ by $\Delta v_g \approx |\omega''|\Delta k$ , so after a time $t$ it has spread by $|\omega''|\Delta k\,t$ : it doubles its width after $t_{\text{sp}} \sim \Delta x^2/|\omega''|$ . A packet of ten deep-water waves of $100\,\mathrm{m}$ ( $\Delta x \approx 1\,\mathrm{km}$ , $\omega'' = -\tfrac14\sqrt{g/k^3} = -50\,\mathrm{m}^{2}/\mathrm{s}$ ) doubles in about six hours; a light pulse of $1\,\mathrm{ns}$ in an optical fibre spreads by tens of picoseconds per kilometre — the limit of the bit rate of long links. The quantum wave packet of Chapter 30 spreads for the same reason.

8.3 The coaxial cable

Proposition 8.9 (Telegrapher’s equations; lossless line)

A coaxial cable (or any two-conductor line) has, per unit length, an inductance $\Lambda$ and a capacitance $\Gamma$ . The voltage $v(x, t)$ between the conductors and the current $i(x, t)$ in the inner one obey the telegrapher’s equations

\frac{\partial v}{\partial x} = -\Lambda\frac{\partial i}{\partial t} , \qquad \frac{\partial i}{\partial x} = -\Gamma\frac{\partial v}{\partial t} ,

hence the d’Alembert equation for $v$ and $i$ with

c = \frac1{\sqrt{\Lambda\Gamma}} , \qquad v = Z_c\,i \ \text{for a wave toward } +x, \quad Z_c = \sqrt{\frac\Lambda\Gamma} ,

$Z_c$ being the characteristic impedance of the line. For a coaxial cable of radii $a < b$ filled with a dielectric of relative permittivity $\varepsilon_r$ , $\Lambda = \dfrac{\mu_0}{2\pi}\ln\dfrac ba$ , $\Gamma = \dfrac{2\pi\varepsilon_0\varepsilon_r}{\ln(b/a)}$ , so $c = c_0/\sqrt{\varepsilon_r}$ (about $2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , two thirds of the speed of light) and $Z_c = \dfrac{60\,\Omega} {\sqrt{\varepsilon_r}}\ln\dfrac ba$ : $50\,\Omega$ or $75\,\Omega$ for the usual cables.

Proof. Model the slice $[x, x + \dd x]$ as a series inductance $\Lambda\dd x$ and a shunt capacitance $\Gamma\dd x$ : the voltage drop along the slice is $\Lambda\dd x\,\partial_ti$ , and the current lost into the capacitance is $\Gamma\dd x \,\partial_tv$ . Cross-differentiate. For $v = f(x - ct)$ , the first equation gives $i = f(x - ct)/\Lambda c = v/\sqrt{\Lambda/\Gamma}$ . The values of $\Lambda$ and $\Gamma$ are those of the cylindrical capacitor and of the coaxial inductor computed in the Year 1 volume. ∎

Left: the equivalent circuit of a slice x of a lossless line — series inductance, shunt capacitance — from which the telegrapher’s equations follow. Right: a coaxial cable in section; its and depend only on the ratio of radii and on the dielectric. — Left: the equivalent circuit of a slice $\dd x$ of a lossless line — series inductance, shunt capacitance — from which the telegrapher’s equations follow. Right: a coaxial cable in section; its $\Lambda$ and $\Gamma$ depend only on the ratio of radii and on the dielectric.

A coaxial cable: inner core, dielectric, braided outer conductor — a transmission line whose inductance and capacitance per metre set its speed and its characteristic impedance.

Proposition 8.10 (Lossy line; the distortionless condition)

With a series resistance $R$ and a shunt conductance $G$ per unit length, the equations become $\partial_xv = -\Lambda\partial_ti - Ri$ , $\partial_xi = -\Gamma\partial_tv - Gv$ , and the dispersion relation $\underline k^2 = (\Lambda\omega - \iu R)(\Gamma\omega - \iu G)$ : the line is dispersive and attenuating. Two limits: (i) $R/\Lambda = G/\Gamma$ (Heaviside’s condition, reached by adding inductance): $\underline k = \omega\sqrt{\Lambda\Gamma} - \iu\sqrt{RG}$ , every frequency travels at the same speed with the same attenuation — the line attenuates but does not distort. (ii) $\Lambda \approx 0$ , $G \approx 0$ (the first submarine cables): $\partial_x^2v = R\Gamma\,\partial_tv$ , a diffusion equation — a pulse does not propagate but smears, reaching a distance $L$ after a time $\sim R\Gamma L^2$ (Kelvin’s law of squares), a hundred times longer for a cable ten times longer.

Proof. Insert $\eu^{\iu(\omega t - kx)}$ : $-\iu kv = -(\iu\omega\Lambda + R)i$ , $-\iu ki = -(\iu \omega\Gamma + G)v$ ; multiply. Under Heaviside’s condition the product is a perfect square. The diffusion limit drops the $\iu\omega\Lambda$ and $G$ terms. ∎

Example 8.11 (Three cables)

The 1858 transatlantic cable: a single copper wire in gutta-percha, $R \approx 3\,\Omega/\mathrm{km}$ , $\Gamma \approx 0.3\,\text{µ}\mathrm{F}/\mathrm{km}$ , $L = 3000\,\mathrm{km}$ : $R\Gamma = 9 \times 10^{-10}\,\mathrm{s}/\mathrm{m}^{2}$ and $R\Gamma L^2 \approx 8\,\mathrm{s}$ per dot: a word a minute, and the operators’ attempts to force the signal with $2\,\mathrm{kV}$ destroyed the insulation within weeks. A telephone line loaded with coils every $2\,\mathrm{km}$ to satisfy Heaviside’s condition (1900): clear speech over hundreds of kilometres. A modern $50\,\Omega$ coaxial cable: $c = 2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , $\Lambda = 0.25\,\text{µ}\mathrm{H}/\mathrm{m}$ , $\Gamma = 100\,\mathrm{pF}/\mathrm{m}$ , and an attenuation, due to the skin effect in the conductors (Chapter 14), growing as $\sqrt f$ : a few decibels per hundred metres at $100\,\mathrm{MHz}$ .

Method 8.12 (Reading a dispersion relation)

(1) Write the wave equation of the medium in complex notation and solve for $\underline k(\omega)$ . (2) Real $k$ : compute $v_\varphi = \omega/k$ and $v_g = \dd\omega/\dd k$ (or $1/(\dd k/\dd\omega)$ ); compare: dispersive or not, normal ( $v_g < v_\varphi$ ) or anomalous. (3) Complex $k$ : split into propagation ( $k'$ ) and attenuation ( $k''$ , length $1/k''$ ); purely imaginary means evanescent, no transport. (4) A cut-off frequency separates a propagating band from an evanescent one. (5) For a signal, reason on the envelope at $v_g$ , never on the crests.

8.4 Exercises

Exercise 8.1 ★

Deep-water waves, $\omega^2 = gk$ . (a) Phase velocity, group velocity and period of a $100\,\mathrm{m}$ swell. (b) A swell of period $14\,\mathrm{s}$ : wavelength, $v_\varphi$ , $v_g$ ; time for its energy to cross $5000\,\mathrm{km}$ . (c) Why does the chop of a $2\,\mathrm{s}$ period not get there at all?

Solution

Solution of Exercise 8.1.

(a) $k = 0.063\,\mathrm{rad}/\mathrm{m}$ , $\omega = \sqrt{gk} = 0.79\,\mathrm{rad}/\mathrm{s}$ , $T = 8.0\,\mathrm{s}$ ; $v_\varphi = 12.5\,\mathrm{m}/\mathrm{s}$ , $v_g = 6.3\,\mathrm{m}/\mathrm{s}$ . (b) $\omega = 0.45\,\mathrm{rad}/\mathrm{s}$ , $k = \omega^2 /g = 0.021\,\mathrm{rad}/\mathrm{m}$ , $\lambda = 310\,\mathrm{m}$ ; $v_\varphi = gT/2\pi = 22\,\mathrm{m}/\mathrm{s}$ , $v_g = 11\,\mathrm{m}/\mathrm{s}$ ; $5 \times 10^6/11 = 4.6 \times 10^{5}\,\mathrm{s} = 5.3$ days. (c) $\lambda = 6\,\mathrm{m}$ , $v_g = 1.6\,\mathrm{m}/\mathrm{s}$ : it would need five weeks, and short waves are damped by viscosity and broken by the wind long before.

Exercise 8.2 ★

A medium obeys $\omega^2 = \omega_c^2 + c^2k^2$ with $\omega_c = 2\pi \times 1 \times 10^{7}\,\mathrm{rad}/\mathrm{s}$ and $c = 3 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ . For $f = 20\,\mathrm{MHz}$ : $k$ , $v_\varphi$ , $v_g$ , and their product. For $f = 5\,\mathrm{MHz}$ : $\kappa$ and the penetration depth.

Solution

Solution of Exercise 8.2.

$\omega = 1.26 \times 10^{8}\,\mathrm{rad}/\mathrm{s}$ , $\omega_c = 6.3 \times 10^{7}\,\mathrm{rad}/\mathrm{s}$ : $k = \sqrt{\omega^2 - \omega_c^2}/c = 0.36\,\mathrm{rad}/\mathrm{m}$ , $v_\varphi = 3.5 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , $v_g = c^2k/\omega = 2.6 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , product $c^2$ . At $5\,\mathrm{MHz}$ , $\omega < \omega_c$ : $\kappa = \sqrt{\omega_c^2 - \omega^2}/c = 0.18\,\mathrm{m}^{-1}$ , depth $5.5\,\mathrm{m}$ .

Exercise 8.3 ★

Two tuning forks at $440\,\mathrm{Hz}$ and $444\,\mathrm{Hz}$ : beat frequency, and the period of the loudness variations. Two water waves of wavelengths $100\,\mathrm{m}$ and $110\,\mathrm{m}$ : frequencies, wavelength and speed of the envelope; compare with $v_g$ at $105\,\mathrm{m}$ .

Solution

Solution of Exercise 8.3.

Beat at $4\,\mathrm{Hz}$ , loudness period $0.25\,\mathrm{s}$ . Water: $f = 0.125$ and $0.119\,\mathrm{Hz}$ ; envelope $\cos(\delta\omega t - \delta kx)$ with $\delta k = 2.9 \times 10^{-3}\,\mathrm{rad}/\mathrm{m}$ (beats $1.1\,\mathrm{km}$ apart) and speed $\delta\omega/\delta k = 6.4\,\mathrm{m}/\mathrm{s}$ — the group velocity at $105\,\mathrm{m}$ , $\tfrac12\sqrt{g\lambda/2\pi} = 6.4\,\mathrm{m}/\mathrm{s}$ .

Exercise 8.4 ★

A coaxial cable: inner conductor $0.9\,\mathrm{mm}$ in diameter, outer $3.0\,\mathrm{mm}$ , polyethylene ( $\varepsilon_r = 2.3$ ). $\Lambda$ , $\Gamma$ , wave speed, characteristic impedance, delay per metre; length of cable equivalent to a $10\,\mathrm{ns}$ delay. Same cable with air instead of polyethylene.

Solution

Solution of Exercise 8.4.

$\ln(b/a) = 1.20$ : $\Lambda = 0.24\,\text{µ}\mathrm{H}/\mathrm{m}$ , $\Gamma = 106\,\mathrm{pF}/\mathrm{m}$ , $c = 1/ \sqrt{\Lambda\Gamma} = 2.0 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , $Z_c = 48\,\Omega$ , $5.0\,\mathrm{ns}/\mathrm{m}$ ; $10\,\mathrm{ns}$ $\leftrightarrow$ $2.0\,\mathrm{m}$ . Air: $\Gamma = 46\,\mathrm{pF}/\mathrm{m}$ , $c = c_0$ , $Z_c = 72\,\Omega$ .

Exercise 8.5 ★★

The stiff string. A real piano string obeys $\mu\partial_t^2y = T\partial_x^2y - EI\,\partial_x^4y$ ( $EI$ the bending stiffness). (a) Dispersion relation, phase and group velocities. (b) For the A $_4$ string of Chapter 6 ( $T = 685\,\mathrm{N}$ , $\mu = 6.1\,\mathrm{g}/\mathrm{m}$ , $EI = 9.8 \times 10^{-3}\,\mathrm{N}\,\mathrm{m}^{2}$ ), relative increase of $v_\varphi$ for the eighth harmonic ( $k = 8\pi/L$ , $L = 0.38\,\mathrm{m}$ ); check against the inharmonicity found there. (c) Is the dispersion normal or anomalous? What does a sharp pluck look like after a few round trips?

Solution

Solution of Exercise 8.5.

(a) $\omega^2 = c^2k^2 + (EI/\mu)k^4$ ; $v_\varphi = \sqrt{c^2 + (EI/\mu)k^2}$ ; $v_g = (c^2k + 2(EI/\mu)k^3)/\omega$ . (b) $c^2 = 1.12 \times 10^{5}\,\mathrm{m}^{2}/\mathrm{s}^{2}$ , $k = 66\,\mathrm{rad}/\mathrm{m}$ , $(EI/\mu)k^2 = 7.0 \times 10^{3}\,\mathrm{m}^{2}/\mathrm{s}^{2}$ : $v_\varphi/c = \sqrt{1.063} = 1.031$ , $+3.1\%$ — the $53$ cents of Chapter 6. (c) $v_g > v_\varphi$ : anomalous; the high harmonics run ahead and the attack of a pluck turns into a brief descending chirp.

Exercise 8.6 ★★

Damped string. $\mu\partial_t^2y + \alpha\partial_ty = T\partial_x^2y$ . (a) Derive the dispersion relation. (b) For $\alpha \ll \mu\omega$ , show $k' \approx \omega/c$ and $k'' \approx \alpha/2Z$ ; attenuation in decibels per metre. (c) Numbers: $\mu = 5\,\mathrm{g}/\mathrm{m}$ , $T = 50\,\mathrm{N}$ , $\alpha = 0.02\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}$ , $f = 100\,\mathrm{Hz}$ : check the approximation and give $\delta$ . (d) Is the damped string dispersive? In which regime of $\alpha$ would it become so?

Solution

Solution of Exercise 8.6.

(a) $-\mu\omega^2 + \iu\alpha\omega = -Tk^2$ : $\underline k^2 = (\omega^2/c^2)(1 - \iu\alpha/\mu\omega)$ . (b) $\underline k \approx (\omega/c)(1 - \iu\alpha/2\mu\omega)$ : $k' = \omega/c$ , $k'' = \alpha/2\mu c = \alpha/2Z$ ; $8.7k''$ dB/m. (c) $c = 100\,\mathrm{m}/\mathrm{s}$ , $Z = 0.5\,\mathrm{kg}/\mathrm{s}$ , $\mu\omega = 3.1 \gg 0.02$ : $k'' = 0.02\,\mathrm{m}^{-1}$ , $\delta = 50\,\mathrm{m}$ , $0.17\,\mathrm{dB}/\mathrm{m}$ . (d) Not in this limit ( $k' = \omega/c$ ); when $\alpha \gtrsim \mu\omega$ the square root no longer linearizes and $k'$ depends non-linearly on $\omega$ : dispersive (the diffusive limit of the cable).

Exercise 8.7 ★★

Ripples. Short water waves are ruled by surface tension: $\omega^2 = \gamma k^3/\rho$ ( $\gamma = 0.072\,\mathrm{N}/\mathrm{m}$ ); the full relation is $\omega^2 = gk + \gamma k^3/\rho$ . (a) $v_\varphi$ and $v_g$ for pure capillary waves; which is larger? (b) Show that the phase velocity of the full relation is minimal at $\lambda_m = 2\pi\sqrt{\gamma/\rho g}$ and compute $\lambda_m$ and $v_{\min}$ . (c) A boat slower than $v_{\min}$ makes no wake: why? (d) Raindrops on a pond make rings whose crests lag behind the ring: in which regime?

Solution

Solution of Exercise 8.7.

(a) $v_\varphi = \sqrt{\gamma k/\rho}$ , $v_g = \tfrac32v_\varphi$ : the group outruns the crests. (b) $v_\varphi^2 = g/k + \gamma k/\rho$ is minimal at $k_m = \sqrt{\rho g/\gamma}$ : $\lambda_m = 2\pi\sqrt{\gamma/\rho g} = 1.7\,\mathrm{cm}$ , $v_{\min} = (4g\gamma/\rho)^{1/4} = 23\,\mathrm{cm}/\mathrm{s}$ . (c) A wake is a pattern stationary with respect to the boat, made of waves whose phase velocity matches its speed; none is slower than $23\,\mathrm{cm}/\mathrm{s}$ . (d) Capillary regime, $v_g > v_\varphi$ : crests are born at the front of the ring and die at its back.

Exercise 8.8 ★★

Spreading. A packet of initial width $\Delta x_0$ doubles its width after $t_{\text{sp}} \approx \Delta x_0^2/|\omega''(k_0)|$ . (a) A group of deep-water waves, $\lambda = 50\,\mathrm{m}$ , $\Delta x_0 = 500\,\mathrm{m}$ . (b) A pulse of light in a glass fibre: $\omega'' \approx -0.16\,\mathrm{m}^{2}/\mathrm{s}$ ; a $10\,\mathrm{ps}$ pulse ( $\Delta x_0 = 2\,\mathrm{mm}$ ): spreading time and the distance travelled meanwhile; the same for a $1\,\mathrm{ns}$ pulse. Why do long optical links use longer pulses or dispersion-compensating fibre?

Solution

Solution of Exercise 8.8.

(a) $k = 0.126\,\mathrm{rad}/\mathrm{m}$ , $\omega'' = -\tfrac14\sqrt{g/k^3} = -18\,\mathrm{m}^{2}/\mathrm{s}$ : $t_{\text{sp}} = 2.5 \times 10^5/18 = 1.4 \times 10^{4}\,\mathrm{s}$ , four hours. (b) $10\,\mathrm{ps}$ : $t_{\text{sp}} = 4 \times 10^{-6}/0.16 = 25\,\text{µ}\mathrm{s}$ , $5\,\mathrm{km}$ at $2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ ; $1\,\mathrm{ns}$ : $0.04/0.16 = 0.25\,\mathrm{s}$ , $50\,000\,\mathrm{km}$ . Short pulses (high bit rates) spread within kilometres: long links use compensating fibre or moderate rates.

Exercise 8.9 ★★

The submarine cable. $R = 3\,\Omega/\mathrm{km}$ , $\Gamma = 0.3\,\text{µ}\mathrm{F}/\mathrm{km}$ , $\Lambda$ and $G$ negligible. (a) Show that $v$ obeys a diffusion equation and give its diffusivity. (b) Dispersion relation $\underline k(\omega)$ ; phase velocity and attenuation length at $1\,\mathrm{Hz}$ and at $10\,\mathrm{Hz}$ . (c) Time for a signal to "arrive" over $3000\,\mathrm{km}$ ( $\sim R\Gamma L^2$ ); words per minute. (d) Heaviside’s cure: the inductance per kilometre needed for $R/\Lambda = G/\Gamma$ if $G = 1 \times 10^{-7}\,\mathrm{S}/\mathrm{km}$ ; why was it hard to add?

Solution

Solution of Exercise 8.9.

(a) $\partial_x^2v = R\Gamma\,\partial_tv$ , $D = 1/R\Gamma = 1.1 \times 10^{12}\,\mathrm{m}^{2}/\mathrm{s}$ . (b) $\underline k^2 = -\iu R\Gamma\omega$ : $k' = k'' = \sqrt{R\Gamma\omega/2}$ ; $1\,\mathrm{Hz}$ : $k = 1.7 \times 10^{-6}\,\mathrm{m}^{-1}$ , $v_\varphi = 3.7 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ , $\delta = 600\,\mathrm{km}$ ; $10\,\mathrm{Hz}$ : $v_\varphi = 1.2 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ , $\delta = 190\,\mathrm{km}$ . (c) $R\Gamma L^2 = 9 \times 10^{-13} \times 9 \times 10^{12} = 8\,\mathrm{s}$ per dot: about a word a minute. (d) $\Lambda = R\Gamma/G = 9\,\mathrm{H}/\mathrm{km}$ — an absurd inductance for a submarine cable; loading coils worked on land telephone lines, whose $G$ and $R\Gamma$ were different.

Exercise 8.10 ★★★

Energy travels at the group velocity. String on an elastic bed: $\mu\partial_t^2y = T\partial_x^2y - Ky$ , wave $y = A\cos(\omega t - kx)$ with $\omega^2 = \omega_c^2 + c^2k^2$ . (a) Energy density $e = \tfrac12\mu(\partial_ty)^2 + \tfrac12T (\partial_xy)^2 + \tfrac12Ky^2$ and flux $\mathcal P = -T\partial_xy\,\partial_ty$ : check $\partial_te + \partial_x\mathcal P = 0$ . (b) Time averages $\langle e\rangle$ and $\langle\mathcal P\rangle$ for the wave. (c) Show that $\langle\mathcal P\rangle/ \langle e\rangle = c^2k/\omega = v_g$ , not $v_\varphi$ . (d) What happens to the energy flux below the cut-off?

Solution

Solution of Exercise 8.10.

(a) $\partial_te = \dot y(\mu\ddot y + Ky) + Ty'\dot y' = \dot yTy'' + Ty'\dot y' = \partial_x (Ty'\dot y) = -\partial_x\mathcal P$ . (b) $\langle e\rangle = \tfrac14A^2(\mu\omega^2 + Tk^2 + K) = \tfrac12\mu\omega^2A^2$ (using $\mu\omega^2 = Tk^2 + K$ ); $\langle\mathcal P\rangle = \tfrac12Tk\omega A^2$ . (c) Ratio $Tk/\mu\omega = c^2k/\omega = v_g$ . (d) $y = A\eu^{-\kappa x}\cos\omega t$ : $\mathcal P \propto \sin2\omega t$ , zero mean — no energy crosses the medium.

Exercise 8.11 ★★★

Swell and tsunami. Water waves of wavelength $\lambda$ on a depth $h$ obey $\omega^2 = gk\tanh kh$ (admitted). (a) Recover the deep-water and the shallow-water ( $kh \ll 1$ ) limits; speed of shallow-water waves; are they dispersive? (b) A tsunami of $\lambda = 200\,\mathrm{km}$ on $4\,\mathrm{km}$ of ocean: which limit, speed, period; time to cross $8000\,\mathrm{km}$ . (c) Its amplitude offshore is $0.5\,\mathrm{m}$ : energy per unit length of crest (admit $E = \tfrac12\rho gA^2$ per unit area) for a $1000\,\mathrm{km}$ front; what happens when the depth falls to $10\,\mathrm{m}$ , if the energy flux $Ev_g$ is conserved (Green’s law, $A \propto h^{-1/4}$ )? (d) A $100\,\mathrm{m}$ swell arriving on a beach: at what depth does it start to feel the bottom, and why do its crests turn parallel to the shore?

Solution

Solution of Exercise 8.11.

(a) $\tanh kh \to 1$ : $\omega^2 = gk$ ; $kh \ll 1$ : $\omega = \sqrt{gh}\,k$ , speed $\sqrt{gh}$ , non-dispersive. (b) $kh = 0.13$ : shallow; $c = \sqrt{9.81 \times 4000} = 198\,\mathrm{m}/\mathrm{s}$ , $T = 17\,\mathrm{min}$ ; $8000\,\mathrm{km}$ in $11\,\mathrm{h}$ . (c) $\tfrac12\rho gA^2 = 1.2\,\mathrm{kJ}/\mathrm{m}^{2}$ over $200 \times 1000$ km $^2$ : $2.5 \times 10^{14}\,\mathrm{J}$ . Green: $A \propto h^{-1/4}$ , $(4000/10)^{1/4} = 4.5$ : $2.2\,\mathrm{m}$ , before steepening and run-up. (d) At $h \approx \lambda/2 = 50\,\mathrm{m}$ ; the part of a crest in shallower water slows down, so the crest swings round parallel to the depth contours — refraction.

Exercise 8.12 ★★★

A pulse on a line. A $50\,\Omega$ cable of length $\ell = 100\,\mathrm{m}$ ( $c = 2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ ) is fed at $x = 0$ by a generator of internal resistance $50\,\Omega$ and ended at $x = \ell$ by a load $R_L$ . (a) Show that a wave arriving at the end is reflected with the voltage coefficient $r = (R_L - Z_c)/(R_L + Z_c)$ (write $v$ and $i$ as incident plus reflected waves and impose Ohm’s law in the load). (b) Values for $R_L = 50$ , $0$ , $\infty$ , $150\,\Omega$ . (c) A $1\,\mathrm{ns}$ pulse of $1\,\mathrm{V}$ is sent: what the oscilloscope at the generator shows, and when, for each load; why the generator’s $50\,\Omega$ matter. (d) A fault in a buried cable returns an echo after $1.8\,\text{µ}\mathrm{s}$ : where is it? (Time-domain reflectometry.)

Solution

Solution of Exercise 8.12.

(a) At $x = \ell$ : $v = v_i + v_r$ , $i = (v_i - v_r)/Z_c$ , $v = R_Li$ : $r = (R_L - Z_c)/(R_L + Z_c)$ . (b) $0$ , $-1$ , $+1$ , $0.5$ . (c) The line’s input looks like $Z_c$ , so the scope sees $0.5\,\mathrm{V}$ at $t = 0$ , then the echo $0.5r$ V at $2\ell/c = 1\,\text{µ}\mathrm{s}$ : nothing, $-0.5\,\mathrm{V}$ , $+0.5\,\mathrm{V}$ , $+0.25\,\mathrm{V}$ ; the matched generator absorbs the echo, so there is no second one. (d) $\tfrac12 \times 2 \times 10^8 \times 1.8 \times 10^{-6} = 180\,\mathrm{m}$ .

8.5 Problem: The cable and the sea

Problem 8.1

Weekend problem — dispersion at work: the telegraph cable that smeared the dots, the swell that announces a storm, the tsunami that crosses an ocean, and the packet that spreads

Part I — The telegraph cable. The 1866 transatlantic cable: $R = 2.0\,\Omega/\mathrm{km}$ , $\Gamma = 0.25\,\text{µ}\mathrm{F}/\mathrm{km}$ , $\Lambda = 0.5\,\mathrm{mH}/\mathrm{km}$ , $G$ negligible, length $L = 3500\,\mathrm{km}$ .

Write the telegrapher’s equations with $R$ and derive the dispersion relation $\underline k^2 = \Gamma\omega(\Lambda\omega - \iu R)$ .
At what frequency does $\Lambda\omega$ equal $R$ ? Conclude that for telegraph frequencies (a few hertz) the line is in the diffusive regime.
In that regime, $\underline k = \sqrt{-\iu R\Gamma\omega}$ : give $k'$ and $k''$ (recall $\sqrt{-\iu} = (1 - \iu)/\sqrt2$ ); phase velocity and attenuation length at $1\,\mathrm{Hz}$ ; at $4\,\mathrm{Hz}$ .
Attenuation in decibels over the whole cable at $1\,\mathrm{Hz}$ and at $4\,\mathrm{Hz}$ . Which frequency survives? What does that do to a sharp dot?
Kelvin’s estimate of the transit time, $R\Gamma L^2$ ; the practical rate of the cable in words per minute if a word is about $10$ dots and dashes and the dots must not overlap.
Heaviside proposed to raise $\Lambda$ : what value makes $\Lambda\omega = R$ at $4\,\mathrm{Hz}$ , and what would the speed and attenuation then be (take $G = 0$ , so the condition $R/\Lambda = G/\Gamma$ cannot be met exactly — use the general relation at that frequency)?
A modern optical fibre carries $1 \times 10^{10}\,\mathrm{bit}/\mathrm{s}$ : compare with question 5, in powers of ten.

Part II — The swell. Deep water: $\omega^2 = gk$ .

Phase and group velocity as functions of the period $T$ .
A storm $4000\,\mathrm{km}$ from a coast generates waves of periods from $6\,\mathrm{s}$ to $18\,\mathrm{s}$ : arrival time of each extreme; duration of the swell event.
Show that at a coast at distance $D$ the arriving period obeys $1/T = gt/4\pi D$ , $t$ counted from the storm: $1/T$ grows linearly in time. A buoy records $T = 16\,\mathrm{s}$ one day at noon and $T = 12\,\mathrm{s}$ exactly $24\,\mathrm{h}$ later: distance of the storm.
Energy per unit area of a swell of amplitude $A$ is $\tfrac12\rho gA^2$ ; power crossing a line of $1\,\mathrm{m}$ of crest, for $A = 1.5\,\mathrm{m}$ , $T = 12\,\mathrm{s}$ . (Use the group velocity — why?)
A wave-energy farm of $1\,\mathrm{km}$ of front with $30\%$ efficiency: electrical power; compare with a wind turbine ( $3\,\mathrm{MW}$ ).
A boat at $5\,\mathrm{m}/\mathrm{s}$ on deep water leaves a stationary wake behind it: which wavelength has a phase velocity equal to the boat’s speed and so follows it? Why does the pattern trail in a V, narrower than the crests’ own spreading would suggest (think of $v_g = v_\varphi/2$ )?
Watching one group of swell from a cliff, describe what an individual crest does during the time the group passes a fixed buoy; how many crests does a group of duration $2\,\mathrm{min}$ contain at $T = 12\,\mathrm{s}$ , and how many crest "lifetimes" does that represent?

Part III — The tsunami. $\omega^2 = gk\tanh kh$ ; Pacific depth $h = 4.0\,\mathrm{km}$ .

An earthquake lifts a sea-floor patch of $100\,\mathrm{km}$ by $2\,\mathrm{m}$ ; the wave has $\lambda \approx 200\,\mathrm{km}$ : check that it is a shallow-water wave and give its speed, period and the time to reach a coast $6000\,\mathrm{km}$ away.
Is it dispersive? Compare with the swell of Part II: does it arrive before or after the swell would? Which arrives as a single pulse?
Its height offshore is $0.6\,\mathrm{m}$ : energy per square metre; total energy of a $500\,\mathrm{km}$ -long front spread over one wavelength; compare with a $1\,\mathrm{Mt}$ explosion ( $4.2 \times 10^{15}\,\mathrm{J}$ ).
Green’s law: height when the depth falls to $20\,\mathrm{m}$ near the shore (energy flux $\tfrac12\rho gA^2v_g$ conserved, $v_g = \sqrt{gh}$ ); speed and wavelength there; why the wave then steepens and breaks like a wall.
A ship at sea does not notice the tsunami passing: why?
Tide gauges $6000\,\mathrm{km}$ apart record the wave: what the travel time gives, and why the measured speed is slightly less than $\sqrt{gh}$ over an ocean of varying depth.

Part IV — The spreading packet.

For deep water, compute $\omega''(k) = \dd^2\omega/\dd k^2$ and its value for $\lambda = 100\,\mathrm{m}$ .
A packet of $10\,$ such waves ( $\Delta x_0 \approx 1\,\mathrm{km}$ ) is released: spreading time $\Delta x_0^2/|\omega''|$ ; distance travelled by the packet meanwhile.
Explain in words why a packet with a narrower spectrum (longer train) spreads more slowly, and why the tsunami of Part III barely spreads at all.
The same estimate for an electron of wavelength $0.1\,\mathrm{nm}$ localized within $1\,\mathrm{nm}$ , with $\omega = \hbar k^2/2m$ ( $\hbar = 1.05 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ , $m = 9.1 \times 10^{-31}\,\mathrm{kg}$ ): spreading time and the distance travelled — the first contact with the quantum wave packet of Chapter 30.
Sum up in a table: for the cable, the swell, the tsunami and the electron, the dispersion relation, $v_\varphi$ , $v_g$ and whether the medium spreads or merely attenuates.

Solution

Solution of Problem 8.1.

1. $\partial_xv = -\Lambda\partial_ti - Ri$ , $\partial_xi = -\Gamma\partial_tv$ ; with $\eu^{\iu(\omega t - kx)}$ : $\iu kv = (\iu\omega\Lambda + R)i$ , $\iu ki = \iu\omega\Gamma v$ , hence $k^2 = \omega\Gamma(\omega\Lambda - \iu R)$ .

2. $\omega = R/\Lambda = 4000\,\mathrm{rad}/\mathrm{s}$ , $640\,\mathrm{Hz}$ : at a few hertz $\Lambda\omega \ll R$ , diffusive.

3. $k' = k'' = \sqrt{R\Gamma\omega/2}$ with $R\Gamma = 5 \times 10^{-13}\,\mathrm{s}/\mathrm{m}^{2}$ : $1\,\mathrm{Hz}$ : $1.25 \times 10^{-6}\,\mathrm{m}^{-1}$ , $v_\varphi = 5 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ , $\delta = 800\,\mathrm{km}$ ; $4\,\mathrm{Hz}$ : $2.5 \times 10^{-6}\,\mathrm{m}^{-1}$ , $1 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ , $400\,\mathrm{km}$ .

4. $8.7L/\delta$ : $38\,\mathrm{dB}$ at $1\,\mathrm{Hz}$ , $76\,\mathrm{dB}$ at $4\,\mathrm{Hz}$ : only the slowest components survive; a sharp dot arrives as a long, low hump.

5. $R\Gamma L^2 = 5 \times 10^{-13} \times 1.2 \times 10^{13} = 6\,\mathrm{s}$ : a dot every six seconds, a word a minute (the real cable, with sensitive mirror galvanometers, managed a few).

6. $\Lambda = R/\omega = 80\,\mathrm{mH}/\mathrm{km}$ . Then $k^2 = \Gamma R\omega(1 - \iu)$ , $k = \sqrt{\Gamma R\omega} \,2^{1/4}\eu^{-\iu\pi/8}$ : $k' = 3.9 \times 10^{-6}\,\mathrm{m}^{-1}$ , $k'' = 1.6 \times 10^{-6}\,\mathrm{m}^{-1}$ : $v_\varphi = 6.4 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ , $\delta = 620\,\mathrm{km}$ , $49\,\mathrm{dB}$ over the cable instead of $76\,\mathrm{dB}$ .

7. A dot every $6\,\mathrm{s}$ is about $0.2\,\mathrm{bit}/\mathrm{s}$ : the fibre is $10^{10}$ to $10^{11}$ times faster.

8. $v_\varphi = gT/2\pi$ , $v_g = gT/4\pi$ .

9. $18\,\mathrm{s}$ : $v_g = 14.1\,\mathrm{m}/\mathrm{s}$ , $4 \times 10^6/14.1 = 3.3\,\mathrm{days}$ ; $6\,\mathrm{s}$ : $4.7\,\mathrm{m}/\mathrm{s}$ , $9.9\,\mathrm{days}$ : the swell lasts more than six days.

10. $t = D/v_g = 4\pi D/gT$ , so $1/T = gt/4\pi D$ . $\Delta(1/T) = 1/12 - 1/16 = 0.0208\,\mathrm{Hz}$ in $86\,400\,\mathrm{s}$ : $g/4\pi D = 2.4 \times 10^{-7}\,\mathrm{s}^{-2}$ , $D = 3200\,\mathrm{km}$ .

11. $\tfrac12\rho gA^2 = 11\,\mathrm{kJ}/\mathrm{m}^{2}$ ; the energy moves at $v_g = gT/4\pi = 9.4\,\mathrm{m}/\mathrm{s}$ : $103\,\mathrm{kW}$ per metre of crest.

12. $10^3 \times 103 \times 0.3 = 31\,\mathrm{MW}$ : ten wind turbines.

13. $\lambda = 2\pi U^2/g = 16\,\mathrm{m}$ . The energy of each wave component travels at half its phase speed, so the pattern is confined to a wedge behind the boat (half-angle $19.5{}^{\circ}$ , whatever its speed).

14. A crest is born at the rear of the group, overtakes it (crests move at $2v_g$ relative to the water, $v_g$ relative to the group) and dies at the front. The group is $120 \times 9.4 = 1.1\,\mathrm{km}$ long, five wavelengths ( $\lambda = gT^2/2\pi = 225\,\mathrm{m}$ ); ten crests pass the buoy, each having lived about two minutes.

15. $kh = 0.13$ : shallow. $c = 198\,\mathrm{m}/\mathrm{s}$ , $T = \lambda/c = 17\,\mathrm{min}$ , $8.4\,\mathrm{h}$ .

16. Non-dispersive: all components at $198\,\mathrm{m}/\mathrm{s}$ , arriving together as one pulse (the swell, at $\sim10\,\mathrm{m}/\mathrm{s}$ , sorted by period, would take a week).

17. $\tfrac12\rho gA^2 = 1.8\,\mathrm{kJ}/\mathrm{m}^{2}$ over $500 \times 200$ km $^2$ : $1.8 \times 10^{14}\,\mathrm{J}$ , $4\%$ of a megatonne.

18. $A = 0.6(4000/20)^{1/4} = 2.3\,\mathrm{m}$ ; $c = \sqrt{gh} = 14\,\mathrm{m}/\mathrm{s}$ , $\lambda = cT = 14\,\mathrm{km}$ . The crest, in deeper water than the trough ahead of it, travels faster and catches up: the front steepens into a bore.

19. A rise of $0.6\,\mathrm{m}$ over a quarter of an hour, on a slope of $10^{-5}$ : nothing to feel.

20. The mean speed, $\sqrt{gh}$ for the mean depth; since $\sqrt h$ is concave, shallow stretches slow the wave more than deep ones speed it up, and the path average is below $\sqrt{g\bar h}$ .

21. $\omega' = \tfrac12\sqrt{g/k}$ , $\omega'' = -\tfrac14\sqrt{g/k^3}$ ; $k = 0.063\,\mathrm{rad}/\mathrm{m}$ : $\omega'' = -50\,\mathrm{m}^{2}/\mathrm{s}$ .

22. $t_{\text{sp}} = 10^6/50 = 2 \times 10^{4}\,\mathrm{s}$ , $5.6\,\mathrm{h}$ ; the packet has moved $v_gt = 6.25 \times 2 \times 10^4 = 125\,\mathrm{km}$ .

23. A long train contains a narrow band of $k$ , whose group velocities nearly coincide: the components stay together longer. The tsunami’s medium is non-dispersive: equal speeds, no spreading.

24. $\omega'' = \hbar/m = 1.2 \times 10^{-4}\,\mathrm{m}^{2}/\mathrm{s}$ : $t_{\text{sp}} = 10^{-18}/1.2 \times 10^{-4} = 9 \times 10^{-15}\,\mathrm{s}$ ; $v_g = \hbar k/m = 7.3 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ : $60\,\mathrm{nm}$ .

25. Cable: $k^2 = -\iu R\Gamma\omega$ , $v_\varphi = \sqrt{2\omega/R\Gamma}$ , no clean $v_g$ , attenuates and smears. Swell: $\omega^2 = gk$ , $v_\varphi = 2v_g$ , spreads. Tsunami: $\omega = \sqrt{gh}\,k$ , $v_\varphi = v_g$ , neither. Electron: $\omega = \hbar k^2/2m$ , $v_g = 2v_\varphi$ , spreads.

Terms defined in this chapter

See all 393 terms in the glossary