Maxwell: the displacement current, and light

Idea. Electrostatics and magnetostatics each define one constant, and together they contain a speed. That speed is a prediction before it is a coincidence.

1. \([\varepsilon_0] = \mathrm{C^2N^{-1}m^{-2}}\) and \([\mu_0] = \mathrm{N\,A^{-2}} = \mathrm{N\,s^2C^{-2}}\). Multiplying, the newtons and coulombs cancel: \[ [\varepsilon_0\mu_0] = \mathrm{C^2N^{-1}m^{-2}}\cdot\mathrm{N\,s^2C^{-2}} = \mathrm{s^2\,m^{-2}} . \]

2. \(\varepsilon_0\mu_0 = 1.112650\times10^{-17}\ \mathrm{s^2m^{-2}}\), so \(1/\sqrt{\varepsilon_0\mu_0} = 2.99792\times10^8\ \mathrm{m\,s^{-1}}\).

3. \([\mu_0/\varepsilon_0] = \mathrm{N\,s^2C^{-2}}/\mathrm{C^2N^{-1}m^{-2}} = \mathrm{N^2s^2m^2C^{-4}}\), so \([Z_0] = \mathrm{N\,m\,s\,C^{-2}} = \mathrm{J\,s\,C^{-2}} = \mathrm{V\,C\,s\,C^{-2}} = \mathrm{V\,s\,C^{-1}} = \mathrm{V\,A^{-1}} = \Omega\). Numerically \(Z_0 = \sqrt{1.25663706\times10^{-6}/8.85418781\times10^{-12}} = 376.730\ \Omega\). For a plane wave \(|\vec H| = |\vec B|/\mu_0 = |\vec E|/(\mu_0c)\), so \(|\vec E|/|\vec H| = \mu_0c = \mu_0/\sqrt{\varepsilon_0\mu_0} = \sqrt{\mu_0/\varepsilon_0} = Z_0\).

4. \(3.1074/2.99792 = 1.0365\): 3.65 per cent high.

Check. The units in part 3 are those of a resistance, and the value is familiar to engineers: an antenna matched to free space presents about \(377\ \Omega\). Limit: if magnetism were absent (\(\mu_0\to0\)), \(Z_0\to0\) and \(c\to\infty\) — the electrostatic world, with instantaneous action ✓.

Idea. Quasi-statics is the first term of an expansion in \(\omega s/c\). Each application of the two induction laws brings one more factor \((\omega s/c)^2\), and the series they build is a Bessel function.

1. Ampère–Maxwell through the disc of radius \(s\), with no conduction current in the gap: \(2\pi sB_\phi = \mu_0\varepsilon_0\pi s^2\partial_tE_z\), so \[ B_\phi(s,t) = \frac{s}{2c^2}\,\partial_tE_z = -\frac{\omega s}{2c^2}\,\hat E\sin\omega t . \]

2. In cylindrical coordinates \((\nabla\times\vec E)_\phi = \partial_zE_s - \partial_sE_z = -\partial_sE_z\). Faraday: \(\partial_sE_z = \partial_tB_\phi = \frac{s}{2c^2}\partial_t^2E_z\). With \(\partial_t^2E_z = -\omega^2\hat E\cos\omega t\) at leading order, integrate from the axis: \[ E_z(s,t) = \hat E\cos\omega t - \frac{\omega^2s^2}{4c^2}\hat E\cos\omega t = \hat E\cos\omega t\Big(1 - \frac{\omega^2s^2}{4c^2}\Big). \]

3. At the rim the relative correction is \((\omega a/c)^2/4\). At 50 Hz, \(\omega a/c = 2\pi(50)(0.05)/2.998\times10^8 = 5.24\times10^{-8}\) and the correction is \(6.9\times10^{-16}\). At 1 GHz, \(\omega a/c = 1.048\) and the correction is \(0.2745\): the "correction" is a quarter of the answer, and the quasi-static picture has failed.

4. Curl Faraday and use Ampère–Maxwell in the gap: \(\nabla^2\vec E = c^{-2}\partial_t^2\vec E\). For \(E_z = f(s)\cos\omega t\) the Laplacian in cylindrical coordinates gives \(f'' + f'/s = -k^2f\), which is Bessel's equation of order \(0\) in the variable \(ks\) (unit 09). The solution regular at \(s = 0\) with \(f(0) = \hat E\) is \(f = \hat E J_0(ks)\), and \(J_0(x) = 1 - x^2/4 + x^4/64 - \dots\), whose first two terms are part 2. At 1 GHz, \(J_0(1.048) = 0.7437\), against \(1 - 0.2745 = 0.7255\) from part 2 and \(0.7443\) with the \(x^4/64\) term added.

Check. Dimensions: \([sB_\phi] \) and \([s^2\partial_tE/c^2]\) agree, both \(\mathrm{T\,m}\); \(\omega^2s^2/c^2\) is dimensionless ✓. Limit \(\omega\to0\): no correction, the steady-current capacitor of the deck ✓. The rim field at 50 Hz: with \(\hat E\) such that the plate current amplitude is \(1\) A, \(B_\phi(a) = 4.00\ \mu\)T in amplitude — the deck's number, now oscillating.

Idea. The divergence of a curl vanishes, so taking the divergence of an evolution equation gives an equation for how the corresponding constraint changes. Charge conservation is exactly what makes Gauss's constraint stationary.

1. Clairaut (all fields \(C^2\)) lets \(\partial_t\) pass through \(\nabla\cdot\). First, \(\partial_t(\nabla\cdot\vec B) = -\nabla\cdot(\nabla\times\vec E) = 0\), so \(\nabla\cdot\vec B\) is constant in time, hence \(0\). Second, let \(C = \varepsilon_0\nabla\cdot\vec E - \rho\). Then \[ \partial_tC = \nabla\cdot(\varepsilon_0\partial_t\vec E) - \partial_t\rho = \nabla\cdot\big(\mu_0^{-1}\nabla\times\vec B - \vec J\big) - \partial_t\rho = -\nabla\cdot\vec J - \partial_t\rho = 0 , \] so \(C\) is constant in time, hence \(0\). \(\square\)

2. Divergence of Ampère–Maxwell: \(0 = \mu_0\nabla\cdot\vec J + \mu_0\varepsilon_0\partial_t\nabla\cdot\vec E = \mu_0(\nabla\cdot\vec J + \partial_t\rho)\), using Gauss in the last step. \(\square\)

3. Divergence of \(\nabla\times\vec B = \mu_0\vec J\) gives \(\nabla\cdot\vec J = 0\). Charge conservation — which holds for any physical sources, and which the problem asks us to respect — then gives \(\partial_t\rho = 0\) everywhere. So if \(\partial_t\rho\ne0\) at some point, no \(C^2\) field \(\vec B\) can satisfy the equation there: the system has no solution. \(\square\) (Without assuming continuity, the system forces \(\nabla\cdot\vec J = 0\), which is equally false on a charging plate.)

4. Six: the three components of each curl equation. The two divergence equations contain no time derivative, and part 1 shows they are preserved by the six evolution equations once they hold initially, so they restrict the initial data rather than competing with the evolution.

Check. On the deck's capacitor: \(\nabla\cdot\vec B = 0\) for \(\vec B = B(s)\hat\phi\) ✓ (a purely azimuthal field depending on \(s\) has zero divergence), and \(\varepsilon_0\nabla\cdot\vec E = 0 = \rho\) in the gap for uniform \(\vec E\) ✓, at every instant, as part 1 promises.

Idea. The energy dissipated in the wire does not travel along it; it arrives from the surrounding space, radially inward, as Poynting found in 1884.

1. \(R = 1.68\times10^{-8}\times1.00/(\pi\times10^{-6}) = 5.348\times10^{-3}\ \Omega\), so \(V = IR = 5.348\) mV and \(E = V/L = 5.348\times10^{-3}\ \mathrm{V\,m^{-1}}\) along \(\hat z\), the direction of the current. Ampère: \(B = \mu_0I/2\pi b = 2.000\times10^{-4}\) T along \(\hat\phi\). Since \(\hat z\times\hat\phi = -\hat s\), \(\vec S = -\frac{EB}{\mu_0}\hat s\): radially inward, with magnitude \(EB/\mu_0 = (5.348\times10^{-3})(2.000\times10^{-4})/(1.2566\times10^{-6}) = 0.8511\ \mathrm{W\,m^{-2}}\).

2. Area \(2\pi bL = 6.283\times10^{-3}\ \mathrm{m^2}\); power in \(= 0.8511\times6.283\times10^{-3} = 5.348\) mW. And \(I^2R = 5.348\) mW ✓.

3. \(|\vec S|\cdot2\pi bL = \frac{E}{\mu_0}\cdot\frac{\mu_0I}{2\pi b}\cdot2\pi bL = EIL = VI\), independent of \(b\) and of the material. Poynting's theorem for the wire's volume with steady fields (\(\partial_tu = 0\)) says the inward flux equals \(\int\vec J\cdot\vec E = JE\cdot\pi b^2L = IEL\) ✓ — the same statement. \(\square\)

Check. Dimensions: \([EI L] = \mathrm{V\,m^{-1}\,A\,m} = \mathrm W\) ✓. Limit: a perfect conductor has \(E = 0\), so \(\vec S\) has no radial component and nothing enters: no dissipation ✓. The end faces carry no flux because \(\vec S\) is radial.

Idea. The stress tensor needs only the total field on a surface, not the split into "the field of the other half" and "the field of this half", which here would be hard.

1. Static fields, so \(\vec g\) plays no part and the force on the charges inside is \(\oint T(\hat n,\cdot)\,\mathrm dA\). On the disc \(\vec E = \vec 0\), so \(T = 0\) there (the thin annulus between \(R\) and \(R + \epsilon\) contributes nothing as \(\epsilon\to0\)). On the hemisphere, with \(E = Q/4\pi\varepsilon_0R^2\): \[ F_z = \int_0^{2\pi}\!\!\int_0^{\pi/2}\frac{\varepsilon_0}{2}E^2\cos\theta\,R^2\sin\theta\,\mathrm d\theta\,\mathrm d\phi = \frac{\varepsilon_0}{2}E^2\cdot\pi R^2 = \frac{Q^2}{32\pi\varepsilon_0R^2}. \] By symmetry \(F_x = F_y = 0\). The force is along \(+\hat z\): the halves repel.

2. \(F = (10^{-6})^2/(32\pi\times8.854\times10^{-12}\times0.0100) = 0.1123\) N.

3. In an orthonormal basis, \(T_{ii} = \varepsilon_0(E_i^2 - \frac12|\vec E|^2) + \mu_0^{-1}(B_i^2 - \frac12|\vec B|^2)\), and summing over \(i\): \(\varepsilon_0(|\vec E|^2 - \frac32|\vec E|^2) + \mu_0^{-1}(|\vec B|^2 - \frac32|\vec B|^2) = -u\). Under a change of orthonormal basis \(T' = R^\top TR\) with \(R\) orthogonal, and \(\operatorname{tr}(R^\top TR) = \operatorname{tr}(TRR^\top) = \operatorname{tr}T\). So the trace is a property of the bilinear form, not of the basis — as it must be, since \(u\) is. \(\square\)

Another way

The outward pressure on a charged conducting-style surface is \(\sigma\cdot\frac12(E_\mathrm{out} + E_\mathrm{in}) = \sigma^2/2\varepsilon_0\), with \(\sigma = Q/4\pi R^2\) (the charge feels the average of the fields on its two sides). Its \(z\)-component over the northern hemisphere: \(\frac{\sigma^2}{2\varepsilon_0}\int\cos\theta\,\mathrm dA = \frac{\sigma^2}{2\varepsilon_0}\pi R^2 = \frac{Q^2}{32\pi\varepsilon_0R^2}\) ✓.

Check. Dimensions: \([Q^2/\varepsilon_0R^2]\) is Coulomb's force ✓. Scaling: halving \(R\) quadruples the force, as for two point charges ✓. Compared with two point charges \(Q/2\) at the hemispheres' centroids (\(R/2\) from the centre, so \(R\) apart), \(\frac{(Q/2)^2}{4\pi\varepsilon_0R^2} = \frac{Q^2}{16\pi\varepsilon_0R^2}\): twice as large, plausibly, since the real charge is spread and much of it is far from the other half.

Idea. A plane wave is fixed by one amplitude; the polarisation ellipse is the singular-value decomposition of a \(2\times2\) matrix.

1. \(E_0 = \sqrt{2S/\varepsilon_0c} = \sqrt{2\times1361/(8.854\times10^{-12}\times2.998\times10^8)} = 1013\ \mathrm{V\,m^{-1}}\). \(B_0 = E_0/c = 3.38\ \mu\)T. Momentum flux \(S/c\): on a black surface \(4.54\ \mu\)Pa; on a mirror, where the momentum is reversed, twice that, \(9.08\ \mu\)Pa.

2. \(\vec a\cdot\vec b = 1\ne0\), so the ellipse is neither a segment (\(\vec a\not\parallel\vec b\)) nor a circle. The ellipse \(\{L(\cos\theta,\sin\theta)\}\) has semi-axes equal to the singular values of \(L\), the square roots of the eigenvalues of \(LL^\top = \begin{pmatrix}1&1\\1&2\end{pmatrix}\): \(\lambda = \frac{3\pm\sqrt5}{2} = 2.618,\ 0.382\), so the semi-axes are \(1.618\) and \(0.618\) V m\(^{-1}\) — the golden ratio and its reciprocal. The major axis is the eigenvector for \(2.618\): \((1-2.618)v_1 + v_2 = 0\), \(v_2 = 1.618v_1\), at \(\arctan1.618 = 58.28°\) from \(\hat x\). Sense: at \(t = 0\) \(\vec E = (1,1)\), a quarter-period later \((0,1)\): turning from \(45°\) towards \(90°\), anticlockwise about \(+\hat z\).

Check. The product of the semi-axes is \(|\det L| = |1\cdot1 - 0\cdot1| = 1 = 1.618\times0.618\) ✓, and the sum of their squares is \(|\vec a|^2 + |\vec b|^2 = 3 = 2.618 + 0.382\) ✓. Energy: \(\langle u\rangle = \frac{\varepsilon_0}2\vec E_0\cdot\overline{\vec E_0} = \frac{\varepsilon_0}{2}\cdot3\), which is the ellipse's mean squared radius times \(\varepsilon_0\) ✓.

Idea. The gauge condition changes by \(-\Box\chi\), so reaching it means solving a wave equation, and keeping it means \(\chi\) solves the free one. For a plane wave the remaining freedom removes \(\varphi\) and the longitudinal part of \(\vec A\), leaving two polarisations.

1. Under the transformation, \(L\mapsto L + \nabla^2\chi - c^{-2}\partial_t^2\chi = L - \Box\chi\). By the deck's retarded-solution theorem (with \(f = L\), which is \(C^2\) and has the required support), \(\chi = G_\mathrm{ret} * L\) is \(C^2\) and solves \(\Box\chi = L\). With this \(\chi\) the new \(L\) is \(0\). \(\square\) (The new potentials are \(C^1\) at least, enough for the fields.)

2. Starting from \(L = 0\), the new \(L\) is \(-\Box\chi\), which is zero iff \(\Box\chi = 0\). \(\square\)

3. Lorenz condition on the amplitudes: \(i\vec k\cdot\vec A_0 - i\omega\varphi_0/c^2 = 0\), i.e. \(\vec k\cdot\vec A_0 = \omega\varphi_0/c^2\). With \(\chi = \operatorname{Re}(\chi_0e^{i\theta})\), \(\partial_t\chi\) has amplitude \(-i\omega\chi_0\) and \(\nabla\chi\) has amplitude \(i\vec k\chi_0\), so \(\varphi_0\mapsto\varphi_0 + i\omega\chi_0\), \(\vec A_0\mapsto\vec A_0 + i\vec k\chi_0\). Choose \(\chi_0 = i\varphi_0/\omega\): then \(\varphi_0' = \varphi_0 - \varphi_0 = 0\) and \(\vec A_0' = \vec A_0 - \vec k\varphi_0/\omega\). Then \[ \vec k\cdot\vec A_0' = \frac{\omega\varphi_0}{c^2} - \frac{|\vec k|^2\varphi_0}{\omega} = \frac{\varphi_0}{\omega c^2}\big(\omega^2 - c^2|\vec k|^2\big) = 0 . \]

4. With \(\varphi = 0\) and \(\vec k\cdot\vec A_0' = 0\), \(\vec E_0 = i\omega\vec A_0'\) and \(\vec B_0 = i\vec k\times\vec A_0'\). The map \(\vec A_0'\mapsto\vec E_0\) is injective, so distinct \(\vec A_0'\) in the plane \(\vec k^\perp\subset\mathbb C^3\) give distinct fields, and every transverse \(\vec E_0\) arises. The plane is two-dimensional: two polarisations. Of the four components of \((\varphi,\vec A_0)\), the Lorenz condition removed one and the residual gauge another.

Check. Dimensions: \([\chi_0] = [\varphi_0]\,\mathrm T\) ✓ since \(\varphi\mapsto\varphi - \partial_t\chi\). The fields are unchanged, as a gauge transformation must leave them: \(\vec E_0 = -i\vec k\varphi_0 + i\omega\vec A_0\) before, and \(i\omega(\vec A_0 - \vec k\varphi_0/\omega) = i\omega\vec A_0 - i\vec k\varphi_0\) after ✓.

Idea. Uniform motion makes everything a function of the comoving coordinates, and the wave operator becomes a stretched Laplacian. The Coulomb-gauge potential is instantaneous; the field is not, because the gauge function carries the difference.

1. Let \(T_{s'}\) translate spacetime by \((vs', 0, 0, s')\). The source satisfies \(\rho\circ T_{s'} = \rho\), and the retarded integral \(\varphi(\vec x,t) = \frac1{4\pi\varepsilon_0}\int\rho(\vec y, t - |\vec x - \vec y|/c)/|\vec x - \vec y|\) commutes with translations (substitute \(\vec y\mapsto\vec y + vs'\hat x\)). So \(\varphi(x + vs', y, z, t + s') = \varphi(x,y,z,t)\) for all \(s'\): \(\varphi\) depends on \(x\) and \(t\) only through \(x - vt\). The same for \(\vec A\).

2. On \(f(\xi,y,z)\), \(c^{-2}\partial_t^2f = (v^2/c^2)f_{\xi\xi}\), so \(\Box f = -\big((1 - v^2/c^2)f_{\xi\xi} + f_{yy} + f_{zz}\big) = -(\gamma^{-2}f_{\xi\xi} + f_{yy} + f_{zz})\). The equation \(\Box f = \rho/\varepsilon_0\) becomes \(\gamma^{-2}f_{\xi\xi} + f_{yy} + f_{zz} = -\frac q{\varepsilon_0}\delta(\xi)\delta(y)\delta(z)\). Put \(u = \gamma\xi\): \(\gamma^{-2}f_{\xi\xi} = f_{uu}\) and \(\delta(\xi) = \delta(u/\gamma) = \gamma\,\delta(u)\), so \[ f_{uu} + f_{yy} + f_{zz} = -\frac{q\gamma}{\varepsilon_0}\delta(u)\delta(y)\delta(z), \] Poisson's equation for a point charge \(q\gamma\) in the variables \((u,y,z)\). The solution decaying at infinity (unit 23) is \(f = q\gamma/4\pi\varepsilon_0\sqrt{u^2 + y^2 + z^2}\), which is the stated \(\varphi_\mathrm L\). Each component of \(\vec A\) obeys the same equation with \(\rho/\varepsilon_0\) replaced by \(\mu_0J_i\), and \(\mu_0\vec J = \mu_0\varepsilon_0v\hat x\cdot\rho/\varepsilon_0\), so \(\vec A_\mathrm L = (v/c^2)\varphi_\mathrm L\hat x\). Lorenz: \(\nabla\cdot\vec A_\mathrm L = \frac v{c^2}\partial_\xi\varphi_\mathrm L\) and \(c^{-2}\partial_t\varphi_\mathrm L = -\frac v{c^2}\partial_\xi\varphi_\mathrm L\); the sum is \(0\) ✓.

3. \(\partial_t\chi = -v\partial_\xi\chi = \frac{q}{4\pi\varepsilon_0}\Big[\frac{\gamma/s}{\sqrt{1 + \gamma^2\xi^2/s^2}} - \frac{1/s}{\sqrt{1 + \xi^2/s^2}}\Big] = \frac{q}{4\pi\varepsilon_0}\Big[\frac{\gamma}{\sqrt{s^2 + \gamma^2\xi^2}} - \frac{1}{\sqrt{s^2 + \xi^2}}\Big] = \varphi_\mathrm L - \varphi_\mathrm C\), so \(\varphi_\mathrm L - \partial_t\chi = \varphi_\mathrm C\) ✓. For the divergence, let \(D = \nabla^2\chi + \nabla\cdot\vec A_\mathrm L = \nabla\cdot\vec A_\mathrm C\). Then \[ \partial_tD = \nabla^2(\varphi_\mathrm L - \varphi_\mathrm C) - c^{-2}\partial_t^2\varphi_\mathrm L = -\Box\varphi_\mathrm L - \nabla^2\varphi_\mathrm C = -\frac\rho{\varepsilon_0} + \frac\rho{\varepsilon_0} = 0, \] using the Lorenz condition \(\partial_t\nabla\cdot\vec A_\mathrm L = -c^{-2}\partial_t^2\varphi_\mathrm L\). Since \(D\) is also a function of \((\xi,y,z)\), \(\partial_tD = -v\partial_\xi D = 0\) makes \(D\) independent of \(\xi\). As \(\xi\to-\infty\) at fixed \(s \gt 0\), \(\operatorname{arsinh}w = -\ln(2|w|) + o(1)\) for \(w\to-\infty\), so the bracket tends to \(-\ln(2\gamma|\xi|/s) + \ln(2|\xi|/s) + \ln\gamma = 0\), and \(\chi\) and its derivatives tend to \(0\), as does \(\nabla\cdot\vec A_\mathrm L\). So \(D\equiv0\): Coulomb gauge ✓.

4. From the last rung, with \(\partial_\xi\varphi_\mathrm L = -\frac{q\gamma\cdot\gamma^2\xi}{4\pi\varepsilon_0(\gamma^2\xi^2 + s^2)^{3/2}}\) and similarly in \(y, z\): \[ \vec E = \frac{q\gamma\,(\xi,\ y,\ z)}{4\pi\varepsilon_0(\gamma^2\xi^2 + y^2 + z^2)^{3/2}},\qquad \vec B = \frac{\vec v\times\vec E}{c^2}. \] \(\vec E\) points away from the charge's present position. For \(\vec B\): \(\nabla\times(\frac v{c^2}\varphi_\mathrm L\hat x) = -\frac v{c^2}\hat x\times\nabla\varphi_\mathrm L\), and \(\hat x\times\vec E = -\hat x\times\nabla\varphi_\mathrm L\) because \(\partial_t\vec A_\mathrm L\parallel\hat x\); so \(\vec B = \frac v{c^2}\hat x\times\vec E\). The Coulomb-gauge fields are identical because the two pairs of potentials differ by the gauge transformation of part 3, and the fields are gauge-invariant (the deck's theorem). They are causal because the Lorenz-gauge potentials are retarded integrals, which at \((\vec x,t)\) use the source only on the past light cone. That the field points from the present position is not a signal from the present: for uniform motion the past determines the present, and the field is built from the past. Had the charge been stopped abruptly, the field outside the sphere of radius \(c\times\)(time since stopping) would still point from where it would have been.

5. At \(\xi = 0\), \(s = r\): \(|\vec E| = q\gamma/4\pi\varepsilon_0r^2\). At \(s = 0\), \(|\xi| = r\): \(|\vec E| = q\gamma r/4\pi\varepsilon_0\gamma^3r^3 = q/4\pi\varepsilon_0\gamma^2r^2\). Ratio \(\gamma^3 = 1.25^3 = 1.953\): the field is compressed towards the plane perpendicular to the motion.

Check. \(v\to0\): \(\gamma\to1\), \(\varphi_\mathrm L = \varphi_\mathrm C\), \(\vec A\to\vec 0\); the bracket in \(\chi\) is \(O(\gamma - 1) = O(v^2/c^2)\), so \(\chi = O(v)\to0\), and \(\vec E\) is Coulomb's ✓. Dimensions: \([q/\varepsilon_0v] = \mathrm{V\,m}/(\mathrm{m\,s^{-1}}) = \mathrm{V\,s}\), so \(\partial_t\chi\) is in volts, like \(\varphi\) ✓. Gauss's law: the flux of \(\vec E\) through any sphere about the present position is \(q/\varepsilon_0\) — it must be, as \(\vec E\) solves Maxwell's equations.

Idea. Transforming in time turns the wave equation into the Helmholtz equation, which has an outgoing and an incoming spherical wave. Causality — holomorphy in the upper half-plane — keeps the outgoing one.

1. Under the time transform \(\partial_t\mapsto-i\omega\), so \(c^{-2}\partial_t^2\mapsto-\omega^2/c^2 = -k^2\), and \(\widehat{\delta(t)} = 1\). So \(-k^2\hat G - \nabla^2\hat G = \delta^{(3)}\), i.e. \((\nabla^2 + k^2)\hat G = -\delta^{(3)}\).

2. Let \(f\) be smooth and radial. As distributions, \(\nabla^2(f\Gamma) = f\nabla^2\Gamma + 2\nabla f\cdot\nabla\Gamma + \Gamma\nabla^2f\) with \(\Gamma = 1/4\pi r\); the product rule holds because \(\Gamma\) and \(\nabla\Gamma\) are locally integrable (integrate by parts against a test function, cutting out a small ball as in unit 23). With \(f\nabla^2\Gamma = -f(0)\delta\), \(\nabla f\cdot\nabla\Gamma = -f'/4\pi r^2\) and \(\nabla^2f = f'' + 2f'/r\): \(\nabla^2(f\Gamma) = -f(0)\delta + f''/4\pi r\). For \(f = \cos kr\) (smooth: \(\cos kr = \sum(-1)^nk^{2n}r^{2n}/(2n)!\) is a power series in \(|\vec x|^2\)): \((\nabla^2 + k^2)\frac{\cos kr}{4\pi r} = -\delta + \frac{-k^2\cos kr + k^2\cos kr}{4\pi r} = -\delta\). And \(g = \sin(kr)/r\) is smooth with \(rg = \sin kr\), so \(\nabla^2g = \frac1r(rg)'' = -k^2g\) for \(r \gt 0\), hence everywhere by continuity: \((\nabla^2 + k^2)g = 0\). So \(\frac{e^{\pm ikr}}{4\pi r} = \frac{\cos kr}{4\pi r}\pm\frac{i}{4\pi}\frac{\sin kr}{r}\) both satisfy the equation. \(\square\)

3. Unit 20: \(G(\vec x,\cdot)\) vanishes for \(t \lt 0\) exactly when its transform extends holomorphically to \(\operatorname{Im}\omega \gt 0\) and is suitably bounded there. For \(\omega = \omega_1 + i\omega_2\) with \(\omega_2 \gt 0\), \(|e^{i\omega r/c}| = e^{-\omega_2r/c}\le1\), while \(|e^{-i\omega r/c}| = e^{\omega_2r/c}\) grows without bound. So \(\hat G_\mathrm{ret} = e^{i\omega r/c}/4\pi r\), the outgoing wave (with \(e^{-i\omega t}\), its phase is constant on \(r = ct + \)const).

4. \(G(\vec x,t) = \frac{1}{4\pi r}\cdot\frac1{2\pi}\int e^{-i\omega(t - r/c)}\,\mathrm d\omega = \frac{\delta(t - r/c)}{4\pi r}\), by unit 20's \(\hat1 = 2\pi\delta\) (shifted). The \(i0\) prescription is now in the choice of sign: evaluating \(e^{\pm ikr}\) at \(k = (\omega + i\varepsilon)/c\) and letting \(\varepsilon\to0^+\) singles out \(e^{+ikr}\), the only one that stays bounded. It is the same prescription seen from a different variable.

Check. \(\omega = 0\): \(\hat G = 1/4\pi r\), the Coulomb Green's function, which is also \(\int G\,\mathrm dt\) ✓. The incoming choice gives \(\delta(t + r/c)/4\pi r\), the deck's \(G_\mathrm{adv}\) ✓. Dimensions: \([\hat G] = [G]\,\mathrm T = \mathrm L^{-1}\) ✓.

Idea. At the centre a spherical pulse is the average over the sphere that is arriving now. The inward-moving half of the pulse converges and grows; once every sphere about the centre has left the support, nothing is left — not a tail, exactly nothing.

1. \(M_r[g](\vec 0) = g\) evaluated at radius \(r\): \(1 - r^2/R^2\) for \(r\le R\), \(0\) for \(r \gt R\).

2. With \(h = 0\), \(u(\vec 0,t) = \partial_t\big(t\,M_{ct}[g](\vec 0)\big)\). For \(0\le t \lt R/c\): \(\partial_t\big(t - c^2t^3/R^2\big) = 1 - 3c^2t^2/R^2\). For \(t \gt R/c\): \(\partial_t(0) = 0\).

3. \(u(\vec 0,0) = 1\) ✓ (the initial datum at the centre). \(u(\vec 0, R/2c) = 1 - \frac34 = 0.25\). As \(t\to R/c^-\): \(1 - 3 = -2\). For \(t \gt R/c\): exactly \(0\).

4. The initial bump splits into an outgoing and an ingoing spherical wave. The ingoing one converges on the centre, and its amplitude grows like \(1/r\) as it focuses; it arrives with the sign reversed (the \(\partial_t\) in Kirchhoff's formula differentiates the profile, and the steep inner edge of the average gives a large negative derivative). The value \(-2\) is \(-2\) times the initial height. The jump from \(-2\) to \(0\) at \(t = R/c\) comes from the kink of \(g\) at \(|\vec x| = R\), where \(g'\) jumps from \(-2/R\) to \(0\): the formula differentiates it. After \(t = R/c\) every sphere about the centre of radius \(ct \gt R\) lies outside the support of the data, and by the strong Huygens principle \(u(\vec 0, t)\) depends on the data on that sphere only. So it is \(0\).

Check. Dimensions: \(u\) has \(g\)'s units, and \(c^2t^2/R^2\) is dimensionless ✓. A second route: for radial solutions \(ru\) solves the one-dimensional wave equation (Darboux's identity), so \(ru = \frac12[F(r + ct) + F(r - ct)]\) with \(F(\sigma) = \sigma g(|\sigma|)\), an odd extension. Then \(u(0,t) = \lim_{r\to0}\frac{F(r+ct) + F(r-ct)}{2r} = F'(ct) = \frac{\mathrm d}{\mathrm d\sigma}\big[\sigma(1 - \sigma^2/R^2)\big]_{\sigma = ct} = 1 - 3c^2t^2/R^2\) ✓. (The data are only \(C^1\) at \(|\vec x| = R\), so this is a solution in the weak sense there; away from the sphere \(ct = R\) it is classical.)

Idea. A source uniform along a line is a two-dimensional source. Each point of the line sends a sharp spherical shell, but the shells from farther points arrive later, so the sum at a given point never stops arriving.

1. From the last rung, \(\delta(t - \sigma/c) = c\,\delta(\sigma - ct)\) picks out \(\sigma = ct\), which lies in the range \((\varrho, \infty)\) iff \(ct \gt \varrho\): \[ G_2(\varrho,t) = \frac{c}{2\pi\sqrt{c^2t^2 - \varrho^2}}\,H(ct - \varrho). \] Descent is legitimate because a \(z\)-independent source gives a \(z\)-independent retarded solution, on which \(\Box\) is the two-dimensional operator.

2. With \(w = c^2t^2 - \varrho^2\), \(f = w^{-1/2}\): \(f_t = -c^2tw^{-3/2}\), \(f_{tt} = -c^2w^{-3/2} + 3c^4t^2w^{-5/2}\); \(f_x = xw^{-3/2}\), \(f_{xx} = w^{-3/2} + 3x^2w^{-5/2}\), and the same in \(y\). So \[ c^{-2}f_{tt} - f_{xx} - f_{yy} = -w^{-3/2} + 3c^2t^2w^{-5/2} - 2w^{-3/2} - 3\varrho^2w^{-5/2} = -3w^{-3/2} + 3w\cdot w^{-5/2} = 0. \] So \(G_2\) is a nonzero classical solution throughout the interior \(ct \gt \varrho\). Its support is the whole solid cone; the solution of a 2D problem with data near the origin is therefore nonzero at \(\vec x\) for all \(t \gt |\vec x|/c\), not only near \(t = |\vec x|/c\). The strong principle fails. \(\square\)

3. With \(a = \sqrt{c^2t^2 - x^2}\) (when \(ct \gt |x|\)): \(\int G_2\,\mathrm dy = \frac c{2\pi}\int_{-a}^a\frac{\mathrm dy}{\sqrt{a^2 - y^2}} = \frac c{2\pi}\cdot\pi = \frac c2\), and \(0\) when \(ct \lt |x|\). So \(G_1 = \frac c2H(ct - |x|)\). Convolving with an initial velocity \(h\), with the deck's normalisation \(u = c^{-2}G_1 * h\): \(u = \frac1{2c}\int_{x - ct}^{x + ct}h\) — d'Alembert's formula ✓. In one dimension the tail does not even decay.

4. \(G_2\propto(c^2t^2 - \varrho^2)^{-1/2}\): at \(ct = 2\varrho\), \((3\varrho^2)^{-1/2}\); at \(ct = 10\varrho\), \((99\varrho^2)^{-1/2}\); ratio \(\sqrt{3/99} = 0.174\). As \(t\to\infty\), \(G_2\sim\frac{c}{2\pi ct} = \frac1{2\pi t}\): a tail decaying like \(1/t\), for ever.

Check. Dimensions: \([G_2] = \mathrm T^{-1}\) (the 2D delta has \(\mathrm T^{-1}\mathrm L^{-2}\) and the operator \(\mathrm L^{-2}\)), and \(c/\sqrt{c^2t^2 - \varrho^2}\) is \(\mathrm T^{-1}\) ✓. The three Green's functions in 1, 2, 3 dimensions: a step, a decaying tail, a spike — only three dimensions has sharp signals.

Idea. Differentiating a retarded function in space is, to leading order in \(1/r\), differentiating it in time times \(-\hat r/c\). The \(1/r\) fields carry a finite power to infinity.

1. \(\vec A(\vec x,t) = \frac{\mu_0}{4\pi r}q\vec v(t - r/c)\).

2. With \(\vec F = \frac{\mu_0q}{4\pi}\vec v\): \(\nabla t_\mathrm r = -\hat r/c\), so \(\vec B = -\frac{\mu_0q}{4\pi cr}\hat r\times\vec a + \nabla\frac1r\times\vec F = \frac{\mu_0q}{4\pi cr}\vec a\times\hat r + O(r^{-2})\). Dropped: the \(\nabla(1/r)\) term, of order \(\mu_0qv/r^2\), smaller than the kept one by \(cv/(ar)\sim cT/r\) (with \(T\) the time scale of the motion); and the corrections of the dipole approximation itself.

3. \(\vec E\perp\vec B\), \(|\vec E| = c|\vec B|\), and \(\vec E\times\vec B = c(\vec B\times\hat r)\times\vec B = c|\vec B|^2\hat r\) (as \(\vec B\perp\hat r\)). So \(\vec S = \frac{c}{\mu_0}|\vec B|^2\hat r = \frac{\mu_0q^2a^2\sin^2\theta}{16\pi^2cr^2}\hat r\), with \(\theta\) the angle between \(\vec a\) and \(\hat r\). Over a sphere, \(\int\sin^2\theta\,\mathrm d\Omega = \frac{8\pi}3\): \[ P = \frac{\mu_0q^2a^2}{6\pi c} = \frac{q^2a^2}{6\pi\varepsilon_0c^3}. \]

4. \(a = evB/m_e = (1.602\times10^{-19})(10^6)(1)/(9.109\times10^{-31}) = 1.759\times10^{17}\ \mathrm{m\,s^{-2}}\) (the orbit has radius \(m_ev/eB = 5.69\ \mu\)m). \(P = e^2a^2/6\pi\varepsilon_0c^3 = 1.77\times10^{-19}\) W. The kinetic energy is \(\frac12m_ev^2 = 4.55\times10^{-19}\) J (2.84 eV). Since \(P\propto a^2\propto v^2\propto\) kinetic energy, the energy decays exponentially with time constant \(\frac12m_ev^2/P = 2.58\) s.

Check. Dimensions: \([\mu_0q^2a^2/c] = \mathrm{N\,A^{-2}\,C^2\,m^2s^{-4}/(m\,s^{-1})} = \mathrm{N\,m\,s^{-1}} = \mathrm W\) ✓. Limit: \(\vec a = \vec 0\) gives \(P = 0\): a uniformly moving charge does not radiate (compare P8, whose fields fall as \(1/r^2\)) ✓. The time constant is independent of \(v\) — \(\tau = 3\pi\varepsilon_0m_e^3c^3/(e^4B^2)\) — so it applies to every non-relativistic electron in 1 T.

Idea. An orbiting charge is accelerated, so it radiates; the energy comes from the orbit, which shrinks, which increases the acceleration. The spiral ends at the proton in about sixteen picoseconds.

1. \(a = e^2/4\pi\varepsilon_0m_er^2\) (centripetal, from Coulomb), \(m_ev^2 = e^2/4\pi\varepsilon_0r\), and \(E = \frac12m_ev^2 - \frac{e^2}{4\pi\varepsilon_0r} = -\frac{e^2}{8\pi\varepsilon_0r}\).

2. Solving the last rung for \(\dot r\): \(\dot r = -\frac{8\pi\varepsilon_0r^2}{e^2}\cdot\frac{e^2}{6\pi\varepsilon_0c^3}\cdot\frac{e^4}{16\pi^2\varepsilon_0^2m_e^2r^4} = -\frac43c\Big(\frac{e^2}{4\pi\varepsilon_0m_ec^2}\Big)^2\frac1{r^2} = -\frac43\frac{r_e^2c}{r^2}\).

3. Separate: \(r^2\,\mathrm dr = -\frac43r_e^2c\,\mathrm dt\), so \(\frac{a_0^3}{3} = \frac43r_e^2c\,t\) and \[ t_\mathrm{collapse} = \frac{a_0^3}{4r_e^2c}. \] With \(r_e = 2.8179\times10^{-15}\) m and \(a_0 = 5.2918\times10^{-11}\) m: \(t = (5.2918\times10^{-11})^3/(4\times(2.8179\times10^{-15})^2\times2.9979\times10^8) = 1.56\times10^{-11}\) s.

4. At \(a_0\), \(v/c = \sqrt{e^2/4\pi\varepsilon_0m_ea_0}/c = 7.30\times10^{-3} = 1/137\): non-relativistic ✓. The initial power is \(4.67\times10^{-8}\) W and the period \(T_0 = 1.52\times10^{-16}\) s, so the energy lost per orbit is \(PT_0/|E| = 4.67\times10^{-8}\times1.52\times10^{-16}/2.18\times10^{-18} = 3.3\times10^{-6}\) of the binding energy: the orbit is circular to excellent accuracy ✓. Number of orbits: the period scales as \(r^{3/2}\), so \(N = \int\mathrm dt/T(r) = \int_0^{a_0}\frac{3r^2}{4r_e^2c}\frac{\mathrm dr}{T_0(r/a_0)^{3/2}} = \frac{a_0^3}{2r_e^2cT_0} = \frac{2t_\mathrm{collapse}}{T_0} = 2.05\times10^5\). The approximations fail only at the very end, as \(v\to c\) — too late to matter.

5. Classical mechanics and classical electrodynamics together predict that every atom collapses in \(10^{-11}\) s, and atoms do not: the theory is wrong about matter, and the repair (quantum mechanics) lies outside this course, which records the failure and does not attempt it.

Check. Dimensions: \([a_0^3/r_e^2c] = \mathrm{m^3/(m^2\,m\,s^{-1})} = \mathrm s\) ✓. Scaling: a larger orbit lasts much longer, as \(a_0^3\) — a classical "atom" of radius 1 cm would last \(1.05\times10^{14}\) s, about three million years. The Larmor power at \(a_0\) against the binding energy \(13.6\) eV gives \(|E|/P = 4.7\times10^{-11}\) s, the right order ✓.