The crisis: Maxwell's equations do not survive a Galilean boost

Idea. A Galilean boost leaves accelerations and all differences of positions and velocities unchanged; a law built only from those is invariant, and a law containing an absolute velocity is secretly about a second body.

1. \(\dot{\vec x}_i' = \dot{\vec x}_i - \vec v\) and \(\ddot{\vec x}_i' = \ddot{\vec x}_i\). Also \(\vec x_1' - \vec x_2' = \vec x_1 - \vec x_2\), since the \(\vec vt\) cancels. So \(m_1\ddot{\vec x}_1' = -Gm_1m_2(\vec x_1' - \vec x_2')/\lVert\vec x_1' - \vec x_2'\rVert^3\): the same equation.

2. Substituting, \(m\ddot{\vec x}' = m\vec g - b(\dot{\vec x}' + \vec v)\): a new constant force \(-b\vec v\) appears. The drag is really \(-b(\dot{\vec x} - \vec w)\), where \(\vec w\) is the velocity of the air, and "still air" silently set \(\vec w = \vec 0\) in the first frame. In the new frame the air moves at \(\vec w' = -\vec v\) and the law reads \(m\ddot{\vec x}' = m\vec g - b(\dot{\vec x}' - \vec w')\) — the same form, once the air is admitted as a body. The principle of relativity is about the laws for the whole system, not about one body in a medium that picks out a frame.

3. \(g(h(t,\vec x)) = \big(t + \tau_2 + \tau_1,\ R_1(R_2\vec x - \vec v_2t + \vec a_2) - \vec v_1(t + \tau_2) + \vec a_1\big)\), so \[ g\circ h = \big(R_1R_2,\ \ R_1\vec v_2 + \vec v_1,\ \ R_1\vec a_2 - \vec v_1\tau_2 + \vec a_1,\ \ \tau_1 + \tau_2\big). \] For pure boosts this is \((I, \vec v_1 + \vec v_2, \vec 0, 0)\): closed, with identity \(\vec v = \vec 0\), inverse \(-\vec v\), and \(\vec v_1 + \vec v_2 = \vec v_2 + \vec v_1\). The map \(\vec v \mapsto (I, \vec v, \vec 0, 0)\) is therefore an isomorphism from \((\mathbb R^3, +)\).

Check. Dimensions in part 2: \(b\) is \(\mathrm{kg\,s^{-1}}\), so \(b\vec v\) is a force ✓. Special case in part 3: \(R_1 = R_2 = I\), \(\vec v_i = \vec 0\) gives translations composing by addition, \((\vec a_1 + \vec a_2, \tau_1 + \tau_2)\) ✓. Note the term \(-\vec v_1\tau_2\): a boost and a time translation do not commute, which is why the full group is not a direct product.

Idea. A Galilean boost only changes \(\partial_t\), by \(-\vec v\cdot\nabla'\). Everything then follows from squaring, and the result is the ether picture: light moves at \(c\) relative to the old frame, so at \(c\hat k - \vec v\) relative to the new one.

1. With \(u(\vec x, t) = u'(\vec x - \vec vt, t)\), the chain rule gives \(\partial_{x_i}u = \partial_{x_i'}u'\) and \(\partial_tu = \partial_{t'}u' - \sum_iv_i\,\partial_{x_i'}u'\). So \(\nabla = \nabla'\) and \(\partial_t = \partial_{t'} - \vec v\cdot\nabla'\), constant coefficients, and \[ \partial_t^2 = \partial_{t'}^2 - 2\,\partial_{t'}(\vec v\cdot\nabla') + (\vec v\cdot\nabla')^2 \] (Clairaut lets \(\partial_{t'}\) and \(\vec v\cdot\nabla'\) commute). Subtract \(c^2\nabla'^2\).

2. On \(e^{\mathrm i(\vec k\cdot\vec x' - \omega t')}\): \(\partial_{t'}\to-\mathrm i\omega\), \(\nabla'\to\mathrm i\vec k\), so \(\partial_{t'} - \vec v\cdot\nabla' \to -\mathrm i(\omega + \vec v\cdot\vec k)\) and the operator gives \(-(\omega + \vec v\cdot\vec k)^2 + c^2\lVert\vec k\rVert^2\). Hence \[ \omega = \pm c\lVert\vec k\rVert - \vec v\cdot\vec k,\qquad \nabla_{\vec k}\,\omega = \pm c\,\hat k - \vec v . \] The phase speed in the direction \(\hat k\) is \(c - \vec v\cdot\hat k\); the energy moves at \(c\hat k - \vec v\).

3. We need \((c\hat k - \vec v)\cdot\vec v = 0\), i.e. \(c\,\hat k\cdot\vec v = v^2\). Then \(\lVert c\hat k - \vec v\rVert^2 = c^2 - 2c\,\hat k\cdot\vec v + v^2 = c^2 - v^2\), so the crossing speed is \(\sqrt{c^2 - v^2} = c/\gamma\): the wave front must be aimed slightly upwind. For \(v = 3.0\times10^4\ \mathrm{m\,s^{-1}}\), \[ 1 - \sqrt{1 - \beta^2} = 5.007\times10^{-9}, \] about \(1.50\ \mathrm{m\,s^{-1}}\) slower than \(c\). This is exactly the slides' \(t_\perp = (2L/c)\gamma\): a round trip of \(2L\) at speed \(c/\gamma\).

Check. Dimensions: \(\vec v\cdot\nabla'\) has the dimensions of \(\partial_{t'}\) ✓. Limit \(\vec v\to\vec 0\): \(\omega = \pm c\lVert\vec k\rVert\) ✓. Special case \(\hat k \parallel \vec v\): group velocity \(c - v\) downwind, matching the slides' worked example 1 ✓. And \(1 - \sqrt{1-\beta^2} \approx \tfrac12\beta^2 = 5.007\times10^{-9}\) ✓.

Idea. A boost changes the heat equation only in a first-order term, which a multiplier can absorb. The wave equation's damage is second-order, in the symbol, where a multiplier cannot reach.

1. Using the derivatives in rung 3, \[ u_t - ku_{xx} = e^{\alpha x+\beta t}\Big[(w_t - kw_{\xi\xi}) - (v + 2k\alpha)\,w_\xi + (\beta - k\alpha^2)\,w\Big]. \] If \(w_t = kw_{\xi\xi}\), this vanishes for all such \(w\) exactly when \(\alpha = -v/2k\) and \(\beta = k\alpha^2 = v^2/4k\): \[ (T_vw)(x,t) = e^{-vx/2k + v^2t/4k}\,w(x - vt, t). \] (Necessity: \(w = 1\) and \(w = \xi\) are solutions, and force the two brackets to vanish.)

2. \((T_{v_2}T_{v_1}w)(x,t) = e^{-v_2x/2k + v_2^2t/4k}\,(T_{v_1}w)(x - v_2t, t) = e^{-v_2x/2k + v_2^2t/4k}\,e^{-v_1(x - v_2t)/2k + v_1^2t/4k}\,w(x - v_2t - v_1t, t)\). The exponent is \(-(v_1 + v_2)x/2k + (v_1^2 + 2v_1v_2 + v_2^2)t/4k = -(v_1+v_2)x/2k + (v_1+v_2)^2t/4k\), so the product is \(T_{v_1+v_2}w\). \(\square\)

3. Suppose \(u = m\,w(\xi,t)\) solves the wave equation for every solution \(w\) of \(w_{tt} = c^2w_{\xi\xi}\). Expanding with the product and chain rules, the terms with two derivatives of \(w\) are \(m\big(w_{tt} - 2v\,w_{\xi t} + v^2w_{\xi\xi}\big) - c^2m\,w_{\xi\xi}\); all others carry at most one derivative of \(w\), multiplied by derivatives of \(m\). Take \(w = e^{\lambda(\xi - ct)}\) with \(\lambda \gt 0\): then \(w_{tt} = c^2\lambda^2w\), \(w_{\xi t} = -c\lambda^2w\), \(w_{\xi\xi} = \lambda^2w\), and the second-derivative terms add up to \(m(2vc + v^2)\lambda^2w\), while the rest are \((\text{bounded near a point})\times\lambda w\) or \(\times w\). Dividing \(u_{tt} - c^2u_{xx} = 0\) by \(\lambda^2w\) and letting \(\lambda\to\infty\) gives \(m(2vc + v^2) = 0\). With \(w = e^{\lambda(\xi + ct)}\) the same argument gives \(m(v^2 - 2vc) = 0\). Since \(m \ne 0\), \(v^2 + 2vc = v^2 - 2vc = 0\), so \(v = 0\). \(\square\)

Check. Dimensions: \(k\) is \(\mathrm{m^2\,s^{-1}}\), so \(vx/k\) and \(v^2t/k\) are pure numbers ✓. Limit \(v\to0\): \(T_0 = \) identity ✓. Test on a solution: \(w = e^{-kq^2t}\cos(qx)\) gives a \(T_vw\) that, by part 1, is a heat solution drifting at speed \(v\) while decaying — the heat pattern carried along by a uniformly moving medium, as it should be.

Another way

Part 3 is the slides' symbol argument: a multiplier changes only first- and zeroth-order terms, and the boosted symbol \(\tau^2 - 2v\tau\xi - (c^2 - v^2)\xi^2\) is not a multiple of \(\tau^2 - c^2\xi^2\) for \(v\ne0\). The plane-wave computation above is that argument unpacked: large \(\lambda\) isolates the symbol.

Idea. Precomposition is contravariant; the inverse repairs that. A set of maps preserving a set exactly is closed under inverses automatically; maps preserving it only one way are not.

1. \(\varphi_g\circ\varphi_h(y) = 2(y+1) = 2y + 2\) and \(\varphi_h\circ\varphi_g(y) = 2y + 1\), so \(u\circ\varphi_g\circ\varphi_h = 2y + 2\) and \(u\circ\varphi_h\circ\varphi_g = 2y + 1\). Define \(A_g u = u\circ\varphi_g\). Then \(A_h(A_gu) = (u\circ\varphi_g)\circ\varphi_h = u\circ(\varphi_g\circ\varphi_h) = u\circ\varphi_{gh}\) (with \(\varphi_{gh} = \varphi_g\circ\varphi_h\)), i.e. \(A_hA_g = A_{gh}\): order reversed. In the example, \(A_gA_h u = u\circ\varphi_h\circ\varphi_g = 2y + 1\) while \(A_{gh}u = 2y + 2\) — they differ. With \(B_gu = u\circ\varphi_g^{-1}\): \(B_gB_hu = u\circ\varphi_h^{-1}\circ\varphi_g^{-1} = u\circ(\varphi_g\circ\varphi_h)^{-1} = B_{gh}u\). \(\square\) In the example, \(\varphi_{gh}^{-1}(y) = y/2 - 1\), and \(B_gB_hu(y) = \varphi_h^{-1}(\varphi_g^{-1}(y)) = y/2 - 1\) ✓.

2. Let \(K = \{g : g\cdot S = S\}\). The identity is in \(K\). If \(g, h\in K\), then \((gh)\cdot S = g\cdot(h\cdot S) = g\cdot S = S\). And from \(g\cdot S = S\), applying \(g^{-1}\) to both sides gives \(S = g^{-1}\cdot S\). \(\square\)

3. Here \(S = \{u : u = 0 \text{ on } (0,1)\}\). With \((g\cdot u)(y) = u(y/2)\), a solution becomes a function vanishing on \((0,2) \supset (0,1)\), so \(g\cdot S\subseteq S\). But \(g^{-1}\cdot u(y) = u(2y)\) vanishes only on \((0,\tfrac12)\): for \(u\) equal to \(0\) on \((0,1]\) and to \(y - 1\) beyond, \(g^{-1}\cdot u\) is \(2y - 1 \ne 0\) at \(y = 0.75\). So \(g^{-1}\notin\{g : g\cdot S\subseteq S\}\), and that set is not a group. \(\square\)

Check. Part 3 is consistent with the slides' lemma: for a group \(G\) with \(G\cdot S\subseteq S\), equality is automatic; here \(\{g^n : n\ge0\}\) is only a semigroup. Special case: for the wave operator and linear maps the two notions agree, because \(S\) is preserved exactly by each symmetry found on the slides.

Idea. The first-order terms cancel on a round trip, leaving a path difference \(L\beta^2\); rotation doubles it. The contraction makes the round trip exactly isotropic.

1. From the slides, \(c(t_\parallel - t_\perp) = 2L(\gamma^2 - \gamma) = L\beta^2 + O(\beta^4)\). Turning through \(90°\) swaps the arms, so the pattern moves by \[ \Delta N = \frac{2L\beta^2}{\lambda} = \frac{2\times11\times1.0\times10^{-8}}{5.5\times10^{-7}} = 0.40 . \]

2. \(\Delta N\propto v^2\), so \(\Delta N \lt 0.01\) requires \(v \lt 30\sqrt{0.01/0.40} = 4.7\ \mathrm{km\,s^{-1}}\) — below the Earth's orbital speed by a factor of six, and the orbital velocity alone reverses every six months.

3. \(L(1 - 1/\gamma) = L\big(\tfrac12\beta^2 + O(\beta^4)\big) = 11\times5.0\times10^{-9} = 5.5\times10^{-8}\) m: 55 nm, a tenth of a wavelength. The round trip loses twice that, \(1.1\times10^{-7}\) m \(= L\beta^2\), which is exactly the path difference in part 1.

4. Outbound, the light must cover the arm plus the distance the far mirror moves: \(c^2t_1^2 = \lVert\ell\hat e + \vec vt_1\rVert^2 = \ell^2 + 2\ell vt_1\cos\varphi + v^2t_1^2\), so \((c^2 - v^2)t_1^2 - 2\ell v\cos\varphi\,t_1 - \ell^2 = 0\) and \(t_1 = \dfrac{\ell v\cos\varphi + \ell\sqrt{v^2\cos^2\varphi + c^2 - v^2}}{c^2 - v^2}\). Back, \(\cos\varphi \to -\cos\varphi\). Adding, \[ t_1 + t_2 = \frac{2\ell\sqrt{c^2 - v^2\sin^2\varphi}}{c^2 - v^2} = \frac{2\gamma^2}{c}\sqrt{\ell^2 - \beta^2\ell^2\sin^2\varphi}. \] For the contracted arm, \(\ell^2 = L^2\cos^2\theta/\gamma^2 + L^2\sin^2\theta\) and \(\ell^2\sin^2\varphi = L^2\sin^2\theta\), so \(\ell^2 - \beta^2\ell^2\sin^2\varphi = L^2\cos^2\theta/\gamma^2 + L^2\sin^2\theta(1 - \beta^2) = L^2/\gamma^2\). Hence \(t_1 + t_2 = (2\gamma^2/c)(L/\gamma) = 2L\gamma/c\), independent of \(\theta\). \(\square\)

Check. Dimensions: \(L\beta^2/\lambda\) is a number ✓. Special cases of part 4: \(\theta = 90°\) gives \(\ell = L\), \(\varphi = 90°\) and \(t = (2L/c)\gamma\), the slides' \(t_\perp\) ✓; \(\theta = 0\) gives \(\ell = L/\gamma\), \(t = (2L/\gamma c)\gamma^2\) ✓. With the exact \(c\) the shift in part 1 is \(0.4006\).

Idea. Each experiment gives one equation; a light-only experiment can only see ratios that a common rescaling of lengths and times leaves alone, and a material clock is needed to see the rest.

1. Equal rest lengths \(L\): \(t_\parallel = \frac{2L\ell_\parallel}{c}\gamma^2\), \(t_\perp = \frac{2L\ell_\perp}{c}\gamma\). No shift on rotating means \(t_\parallel = t_\perp\), i.e. \(\ell_\parallel\gamma = \ell_\perp\), \(\ell_\parallel/\ell_\perp = 1/\gamma\).

2. Given part 1, an arm of rest length \(L_i\) in either direction takes \(\frac{2L_i\ell_\perp\gamma}{c}\) of ether time and \(\tau\ell_\perp\gamma\frac{2L_i}{c}\) of lab time. Unequal arms make the difference \(\tau\ell_\perp\gamma\cdot2(L_1 - L_2)/c\) nonzero, and the Earth's speed \(v\) changes over the year; a null result means \(\tau\ell_\perp\gamma\) is constant, hence equal to its value \(1\) at \(v = 0\): \(\tau\ell_\perp = 1/\gamma\).

3. Crests leave the atom at intervals \(1/(\tau\nu_0)\) of lab time. Forwards, consecutive crests are separated by \((c - v)/(\tau\nu_0)\), so \(\lambda_f = \lambda_0(1 - \beta)/\tau\); backwards \(\lambda_b = \lambda_0(1 + \beta)/\tau\). The mean is \(\lambda_0/\tau\), with the first-order Doppler terms cancelling.

4. MM: \(\ell_\parallel/\ell_\perp\) is \(1/\gamma\) for all three ✓. KT, \(\tau\ell_\perp\): Lorentz \(1/\gamma\) ✓; FitzGerald–Lorentz \(1\) ✗; Voigt \(\gamma\cdot\gamma^{-2} = 1/\gamma\) ✓. IS, mean shift \(\lambda_0(1/\tau - 1)\) with \(\lambda_0 = 486.13\) nm and \(\gamma - 1 = 1.2500\times10^{-5}\):

• Lorentz: \(\lambda_0(\gamma - 1) = 0.00608\) nm \(= 6.08\) pm;
• FitzGerald–Lorentz: \(0\);
• Voigt: \(\lambda_0(\gamma^2 - 1) = 0.01215\) nm \(= 12.15\) pm, twice Lorentz's.

Ives and Stilwell found the first. So MM alone admits all three; KT removes FitzGerald–Lorentz; IS removes Voigt, and with the three equations \(\ell_\parallel = 1/\gamma\), \(\ell_\perp = 1\), \(\tau = 1/\gamma\) are forced.

Check. Dimensions: each condition is a ratio of pure numbers ✓. Limit \(v\to0\): all three conditions read \(1 = 1\) ✓. Size: the individual Doppler shifts are \(\pm\lambda_0\beta\gamma = \pm2.43\) nm (forward line at 483.71 nm, backward at 488.57 nm), four hundred times the effect sought, which is why the mean is measured. Note that the one-parameter family \(\ell_\parallel = 1/(s\gamma)\), \(\ell_\perp = 1/s\), \(\tau = s/\gamma\) passes MM and KT for every \(s\): light-only round trips are blind to a common dilation.

Idea. Four coefficients must match \(\lambda\) times the target: that is three equations after eliminating \(\lambda\), in four unknowns with \(b\) tied to \(a\) by \(v\). The transverse terms make \(\lambda\) a square.

1. The requirement is \[ e^2 - c^2d^2 = \lambda,\quad be - c^2ad = 0,\quad b^2 - c^2a^2 = -c^2\lambda,\quad s^2 = \lambda . \] With \(b = -va\) and \(a \ne 0\), the second gives \(d = -ve/c^2\). The first becomes \(e^2(1 - \beta^2) = \lambda\), the third \(a^2(v^2 - c^2) = -c^2\lambda\), i.e. \(a^2(1 - \beta^2) = \lambda\). So \(a^2 = e^2\) and \(\lambda = s^2 \gt 0\), which forces \(1 - \beta^2 \gt 0\). Continuity from the identity picks \(a = e \gt 0\), and \(a = s/\sqrt{1 - \beta^2} = s\gamma\): \[ x' = s\gamma(x - vt),\quad t' = s\gamma(t - vx/c^2),\quad y' = sy,\quad z' = sz,\qquad |v| \lt c . \] The speed \(c\) came from the operator; nothing was said about light.

2. The operator's symbol matrix is \(A = \mathrm{diag}(1, -c^2, -c^2, -c^2)\) in the order \((t, x, y, z)\), and by unit 18 a linear map with Jacobian \(J\) carries it to \(JAJ^{\mathsf T}\). If this equals \(\lambda A\) with \(\lambda \lt 0\), then \(\lambda A\) has one negative and three positive eigenvalues, while \(JAJ^{\mathsf T}\) has, by Sylvester's law of inertia, the same counts as \(A\): one positive, three negative. Contradiction; and \(\lambda\ne0\) since \(J\) is invertible. So \(\lambda \gt 0\). In part 1, \(\lambda = a^2(1-\beta^2) \gt 0\) gives \(|v| \lt c\). \(\square\)

3. In 1+1, drop \(s^2 = \lambda\): then \(\lambda = a^2(1 - \beta^2)\) may be negative. For \(v = 2c\) and \(a = e = 1\): \(x' = x - 2ct\), \(t' = t - 2x/c\), and \(\lambda = 1 - 4 = -3\): \(\partial_t^2 - c^2\partial_x^2 = -3(\partial_{t'}^2 - c^2\partial_{x'}^2)\). Check the cross term: \(be - c^2ad = -2c + c^2\cdot2/c = 0\) ✓. The signature \((1,1)\) is its own negative, so Sylvester says nothing.

Check. Dimensions: \(s, \gamma\) pure numbers, \(vx/c^2\) a time ✓. Limit \(v\to0\): a dilation by \(s(0)\), which must be \(1\) ✓. Special case \(s = 1\): \(\lambda = 1\), the Lorentz boost, which preserves the operator exactly ✓.

Idea. The symbol is a quadratic form of signature \((1,1)\); the linear symmetries of \(P\) are the maps preserving that form up to a factor. In null coordinates the form is \(\xi\eta\) (up to sign), and the answer is visible: scale \(\xi\) and \(\eta\) separately, or swap them.

1. Group: if \(MAM^{\mathsf T} = A = NAN^{\mathsf T}\), then \((MN)A(MN)^{\mathsf T} = MAM^{\mathsf T} = A\); and \(M^{-1}AM^{-\mathsf T} = M^{-1}(MAM^{\mathsf T})M^{-\mathsf T} = A\). Equations: \(p^2 - q^2 = 1\), \(pr - qm = 0\), \(r^2 - m^2 = -1\). The first gives \(|p|\ge1\), the third \(|m|\ge1\), and taking determinants \(-(\det M)^2 = -1\), so \(\det M = \pm1\). The functions \(\operatorname{sign}p\) and \(\det M\) are continuous with values in \(\{\pm1\}\), so they are constant on components: there are at least four. Exactly four: put \(v = -q/p\in(-1,1)\); then \(p^2(1 - v^2) = 1\), so \(p = \pm\gamma\), \(q = \mp\gamma v\). From \(pr = qm\), \(r = -vm\), and \(r^2 - m^2 = -1\) gives \(m = \pm\gamma\), \(r = \mp\gamma v\) (signs of \(m\) and \(r\) together). So every \(M\) is \(\gamma\begin{pmatrix}\epsilon_1 & -\epsilon_1v\\ -\epsilon_2v & \epsilon_2\end{pmatrix}\) with \(\epsilon_i = \pm1\), and \(\det M = \epsilon_1\epsilon_2\). For each of the four sign pairs, the matrices form a copy of the interval \(|v|\lt1\), which is connected. With \(\epsilon_1 = \epsilon_2 = 1\): the boosts. \(\square\)

2. Taking determinants of \(JAJ^{\mathsf T} = \lambda A\): \(-(\det J)^2 = -\lambda^2\), so \(\lambda\ne0\). If \(\lambda\gt0\), \(M = J/\sqrt\lambda\) satisfies \(MAM^{\mathsf T} = A\), and \(J = \sqrt\lambda\,M\). If \(\lambda\lt0\), since \(SAS^{\mathsf T} = -A\), \(M = JS/\sqrt{|\lambda|}\) has \(MAM^{\mathsf T} = J(-A)J^{\mathsf T}/|\lambda| = A\), and \(J = \sqrt{|\lambda|}\,MS\) since \(S^2 = I\). \(\square\)

3. A boost \(t' = \gamma(t - vx)\), \(x' = \gamma(x - vt)\) gives \(\xi' = x' - t' = \gamma(1 + v)(x - t) = \gamma(1+v)\,\xi\) and \(\eta' = \gamma(1 - v)\,\eta\): diagonal, with \(p = \sqrt{(1+v)/(1-v)}\) and \(q = \sqrt{(1-v)/(1+v)} = 1/p\). Since \(P = -4\partial_\xi\partial_\eta\) (part 4), a diagonal map \((\xi,\eta)\mapsto(p\xi, q\eta)\) gives \(\partial_\xi\partial_\eta = pq\,\partial_{\xi'}\partial_{\eta'}\), so \(\lambda = pq\). The diagonal maps are therefore: dilations times boosts (\(p, q \gt 0\)); those times \(-I = PT\), which sends \((\xi,\eta)\) to \((-\xi,-\eta)\); and, with \(pq \lt 0\), those times the swap \(S : (t,x)\mapsto(x,t)\), which sends \(\xi = x - t\) to \(-\xi\) and fixes \(\eta\) — the \(\lambda \lt 0\) case. The reflections \(x\mapsto-x\) and \(t\mapsto-t\) instead exchange the null directions, sending \((\xi,\eta)\) to \((-\eta,-\xi)\) and \((\eta,\xi)\).

4. In null coordinates \(\partial_x = \partial_\xi + \partial_\eta\), \(\partial_t = -\partial_\xi + \partial_\eta\), so \(P = \partial_t^2 - \partial_x^2 = -4\partial_\xi\partial_\eta\). For \(\xi' = F(\xi)\), \(\eta' = G(\eta)\): \(\partial_\xi = F'(\xi)\partial_{\xi'}\) and \(\partial_\eta = G'(\eta)\partial_{\eta'}\), and because \(F'\) does not depend on \(\eta\), \(\partial_\xi\partial_\eta = F'(\xi)G'(\eta)\,\partial_{\xi'}\partial_{\eta'}\) with no first-order term. So \(\partial_\xi\partial_\eta u = 0 \iff \partial_{\xi'}\partial_{\eta'}u' = 0\). Such a map is linear exactly when \(F(\xi) = p\xi\) and \(G(\eta) = q\eta\): the diagonal maps of part 3. Every other choice of \(F, G\) is a nonlinear symmetry — the conformal group of 1+1, infinite-dimensional.

Check. Special case \(v = 0\): \(p = q = 1\) ✓. Composition: \(p(u)p(v) = \sqrt{\frac{(1+u)(1+v)}{(1-u)(1-v)}}\), and this equals \(p(w)\) for \(w = (u+v)/(1+uv)\), since \(\frac{1+w}{1-w} = \frac{1 + u + v + uv}{1 - u - v + uv}\) ✓ — the composition law in the form "Doppler factors multiply". Number: \(v = 0.6\) gives \(p = 2\), \(q = \tfrac12\).

Idea. The composition law is matrix multiplication; Fresnel's coefficient is its first-order term; the fringe count is proportional to \(f\), so the experiment reads \(f\) off directly.

1. With \(\Lambda(v) = \gamma_v\begin{pmatrix}1 & -v\\ -v/c^2 & 1\end{pmatrix}\) on \((x,t)\), \[ \Lambda(u)\Lambda(v) = \gamma_u\gamma_v\begin{pmatrix}1 + uv/c^2 & -(u+v)\\ -(u+v)/c^2 & 1 + uv/c^2\end{pmatrix} = K\begin{pmatrix}1 & -w\\ -w/c^2 & 1\end{pmatrix}, \] \(K = \gamma_u\gamma_v(1 + uv/c^2)\). Since \((1 + uv/c^2)^2 - (u+v)^2/c^2 = (1 - u^2/c^2)(1 - v^2/c^2)\), \(K^2(1 - w^2/c^2) = \gamma_u^2\gamma_v^2(1 - u^2/c^2)(1 - v^2/c^2) = 1\), so \(K = \gamma_w\) (both positive). \(\square\)

2. \(u_{\text{lab}} = \dfrac{c/n + w}{1 + w/nc} = \dfrac cn + \Big(1 - \dfrac1{n^2}\Big)w - \dfrac{w^2}{nc}\Big(1 - \dfrac1{n^2}\Big) + O\Big(\dfrac{w^3}{c^2}\Big)\). For water \(f = 1 - 1/1.333^2 = 0.4372\); at \(w = 7.059\ \mathrm{m\,s^{-1}}\) the drag is \(3.086\ \mathrm{m\,s^{-1}}\) and the second-order term \(-5.5\times10^{-8}\ \mathrm{m\,s^{-1}}\).

3. \(\Delta N = \dfrac{4n^2Lfw}{\lambda c}\), and with the numbers \(4n^2Lw/(\lambda c) = 4\times1.7769\times1.4875\times7.059/(5.26\times10^{-7}\times2.998\times10^8) = 0.4733\):

• relativity, \(f = 0.4372\): \(\Delta N = 0.207\);
• Hertz, \(f = 1\): \(\Delta N = 0.473\);
• stationary ether, \(f = 0\): \(\Delta N = 0\).

The three differ by far more than a tenth of a fringe, so the experiment discriminates cleanly.

Check. Dimensions: \(Lw/(\lambda c)\) is \(\mathrm{m\cdot m\,s^{-1}/(m\cdot m\,s^{-1})}\), a number ✓. Limits: \(n\to1\) gives \(f\to0\) (vacuum is not dragged) ✓; \(u = c\) in part 1 gives \(w = c\) ✓. The approximation in the hint drops \(f^2w^2\) against \(c^2/n^2\), a relative error of \(10^{-16}\).

Idea. Both effects are first order in a velocity ratio; the second is multiplied by a huge distance, which is why a tiny source dependence would be glaring.

1. \(v/c = 2.978\times10^4/2.99792458\times10^8 = 9.934\times10^{-5}\), and \(\arctan(9.934\times10^{-5}) = 9.934\times10^{-5}\) rad \(= 20.49''\).

2. \(\Delta t \approx 2Du/c^2 = 2\times9.4607\times10^{18}\times10^5/(2.998\times10^8)^2 = 2.105\times10^7\) s \(= 243.7\) days. That is sixty orbital periods: light from sixty different orbital phases would arrive together, and the star's Doppler curve would be unrecognisable. Observed spectroscopic binaries follow Kepler cleanly.

3. With \(c + ku\), \(\Delta t \approx 2Dku/c^2 = k\times2.105\times10^7\) s. Requiring \(\Delta t \lt 0.01\times4\times86\,400 = 3456\) s gives \(k \lt 1.6\times10^{-4}\). Brecher's bound is five orders of magnitude tighter; with \(k = 2\times10^{-9}\) the delay in part 2 would be \(0.042\) s.

Check. Dimensions: \(Du/c^2\) is a time ✓. Limit \(u\to0\): no delay ✓. The exact \(2Du/(c^2 - u^2)\) differs from the approximation by a factor \(1 + 1.1\times10^{-7}\): invisible at four figures.

Idea. Galilean mechanics fixes how the fields would have to transform; with that rule the curl-of-\(\vec E\) half of Maxwell survives and the displacement current does not, with a residual of first order in \(v/c\).

1. The particle's velocity is \(\vec u - \vec v\) in the new frame, and force is frame-independent in Galilean mechanics, so \(\vec E' + (\vec u - \vec v)\times\vec B' = \vec E + \vec u\times\vec B\) for all \(\vec u\). At \(\vec u = \vec v\): \(\vec E' = \vec E + \vec v\times\vec B\). Substituting back: \(\vec u\times\vec B' - \vec v\times\vec B' = \vec u\times\vec B - \vec v\times\vec B\), i.e. \((\vec u - \vec v)\times(\vec B' - \vec B) = \vec 0\) for all \(\vec u\); choosing \(\vec u - \vec v\) along each axis in turn forces \(\vec B' = \vec B\).

2. \(\nabla'\cdot\vec B' = \nabla\cdot\vec B = 0\). And, expanding \(\nabla\times(\vec v\times\vec B)\) by unit 23's product rule for curls with \(\vec v\) constant, \(\nabla'\times\vec E' + \partial_{t'}\vec B' = \nabla\times\vec E + \vec v(\nabla\cdot\vec B) - (\vec v\cdot\nabla)\vec B + \partial_t\vec B + (\vec v\cdot\nabla)\vec B = (\nabla\times\vec E + \partial_t\vec B) + \vec v(\nabla\cdot\vec B) = \vec 0\). \(\square\)

The second identity, proved. For constant \(\vec a\), the \(x\)-component of \(\vec a\times(\nabla\times\vec E)\) is \(a_y(\partial_xE_y - \partial_yE_x) - a_z(\partial_zE_x - \partial_xE_z) = a_y\partial_xE_y + a_z\partial_xE_z - (a_y\partial_y + a_z\partial_z)E_x\). Adding and subtracting \(a_x\partial_xE_x\) makes this \(\partial_x(\vec a\cdot\vec E) - (\vec a\cdot\nabla)E_x\); the other components go the same way cyclically. So \(\vec a\times(\nabla\times\vec E) = \nabla(\vec a\cdot\vec E) - (\vec a\cdot\nabla)\vec E\). \(\square\) (Unit 24 proves the case \(\vec a = \vec E\); the constant-\(\vec a\) case is this one line.)

3. Subtracting as in rung 3, \[ \nabla'\times\vec B' - \tfrac1{c^2}\partial_{t'}\vec E' = -\tfrac1{c^2}\Big[(\vec v\cdot\nabla)\vec E + \vec v\times\partial_t\vec B + (\vec v\cdot\nabla)(\vec v\times\vec B)\Big]. \] With \(\partial_t\vec B = -\nabla\times\vec E\) and the identity just proved, \(\vec v\times\partial_t\vec B = -\nabla(\vec v\cdot\vec E) + (\vec v\cdot\nabla)\vec E\), which gives the stated form. \(\square\)

4. With \(\varphi = kx - \omega t\): \((\vec v\cdot\nabla)\vec E = -vkE_0\sin\varphi\,\hat y\); \(\vec v\cdot\vec E = 0\); \(\vec v\times\vec B = -(vE_0/c)\cos\varphi\,\hat y\), so \((\vec v\cdot\nabla)(\vec v\times\vec B) = (v^2kE_0/c)\sin\varphi\,\hat y\). The residual is \[ -\frac1{c^2}\Big[-2vkE_0 + \frac{v^2kE_0}{c}\Big]\sin\varphi\,\hat y = \frac{E_0k}{c}\big(2\beta - \beta^2\big)\sin\varphi\,\hat y . \] For \(\beta = 1.0\times10^{-4}\), \(2\beta - \beta^2 = 2.0\times10^{-4}\).

Check. Dimensions: \(E/c\) has the units of \(B\) and \(v/c\) is a number, so each term on the right is (tesla)/(metre), the units of \(\nabla\times\vec B\) ✓. Limit \(\vec v\to\vec 0\): residual zero ✓. Consistency: the same factor \(2\beta - \beta^2\) appears in the slides' test of the wave operator on the boosted wave, as it must, since that operator comes from these equations.

Idea. Voigt's map is a Lorentz boost followed by the dilation \(1/\gamma\). Dilations are symmetries of the wave operator, so Voigt passes that test; but a velocity-dependent dilation spoils reciprocity and closure.

1. \(\gamma^{-1}\Lambda(v)\) sends \(x\mapsto\gamma^{-1}\gamma(x - vt) = x - vt\), \(t\mapsto t - vx/c^2\), \(y\mapsto y/\gamma\): this is \(V(v)\). \(\Lambda(v)\) preserves the operator exactly, and a dilation \(y\mapsto sy\) (all four coordinates) gives \(\partial_y = s\,\partial_{y'}\), multiplying the operator by \(s^2\). With \(s = 1/\gamma\) the factor is \(\gamma^{-2}\). \(\square\)

2. \(V(v)^{-1} = \gamma\,\Lambda(v)^{-1} = \gamma\,\Lambda(-v)\): \(x = \gamma^2(x' + vt')\), \(t = \gamma^2(t' + vx'/c^2)\), \(y = \gamma y'\). But \(V(-v) = \gamma^{-1}\Lambda(-v)\). They differ by the factor \(\gamma^2\ne1\) for \(v\ne0\): each frame would find the other's transverse rulers shortened by \(1/\gamma\), and the two statements are incompatible with treating the frames alike.

3. \(V(u)V(v) = \gamma_u^{-1}\gamma_v^{-1}\Lambda(u)\Lambda(v) = \gamma_u^{-1}\gamma_v^{-1}\Lambda(w) = \dfrac{\gamma_w}{\gamma_u\gamma_v}\,V(w) = (1 + uv/c^2)\,V(w)\). This is not a Voigt map: the \(x'\)-row of \((1 + uv/c^2)V(w)\) is \((1 + uv/c^2)(x - wt)\), which fixes the velocity as \(w\), and then its coefficient of \(x\) is \(1 + uv/c^2\) rather than Voigt's \(1\). For \(u = v = c/2\): \(w = 0.8c\) and the factor is \(1.25\). \(\square\)

4. A rod along \(x'\) of rest length \(L\): at fixed \(t\), \(\Delta x' = \Delta x\), so \(\ell_\parallel = 1\). Across: \(\Delta y = \gamma\,\Delta y'\), so \(\ell_\perp = \gamma\). A clock at \(x' = 0\), i.e. \(x = vt\), reads \(t' = t(1 - \beta^2) = t/\gamma^2\), so \(\tau = 1/\gamma^2\). These are P6's Voigt values: MM and KT pass, Ives–Stilwell fails by a factor of two.

Check. Dimensions: all factors pure numbers ✓. Limit \(v\to0\): \(V(0) = I\) ✓, and \(1 + uv/c^2\to1\) as \(c\to\infty\), where Voigt's maps become Galileo's, which do form a group ✓. This is Poincaré's point: the changes of frame must form a group, and within the family \(s(v)\Lambda(v)\) the group property together with isotropy, \(s(-v) = s(v)\), leaves only \(s = 1\) — the slides' reciprocity argument.