Part C · the analytic toolkit · spine

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

The chain rule alone shows the wave operator breaks under x ↦ x − vt; asking which substitution does not break it produces the Lorentz transformation, with c supplied by the equation rather than by a postulate.

needs 24, 03~2 h slides + 4 h problemsproblems: ./mr problems 25

Unit 03 said "an inertial frame" as if there were one. There are infinitely many: any frame moving uniformly, without rotating, relative to an inertial frame is another, and Newton's laws cannot tell them apart. That is Galilean relativity, and unit 03 used it without naming it. Unit 24 then produced a theory — Maxwell's — that contains a speed, $c = 1/\sqrt{\varepsilon_0\mu_0}$, with no frame attached. Something has to give, and this unit finds out what, using nothing more than the chain rule of unit 03 and the principal symbol of unit 18.

The unit has two halves that must be kept apart. One is mathematics: what it means for a system of partial differential equations to be covariant under a group, a proof that the wave operator is not Galilean-covariant, and a derivation of the substitutions under which it is. The other is history and experiment: the thirty years (roughly 1875–1905) in which physicists tried to keep Galileo and Maxwell both, the three ways out they tried, and the experiments that shut each one. The mathematics is complete and short. The experiments are what make it matter.

How to read this deck

One computation runs through it: the one-dimensional wave operator pushed through a change of frame. Section 1 says what "the same law in every frame" means precisely. Section 2 pushes the operator through a Galilean boost and finds terms that should not be there. Section 3 is the history of trying to live with those terms. Section 4 asks which linear substitution removes them and gets the Lorentz boost. Section 5 shows the substitutions form a group, and matches each free function in the answer to the experiment that fixes it.

Every number quoted was computed. Proofs are on the slides where they are short; the longer ones are in these notes under Full proof, and anything imported is labelled cited with where its proof lives. Terms with a dotted underline are glossary entries — click one, or press G. N opens the notes, R stacks the deck into one readable page.

25 · overview

One operator, two substitutions

Every component of a light wave obeys $u_{tt} = c^2u_{xx}$. Galileo's $x' = x - vt$, $t' = t$ breaks it. Which linear change of frame does not?

the route, in four steps

1. The chain rule turns $\partial_t^2 - c^2\partial_x^2$ into a combination of $\partial_{t'}^2$, $\partial_{x'}\partial_{t'}$, $\partial_{x'}^2$ whose coefficients are read off the map. Galileo's leaves $-2v\,\partial_{x'}\partial_{t'}$, which no redefinition of the field removes.

2. Demand no cross term and the right ratio of the other two. The cross term dies only if $t'$ depends on $x$: $\ x' = A(x - vt)$, $t' = A(t - vx/c^2)$.

3. With $y' = sy$, $z' = sz$, the transverse terms force $A^2(1 - v^2/c^2) = s^2 \gt 0$: so $|v| \lt c$, and one scale $s(v)$ is left free.

4. The inverse must be the same rule at $-v$, and space is isotropic: so $s = 1$ — the Lorentz boost, whose invariant speed is the $c$ in the operator.

Galileo tilts only the time axis. The map that saves the operator also tilts $t' = 0$ (dashed), mirroring $x = vt$ in the light line.

getting around

Dotted: a glossary term, click it or press G. Solid: the running example, on hover. N: the notes.

The plan. 1 covariance · 2 the boost fails · 3 three ways out, and their experiments · 4 the right substitution · 5 a group.

The route, step by step. Step 1 is How an operator moves under a linear change of frame (the coefficient formula) and Worked: the wave operator, boosted by Galileo; that the cross term cannot be removed even by multiplying the field is No speed and no multiplier removes the cross term, and What survives, and the term that breaks finds where it sits in Maxwell's equations. Step 2 is The question, posed exactly and Worked: solving the two conditions. Step 3 is Three dimensions: transverse lengths, and $|v| \lt c$. Step 4 is Fixing $s$: reciprocity and isotropy and The Lorentz boost. Section 3, between steps 1 and 2, is the history: what physicists tried before taking step 2.

Why this example. Unit 24 showed every Cartesian component of $\vec E$ and $\vec B$ in vacuum obeys the wave equation, so if Maxwell's equations looked the same in two frames, the wave operator would have to. The one-dimensional operator carries the whole difficulty — the transverse directions add one condition in section 4 and nothing else — and it is small enough that every computation can be done by hand, which is the point: nothing here needs more than the chain rule. The closing slide of section 5 returns to it with both answers side by side.

The figure. The dashed line is the Lorentz boost's line of simultaneity $t' = 0$, i.e. $t = vx/c^2$: it is the mirror image of the worldline $x = vt$ in the light line $x = ct$ (swap $x$ and $ct$). Galileo keeps it horizontal. That single tilt — that "at the same time" depends on the frame — is what kills the cross term, and unit 26 builds everything from it.

The problem sheet follows the same order: P1–P4 are covariance (Newton, the wave operator, a Galilean-covariant equation that needs a multiplier, and the group lemma); P5 and P10 are the experimental numbers; P6 is "which experiment pins which"; P7–P9 derive and use the boost; P11 is Galilean electromagnetism; P12 is Voigt's transformation.

Units and constants: SI; $c = 2.99792458\times10^8\ \mathrm{m\,s^{-1}}$, which is $1/\sqrt{\varepsilon_0\mu_0}$ with unit 24's constants to the digits quoted there.

section 1

Galilean relativity, named at last

Newton's laws cannot tell uniformly moving frames apart. Saying precisely what that means — for particles, and then for fields — is the definition this unit is built on.

25 · covariance

The Galilean group, and why Newton cannot see it

the Galilean group

The maps of spacetime $\mathbb R\times\mathbb R^3$ $$(t, \vec x) \;\mapsto\; (t + \tau,\ R\vec x - \vec vt + \vec a),$$ $R \in SO(3)$ (unit 03), $\vec v, \vec a \in \mathbb R^3$, $\tau \in \mathbb R$: ten parameters. The case $R = I$, $\vec a = \vec 0$, $\tau = 0$ is a Galilean boost — the frame whose origin moves at $\vec v$.

It is a group: the composite of two such maps is another, and each has an inverse of the same form (notes).

theorem · Newton is boost-invariant

If $m_i\ddot{\vec x}_i = \sum_{j\ne i}\vec F_{ij}(\vec x_i - \vec x_j,\ \dot{\vec x}_i - \dot{\vec x}_j)$ holds in one frame, the same equations hold in every boosted frame.

Proof. $\vec x_i' = \vec x_i - \vec vt$ gives $\ddot{\vec x}_i' = \ddot{\vec x}_i$, and differences of positions and of velocities are unchanged. $\square$

the principle of relativity

The laws of physics take the same form in every inertial frame. Newton's mechanics satisfies it for the Galilean group, which is how unit 03 could say "an" inertial frame.

Full proof that it is a group. Write $g = (R, \vec v, \vec a, \tau)$ for the map above. Then $g_1g_2$ sends $(t,\vec x)$ to $(t + \tau_2 + \tau_1,\ R_1(R_2\vec x - \vec v_2t + \vec a_2) - \vec v_1(t + \tau_2) + \vec a_1)$, which is $(R_1R_2,\ R_1\vec v_2 + \vec v_1,\ R_1\vec a_2 - \vec v_1\tau_2 + \vec a_1,\ \tau_1 + \tau_2)$ — of the same form. The identity is $(I, \vec 0, \vec 0, 0)$, and solving $t' = t + \tau$, $\vec x' = R\vec x - \vec vt + \vec a$ for $(t, \vec x)$ gives $\vec x = R^{\!\top}\vec x' + R^{\!\top}\vec v(t' - \tau) - R^{\!\top}\vec a$, again of the form. Associativity is inherited from composition of maps. $\square$ Note that velocities add: boosting by $\vec v_2$ then $\vec v_1$ is boosting by $\vec v_1 + \vec v_2$. Section 5 finds that this is the one piece of the group that has to change.

What the theorem needs. The forces must depend on relative positions and relative velocities only; masses must be the same in both frames; and time must be the same time, $t' = t$ — that last one is so natural that nobody before 1905 stated it as a hypothesis. For rotations as well as boosts, $\vec F_{ij}$ must also be rotation-equivariant, $\vec F(R\vec r, R\vec w) = R\vec F(\vec r, \vec w)$, which gravity and every central force are.

What it forbids. A force that depends on the absolute velocity of a particle — air resistance, say — is not an exception: it depends on the velocity relative to the air, and the air is one of the bodies. The Lorentz force $q(\vec E + \vec u\times\vec B)$ of unit 23 is a different matter, and section 2 takes it up.

History. Galileo's ship (Dialogo, 1632): below decks in a smoothly sailing ship, no experiment tells you that you are moving. Newton's Corollary V to the laws of motion (1687) is the same statement. The name "Galilean transformation" is Philipp Frank's, 1909 — after it had already been replaced.

25 · covariance

What "the same law" means for a field

group action on fields

Let a group $G$ act on spacetime $U$ by bijections $y \mapsto \varphi_g(y)$, with $\varphi_{gh} = \varphi_g\circ\varphi_h$, and let $\rho : G \to GL(V)$ be a homomorphism. For a field $u : U \to V$ put $$(g\cdot u)(y) = \rho(g)\,u\big(\varphi_g^{-1}(y)\big).$$ This is an action: $(gh)\cdot u = g\cdot(h\cdot u)$.

covariance

A linear system $Pu = 0$ is $G$-covariant if $g\cdot u$ is a solution whenever $u$ is, for every $g \in G$. The operator $P$ is invariant up to a factor if $P(g\cdot u) = \lambda(g)\,g\cdot(Pu)$ for scalars $\lambda(g) \ne 0$.

proof it is an action

$\big(g\cdot(h\cdot u)\big)(y) = \rho(g)\rho(h)\,u\big(\varphi_h^{-1}\varphi_g^{-1}y\big) = \rho(gh)\,u\big(\varphi_{gh}^{-1}y\big)$. $\square$

Here $y = (t, \vec x)$, $g$ is a change of frame, and $u$ is a scalar field or the six components of $(\vec E, \vec B)$. The matrix $\rho(g)$ says how the components mix: a scalar has $\rho = 1$; for $(\vec E, \vec B)$, working out $\rho$ is part of the problem.

the physics, restated

The principle of relativity for a field theory is: the equations are covariant under the group relating inertial frames. Invariance of $P$ implies covariance, since $Pu = 0$ gives $P(g\cdot u) = \lambda\,g\cdot 0 = 0$.

Why the inverse in $u(\varphi_g^{-1}(y))$. The field $g\cdot u$ should take at the moved point $\varphi_g(y)$ the value $u$ took at $y$. Without the inverse, $u\mapsto u\circ\varphi_g$ reverses the order of composition — it is an action of the opposite group — and the proof above fails. For a single change of frame the distinction is invisible; for a group it is the difference between a homomorphism and an anti-homomorphism.

Covariance versus invariance. Covariance is about solution sets; invariance is about the operator. The factor $\lambda(g)$ is allowed because $Pu = 0$ and $\lambda Pu = 0$ are the same equation. Some books reserve "invariant" for $\lambda = 1$ and $\rho$ trivial and use "covariant" for the rest; the words are not standardised, and this deck always says which it means.

A generalisation the problem sheet needs. One can let $\rho$ depend on the point, $\rho_g(y)$, provided $\rho_{gh}(y) = \rho_g(y)\,\rho_h(\varphi_g^{-1}y)$ so that the action property still holds. P3 finds the heat equation covariant under Galilean boosts only with such a point-dependent multiplier. The theorem of section 2 excludes even these for the wave operator.

25 · covariance

Symmetries of an equation form a group

lemma · the symmetry group of an equation

Let a group $H$ act on fields, and $S$ be the solution set of $Pu = 0$. Then $$\mathrm{Sym}(P) = \{\,g\in H : g\cdot S = S\,\}$$ is a subgroup of $H$. A subgroup $G \le H$ has $G\cdot S \subseteq S$ if and only if $G \le \mathrm{Sym}(P)$.

proof

If $g\cdot S = S$ and $h\cdot S = S$ then $(gh)\cdot S = g\cdot(h\cdot S) = S$, and applying $g^{-1}$ to $g\cdot S = S$ gives $S = g^{-1}\cdot S$. For the second claim: $G\cdot S\subseteq S$ gives, for $g\in G$, $S = g\cdot(g^{-1}\cdot S) \subseteq g\cdot S \subseteq S$. $\square$

So the question "which changes of frame preserve Maxwell's equations?" has a group for its answer. It is not a list of lucky coincidences, and it can be found by computing the symmetries of one operator.

where this is going

Take $H$ = invertible linear maps of the $(t,x)$-plane acting on scalar fields, and $P = \partial_t^2 - c^2\partial_x^2$. Section 2 shows no Galilean boost with $v\ne0$ is in $\mathrm{Sym}(P)$. Section 4 computes $\mathrm{Sym}(P)\cap H$. Section 5 identifies it.

Why this lemma is the historical crux. Lorentz, between 1892 and 1904, found substitutions that made Maxwell's equations look nearly the same in a moving frame, one patch at a time, each justified by a separate physical argument. Poincaré in 1905 noticed that the patches were the elements of a group — which by this lemma they had to be, if they were symmetries at all — and that the group structure itself fixes the one factor Lorentz had left undetermined. Section 5 reproduces that argument.

Where the group lives. "Symmetries of $P$" depends on which $H$ one starts from. Inside all diffeomorphisms of the plane the answer for the 1+1 wave operator is infinite-dimensional (section 5); inside linear maps it is a two-dimensional group. That is not a contradiction — $\mathrm{Sym}(P)\cap H$ is a subgroup of each — but it is why this deck says "linear" every time.

section 2

The wave operator under a Galilean boost

The chain rule, applied once. Terms appear that no choice of speed and no redefinition of the field can remove — and in Maxwell's equations they sit exactly where unit 24 put the displacement current.

25 · the boost fails

How an operator moves under a linear change of frame

Let $(x', t') = (ax + bt,\ dx + et)$ with $ae - bd \ne 0$, and write a scalar field in the new variables: $u(x,t) = u'(x', t')$. The chain rule (unit 03) gives

$$\partial_x = a\,\partial_{x'} + d\,\partial_{t'},\qquad \partial_t = b\,\partial_{x'} + e\,\partial_{t'} .$$

Squaring, with Clairaut to merge the mixed partials,

$$\begin{aligned}\partial_t^2 - c^2\partial_x^2 \;=\;& (e^2 - c^2d^2)\,\partial_{t'}^2 \\ &+ 2(be - c^2ad)\,\partial_{x'}\partial_{t'} \\ &+ (b^2 - c^2a^2)\,\partial_{x'}^2 .\end{aligned}$$

unit 18, in this case

This is unit 18's theorem $\tilde A = JAJ^{\mathsf T}$ for the symbol matrix $A = \mathrm{diag}(1, -c^2)$ in the order $(t, x)$, with $J = \begin{pmatrix} e & d\\ b & a\end{pmatrix}$ the Jacobian of $(t,x)\mapsto(t',x')$. No lower-order terms arise, because the map is linear.

which way the arrows go

$\partial_x$ is not $\frac{1}{a}\partial_{x'}$. A new coordinate depends on both old ones, so each old derivative is a combination of both new ones — the coefficients are the columns of the Jacobian. Getting this backwards is the commonest error in this unit.

Full proof of the display. For a $C^2$ function $u'$, $u(x,t) = u'(ax + bt, dx + et)$. By the chain rule, $\partial_xu = a\,u'_{x'} + d\,u'_{t'}$ and $\partial_tu = b\,u'_{x'} + e\,u'_{t'}$, each evaluated at $(x',t')$; the coefficients are constants, so differentiating again is the same operation applied to $u'_{x'}$ and $u'_{t'}$. Thus $\partial_t^2u = b^2u'_{x'x'} + 2be\,u'_{x't'} + e^2u'_{t't'}$ and $\partial_x^2u = a^2u'_{x'x'} + 2ad\,u'_{x't'} + d^2u'_{t't'}$, using $u'_{x't'} = u'_{t'x'}$ (Clairaut, unit 03, which is why $C^2$ is assumed). Subtract $c^2$ times the second from the first. $\square$

Dimensions. If $x'$ is a length and $t'$ a time, then $a$ and $e$ are dimensionless, $b$ is a speed and $d$ an inverse speed. Each coefficient in the display then has the dimensions of the corresponding coefficient of $P$: $e^2 - c^2d^2$ is dimensionless, $be - c^2ad$ is a speed and $b^2 - c^2a^2$ a speed squared ✓. Special case $a = e = 1$, $b = d = 0$: the identity gives back $P$ ✓.

The warning, once more. The inverse function theorem of unit 04 is what lets the map be reversed; the matrix of $(\partial_x, \partial_t)$ in terms of $(\partial_{x'}, \partial_{t'})$ is the transpose of the Jacobian, not its inverse. Unit 04 planted this as "the derivative is a covector"; unit 18 made it the reason the symbol transforms by $J(\cdot)J^{\mathsf T}$; unit 28 will make it the difference between upper and lower indices.

25 · the boost fails · example

Worked: the wave operator, boosted by Galileo

worked example 1 · the running example

Step 1, the map. $x' = x - vt$, $t' = t$: so $a = 1$, $b = -v$, $d = 0$, $e = 1$.

Step 2, the derivatives. $\partial_x = \partial_{x'}$ and $\partial_t = \partial_{t'} - v\,\partial_{x'}$. The time derivative picks up a space derivative: at fixed $x$, $x'$ runs backwards at rate $v$.

Step 3, the coefficients. $e^2 - c^2d^2 = 1$; $\ be - c^2ad = -v$; $\ b^2 - c^2a^2 = v^2 - c^2$.

Step 4, the result.

$$P = \partial_{t'}^2 \;-\; 2v\,\partial_{x'}\partial_{t'} \;-\; (c^2 - v^2)\,\partial_{x'}^2 .$$

Step 5, what it predicts. Put $u' = e^{\mathrm i(kx' - \omega t')}$: $\ -\omega^2 - 2vk\omega + (c^2 - v^2)k^2 = 0$, so $\omega = (\pm c - v)k$. Waves run at $c - v$ one way and $c + v$ the other.

checks

Dimensions: $v\,\partial_{x'}\partial_{t'}$ and $v^2\partial_{x'}^2$ both have the dimensions of $\partial_{t'}^2$ ✓.
Limit $v \to 0$: $P$ comes back ✓.
Special case: the speeds $c \mp v$ are exactly Galileo's addition of velocities applied to a signal moving at $\pm c$ in the old frame ✓ — the operator knows the rule it was transformed by.

The next slide draws step 5: the two light lines, before and after the boost.

Step 5 in full. With $u' = e^{\mathrm i(kx' - \omega t')}$, each $\partial_{t'}$ brings down $-\mathrm i\omega$ and each $\partial_{x'}$ brings down $\mathrm ik$. So $\partial_{t'}^2 \to -\omega^2$, $\partial_{x'}\partial_{t'} \to (\mathrm ik)(-\mathrm i\omega) = k\omega$, $\partial_{x'}^2 \to -k^2$, and the operator gives $-\omega^2 - 2vk\omega + (c^2 - v^2)k^2$. As a quadratic in $\omega$: $\omega^2 + 2vk\omega - (c^2 - v^2)k^2 = 0$, so $\omega = -vk \pm\sqrt{v^2k^2 + (c^2 - v^2)k^2} = (-v \pm c)k$. The phase velocity $\omega/k$ is $c - v$ or $-(c + v)$ (unit 20's phase velocity; the medium is non-dispersive, so group and phase velocity agree).

Why this is not yet a contradiction. Nothing says the equation must look the same in a moving frame. If there is a preferred frame — the rest frame of whatever light is a wave in — then the boosted operator is simply the correct description in a frame moving through it, and the terms are real physics: an "ether wind". Section 3 is the story of looking for that wind.

Symbol check. In the order $(t,x)$, $J = \begin{pmatrix}1 & 0\\ -v & 1\end{pmatrix}$ and $JAJ^{\mathsf T} = \begin{pmatrix}1 & -v\\ -v & v^2 - c^2\end{pmatrix}$, whose off-diagonal $-v$ is the $-2v$ above (the operator counts the mixed term twice). Its determinant is $v^2 - c^2 - v^2 = -c^2$, unchanged, as $\det J = 1$ requires.

25 · the boost fails

Galileo's boost, drawn: light is not isotropic

The new operator is a perfectly good hyperbolic operator — Sylvester (unit 18) guarantees the type survives, so there are still two families of characteristics. What does not survive is the form: the characteristics are no longer symmetric about the time axis, and in the moving frame light is not isotropic.

For the Earth's orbital speed, $v/c = 1.0\times10^{-4}$: the lean is $10^{-4}$ rad, invisible in a drawing and, it will turn out, invisible to every experiment.

Spacetime diagrams, time up, with $v = 0.4c$ (exaggerated). Left: a flash at the origin spreads at $c$ both ways (dots, one time unit later). Right: the same two light lines in Galileo's coordinates lean to the left — $c - v$ forwards, $c + v$ backwards — away from the dashed lines where isotropic light would run.

What to look for in the figure. Galileo's map $x' = x - vt$, $t' = t$ shears each horizontal line $t = $ const sideways by $vt$ and keeps it horizontal. Every line through the origin keeps its time coordinates and loses $vt$ from its space coordinate, so the light line $x = ct$ becomes $x' = (c - v)t'$ and $x = -ct$ becomes $x' = -(c + v)t'$: the whole picture leans left, rigidly. The light lines are the characteristics of the operator (unit 18), and step 5 of the worked example found the same two speeds from the dispersion relation $\omega = (\pm c - v)k$.

Why "isotropic" is the right word. In one space dimension "every direction" means "left and right". On the right, a flash at the origin is $0.6$ units to the right and $1.4$ to the left one time unit later: the moving observer sees the flash off-centre, and could measure $v$ by timing the two ways. That is the ether-wind measurement section 3 is about.

25 · the boost fails

No speed and no multiplier removes the cross term

theorem

For $v \ne 0$, the Galilean boost is not a symmetry of $P = \partial_t^2 - c^2\partial_x^2$ — not even if the field is also multiplied by any smooth nowhere-zero function $m(x,t)$, and not even up to any smooth nowhere-zero factor $\lambda(x,t)$.

proof

Multiplying $u'$ by $m$ adds only first- and zeroth-order terms (product rule), so it cannot change the second-order part — the principal symbol (unit 18). The boosted symbol is $\tau^2 - 2v\tau\xi - (c^2 - v^2)\xi^2$ in the dual variables $(\tau,\xi)$ of $(t',x')$, and $\lambda(\tau^2 - c^2\xi^2)$ has no $\tau\xi$ term. They are equal only if $v = 0$. $\square$

"The solution set is carried onto itself" is weaker than "the operator is preserved", so the theorem needs one more step: the notes show that for this $P$ the two are equivalent. The slide's claim is therefore about solution sets — about physics.

why the multiplier matters

Galilean covariance with a multiplier is not a vacuous escape. The heat equation $u_t = ku_{xx}$ is Galilean-covariant once the field is multiplied by $e^{-vx/2k + v^2t/4k}$ (problem P3). The wave operator is not, and the obstruction is in its symbol, where no multiplier reaches.

Full proof that the same solution set forces the operator. Suppose a linear change of variables is a symmetry, so that $u$ solves $Pu = 0$ exactly when the transported $u'$ solves $P'u' = 0$, where $P' = \partial_{t'}^2 - c^2\partial_{x'}^2$. Let $Q$ be $P$ rewritten in primed variables by the chain rule, as on the previous slides, so that $Pu = 0 \iff Qu' = 0$; the hypothesis says that $Q$ and $P'$ have the same solutions, and only one direction of that is needed here — every solution of $P'u' = 0$ is annihilated by $Q$. Every $u' = f(x' - ct')$ and every $u' = g(x' + ct')$ with $f, g \in C^2$ solves $P'u' = 0$ (d'Alembert, unit 18). Writing $Q = \alpha\,\partial_{t'}^2 + 2\kappa\,\partial_{x'}\partial_{t'} + \delta\,\partial_{x'}^2$ (the letter $\kappa$, since $\beta$ is $v/c$ in this deck), $Qf(x' - ct') = (\alpha c^2 - 2\kappa c + \delta)f''$ and $Qg(x' + ct') = (\alpha c^2 + 2\kappa c + \delta)g''$. Both vanish for all $f, g$ only if $\alpha c^2 \mp 2\kappa c + \delta = 0$. Subtracting gives $\kappa = 0$; adding gives $\delta = -c^2\alpha$. Hence $Q = \alpha P'$, with $\alpha \ne 0$ since the change of variables is invertible. $\square$

The two conditions are worth reading: right-moving waves at speed $c$ stay right-moving waves at speed $c$, and so do left-moving ones. The characteristic lines $x \pm ct = $ const (unit 18) must map to $x' \pm ct' = $ const. Section 4 imposes exactly these two conditions.

The multiplier, in detail. If $u = m\,u'$ then $Pu = m\,Pu' + (\text{terms with at most one derivative of } u')$. So "multiplied solutions are solutions" forces the principal parts of the two operators to agree up to the factor $m$, and the argument above runs pointwise with $\alpha$ now a function. The theorem's conclusion $v = 0$ follows at any single point.

Why the heat equation escapes. Its symbol in the order $(t,x)$ is $\mathrm{diag}(0, -k)$, and $JAJ^{\mathsf T}$ with $J$ the Galilean Jacobian is again $\mathrm{diag}(0,-k)$: the boost changes only the first-order term $u_t$, which a multiplier can absorb. Degenerate in time, the parabolic symbol cannot see a boost.

25 · the boost fails

How $\vec E$ and $\vec B$ would have to transform

A covariant theory needs $\rho$: how the field components mix. For electromagnetism Galileo's own mechanics supplies it. A charge $q$ with velocity $\vec u$ feels the Lorentz force of unit 23; in the boosted frame it has velocity $\vec u - \vec v$, and Galilean forces are the same in both frames. So for every $\vec u$,

$$q\big(\vec E' + (\vec u - \vec v)\times\vec B'\big) \;=\; q\big(\vec E + \vec u\times\vec B\big).$$

the Galilean field transformation

$$\vec B' = \vec B,\qquad \vec E' = \vec E + \vec v\times\vec B,$$ both evaluated at the same event.

Proof. $\vec u = \vec v$ gives $\vec E' = \vec E + \vec v\times\vec B$. Substituting back leaves $\vec u\times(\vec B' - \vec B) = \vec 0$ for all $\vec u$, so $\vec B' = \vec B$. $\square$

checks

Dimensions: $[vB] = \mathrm{m\,s^{-1}\cdot T} = \mathrm{V\,m^{-1}}$ ✓.
Limit $\vec v \to \vec 0$: the identity ✓.
Special case: in a pure magnetic field, a charge at rest in the moving frame feels $q\,\vec v\times\vec B$ in the old one and $q\vec E'$ in the new; the formula makes them the same force ✓.

Unit 23's handover slide promised that $\vec v\times\vec B$ depends on the frame. This is that dependence made into a rule. It is linear in the fields — a $6\times6$ matrix $\rho(\vec v)$ — which is what the next example needs.

Is $\rho$ a homomorphism? Two boosts: $\vec B'' = \vec B$ and $\vec E'' = \vec E' + \vec v_1\times\vec B' = \vec E + (\vec v_1 + \vec v_2)\times\vec B$, which is the rule for the composite boost $\vec v_1 + \vec v_2$. So yes: the Galilean field transformation is a representation of the boost subgroup on $\mathbb R^6$, and the definitions of section 1 apply.

What was assumed. That force, charge and the acceleration of the particle are the same in both frames — Galilean mechanics. The point of the derivation is that if Galilean relativity held for electromagnetism, this would be the only possible way for the fields to transform. The next two slides check whether Maxwell's equations are covariant with it. They are not, and the failure is located precisely.

Looking ahead. Unit 28 derives the correct rule: $\vec E_\parallel' = \vec E_\parallel$, $\vec E_\perp' = \gamma(\vec E + \vec v\times\vec B)_\perp$, $\vec B_\perp' = \gamma(\vec B - \vec v\times\vec E/c^2)_\perp$. At first order in $v/c$ the electric part is this slide's; the magnetic part gains $-\vec v\times\vec E/c^2$, which for a light wave, where $|\vec E| = c|\vec B|$, is of order $(v/c)|\vec B|$ — first order, not negligible. That is where the failure comes from.

25 · the boost fails · example

Worked: a light wave seen from a moving frame

worked example 2 · Maxwell is not Galilean-covariant

Step 1, a solution. In vacuum (unit 24), $\vec E = E_0\cos(kx - \omega t)\,\hat y$, $\vec B = \frac{E_0}{c}\cos(kx - \omega t)\,\hat z$ with $\omega = ck$.

Step 2, move the argument. At the event $(x', t')$, $x = x' + vt'$ and $t = t'$, so the phase is $kx' - (\omega - kv)t' = kx' - \omega't'$ with $\omega' = k(c - v)$.

Step 3, mix the components. Any invertible $\rho$ — Galileo's, or any other constant one — gives primed components that are constants times $\cos(kx' - \omega't')$, not all zero. With Galileo's: $\vec E' = E_0(1 - \tfrac vc)\cos(\cdot)\,\hat y$.

Step 4, test it. Every component of a vacuum Maxwell field obeys $P' = \partial_{t'}^2 - c^2\partial_{x'}^2$. On $\cos(kx' - \omega't')$:

$$\begin{aligned}P'\cos(\cdot) &= \big(c^2k^2 - \omega'^2\big)\cos(\cdot)\\ &= k^2v(2c - v)\cos(\cdot) \ne 0 .\end{aligned}$$

conclusion

For $0 \lt |v| \lt 2c$ and any constant invertible $\rho$, the boosted light wave is not a vacuum solution. Maxwell's equations are not covariant under Galilean boosts.

checks

Dimensions: $k^2v(2c - v)$ is $\mathrm{s^{-2}}$, as $\partial_{t'}^2$ ✓. Limit $v \to 0$: $0$ ✓. Size: relative to $c^2k^2$ it is $2\beta - \beta^2$, with $\beta = v/c$: $2.0\times10^{-4}$ for the Earth's speed.

The wave in the new frame moves at $c - v$. Maxwell's equations, with their $\varepsilon_0$ and $\mu_0$, allow only $c$.

Step 4 in full. $\partial_{t'}^2\cos(kx' - \omega't') = -\omega'^2\cos$ and $\partial_{x'}^2\cos = -k^2\cos$, so $P'\cos = (-\omega'^2 + c^2k^2)\cos$. With $\omega' = k(c - v)$: $c^2k^2 - k^2(c - v)^2 = k^2(2cv - v^2) = k^2v(2c - v)$.

The edge case $v = 2c$. Then $\omega' = -ck$: the right-moving wave at $c$ becomes a left-moving wave at $c$, which is a solution. To close the loophole, run the same argument on the left-moving wave $\cos(kx + \omega t)$, which becomes $\cos(kx' + k(c + v)t')$ and fails unless $v = 0$ or $v = -2c$. For every $v \ne 0$ at least one of the two waves fails. $\square$ (A boost at $2c$ is not physics, but a theorem should not have a hole in it.)

Why a constant $\rho$ suffices. The primed field at $(x', t')$ is $\rho$ applied to the six numbers $(\vec E, \vec B)(x' + vt', t')$, each of which is a constant times $\cos(kx' - \omega't')$. Since $\rho$ is invertible and the wave is not zero, some primed component is a nonzero constant times $\cos(kx' - \omega't')$, and it fails the test. A point-dependent $\rho$ is excluded by the symbol argument on the previous slide.

The same example, Galileo's $\rho$, component by component. $\vec v\times\vec B = v\hat x\times\frac{E_0}{c}\cos\,\hat z = -\frac{vE_0}{c}\cos\,\hat y$, so $\vec E' = E_0(1 - v/c)\cos(\cdot)\,\hat y$ and $\vec B' = \frac{E_0}{c}\cos(\cdot)\,\hat z$. The ratio $|\vec E'|/|\vec B'| = c - v$, the speed of the wave — consistent with the next slide, where Faraday's law is satisfied and Ampère–Maxwell is not.

25 · the boost fails

What survives, and the term that breaks

theorem · half of Maxwell is Galilean

With $\vec E' = \vec E + \vec v\times\vec B$, $\vec B' = \vec B$ and $\vec x' = \vec x - \vec vt$, $t' = t$: $$\nabla'\!\cdot\vec B' = 0,\qquad \nabla'\times\vec E' + \partial_{t'}\vec B' = \vec 0$$ hold whenever the unprimed equations do.

Proof. $\nabla' = \nabla$ and $\partial_{t'} = \partial_t + \vec v\cdot\nabla$ on fields at a fixed event. Then $\nabla\times(\vec v\times\vec B) = \vec v(\nabla\!\cdot\vec B) - (\vec v\cdot\nabla)\vec B$ (unit 23's product rule for curls, with $\vec v$ constant), and the $(\vec v\cdot\nabla)\vec B$ terms cancel. $\square$

the displacement current does not

In vacuum, on worked example 2's wave, $$\nabla'\times\vec B' - \frac{1}{c^2}\partial_{t'}\vec E' = \frac{E_0k}{c}\,\big(2\beta - \beta^2\big)\sin(kx' - \omega't')\,\hat y \ne \vec 0 .$$ Faraday's law and $\nabla\cdot\vec B = 0$ are Galilean; the term Maxwell added in 1862 — the one that makes light — is the one that is not.

Drop the displacement current and the rest is a consistent Galilean theory, the magnetic limit of electromagnetism (cited: Le Bellac and Lévy-Leblond, 1973). But it has no light in it.

Full proof of the theorem. Define $\vec B'(\vec x', t') = \vec B(\vec x' + \vec vt', t')$. By the chain rule, $\partial_{t'}\vec B' = \partial_t\vec B + (\vec v\cdot\nabla)\vec B$ and $\nabla'$ acts as $\nabla$. So $\nabla'\cdot\vec B' = \nabla\cdot\vec B = 0$. Next, for constant $\vec v$, $\nabla\times(\vec v\times\vec B) = \vec v(\nabla\cdot\vec B) - (\vec v\cdot\nabla)\vec B$: this is unit 23's product rule $\nabla\times(\vec F\times\vec G) = \vec F(\nabla\cdot\vec G) - \vec G(\nabla\cdot\vec F) + (\vec G\cdot\nabla)\vec F - (\vec F\cdot\nabla)\vec G$ with $\vec F = \vec v$ constant, which kills the middle two terms. (Unit 08's $BAC$–$CAB$ is the algebraic identity for three vectors; it does not apply when one of them is $\nabla$.) Hence $\nabla'\times\vec E' + \partial_{t'}\vec B' = \nabla\times\vec E + \vec v(\nabla\cdot\vec B) - (\vec v\cdot\nabla)\vec B + \partial_t\vec B + (\vec v\cdot\nabla)\vec B = (\nabla\times\vec E + \partial_t\vec B) + \vec v(\nabla\cdot\vec B) = \vec 0$. $\square$

The residual, in full. From worked example 2, $\vec B' = \frac{E_0}{c}\cos\varphi'\,\hat z$ and $\vec E' = E_0(1 - \beta)\cos\varphi'\,\hat y$ with $\varphi' = kx' - \omega't'$, $\omega' = k(c - v)$. Then $\nabla'\times\vec B' = -\partial_{x'}B'_z\,\hat y = \frac{E_0k}{c}\sin\varphi'\,\hat y$ and $\frac1{c^2}\partial_{t'}\vec E' = \frac{E_0(1-\beta)\omega'}{c^2}\sin\varphi'\,\hat y = \frac{E_0k}{c}(1 - \beta)^2\sin\varphi'\,\hat y$. The difference is $\frac{E_0k}{c}\big(1 - (1-\beta)^2\big)\sin\varphi'\,\hat y = \frac{E_0k}{c}(2\beta - \beta^2)\sin\varphi'\,\hat y$. The relative size, $2\beta - \beta^2$, is the same $2.0\times10^{-4}$ as worked example 2 found for the wave operator — the same failure seen in the equation it came from. Dimensions: $E_0k/c$ is $\mathrm{T\,m^{-1}}$, as a curl of $\vec B$ ✓.

Citation. M. Le Bellac and J.-M. Lévy-Leblond, "Galilean electromagnetism", Il Nuovo Cimento 14B (1973) 217–234. They show there are exactly two consistent Galilean limits, "electric" and "magnetic"; the magnetic one is this slide's, and it is how the equations of magnetohydrodynamics and of eddy currents in slowly moving conductors manage to be Galilean-covariant. Neither limit has electromagnetic waves.

The historical irony. Unit 24 added the displacement current on purely logical grounds — consistency with charge conservation — and got light as the reward. The same term is what makes electromagnetism incompatible with Galileo. Maxwell could not have had one without the other.

section 3

Three ways out, and the experiments that closed them

If Maxwell and Galileo disagree, either Maxwell holds in only one frame, or light is not what Maxwell says, or Maxwell is wrong. Each was tried, each made a prediction, and each prediction failed.

25 · three ways out

The three responses

response	claim	what it predicts	killed by
an ether	Maxwell holds in one frame only, the rest frame of a medium; elsewhere the boosted operator of worked example 1 is correct	light at $c \mp v$ in the lab, so an ether wind of order $\beta = 10^{-4}$ or $\beta^2 = 10^{-8}$	Michelson–Morley, 1887
emission	light leaves its source at $c$ relative to the source, as a thrown ball would	light from an approaching star arrives early; double stars look scrambled	de Sitter, 1913
new Maxwell	replace Maxwell's equations by Galilean-covariant ones	moving matter drags light with it completely	Fizeau, 1851 — already on the books

the ether

The medium whose rest frame, the ether frame, is the one frame where $Pu = 0$ holds. Light's transverse polarisation (unit 24) made it an elastic solid, stiff enough to carry waves at $3\times10^8\ \mathrm{m\,s^{-1}}$, yet offering no measurable drag to the planets.

a fourth response

Keep Maxwell exactly and give up Galileo. Nobody took it seriously until the other three had failed, because it means changing $t' = t$ — the hypothesis Newton never stated.

Why the ether was not a silly idea. Every wave known before 1860 was a wave in something: sound in air, ripples on water, unit 18's string. The wave equation for the string holds in the rest frame of the string and in no other — worked example 1 applies verbatim to a string on a moving train, and nobody finds that paradoxical. The ether was the string for light. The crisis is only a crisis once experiments fail to find the frame.

The absurd properties. A transverse wave in an elastic solid travels at $\sqrt{G/\rho}$, with $G$ the shear modulus and $\rho$ the density (cited: it is unit 18's $\sqrt{T/\mu}$ in three dimensions, and unit 22 derives it from the stress tensor). For $\sqrt{G/\rho} = 3\times10^8\ \mathrm{m\,s^{-1}}$ the ratio $G/\rho$ must be $9\times10^{16}\ \mathrm{m^2\,s^{-2}}$; for steel it is about $10^7$. Either the ether is astonishingly rigid or astonishingly light, and in either case the Earth passes through it at $30\ \mathrm{km\,s^{-1}}$ without slowing, and it carries no longitudinal wave at all, which an elastic solid always does. Nineteenth-century ether models (Green, MacCullagh, Kelvin) each fixed one of these at the cost of another.

Why the table has three rows. Each row denies one of three premises: that Maxwell's equations are true, that they are true in every inertial frame, that the relation between frames is Galilean. The first two rows deny the second premise in different ways; the third row denies the first. Einstein's move denies the third.

25 · three ways out

Aberration and Fizeau: the ether must stand still, and not quite

stellar aberration · Bradley, 1728

Stars appear displaced towards the direction of the Earth's motion by up to $\arctan(v/c)$, cycling yearly. With $v = 29.78\ \mathrm{km\,s^{-1}}$ that is $9.93\times10^{-5}$ rad $= 20.49''$ — what Bradley measured.

It is rain on a moving car: the telescope must be tilted because it moves while the light crosses it. It needs light to arrive through a medium not dragged along by the Earth — a stationary ether.

So by 1728 the ether, if there is one, does not move with the Earth. The next slide finds that moving water does drag light — but only partly.

In a frame where the starlight falls straight down: while it falls the tube's length, $cT$, the tube moves $vT$. To keep the light on its axis the tube must lean forward by $\alpha$, $\tan\alpha = v/c$. Schematic, with $v/c$ exaggerated to $0.3$; the real $\alpha$ is $20.49''$.

Reading the figure. The dashed tube is where the telescope is when the light enters its top; the solid one, where it is when the light reaches the eyepiece. The ray is vertical in the frame drawn — the frame in which the light's medium is at rest — and the triangle it makes with the tube's sideways motion has legs $cT$ and $vT$. Six months later the Earth moves the other way and the tube must lean the other way: the star traces a yearly ellipse of semi-axis $20.49''$. Note what the figure assumes: the light keeps going straight down inside the moving tube, i.e. the air in the tube does not carry it along. That is the "not dragged" in the slide.

The aberration number. $v/c = 29.78\times10^3/2.99792458\times10^8 = 9.934\times10^{-5}$, and $\arctan(9.934\times10^{-5})$ rad $\times\,(180/\pi)\times3600 = 20.49$ arcseconds. James Bradley found the effect in 1728, and announced it in 1729, while trying to measure stellar parallax (the largest of which, Proxima Centauri's $0.77''$, is a twenty-seventh as large, and none was measured until Bessel, 1838), and his value of the constant was about $20''$. It was the first direct evidence that the Earth moves, and an independent measurement of $c$, after Rømer's of 1676: the angle is $v/c$ with $v$ known from the size of the orbit.

Airy's water telescope, 1871. Filling the telescope tube with water slows the light by a factor $n$, so a naive stationary-ether theory predicts a larger aberration angle. Airy found the angle unchanged. Fresnel's coefficient explains that exactly, to first order in $v/c$: the partially dragged light is carried along by just enough. So by 1871 a stationary ether with Fresnel's drag was consistent with every first-order experiment. What was missing was a reason for the number $1 - 1/n^2$, which the next slide states.

25 · three ways out

Fizeau: moving water drags light, partly

Fizeau, 1851 · Fresnel drag coefficient

Light in water of refractive index $n$, moving at $w$, travels in the lab at $$\frac cn + f\,w,\qquad f = 1 - \frac{1}{n^2}\ \ \text{(Fresnel, 1818)}.$$ For water, $n = 1.333$: $f = 0.437$. Measured by Michelson and Morley in 1886: $0.434 \pm 0.02$.

partial drag

Not $f = 0$ (stationary ether) and not $f = 1$ (dragged ether): moving water drags light by 44%, a number with no mechanism behind it. Section 5 derives it.

Schematic. The two beams share one loop, so every length and mirror is common to both; only the water's motion distinguishes them. The difference of their travel times, $\propto f\,w$, moves the fringes when the flow is reversed.

Fizeau's apparatus. Light split into two beams, sent through two tubes of flowing water, one with the flow and one against, and recombined. The fringe shift is proportional to $f$ (problem P9 computes it for an apparatus of Fizeau's size). Fizeau's 1851 value agreed with Fresnel's within his accuracy; Michelson and Morley's repetition of 1886 (Am. J. Sci. 31, 377) gave $0.434 \pm 0.02$, against Fresnel's $0.437$ for $n = 1.333$.

Why the loop. In the figure each beam passes through both tubes, and the water flows in opposite directions in the two, so one beam is carried along in both and the other held back in both. Sending the two beams round the same loop in opposite senses (Michelson and Morley's arrangement) means the beams cover identical paths through identical glass and air; any drift of the apparatus affects both alike and cancels in the difference. Only $w$ does not cancel.

25 · three ways out · example

Worked: Michelson–Morley, the expected shift

worked example 3 · the ether wind

Setup. A Michelson interferometer: two arms of length $L$ at right angles, one along the wind $v$. Work in the ether frame, where light moves at $c$.

Along. Out at $c - v$ relative to the mirror, back at $c + v$: $t_\parallel = \frac{L}{c - v} + \frac{L}{c + v} = \frac{2L}{c}\gamma^2$, $\ \gamma^2 = \frac{1}{1 - \beta^2}$.

Across. The light must aim upwind, crossing at $\sqrt{c^2 - v^2}$: $t_\perp = \frac{2L}{c}\gamma$.

Difference. $c(t_\parallel - t_\perp) = 2L(\gamma^2 - \gamma) = L\beta^2 + O(\beta^4)$.

Rotate by 90°. The arms swap roles, so the fringes move by $$\Delta N = \frac{2L\beta^2}{\lambda} = \frac{2(11\ \mathrm m)(1.0\times10^{-8})}{5.5\times10^{-7}\ \mathrm m} = 0.40 .$$

checks

Dimensions: $L\beta^2/\lambda$ is a pure number ✓. Limit $v\to0$: $\Delta N \to 0$ ✓. Second order: the first-order terms cancel on the round trip — which is why nothing before 1881 could see the wind.

Here $\beta = v/c$, $\gamma = 1/\sqrt{1 - \beta^2}$ is shorthand for the factor that falls out of the algebra, and $\lambda$ is the wavelength of the light. The next slide draws both round trips, and gives the verdict.

The across arm in full. In the ether frame the mirror moves sideways at $v$ while the light crosses. If the one-way crossing takes time $T$, the light travels $cT$ along the hypotenuse of a right triangle with legs $L$ and $vT$: $c^2T^2 = L^2 + v^2T^2$, so $T = L/\sqrt{c^2 - v^2} = (L/c)\gamma$, and the round trip is $2T$.

The expansion. $\gamma^2 = 1 + \beta^2 + O(\beta^4)$ and $\gamma = 1 + \tfrac12\beta^2 + O(\beta^4)$, so $\gamma^2 - \gamma = \tfrac12\beta^2 + O(\beta^4)$ and $2L(\gamma^2 - \gamma) = L\beta^2$. For $L = 11$ m and $\beta^2 = 1.0\times10^{-8}$ this path difference is $1.1\times10^{-7}$ m $= 0.20\lambda$; rotating the apparatus reverses its sign, so the pattern moves by twice that.

The numbers are Michelson and Morley's own. Their paper: "Considering only the velocity of the earth in its orbit, this would be $2D\,v^2/c^2 = 2D\times10^{-8}$. The distance $D$ was about eleven meters, or $2\times10^7$ wave-lengths of yellow light; hence the displacement to be expected was $0.4$ of a fringe." Eleven metres was reached by folding each arm with mirrors on a sandstone slab floating on mercury, so it could be rotated smoothly without strain. Using the exact $c$ the shift is $0.4006$. (A. A. Michelson and E. W. Morley, Am. J. Sci. 34 (1887) 333.)

Why the season matters. The ether wind in the lab is the vector difference of the Earth's orbital velocity and whatever the solar system's velocity through the ether is. The latter is unknown, but the orbital part reverses every six months, so it cannot cancel the wind at all seasons. Michelson and Morley observed only for a few days in July 1887; a null result in one season could be bad luck. Morley and Miller repeated the experiment at other times of year (1902–05) and found none either.

25 · three ways out

Michelson–Morley, drawn: two round trips

Left: along the motion the light chases a receding mirror at $c - v$, then meets the splitter at $c + v$. Right: across it, in the ether frame, the light must aim ahead to hit a mirror that has moved $vT$; it runs the hypotenuse $cT$, so $c^2T^2 = L^2 + v^2T^2$. Schematic, $v/c$ exaggerated to $0.5$.

Along: $t_\parallel = \frac{2L}{c}\gamma^2$. Across: $t_\perp = \frac{2L}{c}\gamma$. The across trip is quicker by one factor of $\gamma$; rotating the slab $90°$ swaps the arms and should move the fringes by $0.40$.

the null result

Michelson and Morley, Cleveland, July 1887: the shift was "probably less than the fortieth part" of $0.40$ — below $0.01$ fringe, at noon and in the evening. Later repetitions at other seasons agreed.

Worked example 1's operator predicts $c - v$ and $c + v$. This is that operator, tested at second order, and it fails.

What to look for in the figure. Both arms have the same length $L$, and in both the light covers more than $2L$ in the ether frame, because the apparatus moves during the trip. Along the motion the penalty is $\gamma^2$: the slow leg out ($c - v$) costs more than the fast leg back ($c + v$) saves. Across, it is only $\gamma$: the zig-zag is longer than $2L$ by the triangle's hypotenuse, a smaller penalty. The difference, $\gamma^2 - \gamma \approx \tfrac12\beta^2$, is second order because the first-order effects — the $\mp v$ on the way out and back — cancel in each round trip.

Why rotate. Nobody can make two arms equal to a fraction of a wavelength, so a single reading of the fringes says nothing. Rotating the apparatus swaps which arm is "along"; the time difference changes sign, and the fringe pattern should slide by $2L\beta^2/\lambda$ during the turn. That slide was what Michelson and Morley watched for, and it did not come.

25 · three ways out

The scoreboard: no ether fits all three

ether model	aberration	Fizeau, $f = 0.437$	Michelson–Morley
stationary, Fresnel's partial drag	✓	✓ (by fiat)	✗ predicts $0.40$
completely dragged near the Earth (Stokes, 1845)	✗	✗ predicts $f = 1$	✓
stationary + contraction hypothesis	✓	✓ (by fiat)	✓

the contraction hypothesis · FitzGerald 1889, Lorentz 1892

A body moving through the ether at $v$ shortens along its motion by the factor $\sqrt{1 - \beta^2}$. Then $t_\parallel$ becomes $\frac{2L}{c}\gamma^2\cdot\frac1\gamma = \frac{2L}{c}\gamma = t_\perp$: the null result, exactly, at every speed.

a patch, not a mechanism

The third row fits by adjusting one new function per experiment. For the Earth at $30\ \mathrm{km\,s^{-1}}$ the contraction of its diameter is $6.4$ cm. Section 5 finds that both patches — Fresnel's $f$ and this — fall out of the substitution section 4 derives.

Why the dragged ether fails aberration. If the ether near the Earth moves with it, light arriving at the telescope has been carried along and there is nothing to tilt for — unless the ether's flow bends the rays by exactly the right amount on the way in. Stokes (1845) showed that it does if the flow is irrotational. Lorentz pointed out in 1886 that an irrotational flow cannot also match the Earth's velocity at its surface, which is what "dragged" requires; the two demands contradict each other.

The contraction check in full. With the parallel arm's length $L/\gamma$, $t_\parallel = \frac{2(L/\gamma)}{c}\gamma^2 = \frac{2L}{c}\gamma = t_\perp$. This holds for every $v$, not just to order $\beta^2$, so the hypothesis predicts a null result at any speed and any orientation (a rotated arm at angle $\theta$ also works out; P6 does the general bookkeeping).

The Earth's contraction. $\gamma - 1 = 5.007\times10^{-9}$ at $30\ \mathrm{km\,s^{-1}}$, so $1 - 1/\gamma = 5.007\times10^{-9}$ and a diameter of $1.2742\times10^7$ m shortens by $0.0638$ m. No ruler could detect it, since the ruler contracts too — which is the problem with the hypothesis in one sentence.

History. G. F. FitzGerald, Science 13 (1889) 390, a paragraph of one page. H. A. Lorentz, independently, 1892. Lorentz later tried to derive the contraction from the hypothesis that intermolecular forces transform like electromagnetic ones. That idea is closer to the truth than it looks: unit 28 shows the electromagnetic forces do transform that way, and the principle of relativity says every other force must too.

25 · three ways out

Emission theories, and de Sitter's double stars

emission theory · Ritz, 1908

Light leaves its source at $c$ relative to the source, and keeps the source's velocity: Galilean addition applied to light. Michelson–Morley's null result is then automatic, since source and mirrors share one frame.

It keeps Galileo and gives up Maxwell's field theory: the wave equation's solutions travel at the $c$ in the operator, whatever the source did (notes).

de Sitter's argument, 1913

A star in a binary orbit, speed $u$, distance $D$. Light emitted while approaching arrives earlier than light emitted while receding by $$\Delta t = \frac{D}{c - u} - \frac{D}{c + u} \approx \frac{2Du}{c^2}.$$

checks

Dimensions: $Du/c^2$ is $\mathrm{m\cdot m\,s^{-1}/m^2\,s^{-2}} = \mathrm s$ ✓. Limit $u\to0$: no effect ✓.

The expansion. $\frac{D}{c-u} - \frac{D}{c+u} = \frac{2Du}{c^2 - u^2} = \frac{2Du}{c^2}(1 + O(u^2/c^2))$. With the numbers on the slide, the exact difference is $2.1053\times10^7$ s and the approximation agrees to five figures; divided by 86 400 s it is 243.7 days.

The objection, and its answer. Light passing through gas near the star is absorbed and re-emitted, and the re-emitted light could "forget" the source's velocity (the extinction theorem of Ewald and Oseen, cited; Born and Wolf, Principles of Optics, §2.4). So optical binaries were not decisive. K. Brecher (Phys. Rev. Lett. 39 (1977) 1051) used X-ray binaries, whose high-energy photons are not re-emitted by the intervening gas, and bounded any dependence $c + ku$ at $k \lt 2\times10^{-9}$.

Emission theories and Maxwell. An emission theory cannot be a field theory of Maxwell's kind at all: the wave equation's solutions travel at the $c$ in the operator whatever the source did. Ritz's theory replaced the fields by retarded action-at-a-distance with a source-dependent retardation. It is the most radical of the three responses, and the cleanest to kill.

History. W. de Sitter, "Ein astronomischer Beweis für die Konstanz der Lichtgeschwindigkeit", Physikalische Zeitschrift 14 (1913) 429. Note the date: eight years after Einstein. The constancy of $c$ was a postulate in 1905; it became an observation later.

25 · three ways out

De Sitter's binary, drawn: one star, several images

the size

$u = 100\ \mathrm{km\,s^{-1}}$, $D = 1000$ light-years $= 9.46\times10^{18}$ m: $\ \Delta t = 2.11\times10^7$ s $= 244$ days.

A binary with a period of days would show several images of each star at once, and a Doppler curve scrambled beyond recognition. Spectroscopic binaries show clean Keplerian orbits.

The figure takes a period of 100 days, so that the curve can still be drawn. Shorter periods fold it more often, and give more images.

Under an emission theory, light leaves at $c + u_r$, with $u_r = u\sin(2\pi t/100\ \mathrm{d})$ the star's velocity towards us. Arrival time against emission time swings $\pm 122$ days about the dashed line, so it doubles back: on one night (the dashed horizontal line) light from five moments of the orbit arrives together.

How the curve is computed. A circular orbit seen edge-on, speed $u = 100\ \mathrm{km\,s^{-1}}$, period $100$ days, distance $D = 9.46\times10^{18}$ m. Light emitted at time $t$ travels at $c + u_r(t)$ and arrives at $t + D/(c + u_r)$; the figure plots that minus the constant $D/c$, exactly. Since $D/(c + u_r) - D/c \approx -Du_r/c^2$, the swing is $\pm Du/c^2 = \pm 122$ days, half the slide's $244$. The curve turns back wherever its slope $1 - (2\pi Du/c^2P)\cos(2\pi t/P)$ is negative, which for $P = 100$ days is most of the orbit, since $2\pi Du/c^2P = 7.7$.

What a telescope would see. Each crossing of the horizontal line is a moment of the orbit whose light arrives on the same night, each with its own position on the orbit and its own Doppler shift. The star would appear in several places at once, with spectral lines at several velocities, and the images would come and go as the night's line moves up. Under Maxwell, light leaves at $c$ whatever the source does, the curve is the dashed line shifted by the small light-time across the orbit, and one image moves smoothly round.

25 · three ways out

Changing Maxwell, and Maxwell continuing to be right

Hertz, 1890

Replace each $\partial_t$ in Maxwell's equations by the derivative following the moving matter, $\partial_t + \vec w\cdot\nabla$, as unit 22's material derivative would. The equations become Galilean-covariant by construction, in the way the boosted operator of worked example 1 would become $P$ again if $v$ were the local velocity of the medium.

its prediction

Light in a medium is carried completely by it: $f = 1$ in Fizeau's formula. The measured $f$ is $0.434 \pm 0.02$.

Every other test pointed the same way. Maxwell's $c = 1/\sqrt{\varepsilon_0\mu_0}$ from two bench-top constants matched the speed of light (unit 24). Hertz himself produced radio waves in 1887–88 and showed they reflect, refract and polarise as light does, at a finite speed comparable with light's.

where that leaves things in 1900

Maxwell's equations are right. They hold in the lab to $10^{-8}$ though the lab moves. Galilean relativity forbids both at once. Of the four premises — Maxwell, relativity, Galileo, and the unstated $t' = t$ — something in the last two must go.

What Hertz's theory is. H. Hertz, "Über die Grundgleichungen der Elektrodynamik für bewegte Körper", Annalen der Physik 41 (1890) 369. The ether inside matter is assumed to move with the matter, and the field equations are written with the convective derivative. In the rest frame of the matter the equations have Maxwell's form; a Galilean boost carries one rest frame to another, so they are Galilean-covariant. The cost is that a light wave inside moving matter moves with the matter: $c/n + w$, i.e. $f = 1$. Fizeau's 1851 measurement had already excluded that. (Hertz also fails the later experiments of Eichenwald, 1903, on rotating dielectrics.)

Unit 22's material derivative is not a prerequisite of this unit, so only its definition is used here: $\partial_t + \vec w\cdot\nabla$ is the rate of change following a particle moving with velocity field $\vec w$, which is the chain rule applied to $u(\vec x(t), t)$.

"To $10^{-8}$". Michelson–Morley bounds the lab-frame anisotropy of the round-trip speed at the level of $\beta^2/40 \approx 2.5\times10^{-10}$ in the combination they measured — in round numbers, Maxwell's isotropic propagation holds in the moving lab to far better than the $10^{-8}$ an ether wind would disturb it by. Modern cavity versions of the experiment bound the anisotropy below $10^{-17}$ (cited: Herrmann et al., Phys. Rev. D 80 (2009) 105011).

section 4

Which substitution kills the cross terms

Keep Maxwell, keep the principle of relativity, and ask the chain rule which linear change of frame leaves the wave operator alone. Two conditions; one family; $c$ already inside it.

25 · the right substitution

The question, posed exactly

Look for an invertible linear map $(x', t') = (ax + bt,\ dx + et)$ describing a frame whose origin moves at velocity $v$. Then:

the origin moves at $v$: $x' = 0$ exactly when $x = vt$, so $b = -va$;
the law is the same: the map is a symmetry of $P$, which by the previous section's notes means $P = \lambda\,(\partial_{t'}^2 - c^2\partial_{x'}^2)$ for some constant $\lambda \ne 0$.

The second demand is not "Maxwell holds in the new frame by assumption about light". It is the principle of relativity, applied to Maxwell's equations as they stand.

the two conditions

From the coefficient formula of section 2, form-invariance is (i) no cross term and (ii) the right ratio:

$$\begin{aligned}\text{(i)}\quad & be - c^2ad = 0,\\ \text{(ii)}\quad & b^2 - c^2a^2 = -c^2\,(e^2 - c^2d^2).\end{aligned}$$

Equivalently: the two families of characteristics, $x \pm ct = $ const, are each carried to themselves.

Four unknowns $a, b, d, e$; $v$ fixes $b$ in terms of $a$; two conditions leave one free number.

Why linear. A Galilean boost is linear, and the question is which linear map should replace it. Unit 26 derives linearity from the homogeneity of space and time rather than assuming it: a map that treats every point alike must send straight uniform motions to straight uniform motions, and such maps are affine. Here it is an assumption, stated. The nonlinear symmetries of $P$ exist and are discussed in section 5; they are not changes of inertial frame.

Why the characteristics. In null coordinates $\xi = x - ct$, $\eta = x + ct$, the operator is $P = -4c^2\partial_\xi\partial_\eta$ (unit 18's canonical form). A linear map preserves it up to a factor exactly when it sends $\xi$ to a multiple of $\xi'$ and $\eta$ to a multiple of $\eta'$, or swaps them. Condition (i) and (ii) are that statement written in $(x,t)$. The swap is excluded by continuity from the identity (or, in 3+1, by Sylvester — see the transverse slide).

Where $c$ enters. The $c$ in the conditions is the $c$ in $P$, which unit 24 computed as $1/\sqrt{\varepsilon_0\mu_0} = 2.998\times10^8\ \mathrm{m\,s^{-1}}$ from a capacitor and a coil. Nothing about the speed of light as such has been assumed: only that Maxwell's equations, with their constants, hold in both frames.

25 · the right substitution · example

Worked: solving the two conditions

worked example 4 · the running example, repaired

Step 1, use $b = -va$ in (i). $-vae - c^2ad = 0$, and $a \ne 0$ (else $x'$ would not depend on $x$ and the map would send the line $t = 0$ to a point of $x'$), so $$d = -\frac{v}{c^2}\,e .$$

Step 2, use both in (ii). Left: $v^2a^2 - c^2a^2 = -c^2a^2(1 - \beta^2)$. Right: $-c^2(e^2 - v^2e^2/c^2) = -c^2e^2(1 - \beta^2)$. So, for $|v| \ne c$, $$a^2 = e^2 .$$

Step 3, pick the branch. At $v = 0$ the map must be the identity, and $a, e$ depend continuously on $v$ and never vanish, so $a = e \gt 0$. Call it $A(v)$.

Step 4, the family. $$x' = A(v)\,(x - vt),\qquad t' = A(v)\Big(t - \frac{v x}{c^2}\Big),$$ and then $P = A^2(1 - \beta^2)\,\big(\partial_{t'}^2 - c^2\partial_{x'}^2\big)$.

checks

Dimensions: $vx/c^2$ is $\mathrm{m\,s^{-1}\cdot m/m^2\,s^{-2}} = \mathrm s$ ✓.
Limit $c\to\infty$: $t' = At$, $x' = A(x - vt)$; with $A = 1$ this is Galileo ✓.
Light: $x = ct$ gives $x' = A(c - v)t$ and $t' = A(1 - \beta)t$, so $x' = ct'$ ✓. The invariant speed is $c$, in both directions.

One function, $A(v)$, is still free — and $P$ cannot fix it, since a dilation $(x,t)\mapsto(sx, st)$ multiplies $P$ by $s^{-2}$ and is a symmetry.

A name clash. $A(v)$ is a number, the common factor of the map; it is not the symbol matrix $A$ of section 2, which reappears only in the notes of later slides, always named as the symbol matrix.

Step 4's operator in full. With $a = e = A$, $b = -vA$, $d = -vA/c^2$, the coefficient formula gives: $e^2 - c^2d^2 = A^2(1 - v^2/c^2)$; $be - c^2ad = -vA^2 + vA^2 = 0$; $b^2 - c^2a^2 = A^2(v^2 - c^2) = -c^2A^2(1 - v^2/c^2)$. So $P = A^2(1 - \beta^2)(\partial_{t'}^2 - c^2\partial_{x'}^2)$, with $\lambda = A^2(1 - \beta^2)$. $\square$

The other branch. $a = -e$ gives $x' = a(x - vt)$, $t' = -a(t - vx/c^2)$: a time reversal composed with a member of the family. It is a symmetry of $P$, but not one reachable continuously from the identity, and a change of frame does not reverse the direction of time. Similarly $|v| = c$ makes the map singular: $\det = ae - bd = A^2(1 - \beta^2) = 0$.

What about $|v| \gt c$? In one space dimension nothing yet forbids it: then $\lambda = A^2(1 - \beta^2) \lt 0$ and $P \mapsto -|\lambda|P'$, still form-invariant up to a factor. The map swaps the roles of the two characteristic families' orientations, and in 1+1 dimensions $x$ and $ct$ really are interchangeable for the operator. The next slide shows the transverse directions forbid it.

History. Woldemar Voigt, 1887, looking for the transformations that preserve the wave equation for the purpose of treating the Doppler effect, found exactly this family with $A = 1$ — and the transverse scaling of the next slide. It went largely unnoticed for twenty years; Lorentz acknowledged it in The Theory of Electrons (1909).

25 · the right substitution

The solved map, drawn: light runs through the corners

The new coordinates drawn in the old frame, $v = c/2$, with $A = \gamma = 1.155$. Solid: lines of constant $x'$; dashed: lines of constant $ct'$, one unit apart. The light lines $x = \pm ct$ pass exactly through the grid's corners where $x' = \pm ct'$ (dots).

Galileo's map tilts only the lines of constant $x'$; its lines of constant $t'$ stay horizontal. This one tilts both, by the same angle, towards the light line: the $x'$-axis ($t' = 0$, the line $t = vx/c^2$) is the mirror image of the $t'$-axis ($x = vt$) in $x = ct$.

Condition (i) is that mirror symmetry. Condition (ii), $a = e$, spaces the two families alike, so each light line runs along the diagonals of the cells: each family of characteristics goes to itself.

the price

The cross term died because $t'$ now depends on $x$. The new clocks are set differently at different places: "at the same time" is frame-dependent. That is the price, and it is the whole of it.

What to look for in the figure. Follow a dashed line, a line of constant $t'$: it climbs to the right, so events that the new frame calls simultaneous happen later, in the old frame, the further they are in the direction of motion. That slope, $v/c^2$ in $t$ per unit $x$, is the term $-vx/c^2$ in $t'$. Then follow the light line $x = ct$: it crosses the grid exactly through the points $(x', ct') = (1, 1), (2, 2), \dots$ — the light moves one unit of $x'$ per unit of $ct'$, at $c$, in the new frame too.

Which condition does what. With $b = -va$, condition (i) gives $d = -ve/c^2$, so the map is $x' = a(x - vt)$, $t' = e(t - vx/c^2)$: the lines of constant $t'$ have slope $v/c^2$, the mirror image of the worldline's. On the light line $x = ct$ this map gives $x' = a(c - v)t$ and $ct' = e(c - v)t$, so $x' = (a/e)\,ct'$: the light line runs through the corners $(k, k)$ of the grid only if $a = e$, which is condition (ii) after worked example 4's step 2. The grid's overall spacing in the old frame depends on $A$; its directions and the equal spacing of the two families do not, and those are all the wave operator fixes.

25 · the right substitution

Three dimensions: transverse lengths, and $|v| \lt c$

In space, $P = \partial_t^2 - c^2(\partial_x^2 + \partial_y^2 + \partial_z^2)$. Keep the boost along $x$, and let $y' = sy$, $z' = sz$ with a transverse scale factor $s \gt 0$ (isotropy about the $x$-axis makes the two equal). The $(x,t)$ part is as before, and the transverse part becomes $-c^2s^2(\partial_{y'}^2 + \partial_{z'}^2)$. Form-invariance now needs

$$A^2(1 - \beta^2) = \lambda = s^2 .$$

theorem · the family

The boost-shaped linear maps preserving $P$ up to a factor are exactly $$x' = s\gamma(x - vt),\ \ t' = s\gamma\Big(t - \frac{vx}{c^2}\Big),\ \ y' = sy,\ \ z' = sz,$$ with $\gamma = (1 - \beta^2)^{-1/2}$, $|v| \lt c$, and $s = s(v) \gt 0$ free.

proof

$\lambda = s^2 \gt 0$ forces $1 - \beta^2 \gt 0$, i.e. $|v| \lt c$; then $A = s/\sqrt{1 - \beta^2} = s\gamma$ (positive root, as before). Conversely, each such map multiplies $P$ by $s^2$. $\square$

checks

Dimensions: $\gamma$ and $s$ are pure numbers ✓. Limit $v\to0$: $\gamma\to1$, a pure dilation by $s(0)$, which must be $1$ ✓. Special case: $s = 1/\gamma$ gives $A = 1$, worked example 4's Voigt case ✓.

The speed limit $|v| \lt c$ has appeared, and nothing about particles or energy was used: it is a statement about which linear maps a hyperbolic operator with three space directions allows.

Full proof of the transverse condition. With $y' = sy$, $\partial_y = s\,\partial_{y'}$, so $-c^2\partial_y^2 = -c^2s^2\partial_{y'}^2$, and likewise for $z$; the $(x,t)$ part is worked example 4's, $A^2(1-\beta^2)(\partial_{t'}^2 - c^2\partial_{x'}^2)$. The total is $\lambda(\partial_{t'}^2 - c^2\nabla'^2)$ for a single $\lambda$ exactly when $A^2(1 - \beta^2) = s^2$. $\square$

The general reason, by Sylvester. Any linear symmetry of $P$ up to a factor $\lambda$ satisfies $JAJ^{\mathsf T} = \lambda A$ for the symbol matrix $A = \mathrm{diag}(1, -c^2, -c^2, -c^2)$ (here $A$ is the matrix of section 2, not the factor $A(v)$). If $\lambda \lt 0$, the right side has one negative and three positive eigenvalues, the left side (by Sylvester's law of inertia, unit 18) one positive and three negative. They cannot be equal. So $\lambda \gt 0$ for every linear symmetry of the 3+1 wave operator, boost-shaped or not. In 1+1 the signature $(1,1)$ is its own negative, which is why the one-dimensional computation could not see the speed limit.

Why isotropy gives one $s$ for $y$ and $z$. A rotation about the $x$-axis commutes with the boost along $x$ and mixes $y$ with $z$; if the two scale factors differed, conjugating by a quarter-turn would produce a second, different map for the same physical boost. The boost must also not tilt $y$ into $x$ or $t$: that is what "boost-shaped" means, and a reflection $y\mapsto -y$ commuting with the boost forces it.

25 · the right substitution

Fixing $s$: reciprocity and isotropy

reciprocity

Write $L_s(v)$ for the map of the family at velocity $v$, with its scale $s = s(v)$. The principle of relativity, once more: the old frame's origin moves at $-v$ in the new frame (check: $x = 0$ gives $x' = -vt'$), so the inverse map must be the same rule with $-v$: $$L_s(v)^{-1} = L_s(-v).$$

isotropy

A half-turn about the $z$-axis, $(x,y)\mapsto(-x,-y)$, turns the boost at $v$ into the boost at $-v$ without changing its transverse lengths: $s(-v) = s(v)$.

theorem · $s = 1$

Write $L_s(v) = s(v)\,\Lambda(v)$, $\Lambda$ the $s = 1$ map. Direct multiplication gives $\Lambda(v)\Lambda(-v) = I$. Reciprocity then says $s(v)^{-1}\Lambda(-v) = s(-v)\Lambda(-v)$, so $s(v)s(-v) = 1$; isotropy makes it $s(v)^2 = 1$; and $s \gt 0$. $\square$

Voigt fails here

Voigt's transformation, $s = 1/\gamma$, preserves $P$ up to the factor $1/\gamma^2$ — but its inverse is $\gamma\Lambda(-v)$, not $\gamma^{-1}\Lambda(-v)$. The two frames would disagree about which of them had the shrunken rulers.

$\Lambda(v)\Lambda(-v) = I$ in full. In the order $(x, t)$, $\Lambda(v) = \gamma\begin{pmatrix}1 & -v\\ -v/c^2 & 1\end{pmatrix}$ and $\Lambda(-v) = \gamma\begin{pmatrix}1 & v\\ v/c^2 & 1\end{pmatrix}$. The product is $\gamma^2\begin{pmatrix}1 - v^2/c^2 & v - v\\ -v/c^2 + v/c^2 & 1 - v^2/c^2\end{pmatrix} = \gamma^2(1 - \beta^2)I = I$. The transverse block is the identity. $\square$

Why the half-turn and not a mirror. Isotropy is invariance under rotations, $SO(3)$ (unit 03). A reflection $x \mapsto -x$ alone is not a rotation, and physics does distinguish mirror images (the weak interaction does). The half-turn $R$ about $z$ reverses $x$ and $y$ together; conjugating: $R\,L_s(v)R^{-1}$ has $x' = s\gamma(x + vt)$, $t' = s\gamma(t + vx/c^2)$, $y' = sy$ — the boost at $-v$, with the scale $s(v)$. By isotropy it must be $L_s(-v)$, so $s(-v) = s(v)$.

History. This is Poincaré's argument of 1905 ("Sur la dynamique de l'électron", Comptes Rendus 140, 1504, and the long version in Rendiconti del Circolo Matematico di Palermo 21 (1906) 129). Lorentz in 1904 had the family with an undetermined factor $\ell$ — this slide's $s$ — and set $\ell = 1$ by an argument about the electron. Poincaré observed that the transformations must form a group, that the inverse must be the same transformation with $-v$, and that this forces $\ell = 1$. Unit 26 derives the whole transformation this way, without Maxwell, and needs this unit only for the value of the invariant speed.

25 · the right substitution

The Lorentz boost

the Lorentz boost along $x$ $$\boxed{\begin{aligned}x' &= \gamma(x - vt), & y' &= y,\\ t' &= \gamma\Big(t - \frac{vx}{c^2}\Big), & z' &= z\end{aligned}}$$

with the Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$, $|v| \lt c$. It carries $P$ to $P'$ exactly: $\lambda = s^2 = 1$.

checks

Dimensions: $\gamma$ is a number, $vx/c^2$ a time ✓.
Limit $c \to \infty$: $\gamma \to 1$, $vx/c^2 \to 0$: Galileo ✓.
Size at $30\ \mathrm{km\,s^{-1}}$: $\gamma - 1 = 5.0\times10^{-9}$; $\ vx/c^2$ over $x = 1$ km is $3.3\times10^{-10}$ s.
Light: $x = \pm ct \iff x' = \pm ct'$ ✓.

on the running example

$\partial_x = \gamma\big(\partial_{x'} - \tfrac{v}{c^2}\partial_{t'}\big)$, $\ \partial_t = \gamma\big(\partial_{t'} - v\,\partial_{x'}\big)$, and

$$\begin{aligned}\partial_t^2 - c^2\partial_x^2 &= \gamma^2\big[(1 - \beta^2)\partial_{t'}^2 + 0\cdot\partial_{x'}\partial_{t'}\\ &\qquad - c^2(1 - \beta^2)\partial_{x'}^2\big]\\ &= \partial_{t'}^2 - c^2\partial_{x'}^2 .\end{aligned}$$

Worked example 1's $-2v\,\partial_{x'}\partial_{t'}$ is cancelled by $+2v\,\partial_{x'}\partial_{t'}$ from the $-\tfrac{v}{c^2}\partial_{t'}$ in $\partial_x$ — the term Galileo did not have.

The cancellation, term by term. $\partial_t^2 = \gamma^2(\partial_{t'}^2 - 2v\,\partial_{x'}\partial_{t'} + v^2\partial_{x'}^2)$ and $c^2\partial_x^2 = \gamma^2(c^2\partial_{x'}^2 - 2v\,\partial_{x'}\partial_{t'} + \tfrac{v^2}{c^2}\partial_{t'}^2)$. Subtracting, the mixed terms cancel exactly, and what is left is $\gamma^2(1 - v^2/c^2)(\partial_{t'}^2 - c^2\partial_{x'}^2)$, with $\gamma^2(1 - v^2/c^2) = 1$. $\square$

The numbers. For $v = 3.0\times10^4\ \mathrm{m\,s^{-1}}$ and $c = 2.998\times10^8\ \mathrm{m\,s^{-1}}$, $\beta^2 = 1.001\times10^{-8}$ and $\gamma - 1 = \tfrac12\beta^2 + \dots = 5.007\times10^{-9}$. The desynchronisation $vx/c^2$ for $x = 1000$ m is $3\times10^4\times10^3/(8.988\times10^{16}) = 3.34\times10^{-10}$ s. Tiny — but it is first order in $v/c$ (times $x/c$), while contraction and dilation are second order. The relativity of simultaneity is the largest effect, which is why unit 26 treats it as the content and the others as corollaries.

Only one thing was not derived: that the family should be continuous in $v$ and pass through the identity. That is the statement that a frame moving at $v \to 0$ is the same frame, which no one will dispute.

25 · the right substitution

The postulate that became a conclusion

what was assumed

Maxwell's vacuum equations hold in one inertial frame (so each component obeys $Pu = 0$);
the principle of relativity: they hold in the same form in every inertial frame;
frames are related by linear maps, continuous in $v$, with the origin moving at $v$;
isotropy of space.

what was derived

The Lorentz boost; the speed limit $|v| \lt c$; that $c$ is the same in every frame; and its value, $1/\sqrt{\varepsilon_0\mu_0}$, read off the operator.

Einstein's 1905 postulates were the principle of relativity and the constancy of the speed of light. Here the second is a theorem: the principle of relativity, applied to Maxwell's equations as they were already written, has no room for any other invariant speed.

honestly

"Maxwell holds in every frame with the same $\varepsilon_0, \mu_0$" contains the constancy of $c$, so no magic has happened: the postulate has been located inside a principle everyone already believed and an equation everyone already trusted. Unit 26 derives the same group with the invariant speed as an unknown, and cites this slide for its value.

Why this matters for what follows. Unit 26 will derive the Lorentz group from homogeneity, isotropy and the group axioms, and find that they allow exactly one free parameter, an invariant speed $K$, with Galileo as the case $K = \infty$. Something empirical must then fix $K$. With this unit behind it, unit 26 can say "the invariant speed is $c$", because Maxwell's operator is invariant only under the group with $K = c$. Without it, $K = c$ is an assumption about light.

What Einstein's paper did. A. Einstein, "Zur Elektrodynamik bewegter Körper", Annalen der Physik 17 (1905) 891. §1–3 derive the transformation from the two postulates plus an operational definition of simultaneity by light signals; §6 then shows that Maxwell's equations are covariant under it, with $\vec E$ and $\vec B$ mixing as unit 28 derives. This deck runs the logic in the opposite direction: covariance of the wave operator first, the transformation as the answer. The mathematics is the same; the order of discovery was Einstein's.

A caution. The derivation used only the wave operator, which each field component obeys. That shows the Lorentz boost is necessary for Maxwell covariance. That it is sufficient — that with the right $\rho$ the full first-order system is covariant — is a separate fact, proved in unit 28 once $\vec E$ and $\vec B$ are one tensor. Until then it is cited.

section 5

The patches form a group

Lorentz's local time and FitzGerald's contraction were two patches; Poincaré saw they were one group. Its composition law explains Fizeau, its larger cousin is the conformal group, and its free function is fixed by one experiment.

25 · a group

Lorentz's patches, recognised

local time · Lorentz, 1895

In a frame moving through the ether at $v$, use the Galilean position $x' = x - vt$ but the clock reading $$t_{\mathrm{loc}} = t - \frac{vx}{c^2}.$$ Lorentz called it a mathematical aid, not the time.

It is worked example 4 with $A = 1$: to all orders in 1+1 dimensions it kills the cross term, and $P = (1 - \beta^2)(\partial_{t_{\mathrm{loc}}}^2 - c^2\partial_{x'}^2)$. The leftover factor is harmless in 1+1; in 3+1 it clashes with the transverse terms at order $\beta^2$.

patch	in the family	order in $\beta$
local time, 1895	$t' \ni -vx/c^2$	first
contraction, 1889–92	$x' = \gamma(x - vt)$	second
time dilation, Larmor 1897–1900	$t' = \gamma(\cdots)$	second
$\ell = 1$, Poincaré 1905	$s = 1$	—

All four rows are one map. The next slide shows the contraction falling out of it.

Full check of the local-time claim. $x' = x - vt$, $t_{\mathrm{loc}} = t - vx/c^2$ is worked example 4's family with $A = 1$; there, $P = A^2(1 - \beta^2)(\partial_{t'}^2 - c^2\partial_{x'}^2)$, so the factor is $1 - \beta^2$. In 3+1 dimensions with $y' = y$ the transverse part stays $-c^2(\partial_{y'}^2 + \partial_{z'}^2)$ with factor $1$, and the three factors ($1 - \beta^2$, $1 - \beta^2$, $1$) disagree at order $\beta^2$ — precisely what Michelson–Morley measures. (Lorentz's own local time used the moving-frame coordinate, $t - vx'/c^2$; the two agree to first order in $\beta$, which is all his 1895 "theorem of corresponding states" claimed.)

History. H. A. Lorentz, Versuch einer Theorie der electrischen und optischen Erscheinungen in bewegten Körpern (Leiden, 1895); J. Larmor, Aether and Matter (Cambridge, 1900), which has the full transformation including the factor $\gamma$ on time; H. A. Lorentz, Proc. Roy. Neth. Acad. Arts Sci. 6 (1904) 809. None of them regarded $t'$ as what a clock in the moving frame reads. Einstein's contribution was to say that it is.

25 · a group

Contraction, derived: the rod in spacetime

A rod at rest in the new frame, drawn in the old one with $v = 0.6c$, $\gamma = 1.25$. Its ends sweep out the strip between $x' = 0$ and $x' = L$. Cut across at one old time and the strip is $L/\gamma = 0.8L$ wide; cut along a line of constant $t'$ and it is the rod's own length $L$.

one map

A rod at rest in the new frame, $x' \in [0, L]$, occupies $x \in [vt, vt + L/\gamma]$ at old time $t$: contraction by $1/\gamma$, derived rather than hypothesised.

FitzGerald and Lorentz needed the contraction as a separate hypothesis about matter. Here it is a statement about which events count as "the ends, at the same time" — and the two frames answer differently.

The contraction, in full. At a fixed old time $t$, $x' = \gamma(x - vt)$ gives $x = vt + x'/\gamma$, so the endpoints $x' = 0$ and $x' = L$ sit at $vt$ and $vt + L/\gamma$. The rod's length in the old frame, measured by marking both ends at the same old time, is $L/\gamma$. The italicised phrase is where simultaneity enters, and unit 26's pole-and-barn problem is about nothing else.

What to look for in the figure. The two cuts are both "the rod, now", for two observers. The horizontal cut takes both ends at one old time; the slanted one at one new time, along the slope $v/c^2$ of the solved-map slide. They meet the strip at different events, and the strip is narrower along the horizontal. The slanted cut looks longer than $0.8L$ on the page but is not $L$ in page units: lengths along a slanted line are not read off a spacetime diagram with a ruler, which is the subject of unit 26.

25 · a group

Boosts compose to boosts

theorem · the velocity composition law

$\Lambda(u)\,\Lambda(v) = \Lambda(w)$ with $$w = \frac{u + v}{1 + uv/c^2}.$$ With $\Lambda(v)^{-1} = \Lambda(-v)$ and $\Lambda(0) = I$, the boosts along $x$ form a group.

proof

In the order $(x, t)$, $\Lambda(v) = \gamma_v\begin{pmatrix}1 & -v\\ -v/c^2 & 1\end{pmatrix}$. The product has diagonal entries $\gamma_u\gamma_v(1 + uv/c^2)$ and off-diagonal entries $-\gamma_u\gamma_v(u + v)$ and $-\gamma_u\gamma_v(u + v)/c^2$ — the shape of $\Lambda(w)$, with prefactor $\gamma_u\gamma_v(1 + uv/c^2) = \gamma_w$ (notes). $\square$

Together with rotations, these generate the Lorentz group (Poincaré's name, 1905), and by the lemma of section 1 it is a subgroup of the symmetry group, among linear maps, of Maxwell's vacuum equations — the whole of it, once unit 28 supplies the $\rho$ that makes the full system covariant (cited). Adding the four translations gives the ten changes of inertial frame. Unit 26 parametrises it properly.

Full proof. $\Lambda(u)\Lambda(v) = \gamma_u\gamma_v\begin{pmatrix}1 & -u\\ -u/c^2 & 1\end{pmatrix}\begin{pmatrix}1 & -v\\ -v/c^2 & 1\end{pmatrix} = \gamma_u\gamma_v\begin{pmatrix}1 + uv/c^2 & -(u + v)\\ -(u+v)/c^2 & 1 + uv/c^2\end{pmatrix}$. Factor out $K = \gamma_u\gamma_v(1 + uv/c^2)$: the matrix becomes $K\begin{pmatrix}1 & -w\\ -w/c^2 & 1\end{pmatrix}$ with $w = (u+v)/(1 + uv/c^2)$. It remains to show $K = \gamma_w$, i.e. $K^2(1 - w^2/c^2) = 1$. Compute $(1 + uv/c^2)^2 - (u + v)^2/c^2 = 1 + u^2v^2/c^4 - u^2/c^2 - v^2/c^2 = (1 - u^2/c^2)(1 - v^2/c^2)$. So $K^2(1 - w^2/c^2) = \gamma_u^2\gamma_v^2\big[(1 + uv/c^2)^2 - (u+v)^2/c^2\big] = \gamma_u^2\gamma_v^2(1 - u^2/c^2)(1 - v^2/c^2) = 1$. $\square$ That also shows $|w| \lt c$ when $|u|, |v| \lt c$, so the product stays in the family.

Which way round. $\Lambda(v)$ takes coordinates of frame $S$ to those of $S'$, moving at $v$ in $S$; $\Lambda(u)$ then takes $S'$ to $S''$, moving at $u$ in $S'$. So $S''$ moves at $w$ in $S$. For collinear boosts the order does not matter — the formula is symmetric in $u, v$ — but for boosts in different directions it does, and the composite includes a rotation (the Thomas–Wigner rotation).

Where the number $1 + uv/c^2$ comes from. It is the cross term again: the $t'$ of the first boost depends on $x$, so the second boost's $x'$ inherits a piece of the first's time. Rapidity, $\tanh\phi = v/c$, turns the composition law into $\phi_w = \phi_u + \phi_v$; unit 26 proves that and uses it.

25 · a group

Composition, plotted: nothing passes $c$

checks

Dimensions: $uv/c^2$ is a pure number, $w$ a speed ✓.
Limit $c\to\infty$: $w = u + v$, Galileo's addition ✓.
Light: $u = c$ gives $w = (c + v)/(1 + v/c) = c$ for every $v$ ✓.
Number: $u = v = c/2$ gives $w = 0.8c$ ✓.

Galileo's sum crosses $c$ as soon as $u + v \gt c$. The composition law bends every curve over so that it reaches $c$ exactly at $u = c$ — light composed with anything is still light — and never above.

$w = (u + v)/(1 + uv/c^2)$ against $u$, for $v = c/2$ and $v = 0.9c$. The dotted line is Galileo's $u + c/2$; the open circle is where it says $c/2$ plus $c/2$ is $c$, and the dot is the composition law's $0.8c$.

25 · a group · example

Worked: Fizeau's drag, from the composition law

worked example 5 · Fresnel's coefficient, derived

Step 1, the rest frame of the water. There light moves at $u = c/n$: the water is at rest, and its refractive index is a property of the water at rest.

Step 2, the lab. The water moves at $w$ in the lab, so compose: $$u_{\mathrm{lab}} = \frac{c/n + w}{1 + w/(nc)} .$$

Step 3, expand in $w/c$. $\frac{1}{1 + w/nc} = 1 - \frac{w}{nc} + O(w^2/c^2)$, so $$u_{\mathrm{lab}} = \frac cn + w - \frac{w}{n^2} + O\Big(\frac{w^2}{c}\Big) = \frac cn + \Big(1 - \frac1{n^2}\Big)w + \dots$$

Step 4, read off. $f = 1 - 1/n^2$: Fresnel's coefficient, $0.437$ for water, with no ether in sight.

checks

Dimensions: $w/(nc)$ is a pure number ✓.
Limit $n\to1$: $f\to0$ — vacuum is not dragged, and $u_{\mathrm{lab}} = c$ exactly ✓.
Limit $n\to\infty$: $f\to1$ — light stuck in the medium goes where it goes ✓.
Size: at $w = 7.059\ \mathrm{m\,s^{-1}}$, the first-order drag is $3.086\ \mathrm{m\,s^{-1}}$ and the next term $-5.5\times10^{-8}\ \mathrm{m\,s^{-1}}$.

The next slide sets the derived coefficient against the two ether theories it replaces.

Step 3 with the error term. $(c/n + w)(1 - w/nc + w^2/n^2c^2 - \dots) = c/n + w - w/n^2 - w^2/(nc) + w^2/(n^3c) + O(w^3/c^2)$. The second-order term is $-\frac{w^2}{nc}\big(1 - \frac1{n^2}\big)$; for $n = 1.333$ and $w = 7.059\ \mathrm{m\,s^{-1}}$ (the water speed used in problem P9) it is $-5.45\times10^{-8}\ \mathrm{m\,s^{-1}}$, while the first-order drag is $0.4372\times7.059 = 3.086\ \mathrm{m\,s^{-1}}$. Fizeau measured the first-order term; nobody has needed the second.

Why the refractive index belongs to the rest frame. Units 06 and 20 treat $n$ as the response of the medium's charges to a wave, computed with the medium at rest; that is the only frame in which it is a property of the water alone. The composition law is then the only way to get the lab speed. Dispersion adds a small correction (Lorentz, 1895: the light's frequency in the water's frame is Doppler shifted, and $n$ depends on it), which turns the coefficient into $1 - 1/n^2 - (\lambda/n)\,\mathrm dn/\mathrm d\lambda$; Zeeman measured the extra term in 1914–15. The slide ignores it.

History. Max von Laue, 1907, "Die Mitführung des Lichtes durch bewegte Körper nach dem Relativitätsprinzip", Annalen der Physik 23, 989: the paper that made Fizeau's experiment a consequence of relativity, two years after Einstein. It is one of the few experiments that confirms relativity at first order in $v/c$.

25 · a group

Fresnel's coefficient, plotted

A coefficient that took an ether, a partial drag and sixty years of argument is two lines of the group law. It is Galileo's addition, $u + w$, that was wrong for light in water — Hertz's $f = 1$.

three theories, one plot

A stationary ether says light in moving water ignores the water: $f = 0$. A dragged ether, or Hertz's equations, say the water carries it fully: $f = 1$. The composition law says $f = 1 - 1/n^2$ — between the two, near $0$ for thin media and near $1$ for dense ones.

The drag coefficient against refractive index. Water sits on the curve: $1 - 1/1.333^2 = 0.437$, against Michelson and Morley's $0.434 \pm 0.02$ (bar). Both ether lines miss it by far more than the error.

25 · a group

The full symmetry group of the wave operator

theorem · linear symmetries, 1+1

In units with $c = 1$, the invertible linear maps under which $P = \partial_t^2 - \partial_x^2$ is invariant up to a factor are exactly $\mu M$ and $\mu MS$ with $\mu \gt 0$, $M$ in the group $O(1,1)$ of maps preserving $t^2 - x^2$, and $S$ the swap $(t,x)\mapsto(x,t)$.

$O(1,1)$ has four pieces: the boosts, and the boosts composed with $x\mapsto-x$, with $t\mapsto -t$, and with both. The dilations $\mu$ are what $P$ cannot see, and why $s$ needed reciprocity.

The swap $S$ exchanges space and time; it is a symmetry of the 1+1 operator only, and the transverse slide excluded it in 3+1. Allow nonlinear maps and the group grows enormously — the next slide.

Full proof of the 1+1 linear theorem. With $c = 1$ the symbol matrix is $A = \mathrm{diag}(1, -1)$ — the deck keeps the name $A$ for it throughout, since $\xi$ and $\eta$ are already the null coordinates — and a linear map with Jacobian $J$ is a symmetry up to $\lambda$ iff $JAJ^{\mathsf T} = \lambda A$ (the notes of section 2's theorem slide show this is also necessary for the solution set to be carried onto itself). Taking determinants, $(\det J)^2\cdot(-1) = \lambda^2\cdot(-1)$, so $|\det J| = |\lambda|$. If $\lambda \gt 0$, put $M = J/\sqrt\lambda$: then $MAM^{\mathsf T} = A$, i.e. $M \in O(1,1)$ (for any such matrix $MAM^{\mathsf T} = A \iff M^{\mathsf T}AM = A$, since both say $M^{-1} = AM^{\mathsf T}A$). If $\lambda \lt 0$, note $SAS^{\mathsf T} = -A$, so $M = J S/\sqrt{|\lambda|}$ satisfies $MAM^{\mathsf T} = A$ and $J = \sqrt{|\lambda|}\,MS$ (as $S^2 = I$). $\square$

The four pieces of $O(1,1)$. Writing $M = \begin{pmatrix}p & q\\ r & m\end{pmatrix}$, the conditions are $p^2 - q^2 = 1$, $pr - qm = 0$, $r^2 - m^2 = -1$. So $|p| \ge 1$, $|m| \ge 1$, and the signs of $p$ (does time keep its direction?) and of $\det M = \pm1$ are locally constant — four components. The one containing $I$ has $p \ge 1$, $\det M = 1$, and is exactly the boosts $\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}$, $|v| \lt 1$. Unit 26 does the 3+1 version, $O(1,3)$.

25 · a group

The conformal group: relabelling each family of light lines

the conformal group

Allow nonlinear maps. In null coordinates $\xi = x - ct$, $\eta = x + ct$, $P = -4c^2\partial_\xi\partial_\eta$, and for any $C^2$ bijections $F, G$ of $\mathbb R$ with $F', G' \ne 0$, $$\partial_\xi\partial_\eta = F'(\xi)\,G'(\eta)\,\partial_{\xi'}\partial_{\eta'},\quad \xi' = F(\xi),\ \eta' = G(\eta).$$ Solutions go to solutions: in 1+1 the symmetry group is infinite-dimensional.

In 3+1 it is fifteen-dimensional: translations (4), Lorentz (6), dilation (1), and four "special conformal" maps — cited: Bateman and Cunningham, 1909–10. Only the ten Poincaré maps are changes of inertial frame.

Left: light lines $\xi = x - ct$ and $\eta = x + ct$ at even spacing, and a straight worldline. Right: the same lines after $\xi' = F(\xi)$, $\eta' = G(\eta)$ for two increasing functions — still light lines, so still the characteristics, but unevenly spaced; the worldline has bent.

What to look for in the figure. A symmetry of $\partial_\xi\partial_\eta$ only has to send each family of characteristics to itself; it is free to slide the lines of each family about, independently, as long as it keeps their order. That is a function's worth of freedom per family, which is why the group is infinite-dimensional. A linear map can only rescale each family uniformly — the boost multiplies $\xi$ by one constant and $\eta$ by another — and that is why the linear answer was so small. The bent worldline is the reason these maps are not changes of inertial frame: a body moving uniformly in one description would be accelerating in the other.

The functions drawn. $F(\xi) = \xi + 0.3\sin 2.5\xi$ and $G(\eta) = \eta + 0.22\,\eta|\eta|$, in units with $c = 1$; both have positive derivative everywhere, so they are bijections and keep the order of the lines.

The nonlinear symmetries. $\partial_\xi = F'(\xi)\partial_{\xi'}$ and $\partial_\eta = G'(\eta)\partial_{\eta'}$ by the chain rule, and since $F'$ depends only on $\xi$ and $G'$ only on $\eta$, $\partial_\xi\partial_\eta = F'G'\,\partial_{\xi'}\partial_{\eta'}$ with no first-order terms. So $u$ solves $\partial_\xi\partial_\eta u = 0$ iff the transported $u'$ solves $\partial_{\xi'}\partial_{\eta'}u' = 0$ — consistent with d'Alembert: $u = f(\xi) + g(\eta)$ becomes $f(F^{-1}(\xi')) + g(G^{-1}(\eta'))$. These are symmetries, but $\lambda = F'G'$ depends on the point, and they do not send straight uniform motions to straight uniform motions, so they are not changes of inertial frame.

Citation, 3+1. H. Bateman, "The transformation of the electrodynamical equations", Proc. London Math. Soc. (2) 8 (1910) 223; E. Cunningham, "The principle of relativity in electrodynamics and an extension thereof", ibid. 77. They showed Maxwell's equations are covariant under the 15-dimensional conformal group. For the wave operator the field must be multiplied by a conformal weight; Liouville's theorem (every conformal map of $\mathbb R^n$, $n \ge 3$, is a composition of similarities and inversions) is what bounds the group at fifteen dimensions. A proof is in Schottenloher, A Mathematical Introduction to Conformal Field Theory, ch. 1–2, which is written in the language of unit 31's manifolds: come back to it after that unit.

25 · a group

Which experiment pins which

Make the ether theory as general as possible — a test theory. In the frame where light is isotropic, an apparatus moving at $v$ has its lengths along and across the motion scaled by $\ell_\parallel(v), \ell_\perp(v)$, and its clocks run at rate $\tau(v)$. Round trips then give:

experiment	measures	forces
Michelson–Morley, 1887	equal arms, rotated	$\ell_\parallel/\ell_\perp = 1/\gamma$
Kennedy–Thorndike experiment, 1932	unequal arms, watched for months	$\tau\,\ell_\perp = 1/\gamma$
Ives–Stilwell experiment, 1938	moving atoms' frequency	$\tau = 1/\gamma$

on the family

A rod along $x'$ has $\ell_\parallel = 1/(s\gamma)$; across, $\ell_\perp = 1/s$; a clock at $x' = 0$ reads $t' = st/\gamma$, so $\tau = s/\gamma$. Then $\ell_\parallel/\ell_\perp = 1/\gamma$ ✓ and $\tau\ell_\perp = 1/\gamma$ ✓ for every $s$, and $\tau = 1/\gamma$ only for $s = 1$.

The next slide plots the Ives–Stilwell prediction for Lorentz, for Voigt, and for an ether with absolute time.

Full derivation of the three conditions. Work in the frame where light moves at $c$ in every direction. An arm of rest length $L$ along the motion has length $L\ell_\parallel$ there, so its round-trip time is $\frac{2L\ell_\parallel}{c}\gamma^2$ (worked example 3); across the motion, $\frac{2L\ell_\perp}{c}\gamma$. MM: equal arms, and the fringe position must not change on rotation, so $\ell_\parallel\gamma^2 = \ell_\perp\gamma$, i.e. $\ell_\parallel/\ell_\perp = 1/\gamma$. KT: with MM satisfied, both arms take $\frac{2L_i\ell_\perp\gamma}{c}$, and the lab measures this with its own clocks as $\tau\cdot\frac{2L_i\ell_\perp\gamma}{c}$. Kennedy and Thorndike (arms of deliberately unequal length) watched for a change as the Earth's velocity changed over the year, and found none; so $\tau\ell_\perp\gamma$ is independent of $v$, and equal to its value $1$ at $v = 0$. IS: a hydrogen atom moving at $\beta c$ emits at rate $\tau\nu_0$; seen forwards and backwards its wavelengths are $\lambda_0(1 \mp \beta)/\tau$, whose mean is $\lambda_0/\tau$. Ives and Stilwell measured the mean shift, $\lambda_0(1/\tau - 1) = \lambda_0(\gamma - 1) \approx \tfrac12\beta^2\lambda_0$, so $\tau = 1/\gamma$.

MM alone is consistent with several laws. Lorentz ($\ell_\parallel = 1/\gamma$, $\ell_\perp = 1$, $\tau = 1/\gamma$); FitzGerald–Lorentz 1892 with absolute time ($1/\gamma$, $1$, $1$) — killed by KT, since $\tau\ell_\perp = 1$; Voigt ($1$, $\gamma$, $1/\gamma^2$) — passes KT, killed by IS. Problem P6 does this in full.

Citations. H. P. Robertson, "Postulate versus observation in the special theory of relativity", Rev. Mod. Phys. 21 (1949) 378, first showed these three experiments fix the transformation; R. Mansouri and R. U. Sexl, Gen. Rel. Grav. 8 (1977) 497, is the test theory used today, with synchronisation conventions treated properly (this slide uses only round trips, where synchronisation cannot enter). R. J. Kennedy and E. M. Thorndike, Phys. Rev. 42 (1932) 400; H. E. Ives and G. R. Stilwell, J. Opt. Soc. Am. 28 (1938) 215. The slide's IS analysis assumes the lab is at rest in the isotropic frame; the full treatment keeps the lab's own motion and gets the same condition.

25 · a group

Ives–Stilwell: the clock that sees $s$

the moral

Both light-only experiments are blind to $s$, because the wave operator is: dilations are its symmetries. The one experiment with a material clock — time dilation of moving atoms — measures it. Voigt's $s = 1/\gamma$ predicts twice the Ives–Stilwell shift.

An atom moving at $\beta c$ emits at the rate $\tau\nu_0$, and seen forwards and backwards its line sits at $\lambda_0(1 \mp \beta)/\tau$. The mean of the two is shifted by $\lambda_0(1/\tau - 1)$: no first-order Doppler, only the clock.

The mean shift of the H$_\beta$ line, $\lambda_0 = 486.13$ nm, from atoms at speed $\beta c$, for three clock rates: Lorentz's $\tau = 1/\gamma$, Voigt's $\tau = 1/\gamma^2$, and absolute time, $\tau = 1$. At $\beta = 0.005$ (dots) the first two differ by a factor of two. Ives and Stilwell found the Lorentz value.

Numbers. For H$_\beta$, $\lambda_0 = 486.13$ nm, at $\beta = 0.005$: the Lorentz prediction is a mean shift of $0.00608$ nm; Voigt's ($\tau = 1/\gamma^2$) is $0.01215$ nm, twice as large; the Lorentz–FitzGerald ether without time dilation ($\tau = 1$) predicts none. Ives and Stilwell's result agreed with the first.

What to look for in the plot. Both curves grow like $\beta^2$: the mean of a red and a blue shift cancels the first-order Doppler effect, and what survives is second order, $\lambda_0(\gamma - 1) \approx \tfrac12\beta^2\lambda_0$ for Lorentz and $\lambda_0(\gamma^2 - 1) \approx \beta^2\lambda_0$ for Voigt. The two light-only experiments of the previous slide cannot distinguish these curves, because both members satisfy their conditions. A spectrometer resolving a few thousandths of a nanometre can.

25 · a group

The running example, closed

	Galileo	Lorentz
map	$x - vt,\ \ t$	$\gamma(x - vt),\ \ \gamma(t - vx/c^2)$
$\partial_t$	$\partial_{t'} - v\partial_{x'}$	$\gamma(\partial_{t'} - v\partial_{x'})$
$\partial_x$	$\partial_{x'}$	$\gamma(\partial_{x'} - \tfrac{v}{c^2}\partial_{t'})$
cross term	$-2v\,\partial_{x'}\partial_{t'}$	$0$
light speeds	$c - v,\ \ c + v$	$c,\ \ c$
composition	$u + v$	$(u+v)/(1 + uv/c^2)$
$c \to \infty$	—	Galileo

what the running example showed

The chain rule, applied to $P$ under $x' = x - vt$, produces a cross term; Maxwell's equations are therefore not Galilean-covariant, and the failure is exactly where the displacement current sits. Demanding that the cross term vanish, that the ratio be right, and that the frames be reciprocal leaves one answer, whose invariant speed is the $c$ written in $P$.

At $30\ \mathrm{km\,s^{-1}}$ the two columns differ by $\gamma - 1 = 5\times10^{-9}$ and by clocks offset $vx/c^2$. Every nineteenth-century experiment lived in that gap.

Reading the table. The only entry of the Lorentz column not visible in the Galilean one is the $-\frac{v}{c^2}\partial_{t'}$ in $\partial_x$: the new time depends on the old position. Every other difference follows from it. Put $c = \infty$ and it disappears, with it the factor $\gamma$, and the right column becomes the left.

The logical structure of the unit, compressed. Section 1 made "the same law in every frame" a precise statement about a group acting on fields. Section 2 showed the Galilean group is not in the symmetry group of Maxwell's operator — at the level of the symbol, so no redefinition of the fields helps. Section 3 showed that each way of keeping Galileo failed experimentally. Section 4 computed the linear symmetry group of the operator and found the Lorentz boosts. Section 5 checked it was a group, derived the one drag coefficient nobody could explain, and matched its one free function to the experiment that measures it.

What this does not prove. That mechanics must also change. Newton's second law is Galilean-covariant and not Lorentz-covariant, and the principle of relativity says the two groups cannot both be right. Unit 27 rebuilds mechanics for the Lorentz group, and finds why colliders exist.

25 · what next

What this unit hands on

26 Minkowski	derives the boost from homogeneity, isotropy and the group axioms with the invariant speed unknown; this unit's wave operator fixes it at $c$. The symbol $\tau^2 - c^2\|\vec\xi\|^2$ becomes the bilinear form $\eta$.
27 dynamics	Newton's law is Galilean-covariant and so must change; four-vectors are what the Lorentz group acts on.
28 field tensor	the $\rho$ this unit could not find: $\vec E$ and $\vec B$ as one antisymmetric tensor, with sufficiency of the Lorentz boost proved.
30 equivalence	a non-inertial frame is a nonlinear change of coordinates — the kind section 4 excluded — and the metric stops being constant.

the three sentences to keep

A law is the same in every frame when the equations are covariant under the group relating the frames, and the symmetries of an equation always form a group. Maxwell's operator acquires a cross term under a Galilean boost that no speed and no field redefinition removes. The linear maps that preserve it, made reciprocal, are the Lorentz boosts, and their invariant speed is the $c$ in the operator.

What was deliberately left out. The Trouton–Noble experiment (1903), the torque on a moving capacitor, whose null result is explained only once unit 28's stress-energy tensor is in hand. The Sagnac effect, which is a rotating frame and therefore unit 30's. The detailed history of the Lorentz–Einstein priority question, on which the careful account is Darrigol, Electrodynamics from Ampère to Einstein (Oxford, 2000).

Reading. Jackson, Classical Electrodynamics (3rd ed.), §11.1–11.3: the situation before 1900, modern tests of the source-independence of $c$, and the Lorentz transformation with its kinematics. Stop at §11.4, where four-velocity begins: that is unit 27. Pais, Subtle is the Lord (Oxford, 1982), ch. 6–8, for the history, including Voigt, Lorentz and Poincaré. For the magnetic limit, Le Bellac and Lévy-Leblond (1973), cited on the slide on what survives.

25 · checkpoint

Done when

checkpoint

On paper and without notes, you can:

show by direct computation with the chain rule that the wave operator is not invariant under a Galilean boost, exhibiting the cross term, and that it is invariant under the Lorentz boost;
derive the Lorentz boost by demanding form-invariance of the wave operator — the two conditions, the transverse condition, and reciprocity — without assuming the constancy of $c$ as a postulate.

Failing it is unambiguous

A Galilean computation with no cross term (the chain rule went the wrong way); a Lorentz computation that leaves one; a derivation that writes "light moves at $c$ in both frames" as a premise; or a family with $s$ left free and called the answer.

26 Minkowski space, where this group becomes the geometry of a quadratic form · 28 after 27, where Maxwell's equations are shown covariant in full.

Try the checkpoint on a case in neither the deck nor the sheet: a boost along the diagonal direction $\hat n = (\hat x + \hat y)/\sqrt2$. The Galilean map $\vec x' = \vec x - v\hat nt$ gives $\partial_t = \partial_{t'} - v\,\hat n\cdot\nabla'$, so the cross term is $-2v\,\partial_{t'}(\hat n\cdot\nabla')$ and the transformed operator is $\partial_{t'}^2 - 2v\,\partial_{t'}\hat n\cdot\nabla' + v^2(\hat n\cdot\nabla')^2 - c^2\nabla'^2$. Then repeat section 4 with coordinates adapted to $\hat n$: the component along $\hat n$ plays the role of $x$, the two perpendicular ones of $y$ and $z$, and the answer is $\vec x' = \vec x + (\gamma - 1)(\hat n\cdot\vec x)\hat n - \gamma v\hat nt$, $t' = \gamma(t - v\,\hat n\cdot\vec x/c^2)$. Check that it reduces to the slide's boost for $\hat n = \hat x$.