Part II · Special Relativity — Chapter 2.1

The Crisis of 1900

Three facts, each of them solid. Any two of them are comfortable together. All three cannot be true at once, and this chapter is the proof.

Tools you'll need  — Chapter 0.3: the binomial series, and what "to leading order in $\beta^{2}$" means quantitatively. Chapter 0.6 §5.2: the multivariable chain rule for a change of coordinates, which is all that §4 is, ground out. Chapter 0.7 §4 (curl), §3 (divergence), §7 (the Laplacian and the second-derivative identities): all four Maxwell equations are written in that language, and §2 manipulates them there. Chapter 0.8 §7.6: the wave equation, its solutions $f(x-vt)$, and the fact that a wave equation's speed is a property of the medium. Chapter 1.1 §3.3: the third law failing for moving charges, and the loose end that failure left.

1 · Galilean relativity, stated precisely

Let's start by looking closely at an idea that has been central to physics for over four hundred years: Galilean relativity. It is that old, and it is correct. To understand the crisis that sparked modern physics, we first need to separate the core principle of relativity from the specific mathematical transformation we use to calculate it. For centuries these two concepts were treated as the exact same thing. Once we pry them apart, the path forward becomes much clearer. Chapter 2.2 is where they finally come apart for good.

1.1 · The transformation

Imagine two observers, each with their own coordinate system, or "frame", of reference. We will call the stationary frame $S$ and the moving frame $S'$. Let's give them both a measuring tape and a clock, and assume they perfectly agree on how to use them.

To keep the maths simple, we'll align their axes and have them start their stopwatches at $t=0$ exactly when they pass each other. After that moment, the $S'$ frame slides away in the positive $x$ direction at a steady speed $v$, as measured in $S$. This setup is known as standard configuration, and it is the baseline for everything in Part II.

Now, suppose something happens — an "event" — and both observers write down where and when it occurred.

• The observer in frame $S$ records the position as $x$ and the time as $t$.
• Meanwhile, the origin of the moving frame $S'$ has shifted forward by a distance $vt$. Because of this shift, the moving observer will measure the event's position as $x-vt$.

We are also going to make a crucial assumption here, and we put it in explicitly so that we can watch it die later: both observers will record the exact same time $t$. This leads us directly to the classical Galilean transformation:

Let's pause and look at that final equation, $t'=t$. It might look like simple bookkeeping, but it actually contains a massive physical claim: there is a single, universal clock ticking at the exact same rate for everyone in the universe, regardless of how fast they are moving, so that two events either are simultaneous or are not, with no reference to who is asking. Isaac Newton built his mechanics on this idea of absolute time, and he stated it in exactly those terms: absolute time, flowing equably, without relation to anything external. It feels so intuitive to our everyday experience that it is hard to realise it is merely an assumption.

As we dive deeper into relativity, the first three equations of (2.1.1) will survive into Chapter 2.2 with a few tweaks. This universal time equation will completely fall apart.

1.2 · Velocities add

Next, let's see how our observers view motion. Imagine a particle travelling along a path. The stationary observer tracks its position as $x(t)$. Using our transformation (2.1.1), the moving observer tracks its position as $x'(t') = x(t) - vt$.

To find the particle's velocity, we just need to take the derivative. Since we assumed that time is the same for both observers ($t'=t$), differentiating with respect to $t'$ is identical to differentiating with respect to $t$:

This equation tells us something very familiar: velocities add or subtract depending on your point of view. If you throw a ball forward at $u'$ inside a train moving at $v$, a person standing on the platform sees the ball moving at $u'+v$, the speed of the throw plus the speed of the train. Nobody has ever been surprised by this, and for everyday objects it is incredibly accurate.

But notice the hidden mechanics here. The subtraction comes from the shift in space, while the ease of the derivative relies entirely on the assumption that $t'=t$. Chapter 2.2 keeps the first ingredient and destroys the second. When that assumption breaks later on, this rule for adding velocities will have to change too, and (2.1.2) is what changes.

1.3 · Accelerations do not change, so $\vv F=m\vv a$ survives

Now differentiate (2.1.2) once more. The relative velocity $v$ is a constant, since that is what it means for one inertial frame to move uniformly relative to another, so its derivative vanishes:

In three dimensions the same computation on each component gives $\ddot{\vv r}' = \ddot{\vv r}$. Acceleration is a Galilean invariant. Two observers in uniform relative motion disagree about where a particle is and about how fast it is going, and they agree exactly about how it is accelerating.

That single fact is why Newtonian mechanics is compatible with the principle of relativity. Here is the argument. Newton's second law in $S$ reads $\vv F = m\ddot{\vv r}$, and we want to know what it looks like in $S'$, so transform it piece by piece. The mass $m$ is a property of the body and does not depend on who is watching. The acceleration is unchanged by (2.1.3). So the law in $S'$ reads $\vv F = m\ddot{\vv r}'$, and the only remaining question is whether the force is the same in both frames.

For the forces Newtonian physics actually uses, it is. Every one of them depends on the particles' relative positions, and possibly on their relative velocities:

The boost cancels in both differences. Gravity depends on $\abs{\vv r_{i}-\vv r_{j}}$, and so do Coulomb's law and Hooke's law. Even a drag force proportional to the relative velocity of body and fluid is untouched. So $\vv F' = \vv F$ for each of them, and

The law has the same shape in the new frame. Not merely a shape that can be corrected into the old one by adding terms, but the identical shape, with the identical symbols meaning the identical things. Compare Chapter 1.1 §4.2, where writing $\vv F=m\vv a$ in polar coordinates produced two extra terms out of nothing. Nothing like that happens here.

Grind box — the full Galilean group, and what "same form" is allowed to mean

(2.1.1) is one member of a larger family. The complete set of transformations that carry one inertial frame to another, in Newtonian physics, is

$$\vv r' \;=\; R\,\vv r \;-\; \vv v\,t \;+\; \vv d, \qquad t' \;=\; t + s,$$

where $R$ is a constant rotation matrix ($3$ parameters), $\vv v$ a constant boost velocity ($3$), $\vv d$ a constant spatial displacement ($3$) and $s$ a constant time offset ($1$). Ten parameters. They compose and invert among themselves, since the composition of two boosts is a boost with $\vv v_{1}+\vv v_{2}$ and the inverse of a rotation is a rotation. So they form a group, the Galilean group. Chapter 6.1 gives that word a definition. For now the content is just that "change of inertial frame" is a closed operation.

Differentiating twice kills $\vv d$ and $\vv v t$ outright and leaves $\ddot{\vv r}' = R\,\ddot{\vv r}$, so acceleration is not quite invariant under the full group. It rotates, as a vector must. But it rotates the same way $\vv F$ does, so $\vv F = m\ddot{\vv r}$ still maps to $\vv F' = m\ddot{\vv r}'$. That is the precise sense in which "the form is unchanged": both sides of the equation transform identically, so the equality is preserved. Hold onto that formulation. It is the exact criterion Chapter 2.4 turns into the definition of a tensor equation, and it is the reason index notation exists.

What is not allowed. Let the relative velocity depend on time, $\vv r' = \vv r - \vv s(t)$ with $\ddot{\vv s}\neq\vv 0$. Then

$$\ddot{\vv r}' = \ddot{\vv r} - \ddot{\vv s}(t) \qquad\Longrightarrow\qquad m\ddot{\vv r}' = \vv F - m\ddot{\vv s}(t),$$

and an extra term has appeared that is not any interaction between bodies. That is the fictitious force of Chapter 1.1 §1.1, and its presence is exactly why the first law is an existence claim about a restricted class of frames rather than a corollary of the second. The principle of relativity was never a claim about all frames. It is a claim about the inertial ones, and the boost velocity must be constant for the argument of §1.3 to go through.

One more piece of fine print, cheap now and expensive later. Nothing above proves that (2.1.1) is the only transformation with these properties. We derived $\vv F'=m\vv a'$ from the Galilean transformation. We did not show that a theory respecting the principle of relativity has to use it. That gap is the whole of Chapter 2.2, and the reason nobody looked into it for three hundred years is that $t'=t$ did not feel like an assumption.

1.4 · The principle itself

Now separate the two ideas that §1.1–1.3 ran together.

The principle says something about the relationship between frames. It does not say what the transformation between them is. Those are separate questions, and the failure to notice they were separate is the single reason the crisis took forty years to resolve. Written out:

	Claim	Status in 1900
(P)	The laws take the same form in all inertial frames	Universally accepted; verified constantly
(G)	Inertial frames are related by (2.1.1), with $t'=t$	Assumed without comment by everyone

Everybody held both. It had never occurred to anyone that they were different claims, because (G) was not perceived as a claim at all. It was perceived as what the words in (P) meant. The resolution of this chapter's crisis is that (P) is right and (G) is wrong. Keep the pair separate from here on, because the rest of the chapter is about the strain between them.

2 · Maxwell's equations produce a speed

Now the second of the three claims. It has to be derived rather than quoted, because the number that falls out at the end is the whole point.

2.1 · One identity first

The derivation needs the curl of a curl, which Chapter 0.7 §7 did not have occasion to compute. It is worth deriving rather than looking up, because the pattern of the answer is not guessable.

Here $\nabla^{2}\vv A$ means the Laplacian of Chapter 0.7 §7.5 applied to each Cartesian component separately. That definition is only available because the components are Cartesian, a caveat that becomes important in Chapter 3.3 and is harmless here.

Grind box — (2.1.6) by brute force, one component

Write $\vv C = \nabla\times\vv A$, whose components are, from Chapter 0.7 §4,

$$C^{x} = \partial_{y}A^{z}-\partial_{z}A^{y}, \qquad C^{y} = \partial_{z}A^{x}-\partial_{x}A^{z}, \qquad C^{z} = \partial_{x}A^{y}-\partial_{y}A^{x}.$$

Take the $x$-component of $\nabla\times\vv C$ and substitute:

$$\begin{aligned} \big(\nabla\times\vv C\big)^{x} &= \partial_{y}C^{z} - \partial_{z}C^{y}\\[3pt] &= \partial_{y}\big(\partial_{x}A^{y}-\partial_{y}A^{x}\big) \;-\; \partial_{z}\big(\partial_{z}A^{x}-\partial_{x}A^{z}\big)\\[3pt] &= \partial_{y}\partial_{x}A^{y} + \partial_{z}\partial_{x}A^{z} \;-\; \partial_{y}\partial_{y}A^{x} - \partial_{z}\partial_{z}A^{x}. \end{aligned}$$

Now the trick, and it is the only step with any content. Add and subtract the missing term $\partial_{x}\partial_{x}A^{x}$. The first group then completes to a divergence and the second to a Laplacian:

$$\begin{aligned} \big(\nabla\times\vv C\big)^{x} &= \underbrace{\partial_{x}\partial_{x}A^{x} + \partial_{y}\partial_{x}A^{y} + \partial_{z}\partial_{x}A^{z}}_{\textstyle \partial_{x}\left(\nabla\cdot\vv A\right)} \;-\; \underbrace{\partial_{x}\partial_{x}A^{x}+\partial_{y}\partial_{y}A^{x}+\partial_{z}\partial_{z}A^{x}}_{\textstyle \nabla^{2}A^{x}}. \end{aligned}$$

Both regroupings used Clairaut's theorem (Chapter 0.6 §6.1) to swap $\partial_{y}\partial_{x}$ for $\partial_{x}\partial_{y}$, and its hypothesis is continuous second partials. That is the same hypothesis that made $\nabla\cdot(\nabla\times\vv A)=0$ work in Chapter 0.7 §7.2. The $y$ and $z$ components follow by cycling $x\to y\to z\to x$, under which every expression above is invariant. Hence (2.1.6). ✓

Why the answer looks like that. $\nabla\times(\nabla\times\ \cdot\ )$ is a second-order operator, so it must be some combination of $\partial_{i}\partial_{j}$ acting on $A^{k}$. Only two such combinations produce a vector: the gradient of the divergence, and the Laplacian of the vector. The identity says the coefficients are $+1$ and $-1$. In Chapter 3.5 this becomes the statement that the operator $\dd\star\dd$ splits in exactly this way, and the minus sign is the one that makes the electromagnetic wave equation come out with the right relative sign between space and time.

2.2 · Take the curl of Faraday's law

We want an equation for $\vv E$ alone. Faraday's law relates $\vv E$ to $\vv B$, and Ampère–Maxwell relates $\vv B$ back to $\vv E$. So apply one to the other and eliminate $\vv B$. The operation that lets Ampère–Maxwell in is the curl, because the curl of $\vv B$ is precisely what that law tells us. Start from Faraday and hit both sides with $\nabla\times$:

The second step swapped a space derivative with a time derivative. That is Clairaut again. The curl is built from $\partial_{x},\partial_{y},\partial_{z}$, none of which is $\partial_{t}$, so as long as $\vv B$ has continuous second partials the two operations commute. It is a small step and worth naming, because in Chapter 3.3 the analogous swap will fail, and the failure will be the curvature of spacetime.

Now substitute Ampère–Maxwell on the right, with $\nabla\times\vv B = \mu_{0}\epsilon_{0}\, \partial\vv E/\partial t$:

Next expand the left-hand side with (2.1.6), using $\nabla\cdot\vv E=0$, which holds because we are in vacuum with no charge anywhere to source a divergence:

Set the two expressions equal and cancel the overall minus sign:

Look at what has happened. We started with four equations coupling two fields, performed three manipulations, and produced an equation for $\vv E$ by itself with no $\vv B$ in it. And the equation is not a new object. It is the wave equation of Chapter 0.8 §7.6, so let us set the two side by side:

with three spatial dimensions instead of one, and a vector where the string had a scalar. Now read off the coefficient:

Nobody put a speed in. Two constants measured with capacitors and current-carrying wires went in, and a propagation speed came out.

Grind box — the same for $\vv B$, and the plane-wave solution

The magnetic field obeys the identical equation. Run the argument the other way. Take the curl of Ampère–Maxwell:

$$\nabla\times\big(\nabla\times\vv B\big) = \mu_{0}\epsilon_{0}\,\pdv{}{t}\big(\nabla\times\vv E\big) = -\mu_{0}\epsilon_{0}\,\pdv{^{2}\vv B}{t^{2}},$$

using Faraday in the last step. Now expand the left with (2.1.6) and use $\nabla\cdot\vv B=0$, which holds always rather than merely in vacuum, since there are no magnetic charges. So

$$\nabla^{2}\vv B \;=\; \mu_{0}\epsilon_{0}\,\pdv{^{2}\vv B}{t^{2}},$$

the same equation with the same speed. The two fields propagate together, which they must, since they are two aspects of one thing, as Chapter 2.6 makes literal.

A plane wave, and $c=\omega/k$. Try a solution travelling along $x$ with $\vv E$ pointing along $y$, so that $\vv E = \hat{\vv y}\,E_{0}\cos(kx-\omega t)$. Then $\nabla^{2}\vv E = -k^{2}\vv E$ and $\partial^{2}\vv E/\partial t^{2} = -\omega^{2}\vv E$, so (2.1.10) requires

$$-k^{2} = -\mu_{0}\epsilon_{0}\,\omega^{2} \qquad\Longrightarrow\qquad \frac{\omega}{k} = \frac{1}{\sqrt{\mu_{0}\epsilon_{0}}} = c.$$

That is Chapter 0.8's relation between frequency, wavenumber and wave speed, now with the speed fixed by two laboratory constants. Notice that it holds for every $k$. The speed does not depend on the wavelength, so vacuum is non-dispersive and a pulse of any shape travels undistorted. That is Chapter 0.8's $f(x-ct)$, and it is why we see sharp images of distant stars rather than smeared ones.

One honest caveat. (2.1.10) is a consequence of Maxwell's equations, not an equivalent of them. Every solution of Maxwell solves it, and not every solution of it solves Maxwell. Feeding the plane wave back into the original four equations imposes two further conditions we did not need above. First, $\nabla\cdot\vv E=0$ forces $\vv E\perp$ propagation direction, so the wave is transverse. Second, Faraday fixes $\vv B = \hat{\vv x}\times\vv E/c$, so $\vv B$ is perpendicular to both $\vv E$ and the direction of travel, in phase with $\vv E$, and smaller by a factor $c$. None of that is needed for this chapter's argument, and all of it matters in Chapter 2.6.

2.3 · Put the numbers in

Now the moment. Here is where $\epsilon_{0}$ and $\mu_{0}$ come from, because the provenance is the point.

⚑ $\epsilon_{0}$, the permittivity of free space, is fixed by measuring the capacitance of a parallel-plate capacitor. Put a known charge on two plates of known area and separation, measure the voltage, and $\epsilon_{0}$ falls out of $C=\epsilon_{0}A/d$. This is a tabletop electrostatics experiment involving no light, no motion and no waves. Its value is $\epsilon_{0}=8.854\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}$.

⚑ $\mu_{0}$, the permeability of free space, is fixed by measuring the force between two parallel current-carrying wires. Run known currents through them, measure the attraction per unit length, and $\mu_{0}$ falls out of $F/\ell = \mu_{0}I_{1}I_{2}/2\pi d$. This is a tabletop magnetostatics experiment involving no light and no waves either. Its value is $\mu_{0}=4\pi\times10^{-7}\ \mathrm{N\,A^{-2}} = 1.2566\times10^{-6}\ \mathrm{N\,A^{-2}}$.

Multiply them together and take the reciprocal square root:

The measured speed of light, from Fizeau's toothed wheel in 1849 and Foucault's rotating mirror in 1862, was ⚑ about $3.0\times10^{8}\ \mathrm{m\,s^{-1}}$.

3 · The question nobody could answer: $c$ with respect to what?

Now the trouble starts, and it starts with a question that is embarrassing in its simplicity.

(2.1.12) gives a speed. A speed is a rate of change of position, and position is measured relative to something. So here is the question: relative to what is the light going at $c$?

3.1 · Every other wave answers this question easily

Chapter 0.8 §7.6 built the wave equation from a chain of masses and springs and got $v=\sqrt{T/\rho}$. Look at what that formula is made of: the tension and the linear density of the string. Both are properties of the medium, so the speed is a property of the medium. It is therefore the speed relative to the medium. If the string is being reeled in while the wave travels along it, an observer at the side of the room measures something else.

Every wave in nineteenth-century physics was like this, without exception:

Wave	Speed	Relative to
Transverse wave on a string	$\sqrt{T/\rho}$	the string
Sound in air	$\sqrt{\gamma p/\rho}\approx343\ \mathrm{m\,s^{-1}}$	the air
Ripples on water	$\sqrt{g\lambda/2\pi}$ (deep water)	the water
Seismic $S$-waves	$\sqrt{G/\rho}$	the rock

And the consequence is entirely familiar. Stand in a wind of speed $u$ and sound travelling downwind reaches you at $343+u$, upwind at $343-u$. That is (2.1.2) applied to a wave, and it is measured routinely. The speed in the formula is the speed in the rest frame of the medium, and in any other frame you add the medium's velocity.

3.2 · The wave equation itself makes the point

You do not even need the physical picture. The mathematics says it directly. The wave equation is written in some particular set of coordinates $(x,t)$, and its solutions are $f(x-vt)$ and $g(x+vt)$, which are disturbances moving at $\pm v$ in those coordinates. Suppose the equation holds in one coordinate system with speed $v$, and suppose coordinates transform by (2.1.1). Then in a coordinate system moving at $u$ the same disturbance moves at $v-u$. The number $v$ in the equation is therefore attached to one preferred frame, namely the frame in which the equation takes that form.

Apply this to (2.1.10). Maxwell's equations, written as they always are written, hold in some frame. In that frame light goes at $c$ in every direction. In a frame moving at $u$ through it, light should go at $c-u$ in one direction and $c+u$ in the other. So Maxwell's equations appear to single out a preferred frame of reference. That flatly contradicts claim (P) of §1.4, and it does so for reasons that have nothing to do with any experiment.

3.3 · The ether, taken seriously

Given all of §3.1, there was one obvious move, and it was not a stupid one. If light is a wave, it is a wave in something. Name that something the luminiferous ether, and declare Maxwell's equations to hold in the ether's rest frame. The preferred frame of §3.2 is then no longer an embarrassment. It is only the rest frame of a medium, exactly like the air for sound. The principle of relativity survives untouched, because the ether is a physical object, and detecting motion relative to it is no more mysterious than feeling a breeze.

This was not a fudge. It was the only known way for a wave to work, and every other wave anyone had ever studied confirmed it. Refusing to posit a medium in 1880 would have been the strange move, not the sober one. And the hypothesis was productive. It made a sharp prediction: the Earth moves, so there must be an ether wind, so the speed of light must be anisotropic in the laboratory. That prediction is testable, and §5 tests it.

The ether was, however, a demanding object. Its properties were fixed by (2.1.10), and they do not sit comfortably together.

Grind box — what the ether had to be like, quantitatively

It had to be a solid. Light is transverse, since a plane wave has $\vv E$ perpendicular to the direction of travel, as the previous grind box showed. A transverse wave is a shear disturbance, with neighbouring layers of the medium sliding past one another. Fluids do not resist shear, which is why sound in air is purely longitudinal and why the Earth's liquid outer core transmits no $S$-waves. So the ether had to have rigidity. It had to be an elastic solid filling all of space, and the planets had to pass through it without measurable resistance.

And an extraordinarily stiff one. For a transverse wave in an elastic solid the speed is $v=\sqrt{G/\rho}$, with $G$ the shear modulus, which is the same square root as $\sqrt{T/\rho}$ in Chapter 0.8. Setting $v=c$ fixes neither the stiffness nor the density but their ratio:

$$\frac{G}{\rho} \;=\; c^{2} \;=\; 8.99\times10^{16}\ \mathrm{m^{2}\,s^{-2}}.$$

Compare steel: shear modulus $79.3\ \mathrm{GPa}$, density $7850\ \mathrm{kg\,m^{-3}}$, so $G/\rho = 1.01\times10^{7}\ \mathrm{m^{2}\,s^{-2}}$. And indeed $\sqrt{1.01\times10^{7}}=3.18\ \mathrm{km\,s^{-1}}$, which is the shear-wave speed in steel. The ether's stiffness-to-density ratio therefore had to exceed steel's by a factor

$$\frac{8.99\times10^{16}}{1.01\times10^{7}} \;=\; 8.9\times10^{9},$$

nearly ten billion. You may make the density as small as you like to keep the planets moving, but the ratio is not negotiable, and it is the ratio that has to be explained.

And it had to be undetectable in every other way. Transparent, frictionless, incompressible enough to fill the space between the stars, non-interacting with matter except through this one channel, and possessed of no measurable effect on anything except that light goes through it.

Nineteenth-century physicists were entirely aware of this list, and it bothered them. Kelvin called the ether one of the two clouds on the horizon of physics. But an awkward hypothesis that makes a testable prediction is a perfectly respectable scientific object, and the response was the right one: go and measure the wind.

4 · Maxwell's equations are not Galilean invariant

Section 3 argued from physical analogy that Maxwell's equations must pick out a frame. This section proves it, by computation, with no analogy anywhere. It is the hinge of the chapter.

Strip the problem to its skeleton. Take one Cartesian component of $\vv E$ and call it $\phi$. Suppress $y$ and $z$, so that the Laplacian is $\partial^{2}/\partial x^{2}$. Then (2.1.10) becomes the one-dimensional wave equation:

Here is the question. What does this equation look like in a frame moving at $v$ along $x$? Not what do its solutions look like, but what does the equation look like. If the principle of relativity holds with the Galilean transformation, the answer must be $\partial^{2}\phi/\partial x'^{2} = c^{-2}\,\partial^{2}\phi/\partial t'^{2}$, character for character.

4.1 · The chain rule, set up carefully

We have new coordinates $x'=x-vt$ and $t'=t$, and a function $\phi$ that we may regard either as a function of $(x,t)$ or of $(x',t')$. It is one physical field, described two ways. Chapter 0.6 §5.2 gives the rule for converting derivatives: each old derivative becomes a sum over the new ones, weighted by how each new coordinate responds to the old one.

The four partial derivatives of the new coordinates with respect to the old come straight from $x'=x-vt$ and $t'=t$:

Take a moment over $\partial t'/\partial x = 0$. It says that changing where you are does not change what time it is. That is the mathematical form of absolute simultaneity. It is the only place $t'=t$ enters this calculation, and it is the entry that Chapter 2.2 will make nonzero. Now substitute (2.1.16) into (2.1.15):

The spatial derivative is untouched, and the time derivative acquires an extra piece. That is the whole asymmetry, and everything below is bookkeeping. The physical reading of the second relation is worth having. Holding your position fixed in the old frame means drifting backwards at $-v$ in the new one, so the rate of change you measure standing still picks up a term from the motion. It is the convective derivative of fluid mechanics, arrived at without meaning to.

4.2 · Second derivatives, and the term that ruins everything

Grind box — squaring the operators, every step

Space. Apply (2.1.17) twice:

$$\pdv{^{2}}{x^{2}} = \pdv{}{x}\left(\pdv{}{x}\right) = \pdv{}{x'}\left(\pdv{}{x'}\right) = \pdv{^{2}}{x'^{2}}.$$

Nothing to do.

Time. Here we must square a sum of two operators. The one thing to be careful about is that operators need not commute, so $(A+B)^{2}=A^{2}+AB+BA+B^{2}$, and we may not jump to $A^{2}+2AB+B^{2}$ without checking. Write $A=\partial/\partial t'$ and $B=-v\,\partial/\partial x'$:

$$\begin{aligned} \pdv{^{2}}{t^{2}} &= \left(\pdv{}{t'} - v\pdv{}{x'}\right)\left(\pdv{}{t'} - v\pdv{}{x'}\right)\\[5pt] &= \pdv{^{2}}{t'^{2}} \;-\; v\,\pdv{}{t'}\pdv{}{x'} \;-\; v\,\pdv{}{x'}\pdv{}{t'} \;+\; v^{2}\pdv{^{2}}{x'^{2}}. \end{aligned}$$

The two middle terms are equal because of Clairaut's theorem, since mixed partials of a twice-continuously-differentiable function commute. They are also equal because $v$ is a constant, so it passes through $\partial/\partial x'$ without generating anything. Both facts are needed, and both are easy to use without noticing. Hence

$$\pdv{^{2}}{t^{2}} \;=\; \pdv{^{2}}{t'^{2}} \;-\; 2v\,\frac{\partial^{2}}{\partial x'\,\partial t'} \;+\; v^{2}\,\pdv{^{2}}{x'^{2}}.$$

Assemble. Substitute both results into (2.1.14):

$$\pdv{^{2}\phi}{x'^{2}} \;=\; \frac{1}{c^{2}}\left[\pdv{^{2}\phi}{t'^{2}} - 2v\,\frac{\partial^{2}\phi}{\partial x'\,\partial t'} + v^{2}\,\pdv{^{2}\phi}{x'^{2}}\right].$$

Bring the $v^{2}$ term to the left and collect the coefficient of $\partial^{2}\phi/\partial x'^{2}$:

$$\left(1 - \frac{v^{2}}{c^{2}}\right)\pdv{^{2}\phi}{x'^{2}} \;+\; \frac{2v}{c^{2}}\,\frac{\partial^{2}\phi}{\partial x'\,\partial t'} \;=\; \frac{1}{c^{2}}\,\pdv{^{2}\phi}{t'^{2}},$$

which is (2.1.18) below, rearranged.

Here is the result of that grind, written with everything on one side and with $\beta \equiv v/c$, as the convention of Part II requires:

Compare that with what we wanted, which was (2.1.14) with primes on it. Two things have gone wrong, and they are worth naming separately.

(i) A cross term has appeared. The middle term $\propto \partial^{2}\phi/\partial x'\,\partial t'$ has no counterpart in (2.1.14). It is odd in $v$, so reversing the direction of motion changes its sign, and that is precisely how an equation encodes a preferred direction. A term that knows which way is "forwards" cannot appear in an isotropic law.

(ii) The leading coefficient has changed. $1$ has become $1-\beta^{2}$. Even if you could somehow dispose of the cross term, the coefficients no longer match.

4.3 · What the transformed equation is actually saying

(2.1.18) is not wrong. It is the correct description of a wave in a medium, as seen by an observer moving through that medium. It is worth extracting exactly what it says, both to check the algebra and to see what is at stake.

So look for travelling-wave solutions in the new frame: $\phi = f(x'-ut')$ for some speed $u$ and arbitrary shape $f$, exactly as in Chapter 0.8 Problem 4. Each derivative brings down a factor, with $\partial^{2}_{x'}\phi = f''$, $\partial_{x'}\partial_{t'}\phi = -u f''$ and $\partial^{2}_{t'}\phi = u^{2}f''$. Substitute into (2.1.18) and cancel the common $f''$, which is nonzero for any wave worth the name. What is left is a condition on $u$ alone:

That is a quadratic in $u$. Solve it, either by completing the square or with the formula:

Those are exactly the two speeds §3.1 predicted on physical grounds. In the moving frame, light chases off in the $+x'$ direction at $c-v$ and comes back the other way at $c+v$. The algebra of §4.2 and the sound-in-a-wind picture of §3.1 are the same statement.

So (2.1.18) is perfectly consistent physics, for a wave in a medium. Sound obeys the analogous equation and nobody minds, because sound has a medium and its rest frame is not mysterious. The entire question is whether light is like that.

5 · Michelson–Morley

Branch (A) is the one that is directly testable, and testing it produced the most famous experiment in physics. The logic is short. The Earth orbits the Sun at about $30\ \mathrm{km\,s^{-1}}$, so unless the ether happens to be dragged along perfectly, there is an ether wind blowing through the laboratory. Measure the speed of light along the wind and across it, and the two answers should differ.

The difficulty is that they differ by very little. From §4.3 the two one-way speeds are $c\pm v$ with $v/c\approx10^{-4}$. A direct timing measurement would therefore have to resolve one part in $10^{4}$ of a quantity that was not itself known that well. The best absolute determination in 1887 was Michelson's own, from 1879, at $299\,910\pm50\ \mathrm{km\,s^{-1}}$, which is good to about two parts in $10^{4}$. The difference he needed to see was smaller than the error bar on the whole. Michelson's answer was to stop measuring speed and start measuring interference, which compares two light paths against each other and can resolve a hundredth of a wavelength.

5.1 · The instrument

A source of monochromatic light shines on a half-silvered mirror set at $45^{\circ}$, called a beam splitter. Half the light passes through and continues to a mirror at distance $L_{1}$. Half reflects off at right angles and travels to a second mirror at distance $L_{2}$. Both beams return to the splitter, recombine, and go on to a detector. Whether they arrive in phase or out of phase depends on the difference in their travel times, so the recombined beam shows a pattern of interference fringes.

Two facts about this arrangement do the real work. First, the arms are perpendicular, so if there is an ether wind then one arm can lie along it and the other across it. Second, and this is the piece of experimental cunning without which the whole thing is impossible, you cannot know $L_{1}$ and $L_{2}$ to a fraction of a wavelength. The absolute fringe position is therefore uninterpretable. What you can do instead is rotate the whole apparatus by $90^{\circ}$, which exchanges the roles of the two arms without changing either length, and then watch the fringes move. §5.5 shows that the unknown mismatch $L_{1}-L_{2}$ cancels out of that measurement exactly.

5.2 · The arm along the wind

Work in the ether's rest frame, where light travels at $c$ in all directions and the apparatus moves at $v$ in the $+x$ direction. Working in the ether frame is the honest choice, because it is the frame where we are entitled to say light moves at $c$, and that is the hypothesis under test.

Set $t=0$ when a pulse leaves the splitter, and put the splitter at the origin at that instant. Thereafter the splitter is at $vt$ and the far mirror at $L_{1}+vt$, both sliding to the right at $v$.

Outbound. The pulse moves right at $c$, so it is at $ct$, while the mirror is at $L_{1}+vt$. They meet when

The mirror is running away from the light, so the light closes the gap at only $c-v$. Notice that this is not a claim that the light moves at $c-v$. It moves at $c$ in the ether frame, as assumed. The $c-v$ is a closing speed between two objects, and closing speeds may perfectly well exceed $c$ or fall below it without anything moving at that rate.

Return. Now the light travels left at $c$ from the mirror's position while the splitter advances to meet it. By the identical argument the gap closes at $c+v$:

Add them, put over a common denominator, and divide top and bottom by $c^{2}$:

The round trip is slower than the $2L_{1}/c$ it would take with no wind, by a factor $1/(1-\beta^{2})$. That is worth a sentence, because the intuition points the wrong way. The two legs seem as though they ought to average out, one gaining what the other loses. They do not. The reason is that the leg travelled at the lower speed lasts longer, so it gets more weight in the total. Losing time at $c-v$ for a long while beats gaining it at $c+v$ briefly. The same asymmetry is why a round trip against and with a current always takes longer than the same trip in still water, and why your average speed on a there-and-back drive is the harmonic mean rather than the arithmetic one.

5.3 · The arm across the wind, done properly

This is where most treatments go soft, so we will be slow. The naive answer says the light goes straight across at $c$, so that the round trip is $2L_{2}/c$ and the transverse arm is unaffected. That answer is wrong, and it is wrong for a reason that becomes central in Chapter 2.2.

The mirror at the end of the transverse arm is moving sideways. During the time the light is in flight, that mirror slides a distance $vt$ down the $x$-axis. Light that left the splitter travelling straight along $y$ in the ether frame would arrive where the mirror used to be and miss it. To hit the mirror, the light must leave with a velocity that has an $x$-component matching the apparatus's own motion.

Write that as an equation. Let the light's velocity in the ether frame be $\vv u = (u_{x},u_{y})$ with, by hypothesis, $\abs{\vv u}=c$. For the light to stay over the moving arm it needs $u_{x}=v$. Then

The useful component of the light's velocity, the part that gets it across the arm, is not $c$ but $\sqrt{c^{2}-v^{2}}$. The rest of its speed budget is spent keeping up sideways. The arm is $L_{2}$ long and the crossing is covered at $u_{y}$, so

with the factor of $2$ because the return trip is the mirror image of the outbound one and takes the same time.

The same result comes out of a right triangle, and that is the version to carry in your head. In time $t_{1}$ the light travels a straight-line distance $ct_{1}$ in the ether frame, which is the hypotenuse. The apparatus has drifted $vt_{1}$, which is the base. The arm is $L_{2}$, which is the height. Pythagoras:

and $t_{\perp}=2t_{1}$ as before. Same answer, same square root. The square root has the same origin in both derivations: a fixed total speed $c$ has to be shared between "across" and "along", and Pythagoras charges you for the sharing. That triangle is going to reappear in Chapter 2.2 as the derivation of time dilation, with the arm replaced by a light clock, and there the factor $1/\sqrt{1-\beta^{2}}$ will be named $\gamma$.

5.4 · The difference

Take equal arms, $L_{1}=L_{2}=L$, which is what the instrument is built to have and what §5.5 shows we do not actually need. Subtract (2.1.25) from (2.1.23):

That is exact. Now expand for small $\beta$, since $\beta\approx10^{-4}$ and what we want to know is how big the effect is. Chapter 0.3's binomial series gives, with $s=\beta^{2}$,

We want the difference of those two series, so subtract the second from the first term by term. The bracket in (2.1.27) becomes

Only the leading piece matters at $\beta\approx10^{-4}$, so keep the $\tfrac12\beta^{2}$ term and drop everything after it. The round-trip difference is then

Look at what has dropped out. The two arms agree at order $\beta^{0}$ and again at order $\beta^{1}$, and they first differ at order $\beta^{2}$. The effect is second order in $v/c$, and that is the central practical fact about the entire subject. There is no first-order ether-wind effect in a round-trip measurement, because the delay on the way out is compensated to first order by the gain on the way back. Every experiment sensitive only to first order was therefore guaranteed a null result before it was built, and there had been several. Michelson's design was the first with second-order sensitivity, which is why it counts.

And that $\beta^{2}=v^{2}/c^{2}$ is the same suppression factor that made the third law's failure so hard to notice in Chapter 1.1 §3.3. Both are the leading symptom of the same disease.

5.5 · What rotating the apparatus measures

Now the experimental point flagged in §5.1. You cannot measure $\Delta t$ directly. Doing so would require knowing $L_{1}-L_{2}$ to a fraction of $590\ \mathrm{nm}$, which nobody can build. So rotate instead.

Before rotation, with arm 1 along the wind and arm 2 across it, the time difference between the beams is, from (2.1.23) and (2.1.25),

Turn the table through $90^{\circ}$. Arm 1 is now across the wind and arm 2 along it, and neither length has changed. So

Subtract. The terms regroup by denominator rather than by arm, and the arms enter only through their sum:

The unknown mismatch $L_{1}-L_{2}$ has vanished identically. Only $L_{1}+L_{2}$ survives, and that is a quantity you can measure with a ruler. Set $L_{1}=L_{2}=L$ so that $L_{1}+L_{2}=2L$, and compare with (2.1.27). The change on rotation is exactly $2\Delta t$.

Now turn that into something visible. A time difference $\tau$ between the two beams corresponds to an optical path difference $c\tau$, and one whole fringe of movement corresponds to a path difference of one wavelength $\lambda$. So the number of fringes the pattern shifts as the apparatus turns is

Clean, and made of measurable things: an arm length, a wavelength, and the square of the Earth's speed in units of $c$.

5.6 · The numbers, 1887

Michelson and Morley floated a sandstone slab on a bath of mercury so that it could be turned smoothly without flexing, and they folded each light path back and forth between multiple mirrors to lengthen it. ⚑ The effective one-way path was $L\approx11\ \mathrm{m}$, the source was a sodium lamp at $\lambda\approx590\ \mathrm{nm}$, and the assumed velocity was the Earth's orbital speed, $v=30\ \mathrm{km\,s^{-1}}$. Then

Now feed that value of $\beta^{2}$ into (2.1.34), along with the arm length and the wavelength above:

(The exact expression (2.1.34) gives $0.373398$, and the expansion also gives $0.373398$. At $\beta=10^{-4}$ the correction term $\tfrac54\beta^{2}$ of (2.1.29) is one part in $10^{8}$, so the two agree to six figures, as they should.)

About four tenths of a fringe. That is not enormous. But Michelson's interferometer could see a shift of about a hundredth of a fringe, so the predicted effect was roughly forty times the detection threshold.

5.7 · The interactive

ether wind v

v = 30.00 km/s

arm L

L = 11.0 m

Earth's orbital speed

1887 apparatus

t∥ = 73.384101678 ns

t⊥ = 73.384101311 ns

Δt = 3.6743e-16 s = 367.4 as

ΔN = 0.3734 fringes

ΔN / bound = 37.3× — far above the threshold; crosses the bound at v = 4.91 km/s

5.8 · The result

They found nothing.

⚑ Michelson and Morley reported in 1887 that the observed displacement was certainly less than a twentieth of the predicted $0.4$ fringes, and probably less than a fortieth. That is an upper bound of about $0.01$ fringes, which is to say the noise floor of the instrument. They repeated the measurement at different times of day and in different seasons, to catch the Earth's rotation and its orbital motion pointing the apparatus in every possible direction relative to any conceivable ether frame. Nothing.

6 · The rescues, and why they failed

A null result does not by itself dispose of the ether. What it disposes of is the simplest ether: one at rest in some frame, through which the Earth moves freely and which does nothing to the apparatus. Several serious alternatives were proposed. It is worth being specific about them, because they were good physics, and because the last one was very nearly right.

6.1 · Ether drag

Here is the obvious first move. Suppose the Earth drags the local ether along with it, so that near the surface there is no wind at all. The interferometer then sits in still air, so to speak, and sees nothing. Two independent observations rule this out.

Stellar aberration. Point a telescope at a star directly overhead. The Earth is moving sideways at $v$ while the light travels down the tube, so the telescope must be tilted slightly into the direction of motion. Otherwise the eyepiece has moved out from under the light by the time it arrives. If the tube has length $\ell$, the light takes $\ell/c$ to traverse it and the tube advances $v\ell/c$ in that time, so the required tilt is

With $v=29.78\ \mathrm{km\,s^{-1}}$ this comes to $\theta = 9.933\times10^{-5}\ \mathrm{rad}$, and converting to arcseconds by multiplying by $206265$,

⚑ Bradley discovered exactly this in 1728. Every star traces out a small ellipse over the course of a year, of angular radius $20.4955''$, and the effect is the single most reliable confirmation that the Earth moves. Now put that together with drag. If the ether were carried along with the Earth, the light would enter that co-moving ether at the top of the atmosphere and thereafter travel in the telescope's own rest frame. There would then be no aberration at all, or at most a fringe effect at the boundary. The aberration is observed, it has exactly the size $v/c$ predicted for undragged ether, and it is the same for all stars regardless of distance. Complete drag is dead.

Fizeau's water tube. In 1851 Fizeau sent light through a tube of water flowing at speed $u$ and measured how much the water carried the light along with it. Complete drag predicts that the light speed in the lab is $c/n + u$, and no drag predicts $c/n$. He measured neither. ⚑ The light was dragged by a fraction of the water's speed,

which for water ($n=1.333$) is $f=0.437$. Fizeau confirmed that to within a few per cent, and Michelson's much more precise repetition in 1886 returned $0.434\pm0.02$. A partial drag coefficient is a very strange thing for a mechanical medium to have. How does a substance know to transmit exactly $43.7\%$ of its own motion? No ether model ever gave a convincing account of it. (Chapter 2.2 will get (2.1.39) in one line, as the leading term of relativistic velocity addition, with no medium, no drag and no free parameter. It is one of the cleanest confirmations of the whole framework, and it is a nineteenth-century measurement.)

6.2 · The FitzGerald–Lorentz contraction

Now the interesting one. In 1889 FitzGerald, and independently Lorentz in 1892, made a proposal of startling economy: objects moving through the ether contract along their direction of motion by exactly the factor $\sqrt{1-\beta^{2}}$.

Watch what that does. The parallel arm lies along the motion, so its length in the ether frame is not $L$ but $L\sqrt{1-\beta^{2}}$. The transverse arm is perpendicular and is unaffected. Substitute into (2.1.23):

The two round-trip times are equal. Not approximately equal, and not equal to order $\beta^{2}$. They are identically equal, for every $\beta$, because $\sqrt{1-\beta^{2}}/(1-\beta^{2}) = 1/\sqrt{1-\beta^{2}}$ is an algebraic identity. The predicted fringe shift is exactly zero to all orders, and Michelson–Morley is explained.

It is impossible to look at (2.1.40) and not feel that something is being got away with. The contraction factor was chosen for no reason except that it makes the answer come out. It was, in FitzGerald's own framing, a hypothesis with one job.

And yet it is not absurd. Lorentz gave it a physical rationale. Suppose the forces holding matter together are electromagnetic in origin, which by 1892 was a reasonable guess. Suppose also that electromagnetic fields are distorted by motion through the ether. Then the equilibrium spacing of the atoms in a solid might well change when the solid is set in motion. On that reading the contraction is a real, dynamical effect on matter, produced by the ether acting on the electrons in the rod. The rod is genuinely shorter. The ether is genuinely there. It just conspires to be undetectable.

6.3 · Local time

Contraction alone is not quite enough, and Lorentz knew it. He wanted Maxwell's equations to come out with the same form in a moving frame, so that the theory would work at first order in $\beta$ for optical experiments generally and not only for Michelson–Morley. To get that, he found he had to introduce a further substitution, which in his 1895 form reads

He called $t'$ the local time (Ortszeit), and he was explicit that he regarded it as a mathematical convenience with no physical meaning whatever. The real time was $t$, the ether's time, universal and absolute in Newton's sense. The quantity $t'$ was an auxiliary variable, a change of integration variable that made the equations tractable, no more significant than substituting $u=x^{2}$ in an integral.

By 1904 Lorentz had the full set. Written in modern notation, with $\gamma = (1-\beta^{2})^{-1/2}$:

Those are the Lorentz transformations. Chapter 2.2 derives them from scratch and will not use this equation. It is here only so that the historical point can be made without ambiguity.

Grind box — how close was Lorentz, exactly?

Uncomfortably close, and the gap is instructive. Take Lorentz's own 1904 apparatus and ask what it gets right.

Right: the algebra. (2.1.42) is character for character what Chapter 2.2 will derive. The contraction factor, the local time, the group property, the invariance of Maxwell's equations under the substitution: all correct, and all published before 1905.

Right: the experimental predictions, for everything then measured. Michelson–Morley, Fizeau, aberration, the first-order optical experiments: Lorentz's theory reproduces them all.

Wrong: the status of $t'$. Because it is a substitution rather than a time, Lorentz had no account of time dilation as a physical phenomenon, and no reason to expect a moving clock of any construction to run slow. Ask him about an unstable particle and the theory is silent.

Wrong: the asymmetry. In Lorentz's scheme the contraction is real for a rod moving through the ether and not for a rod at rest in it, so the two frames are not equivalent even though no experiment can tell them apart. Einstein's symmetry between the frames is exact, and it is the reason each observer finds the other's rods short. Lorentz's framework cannot make that statement, since on his account one of the two rods is really contracted and the other really is not.

Wrong: the generality. Lorentz's contraction was derived from a hypothesis about electromagnetic forces inside matter. Nothing in it covers gravitational binding, nuclear binding, or the decay rate of a particle. Every one of those obeys the same $\gamma$ anyway, which on Lorentz's account would be an extraordinary coincidence and on Einstein's is a tautology.

The lesson generalises past this episode and is worth carrying: having the right equations is not the same as having the right theory. Two accounts can agree on every equation and every measured number and still differ in what they say the symbols mean. That difference shows up the moment you try to apply the theory somewhere it was not built for. Lorentz's electrodynamics and Einstein's relativity were empirically indistinguishable in 1905. They stopped being indistinguishable the moment anyone measured a particle lifetime.

7 · The two postulates

Everything above is the case for the prosecution. Here is what Einstein proposed instead, stated in the form Chapter 2.2 will use as input.

Notice how little that is. There is no ether, no statement about what light is, no model of matter, no dynamics and no mechanism. Two sentences, one of which was universally accepted already. Chapter 2.2 shows that the pair of them determines the transformation between inertial frames essentially uniquely, and everything in Part II after that is consequences.

7.1 · Postulate 1 is not new

It is §1.4's principle with one word changed: all the laws, not just the mechanical ones. That single word is what rules out branch (A) of §4.3, where mechanics obeyed the principle and electromagnetism did not. We already have (2.1.18), in which Maxwell's equations demonstrably change form under (2.1.1). So asserting Postulate 1 forces you into branch (C), and the Galilean transformation must go.

7.2 · Postulate 2 is the radical one, and it is not a definition

This is where the strangeness lives, and it is worth being precise about why.

Consider a lamp on the front of a train moving at $v$, and a lamp on the platform, flashing as they pass. Postulate 2 says both flashes travel at $c$ relative to the platform and at $c$ relative to the train. Not $c+v$ and $c$. Not $c$ and $c-v$. The same $c$, measured by both observers, for both flashes. (2.1.2), the rule that velocities add, is being denied for light, flatly.

Two objections need answering.

"Isn't this just a convention about how to set clocks?" No. There is a real convention lurking nearby, since synchronising distant clocks requires a rule and any rule involves an assumption about one-way light speed, and Chapter 2.3 will treat that carefully. But Postulate 2 has empirical content independent of any such convention, because the round-trip speed is measurable without synchronising anything: one clock, one mirror, one pulse. The claim that the two-way speed is the same in all frames and independent of the source's motion is a fact about the world, and it can be checked.

"Isn't the source-independence part obvious for a wave?" For a wave in a medium, yes. Sound from a moving whistle travels at $343\ \mathrm{m\,s^{-1}}$ through the air regardless of the whistle, since once the disturbance is launched the medium takes over and forgets the source. That is precisely why the ether was attractive: it delivers source-independence for free. Postulate 2 asserts source-independence with no medium to enforce it, which is a much stranger claim. And it also asserts frame-independence, which no medium theory delivers at all.

So Postulate 2 is an empirical claim, and it must be tested. It has been, exhaustively. Worked example 2 gives the sharpest version: photons emitted by particles moving at $0.99975c$ travel at $c$, not at $1.99975c$, to within a part in $10^{4}$.

8 · Worked examples

9 · Your turn