Galilean Transformations

A firecracker goes off beside the track. You, on the platform, log it at $x = 100\ \mathrm{m}$ , $t = 2\ \mathrm{s}$ . A passenger on a train sliding past at $v = 30\ \mathrm{m/s}$ logs the same flash at $x' = 40\ \mathrm{m}$ , $t' = 2\ \mathrm{s}$ . Neither of you is mistaken; your rulers simply have origins that have drifted apart by $vt = 60\ \mathrm{m}$ . The rule converting one logbook into the other is the Galilean transformation — and for two hundred years it was not thought of as a physical hypothesis at all, merely as bookkeeping so obvious it needed no name. The machinery itself belongs to mechanics. The article Relative Motion and Galilean Transformations (`aq-mech-relative-motion-galilean-transformations`) owns it: the dictionary $\mathbf{r}' = \mathbf{r} - \mathbf{V}t$ with $t' = t$ , the velocity rule $\mathbf{v}' = \mathbf{v} - \mathbf{V}$ , the invariance of acceleration $\mathbf{a}' = \mathbf{a}$ and hence of $\mathbf{F} = m\mathbf{a}$ , the ten-parameter Galilean group, consisting of boosts, rotations, spatial translations, and time translations, and the two quantities every classical observer agrees on — the time interval between any two events, and the distance between simultaneous ones. We recall that by name; we do not re-derive it as if it were new. What this article owns is the structure of the map and the single place it shatters. Read as geometry, the transformation is a shear of the spacetime diagram: worldlines tilt, but the lines of constant time stay stubbornly horizontal, because $t' = t$ . Read as algebra, it is a one-parameter group — two boosts compose by simply adding their velocities. Now set off a flash and let the light sphere expand: in $S$ it obeys $x^2 + y^2 + z^2 = c^2t^2$ , a sphere centred on the origin. Push that through the Galilean map and the sphere in $S'$ comes out centred not on the moving observer but a distance $vt'$ behind her. The transformation cannot keep the speed of light constant in both frames, since $t'=t$ . That offset is the crack the Michelson–Morley Experiment was built to measure, and the reason the clause $t' = t$ had to go.

Key takeaways

The Galilean transformation is the coordinate transformation between two inertial reference frames in uniform relative motion with constant velocities ( $\frac{dv}{dt}=0$ ): $x' = x - vt$ , $y' = y$ , $z' = z$ , and — the decisive clause — $t' = t$ .
$t' = t$ asserts absolute time: one universal clock, so simultaneity is absolute — events simultaneous for one inertial observer are simultaneous for all of them.
The map is linear, and that is forced: free particles move in straight lines for every inertial observer, and only a linear map ( $x'=Ax+b$ ) sends straight worldlines to straight worldlines.
Two boosts compose by adding velocities — $B(v_2)\,B(v_1) = B(v_1 + v_2)$ — a one-parameter group. Differentiating $x'=x-vt$ with time gives galilean velocity addition, $u' = u - v$ .
Drawn on an $x$ – $t$ diagram it is a shear: worldlines tilt, simultaneity lines stay horizontal. Unlike a Galilean transformation, a Lorentz transformation tilts both the worldlines and the lines of simultaneity.
Key idea: the Galilean map preserves time intervals and the distances between simultaneous events — but it does not preserve the light sphere. Substituting $x = x' + vt'$ , $t = t'$ into $x^2+y^2+z^2 = c^2t^2$ gives $(x'+vt')^2 + y'^2 + z'^2 = c^2t'^2$ : a sphere of radius $ct'$ whose centre has slipped to $x' = -vt'$ . Galilean kinematics cannot give light the same speed in two frames.

The core insight

The Galilean transformation is a shear of spacetime: it slides the spatial origin by $vt$ and leaves the clock alone. Because $t' = t$ , all inertial observers share one "now", and every time interval and every distance between simultaneous events survives the map. What does not survive is the spherical light front: push $x^2+y^2+z^2 = c^2t^2$ through and the sphere reappears in $S'$ displaced by $vt'$ , so light comes out at $c - v$ forwards and $c + v$ backwards. To preserve the same speed of light for every inertial observer, the assumption $t'=t$ must be abandoned, leading to the Lorentz transformation.

Where this lives today

Every piece of software that fuses sensor data across moving platforms runs equation (1) millions of times a second. An inertial navigation system transforms accelerometer readings from the vehicle frame into the local level frame; a self-driving car's perception stack maps LiDAR returns into the world frame, correcting for the vehicle's drift $vt$ between scans; a radar tracker converts targets between antenna and ground frames. All use the Galilean map without apology, because at $v/c \sim 10^{-7}$ the shear is the exact answer to every digit their hardware can resolve. And the failure has become an instrument. Modern Lorentz-invariance tests — cryogenic optical resonators on turntables, optical clocks compared as the Earth turns — hunt for exactly the anisotropy of worked example 2, a speed of light that depends on direction, and now bound it at the level of parts in $10^{18}$ . The Galilean transformation is not merely superseded; it is one of the most precisely falsified equations in physics.

The physical picture

Two observers, one event. The platform's ruler and the train's ruler read from origins that started together and have drifted apart at $v$ ever since, so by time $t$ they disagree by exactly $vt$ . The clocks, though, are assumed to agree perfectly and forever. That asymmetry between the space entry and the time entry is the whole content of the transformation — and the whole target of what comes after it.

Figure

One event, two logbooks — platform frame S and train frame S′ (v = 30 m/s). After t = 2 s the origins have drifted apart by vt = 60 m, so the flash logged at x = 100 m in S is logged at x′ = 100 − 60 = 40 m in S′. Both clocks read 2 s: t′ = t.

Now put time on the vertical axis and watch what the map does to the whole diagram at once. Every line of constant $x'$ — the worldline of a point fixed in the train — tips over to the right, because that point is marching down the track at $v$ . But every line of constant $t$ stays flat, because $t' = t$ leaves the clock untouched. Tilt the verticals, keep the horizontals: that operation is a shear, and it is what a Galilean boost looks like geometrically. The relativistic replacement performs the same tilt on the worldlines and then, scandalously, tilts the horizontals too.

Figure

PUNCHLINE — the Galilean boost is a SHEAR of the x–t diagram. The train's worldlines (blue) tilt over; the lines of constant time (green) stay exactly horizontal, because t′ = t. The event at (x = 100 m, t = 2 s) sits on the x′ = 40 m worldline, so S′ logs (40 m, 2 s). The dashed red line is what the relativistic replacement does instead — it tilts the simultaneity lines too (tilt exaggerated).

The core equations

Two frames in standard configuration: $S'$ slides along the $x$ -axis of $S$ at constant velocity $v$ , the axes stay parallel, and the origins coincide at $t = 0$ . Add the one classical assumption — a single clock for the universe — and the dictionary between the two logbooks is fixed.

x' = x - vt, \qquad y' = y, \qquad z' = z, \qquad t' = t \qquad (1)

where: $(x,y,z,t)$ — an event's coordinates in $S$ ; $(x',y',z',t')$ — the same event's coordinates in $S'$ ; $v$ — the constant speed of $S'$ along $x$ . Physical reading: the transverse coordinates are untouched (nothing moves sideways), the along-motion coordinate is sheared by the drift $vt$ of the origins, and the clock is copied straight across. The clause $t' = t$ is not a definition or a convenience; it is a physical hypothesis about the world, and it is the one relativity discards. Differentiate the first line once with respect to the shared time and you get the velocity rule; differentiate again and the constant $v$ vanishes.

u' = u - v, \qquad\qquad a' = a \qquad (2)

where: $u, a$ — a particle's velocity and acceleration in $S$ ; $u', a'$ — the same in $S'$ . Physical reading: velocities shift by the frame velocity, accelerations do not shift at all. Since classical forces depend only on relative positions and velocities, $\mathbf{F} = m\mathbf{a}$ therefore reads identically in both frames — the mechanics article named above proves this, and Galilean Relativity turns it into a principle. Because $t' = t$ , differences of coordinates transform as simply as the coordinates do, and two quantities survive the map untouched.

\Delta t' = \Delta t, \qquad\qquad |\Delta \mathbf{r}'| = |\Delta \mathbf{r}| \quad\text{whenever}\quad \Delta t = 0 \qquad (3)

where: $\Delta t$ — the time between two events; $|\Delta\mathbf{r}|$ — the distance between two simultaneous events. Physical reading: these are the invariants of Galilean spacetime. Durations are absolute, lengths measured at one instant are absolute, and — the sharpest consequence — simultaneity is absolute: if $\Delta t = 0$ then $\Delta t' = 0$ for every inertial observer, so all share one universal "now". The fine print: the distance between events at different times is not invariant — it shifts by $v\,\Delta t$ , which is the shear again. Perform one boost and then another, and the composite is a third boost. This is what makes the boosts a group, and it is the algebraic face of the same shear.

B(v_2)\,B(v_1) = B(v_1 + v_2) \qquad (4)

where: $B(v)$ — the boost of equation (1) with velocity $v$ . Physical reading: boosts compose by adding their velocities, with identity $B(0)$ and inverse $B(-v)$ — a one-parameter group, closed and commutative. Velocity is an additive label, which is precisely why the addition rule of the previous article is exactly linear, with no ceiling anywhere. And now the equation that breaks it. Let a flash go off at the common origin at $t = 0$ . In $S$ its wavefront is a sphere expanding at $c$ ; substitute the inverse map $x = x' + vt'$ , $t = t'$ and see what $S'$ is obliged to report.

x^2 + y^2 + z^2 = c^2t^2 \quad\xrightarrow{\ \text{Galilean}\ }\quad (x' + vt')^2 + y'^2 + z'^2 = c^2 t'^2 \qquad (5)

where: $c = 2.998\times10^{8}\ \mathrm{m/s}$ — the speed of light. Physical reading: the right-hand equation is a sphere of radius $ct'$ — but its centre sits at $x' = -vt'$ , not at the origin of $S'$ . The moving observer has run out from under her own light front by $vt'$ , and along her $x'$ -axis the wavefront reaches $(c - v)t'$ forwards and $-(c + v)t'$ backwards: light at $c - v$ one way and $c + v$ the other. Form-invariance fails, for exactly one reason — the clause $t' = t$ . This is the crisis that The Constancy of the Speed of Light and Einstein's Two Postulates exist to resolve.

Where the results come from

Derivation AThe transformation, its linearity, and the group law

The goal is to derive equation (1) rather than assert it, and then to expose the two structural facts hiding inside it. We assume only three things: that the frames are inertial, that the origins coincide at $t = 0$ , and that a single universal clock serves both observers. From the third assumption the time line is immediate; from the first, linearity is forced; and from linearity, the composition law $B(v_2)B(v_1) = B(v_1 + v_2)$ falls out, which is the statement that the boosts form a one-parameter group and that velocity is an additive label.

Step 1. Take absolute time at its word: the same clock reading labels the event for both observers, so $t' = t$ . Why this is a hypothesis and not a triviality: it asserts that the clocks can be synchronised once and will stay synchronised regardless of how the frames move. Nothing forces it — it is an empirical claim about the world, and it is the claim that will fail.

Step 2. The origin of $S'$ sits at $x = vt$ in $S$ . A point at $x$ in $S$ is therefore a distance $x - vt$ from that moving origin, and since the axes stay parallel and nothing moves transversely, $x' = x - vt$ , $y' = y$ , $z' = z$ . Why nothing more is needed: with a shared clock the only difference between the two rulers is where their zeros are, and that gap grows at exactly $v$ per second.

Step 3. Check that the map had to be linear in $(x, t)$ . A free particle moves at constant velocity in $S$ , so its worldline is straight, and the principle of inertia demands it be straight in $S'$ too — and only a linear map carries every straight line to a straight line. Why this matters: linearity is a consequence of the frames being inertial, not an aesthetic choice — and it is a constraint the eventual replacement must also satisfy, which is why the Lorentz transformation is linear too.

Step 4. Compose two boosts. Let $S'$ move at $v_1$ relative to $S$ , and $S''$ at $v_2$ relative to $S'$ . Then $x'' = x' - v_2 t' = (x - v_1 t) - v_2 t = x - (v_1 + v_2)t$ , with $t'' = t' = t$ . So the composite is precisely $B(v_1 + v_2)$ . Why this is the group law: the boosts are closed under composition, $B(0)$ is the identity, $B(-v)$ the inverse, and composing simply adds the parameters. Velocity is an additive label with no upper bound — exactly the property the speed of light will refuse to respect. *As a check: composing two boost matrices symbolically returns the single boost $B(v_1+v_2)$ ; the matrix has a constant Jacobian (confirming linearity); $B(0)$ is the identity and $B(-v)$ the inverse; and, for Worked Example 1's numbers (event at $x = 100\ \mathrm{m}$ , $t = 2\ \mathrm{s}$ , with $v_1 = 30\ \mathrm{m/s}$ and $v_2 = 12\ \mathrm{m/s}$ ), the results $x' = 100 - 60 = 40\ \mathrm{m}$ and the two-boost composite $v_1 + v_2 = 42\ \mathrm{m/s}$ follow directly.*

Go deeper · Derivation B · 2 minThe light sphere is not form-invariant — the centre slips by vt′

Derivation BThe light sphere is not form-invariant — the centre slips by vt′

Here is the result this article owns, and the seed of the entire crisis. Equation (3) showed that the Galilean map preserves time intervals and the distances between simultaneous events, which is why classical physics never noticed anything amiss. We now feed the map the one object it cannot digest: a spherical light front. If the speed of light is $c$ in $S$ , the wavefront from a flash at the origin satisfies $x^2 + y^2 + z^2 = c^2t^2$ . Ask what that same surface looks like in $S'$ , and the answer is a sphere whose centre has slid backwards by $vt'$ — so the wavefront cannot be racing away at $c$ in every direction for both observers.

Step 1. Fire a flash at the common origin at $t = 0$ , and let it spread at speed $c$ in $S$ . At time $t$ the wavefront is the set of points at distance $ct$ from the origin: $x^2 + y^2 + z^2 = c^2t^2$ . Why this is the right test object: it is the only surface in physics whose shape encodes a law-given speed. A cricket ball's trajectory says nothing about the frame; a light sphere says everything.

Step 2. Invert the transformation to express $S$ coordinates in terms of $S'$ ones: $x = x' + vt'$ , $y = y'$ , $z = z'$ , $t = t'$ . Substitute into the wavefront equation:

(x' + vt')^2 + y'^2 + z'^2 = c^2 t'^2

Why the substitution is legitimate: the two observers are describing the same physical surface at the same instant — which is a meaningful thing to say precisely because $t' = t$ makes "the same instant" frame-independent.

Step 3. Read the geometry. Written as $\big(x' - (-vt')\big)^2 + y'^2 + z'^2 = (ct')^2$ , this is a sphere of radius $ct'$ centred at $x' = -vt'$ . It is not $x'^2 + y'^2 + z'^2 = c^2t'^2$ ; the two differ by $2vt'x' + v^2t'^2$ , which vanishes only if $v = 0$ . Why the sign is worth pausing on: the centre lags behind the moving observer, at negative $x'$ . She is running forward out from under the light front that was launched from where she used to be.

Step 4. Extract the speeds. Set $y' = z' = 0$ : the wavefront cuts the $x'$ -axis at $(c - v)t'$ and $-(c + v)t'$ , so $S'$ measures light receding at $c - v$ along her motion and $c + v$ against it — a directional difference of $2v$ . Why this is the whole story in one line: light of speed $c$ in some frame, plus a Galilean transformation, is a theory with a preferred frame, detectable by comparing light speeds forwards and backwards. Nature has been asked, repeatedly, and says no.

Step 5. Locate the guilty assumption. Nothing above used dynamics, forces or masses — only the coordinate map — and within the map the offset traces to a single place: $t' = t$ forces the time coordinate to carry no dependence on position, so the only freedom left is the spatial shear, and a shear cannot preserve a sphere's centre. Why this points forward: the repair must let $t'$ depend on $x$ , which is to say it must abandon absolute simultaneity — the business of The Relativity of Simultaneity and The Lorentz Transformation. *As a check: substituting $x = x' + vt'$ , $t = t'$ into $x^2+y^2+z^2-c^2t^2$ gives $(x'+vt')^2 + y'^2 + z'^2 - c^2t'^2$ ; the residual $2vt'x' + v^2t'^2$ is non-zero for $v \neq 0$ ; solving for the centre gives $x' = -vt'$ with radius $ct'$ ; the axis intercepts yield speeds $c - v$ and $c + v$ ; and the centre offsets are $60\ \mathrm{m}$ (at $v = 30\ \mathrm{m/s}$ , $t' = 2\ \mathrm{s}$ ) and $29.8\ \mathrm{km}$ (at Earth's orbital speed, $t' = 1\ \mathrm{s}$ ).*

Limiting cases

Case	Result	Meaning
Two boosts in succession	$B(v_2)B(v_1) = B(v_1+v_2)$	velocities add; the boosts are a one-parameter group with no ceiling
Simultaneous events, $\Delta t = 0$	$\Delta t' = 0$ , $\|\Delta\mathbf{r}'\| = \|\Delta\mathbf{r}\|$	simultaneity and instantaneous distance are absolute — the assumption relativity overturns
Events at different times	$\|\Delta\mathbf{r}'\| \neq \|\Delta\mathbf{r}\|$	the separation shifts by $v\,\Delta t$ ; only simultaneous distances are invariant
A light sphere	centre slips to $x' = -vt'$	form-invariance fails: light runs at $c - v$ forwards, $c + v$ backwards
The formal limit $c \to \infty$	the offset $vt'$ is negligible against the radius $ct'$	the ratio $v/c \approx 10^{-7}$ for a train — why nobody noticed for two centuries

Common trap“t′ = t is just a definition — of course the clocks agree”

It is not a definition. It is the single boldest physical claim in classical mechanics, and it is false. A definition is a free choice with no consequences; $t' = t$ has enormous consequences, every one testable. It says that two observers in relative motion, having once synchronised their clocks, will agree forever on the reading of any event; that a process lasting $\Delta t$ for one lasts $\Delta t$ for all; and — the load-bearing consequence — that if two firecrackers explode simultaneously for you, they explode simultaneously for the passenger, however fast the train. Each is a prediction, and each fails once speeds get large enough to measure the difference. The tell is already visible in equation (5): keeping the clock rigid forces the whole burden of the transformation onto the spatial coordinate, and a pure spatial shear cannot hold a light sphere centred. Take away one habit from this page: the habit of noticing $t' = t$ every time you write it.

Worked examples

1 · Transforming an event, and stacking two boosts

A firecracker explodes at $x = 100\ \mathrm{m}$ , $t = 2\ \mathrm{s}$ in the platform frame $S$ . Frame $S'$ (a train) moves at $v_1 = 30\ \mathrm{m/s}$ along $+x$ , with origins coinciding at $t = 0$ . (a) Give the event's coordinates in $S'$ . (b) A trolley $S''$ rolls along the train's aisle at $v_2 = 12\ \mathrm{m/s}$ . Give the event's coordinates in $S''$ , and the speed of $S''$ over the ground.

Step 1. Apply equation (1) directly. The origins have drifted apart by $v_1 t = (30)(2) = 60\ \mathrm{m}$ , so $x' = 100 - 60 = \boxed{40\ \mathrm{m}}$ , while the clock is copied across: $t' = t = \boxed{2\ \mathrm{s}}$ . The train logs the flash $40\ \mathrm{m}$ ahead of its own origin, at the same instant the platform does.

Step 2. Now boost again, from $S'$ to $S''$ , using the same rule with $v_2 = 12\ \mathrm{m/s}$ and $t'' = t' = 2\ \mathrm{s}$ : $x'' = x' - v_2 t' = 40 - 24 = \boxed{16\ \mathrm{m}}$ .

Step 3. Check with the group law, equation (4). Composing the two boosts should be a single boost at $v_1 + v_2 = 30 + 12 = \boxed{42\ \mathrm{m/s}}$ — the trolley's speed over the ground. Applying that single boost to the original event: $x'' = 100 - (42)(2) = 100 - 84 = 16\ \mathrm{m}$ . It agrees, as the group law guarantees.

Step 4. Note what the check proved. Two successive frame changes equal one frame change with the velocities added — no correction term, no ceiling. Stack enough boosts and the Galilean rule reports a speed greater than $c$ , without a murmur of complaint.

2 · The light sphere refuses to sit still

A flash goes off at the common origin at $t = 0$ . Using the Galilean transformation, locate the wavefront in $S'$ after $t' = 2\ \mathrm{s}$ for a train at $v = 30\ \mathrm{m/s}$ , and then repeat for the Earth in its orbit ( $v = 29.8\ \mathrm{km/s}$ , $t' = 1\ \mathrm{s}$ ). How big is the effect, and which way does it point?

Step 1. Write the transformed wavefront, equation (5): $(x' + vt')^2 + y'^2 + z'^2 = c^2t'^2$ . This is a sphere of radius $R = ct'$ centred at $x'_{\rm c} = \boxed{-vt'}$ — displaced backwards along the direction of motion.

Step 2. Put in the train's numbers. Radius: $R = ct' = (2.998\times10^{8})(2) = \boxed{5.996\times10^{8}\ \mathrm{m}}$ . Centre offset: $|x'_{\rm c}| = vt' = (30)(2) = \boxed{60\ \mathrm{m}}$ — the very same $60\ \mathrm{m}$ that shifted our firecracker from $100\ \mathrm{m}$ to $40\ \mathrm{m}$ . As a fraction of the radius that is $v/c = \boxed{1.0\times10^{-7}}$ — one part in ten million, which is why the shear went unnoticed for two centuries.

Step 3. Now the Earth, ploughing through the supposed ether at its orbital speed. Offset after one second: $vt' = 29.8\ \mathrm{km}$ , against a radius $ct' = 2.998\times10^{8}\ \mathrm{m}$ , a fraction $v/c = \boxed{9.94\times10^{-5}}$ — about one part in ten thousand, and now genuinely within reach of a good interferometer.

Step 4. Convert the offset into the observable. The wavefront cuts the $x'$ -axis at $(c-v)t'$ and $-(c+v)t'$ , so $S'$ should measure light at $c - v = \boxed{2.99770\times10^{8}\ \mathrm{m/s}}$ downwind and $c + v = \boxed{2.99830\times10^{8}\ \mathrm{m/s}}$ upwind — a directional difference of $2v = 59.6\ \mathrm{km/s}$ . That anisotropy is precisely what the Michelson–Morley Experiment went hunting, and precisely what it failed to find. The failure is not the experiment's; it is equation (1)'s.

Physics of the portrait

The portrait here is not a person but an idea, and its provenance is more tangled than its name suggests. Galileo argued the principle — a sealed cabin below decks reveals nothing about the ship's motion — in the Dialogo of 1632, and put motion on a mathematical footing in the Discorsi of 1638. He never wrote $x' = x - vt$ : the coordinate form belongs to the Newtonian tradition, whose absolute time supplies the clause $t' = t$ . The name is retrospective, attached in the twentieth-century relativity literature, which needed a label for the map that the Lorentz transformation had come to replace.

Origins · 1 minGalilean transformation and how this was discovered

Galilean transformation

1632–1687 · from Galileo's ship to Newton's absolute time

The idea has three fathers and no single author. Galileo (1632) supplied the physical content: below decks on a smoothly sailing ship no experiment betrays the ship's motion, so the descriptions given by two uniformly moving observers must be related by some dictionary that leaves the laws alone. Newton (1687) supplied the assumption that fixes which dictionary — absolute time, "flowing equably without relation to anything external" — and it is that assumption, not Galileo's principle, which forces the clause $t' = t$ . The nineteenth century supplied the algebra, writing the map out in coordinates. Only in the twentieth did the transformation acquire Galileo's name, and it acquired it precisely because relativity had produced a rival worth contrasting it with — the point argued in the title of a 2020 American Journal of Physics paper, "Galilei proposed the principle of relativity, but not the 'Galilean transformation'". The moral is worth keeping: Galileo's principle survived 1905 intact and is a cornerstone of relativity to this day. What did not survive was Newton's clock, smuggled in as $t' = t$ .

History

1632

— Galileo's Dialogo: the sealed ship's cabin. Uniform motion is undetectable from within — the relativity principle, stated as physics, not as equations.

1638

— Galileo's Discorsi (Two New Sciences), printed in Leiden, gives uniform and accelerated motion their first systematic mathematical treatment.

1687

— Newton's Principia: absolute time "flows equably without relation to anything external", supplying the clause

t' = t

; Corollary V states the relativity of uniform motion.

1887

— Michelson and Morley find no trace of the directional difference in the speed of light that equation (5) demands.

1905

— Einstein keeps the relativity principle and drops absolute time; the Lorentz transformation replaces the Galilean one.

1908

— Minkowski's Cologne lecture Raum und Zeit makes the contrast geometric: Galilean spacetime carries an absolute slicing into "nows"; relativistic spacetime does not.

2020

— An American Journal of Physics paper argues the historical point in its title: Galilei proposed the principle of relativity, but not the "Galilean transformation".

Going further

Texts. Taylor and Wheeler, Spacetime Physics, 2nd ed., ch. 3 · Rindler, Introduction to Special Relativity, 2nd ed., §§1–2 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.1 · Hartle, Gravity, ch. 4. All relations are derived in the sections above. Key results: the composition law $B(v_2)B(v_1) = B(v_1+v_2)$ with identity $B(0)$ and inverse $B(-v)$ ; the invariance of $\Delta t$ and of $|\Delta\mathbf{r}|$ at equal times, and its failure at unequal times (a shift of $v\Delta t$ ); and the substitution $x = x'+vt'$ , $t = t'$ into $x^2+y^2+z^2 = c^2t^2$ , giving $(x'+vt')^2+y'^2+z'^2 = c^2t'^2$ — a sphere of radius $ct'$ centred at $x' = -vt'$ , residual $2vt'x' + v^2t'^2 \neq 0$ , axis speeds $c \mp v$ . Numbers: $x' = 40\ \mathrm{m}$ , $x'' = 16\ \mathrm{m}$ , $30 + 12 = 42\ \mathrm{m/s}$ ; sphere radius $5.996\times10^{8}\ \mathrm{m}$ , offset $60\ \mathrm{m}$ , $v/c = 1.0\times10^{-7}$ ; Earth's orbital case, offset $29.8\ \mathrm{km}$ , $v/c = 9.94\times10^{-5}$ , light at $2.99770$ and $2.99830\times10^{8}\ \mathrm{m/s}$ ( $2v = 59.6\ \mathrm{km/s}$ ). Constant: $c = 2.998\times10^{8}\ \mathrm{m/s}$ .

Builds on. Inertial Frames of Reference · The Principle of Relativity · Absolute Space and Absolute Time. Leads to. Galilean Relativity · Symmetries of Galilean Spacetime · Electromagnetism and the Failure of Galilean Invariance · The Constancy of the Speed of Light.

If the transformation is just a change of coordinates, how can it be wrong? Coordinates are our choice.

The choice of labels is free; the claim that a particular relabelling connects two physical observers is not. Equation (1) does not merely define $x'$ — it asserts that the number a moving observer's ruler actually reads is $x - vt$ , and that the number her clock actually reads is $t$ . Both are measurable claims about apparatus, and the second is false. The giveaway is that the map has physical consequences: it predicts light travelling at $c - v$ for her. A pure bookkeeping convention could never predict the outcome of an experiment — and this one does, wrongly.

Why does absolute simultaneity follow from t′ = t, and why is that such a big deal?

Take two events with $\Delta t = 0$ in $S$ — simultaneous for you. Since $t' = t$ for each separately, $\Delta t' = \Delta t = 0$ : simultaneous for every inertial observer, whatever their velocity or separation. So the Galilean world has a universal "now", one slicing of spacetime into instants that everybody shares. It matters because almost every classical intuition rests on it — that a length can be measured by noting both ends at once, that a duration belongs to a process rather than to a process-and-an-observer, that "before" and "after" are absolute. Delete the slicing and all three need rewriting: the business of The Relativity of Simultaneity.

Where exactly does the light sphere's offset come from? I can see the algebra but not the picture.

Picture the flash going off at the moment the two origins coincide. The light spreads outwards from that point in space — a definite place on the platform. As the seconds tick, the train observer keeps moving forward, away from that place, at $v$ ; after $t'$ seconds she is $vt'$ ahead of it. But the light sphere is still centred where it was launched. So in her coordinates the centre sits $vt'$ behind her, at $x' = -vt'$ , and the wavefront ahead of her is only $(c - v)t'$ away while the one behind is $(c + v)t'$ away. The algebra of equation (5) is just this sentence written in symbols. The reason it is a crisis is that nature says her forward and backward measurements come out equal — both $c$ .

If two boosts just add, can I stack enough of them to exceed the speed of light?

Within Galilean kinematics, yes — and that is not a paradox but a symptom. The group law $B(v_2)B(v_1) = B(v_1+v_2)$ has no fixed point, no ceiling, nothing distinguished about the value $c$ ; stack seventy-one boosts of $0.02c$ and it reports $1.42c$ without blinking. A theory whose composition law knows nothing about $c$ cannot protect $c$ . That is the light sphere's offset seen from the algebraic side, and it tells you what the replacement must look like: a composition law with $c$ built in as a fixed point.

The offset vt′ is tiny for anything I can build. So why not keep using the Galilean map?

You should keep using it, and everyone does. For a train the offset is one part in $10^{7}$ of the light sphere's radius; for the Earth in orbit, one part in $10^{4}$ . Inertial navigation, radar tracking and robot perception all run on equation (1) and are right to every digit their sensors deliver. The point is not that the map is useless but that it is not exact, and that the inexactness is structural rather than a matter of small numbers. Relativity is the study of a correction that is negligible in the laboratory and decisive for the logic — measured by $\beta = v/c$ , about $10^{-4}$ for the Earth.

What single change turns the Galilean transformation into the Lorentz transformation?

Letting the time coordinate depend on position. In equation (1) the time line is $t' = t$ — flat, position-blind — and it is exactly that flatness which pins the light sphere's centre to $x' = -vt'$ and lets light run at $c \pm v$ . Allow $t'$ to vary with $x$ and the clock at the far end of a moving frame is offset from the one at the near end; the shear then acts on the time axis too, tilting the simultaneity lines that the punchline figure shows lying flat. Everything counterintuitive in relativity descends from that one modification. The details belong to The Lorentz Transformation; here the job was to show why one is needed.