The Principle of Relativity

Shut yourself in a windowless cabin below the waterline of a ship. Watch the flies circle, drip water into a jar, jump about, toss a ball to a friend. Now let the ship get under way, smoothly, on a glassy sea. Galileo's claim — made in 1632, and one of the boldest sentences in physics — is that nothing whatever changes. The flies do not pile up against the stern bulkhead. The drops still land in the jar. You need not jump harder toward the bow than toward the stern. You can watch as long as you like and never learn, from inside the cabin, whether the ship is moving. That is the principle of relativity, and its content is a prohibition: no law of physics may contain your velocity. Not "velocity is hard to measure" — velocity relative to space itself has no meaning at all. Only velocities of bodies relative to other bodies exist. The principle does not say everything looks the same in every frame; it plainly does not, since the ball you drop falls straight down for you and along a curve for the harbourmaster on the quay. It says the laws governing those different-looking motions are identical, so no experiment inside the cabin can single your cabin out. Below we prove the mechanical half of that claim. Newton's second law is form-invariant under a change to a uniformly moving frame: a constant velocity dies under the second derivative, so acceleration is the same for everybody, and forces built from relative positions and velocities never notice the switch. The equations come out identical, symbol for symbol — which is why the flies do not pile up. Two and a half centuries later the principle collided with Maxwell's electromagnetism, which hands light one definite speed and attaches it to no ship at all. Einstein kept the principle, promoted it to all of physics, and gave up something else instead.

Key takeaways

The laws of physics take the same form in every inertial frame: no experiment performed entirely inside a uniformly moving laboratory can reveal that it is moving.
It constrains the form of laws, not the values of quantities. Positions, velocities and kinetic energies differ between frames — the equations they obey do not.
Mechanically it is a theorem, not a hope: under the Galilean boost $x' = x - vt,\ t' = t$ with $v$ constant, $a' = a$ and $\mathbf F' = \mathbf F$ , so $F = ma$ maps to $F' = ma'$ symbol for symbol.
The boosted equations contain no $v$ at all, so absolute uniform velocity is unmeasurable — while acceleration, which does survive, is measurable at once (see Inertial Frames of Reference).
Total momentum transforms as $\mathbf P' = \mathbf P - M\mathbf v$ , a constant shift, so a collision conserving momentum in one inertial frame conserves it in all of them.
Key idea: the principle of relativity is not the claim that everything looks the same in every frame — lengths, speeds and trajectories famously do not. It is the claim that the LAWS are the same in every inertial frame, so no internal experiment can tell you which frame you are in.

The core insight

Uniform motion has no consequences detectable from inside. Change to a frame moving at constant velocity and every equation of mechanics comes back looking exactly as it did, because a constant velocity vanishes under the second time derivative and cancels out of every relative separation. Velocity is a relation between bodies, never a property one body owns — and any proposed law that would let you measure your own speed through empty space is, for that reason alone, wrong.

Where this lives today

The most practical descendant of Galileo's cabin sits in every phone and every airliner: the inertial measurement unit. Its accelerometers and gyroscopes report nothing at all while you cruise at constant velocity — they physically cannot, because uniform velocity has no observable — and everything the instant you turn, brake or climb. That is not a design limitation waiting to be engineered away; it is equation (2) in silicon, and it is why an inertial navigator must integrate acceleration twice to know where it is. The deeper legacy is methodological. The principle is used as a filter: any proposed law must take the same form in every inertial frame, and that demand kills vast classes of candidate theories before an experiment is designed. Modern precision tests hunt for the tiniest breach — comparing optical clocks and nuclear-spin resonances as the Earth turns and swings around the Sun, looking for any direction or velocity that physics secretly prefers. None has been found, at parts in $10^{18}$ and beyond. Four centuries on, the flies in Galileo's cabin still show no sign of piling up against the stern.

The physical picture

Start with the experiment Galileo actually described. A ball is released from the top of a mast on a ship gliding at steady speed. Aristotle's followers — and, honestly, most people asked today — expect the ball to be "left behind": the ship sails on while the ball falls, so it should land back toward the stern. It does not. It lands at the foot of the mast, exactly where it lands when the ship is moored. The ball was already moving with the ship when released, and nothing takes that motion away from it.

The fallacy dies easily. The ball is "left behind" only if releasing it strips away the horizontal motion it already had — only, that is, if the natural state of matter is rest with respect to some privileged background. Take that away and the ball keeps what it has: it moves forward with the ship at $10\ \mathrm{m/s}$ throughout its fall, so the mast base never runs away underneath it. Now the sharper question. What does the harbourmaster on the quay see? He sees the ball leave the masthead already travelling forward, curve down along a parabola, and land metres downrange. The passenger sees a straight vertical drop. Two completely different trajectories — yet both agree the ball lands at the foot of the mast, both find the fall takes $0.495\ \mathrm{s}$ , and both explain what they see with $F = ma$ and the same downward weight.

Figure

The punchline — a ball dropped from h = 1.2 m in a cabin moving at v = 10 m/s. In the cabin it drops straight down (Δx′ = 0); over the ground it flies a 4.95 m parabola. Same fall time t = √(2h/g) = 0.495 s, same landing point on the cabin floor, same law F = ma

Look hard at what the figure is not saying. It is not saying the observers measure the same things: the ground observer clocks the ball at $11.12\ \mathrm{m/s}$ when it lands, the passenger at $4.85\ \mathrm{m/s}$ ; the ground observer sees a $4.95\ \mathrm{m}$ horizontal displacement, the passenger sees zero. What it is saying is that both descriptions come out of the same equation with the same force, and that the physically meaningful event — where the ball meets the floor — is one they agree on. The disagreement about numbers is what the transformation is for; the agreement about form is the principle of relativity.

The core equations

Everything rides on one map between frames. Let $S$ be an inertial frame (the ground) and $S'$ a frame (the cabin) sliding along $x$ at a constant velocity $v$ , origins coinciding at $t = 0$ . Newton's world assumes one further thing that will not survive the century: a single universal time ticking for both.

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

where: $x, t$ — coordinates of an event in $S$ ; $x', t'$ — coordinates of the same event in $S'$ ; $v$ — the constant relative speed. Physical reading: to convert to the moving frame, subtract off how far that frame has travelled — and leave the clocks alone. This is the Galilean transformation, the world as Newton assumed it. Differentiate (1) with respect to the common time, following one particle. The constant $v$ survives the first derivative and dies in the second — and that single fact, that a constant cannot survive being differentiated twice, is the whole mechanical content of relativity.

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

where: $u = \dot x$ — the particle's velocity in $S$ ; $u'$ — its velocity in $S'$ ; $a = \ddot x$ — its acceleration. Physical reading: velocity is a matter of opinion between inertial frames, acceleration a matter of fact — all inertial observers assign the same $a$ to the same particle. Everything follows from this asymmetry. Now the dynamics. Real forces — gravity between two masses, a spring between two ends, contact and friction — depend on how bodies are placed and how they move relative to one another, never on where they sit in absolute space or how fast the laboratory drifts. Write an isolated system that way:

m_i\,\ddot{\mathbf r}_i \;=\; \sum_{j \neq i} \mathbf F_{ij}\big(\mathbf r_i - \mathbf r_j,\; \dot{\mathbf r}_i - \dot{\mathbf r}_j\big) \qquad (3)

where: $m_i$ — the mass of body $i$ , frame-independent in Newtonian mechanics; $\mathbf F_{ij}$ — the force on $i$ from $j$ , a function only of relative separation and relative velocity. Physical reading: under a boost, $\mathbf r_i - \mathbf r_j$ and $\dot{\mathbf r}_i - \dot{\mathbf r}_j$ are unchanged (the $-\mathbf v t$ and $-\mathbf v$ cancel in every difference) while $\ddot{\mathbf r}_i$ is unchanged by (2) — so (3) reproduces itself, with primes on every symbol and $v$ nowhere. The symmetry propagates into the conserved quantities, which is what makes it useful rather than merely elegant. Total momentum is not invariant — every velocity shifts by $-\mathbf v$ — but it shifts by a constant, and a constant cannot spoil a conservation law.

\mathbf P' \;=\; \sum_i m_i (\dot{\mathbf r}_i - \mathbf v) \;=\; \mathbf P - M\mathbf v, \qquad M = \sum_i m_i \qquad (4)

where: $\mathbf P$ — total momentum in $S$ ; $M$ — the fixed total mass. Physical reading: $M\mathbf v$ is the same before and after any collision, so $\Delta\mathbf P' = \Delta\mathbf P$ — momentum conserved in one inertial frame is conserved in every one, and the same argument run on kinetic energy is why an elastic collision is elastic for everybody.

Where the invariance comes from

Derivation ANewton's second law has the same form in every inertial frame

This is the mechanical core of the whole subject and deserves proof rather than assertion: if a law of motion is second-order in time, and its forces are built from relative positions and relative velocities, then changing to a uniformly moving frame leaves it literally unchanged — same symbols, same structure, no leftover $v$ anywhere. Apply the map (1), differentiate twice, and watch what the boost does to each ingredient of $F = ma$ . The payoff is the corollary at the end: absolute uniform velocity is not merely hard to measure, it is not in the theory at all.

Step 1. Boost the coordinates: $\mathbf r\,' = \mathbf r - \mathbf v t$ , $t' = t$ , with $\mathbf v$ constant. Differentiate once, following the particle: $\dot{\mathbf r}\,' = \dot{\mathbf r} - \mathbf v$ . Why: the observers must disagree about velocity — one moves relative to the other — but by a fixed amount that never changes with time.

Step 2. Differentiate again: $\ddot{\mathbf r}\,' = \ddot{\mathbf r}$ , since $\dot{\mathbf v} = \mathbf 0$ . Acceleration is boost-invariant. Why: this is the hinge. A constant velocity survives one derivative and dies in the second — which is why a law written in accelerations can never see it.

Step 3. Transform the forces. Every relative separation is untouched, $\mathbf r_i\,' - \mathbf r_j\,' = \mathbf r_i - \mathbf r_j$ , and likewise every relative velocity — so $\mathbf F_{ij}' = \mathbf F_{ij}$ . Why: the boost shifts everything together, and a force that cares only about differences cannot notice a shift common to all of its arguments.

Step 4. Reassemble: $m_i\,\ddot{\mathbf r}_i\,' = m_i\,\ddot{\mathbf r}_i = \sum_j \mathbf F_{ij} = \sum_j \mathbf F_{ij}'$ — equation (3) with a prime on every symbol. Formally $\partial(\text{eom})/\partial v \equiv 0$ : the boosted equations contain no $v$ , so no solution of them — no mechanical experiment — can depend on $v$ . Therefore no mechanical experiment can detect absolute uniform velocity. Why: this is Galileo's cabin, proved. The flies, the drops and the jumps all solve equations from which $v$ has vanished, so none can report it. *Verified symbolically in verify_aq-sr-principle-of-relativity.py (S1–S3): $u' = u - v$ , $a' = a$ , $\mathbf F' = \mathbf F$ , the primed equation of motion identical to the unprimed one, and $\partial_v(\text{eom}) = 0$ .*

Go deeper · Derivation B · 2 minConservation laws survive the boost, and a collision proves it

Derivation BConservation laws survive the boost, and a collision proves it

Form-invariance of $F = ma$ would be thin if it did not descend into what experimenters actually measure. Nobody in a laboratory checks a differential equation; they check that momentum comes out the same before and after a collision. So we show the boost carries conservation laws with it — a collision conserving momentum on the platform necessarily conserves it on the train, though the two observers write down different velocities, a different total momentum and a different kinetic energy. The mechanism: total momentum is not invariant, but it changes by a term that is the same before and after, and such a term cannot break a conservation law. The same trick recurs throughout physics.

Step 1. Boost every velocity: $\mathbf u_i\,' = \mathbf u_i - \mathbf v$ , by equation (2). Why: a boost acts on each body identically — it does not care which body is which, or whether a collision is in progress.

Step 2. Sum with the masses: $\mathbf P\,' = \sum_i m_i \mathbf u_i - \big(\sum_i m_i\big)\mathbf v = \mathbf P - M\mathbf v$ , equation (4). Why: the boost adds a rigid drift $-\mathbf v$ to the system, and the momentum of a rigid drift is total mass times drift velocity.

Step 3. Take the change across the collision. Mass is conserved and $\mathbf v$ is a constant of the frame, so the offset $M\mathbf v$ is identical before and after: $\Delta\mathbf P\,' = \Delta\mathbf P$ , hence $\Delta\mathbf P = \mathbf 0 \iff \Delta\mathbf P\,' = \mathbf 0$ . Why: a fixed bias on the scales cancels out of any difference — it can never turn a conserved quantity into a non-conserved one.

Step 4. Repeat for kinetic energy: $T' = \sum \tfrac12 m_i(\mathbf u_i - \mathbf v)^2 = T - \mathbf v\cdot\mathbf P + \tfrac12 M v^2$ . The extra terms are built from $\mathbf P$ and $M$ , neither changed by the collision, so $\Delta T' = \Delta T$ — elastic in one inertial frame, elastic in all. Why: kinetic energies are wildly frame-dependent ( $9\ \mathrm{J}$ in one frame, $16.5\ \mathrm{J}$ in another), yet whether energy is conserved is not a matter of opinion.

Step 5. Note the boundary. Every step used $\dot{\mathbf v} = \mathbf 0$ . Let the frame accelerate at $\mathbf A$ and the argument collapses at Step 1: $\mathbf a\,' = \mathbf a - \mathbf A$ , free bodies swerve, momentum is not conserved in the frame at all. Why: the principle speaks about <strong>uniform</strong> motion, and this is exactly where its authority stops — the subject of Inertial Frames of Reference. *Verified in verify_aq-sr-principle-of-relativity.py (S4, S5, N4–N6): $\mathbf P' = \mathbf P - M\mathbf v$ and $\Delta\mathbf P' = \Delta\mathbf P$ symbolically, plus both frame audits of the collision.*

Limiting cases

Case	Result	Meaning
Uniform boost, $v$ constant	$a' = a$ , $\mathbf F' = \mathbf F$ ; no $v$ in the equations	the principle holds exactly — no mechanical experiment finds $v$ (Galileo's cabin)
Frame accelerating at $\mathbf A$	$\mathbf a\,' = \mathbf a - \mathbf A$ ; free bodies swerve	the principle does not apply; the acceleration is measurable from inside — see Inertial Frames of Reference
Rotating frame, $\boldsymbol\Omega \neq 0$	centrifugal and Coriolis terms appear	also non-inertial; rotation, like acceleration, is detectable without looking out
Low speed, $\beta = v/c \to 0$ (the cabin: $\beta = 3.3\times10^{-8}$ )	the Galilean map (1) is the correct realisation	the Newtonian world — velocities add, time is universal
$v$ comparable to $c$	the principle stands; the group implementing it cannot be Galilean	the transformation must be replaced — see Einstein's Two Postulates and The Lorentz Transformation
Gravity, extended regions	no global inertial frame exists	the principle survives only locally, in freely falling frames — see The Equivalence Principle

Read the table downward and you have the plot of the whole SR track. The principle never fails; what fails is one particular way of realising it — the Galilean transformation with its universal time — and only when light gets involved.

Common trap“Same laws in every frame” does not mean “same numbers in every frame”

This misreading wrecks more first encounters with relativity than any other, and it arrives as a triumphant objection: if all frames are equivalent, how can the ground observer see the ball fly 4.95 m sideways while the passenger sees it drop straight down? Neither is wrong. Positions, velocities, trajectories, momenta and kinetic energies are all frame-dependent, and they are supposed to be: worked example 2 gets $6\ \mathrm{kg\,m/s}$ of momentum in one inertial frame and $-9\ \mathrm{kg\,m/s}$ in another, both correct. What the principle asserts is invariance of the laws — both observers write $F = ma$ with the same $F$ , and both find momentum conserved. The transformation exists precisely because the numbers differ. So when relativity later announces that even lengths and elapsed times are frame-dependent, that is no violation of the principle — it is the same lesson, one level deeper.

Worked examples

1 · The ball dropped in a moving train

A passenger on a train moving at a constant $v = 10\ \mathrm{m/s}$ drops a ball from height $h = 1.2\ \mathrm{m}$ . Where does it land, according to the passenger and to a signaller beside the track? Take $g = 9.81\ \mathrm{m/s^2}$ .

Step 1. Work in the train frame $S'$ — inertial, because $v$ is constant. The ball starts at rest there and the only force is its weight, so $a_y' = -g$ , $a_x' = 0$ , and $h = \tfrac12 g t^2$ gives $t = \sqrt{2h/g} = \sqrt{2 \times 1.2 / 9.81}$ .

Step 2. Evaluate: $t = \boxed{0.495\ \mathrm{s}}$ . Horizontally the ball had zero velocity and there is no horizontal force, so $\Delta x' = \boxed{0}$ — it lands exactly at the dropper's feet. The train's speed appears nowhere in this calculation, which is the principle doing its work.

Step 3. Now the signaller's frame $S$ . The ball leaves the hand already moving forward at $10\ \mathrm{m/s}$ and keeps that horizontal velocity, so $x(t) = 10t$ , $y(t) = 1.2 - \tfrac12(9.81)t^2$ — a parabola. The vertical equation is identical, so the fall time is again $0.495\ \mathrm{s}$ .

Step 4. In that time the ball travels $\Delta x = vt = 10 \times 0.4946 = \boxed{4.95\ \mathrm{m}}$ downrange — and the train, carrying the passenger's feet, travelled the same $4.95\ \mathrm{m}$ . The ball still lands at the passenger's feet.

Step 5. Compare the ledgers. Landing speed in the train: $gt = 4.85\ \mathrm{m/s}$ , straight down. Over the ground: $\sqrt{10^2 + 4.85^2} = 11.12\ \mathrm{m/s}$ , slanted. The observers agree on the event and disagree on almost every number describing how it got there — and both used $F = ma$ with the identical force $-mg$ .

2 · One elastic collision, two inertial frames

A $2.0\ \mathrm{kg}$ trolley at $3.0\ \mathrm{m/s}$ hits a stationary $1.0\ \mathrm{kg}$ trolley elastically, head-on. Check momentum and kinetic energy (a) in the laboratory and (b) in the frame of a truck driving past at $V = 5.0\ \mathrm{m/s}$ .

Step 1. Solve in the lab: $w_1 = \dfrac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2} = \dfrac{(2-1)(3)}{3} = 1.0\ \mathrm{m/s}$ and $w_2 = \dfrac{(m_2-m_1)u_2 + 2m_1u_1}{m_1+m_2} = \dfrac{2(2)(3)}{3} = 4.0\ \mathrm{m/s}$ — the heavy trolley slows to $1\ \mathrm{m/s}$ , the light one is kicked to $4\ \mathrm{m/s}$ .

Step 2. Audit the lab frame. Momentum: $2(3) + 1(0) = 6.0$ before, $2(1) + 1(4) = 6.0\ \mathrm{kg\,m/s}$ after. Kinetic energy: $\tfrac12(2)(9) = 9.0\ \mathrm{J}$ before, $\tfrac12(2)(1) + \tfrac12(1)(16) = 9.0\ \mathrm{J}$ after. Both conserved: $p = 6.0\ \mathrm{kg\,m/s}$ , $T = 9.0\ \mathrm{J}$ .

Step 3. Boost to the truck: subtract $5.0\ \mathrm{m/s}$ from every velocity. Before: $u_1' = -2.0$ , $u_2' = -5.0\ \mathrm{m/s}$ . After: $w_1' = -4.0$ , $w_2' = -1.0\ \mathrm{m/s}$ . From the truck, both trolleys move backwards throughout — a completely different-looking event.

Step 4. Audit the truck frame. Momentum: $2(-2) + 1(-5) = -9.0$ before, $2(-4) + 1(-1) = -9.0\ \mathrm{kg\,m/s}$ after. Kinetic energy: $\tfrac12(2)(4) + \tfrac12(1)(25) = 16.5\ \mathrm{J}$ before, $\tfrac12(2)(16) + \tfrac12(1)(1) = 16.5\ \mathrm{J}$ after. Both conserved: $p' = -9.0\ \mathrm{kg\,m/s}$ , $T' = 16.5\ \mathrm{J}$ .

Step 5. Check the theory: equation (4) predicts $P' = P - MV = 6 - (3)(5) = -9\ \mathrm{kg\,m/s}$ , exactly the audited value, with the same offset before and after — which is why conservation survived. Every number changed; every law held. (Boost instead by $V = P/M = 2.0\ \mathrm{m/s}$ and you land in the centre-of-mass frame, where $p' = 0$ before and after.)

Physics of the portrait

Galileo never wrote $a' = a$ . He did something harder: he saw that the question "is the ship moving?" has no meaning for anyone locked below its decks — and he saw it while the fate of the Copernican system hung on that point. The stock objection to a moving Earth was mechanical: if the Earth spun eastward at hundreds of metres per second, a stone dropped from a tower would land far to the west, and a cannon fired east would out-range one fired west. Galileo's reply was not to measure anything, but to notice that every such argument assumes motion is something a body can be robbed of by being let go — and the sealed cabin proves it is not.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa · Padua · Florence

Galileo Galilei was born in Pisa in 1564 and made his name in Padua, where he built telescopes and found Jupiter's four large moons in 1610, before trading a secure mathematics chair for a risky court appointment in Florence. The cabin argument appears in the Dialogo sopra i due massimi sistemi del mondo of 1632, spoken by the character Salviati, and it repays reading in the original: the fish in their bowl, the flies, the drops from a hanging bottle, the jumps toward bow and stern — all exactly as before, and from none of them could you tell whether the ship was moving or standing still. He states no transformation law; he states an invariance, and uses it to dismantle the anti-Copernican mechanics of his day. His ship is a piece of theoretical physics: an ideal, frictionless, windowless system built to isolate a symmetry. The Dialogo was placed on the Index, and in 1633 the Inquisition condemned him; he spent his last years under house arrest at Arcetri, where, going blind, he wrote the Two New Sciences.

History

1632

— Galileo publishes the Dialogo sopra i due massimi sistemi del mondo; the sealed ship's-cabin argument shows that no mechanical experiment below decks reveals uniform motion.

1687

— Newton's Principia, Corollary V to the Laws of Motion: the motions of bodies in a given space are the same among themselves whether that space is at rest or moves uniformly forwards in a straight line without circular motion.

1865

— Maxwell's A Dynamical Theory of the Electromagnetic Field fixes the speed of light from electric and magnetic constants alone, with no frame attached.

1904

— Poincaré, at the St Louis Congress of Arts and Science, names it "the principle of relativity" and raises it to a general guiding principle of physics.

1905

— Einstein, in "Zur Elektrodynamik bewegter Körper", takes the principle as a postulate and extends it from mechanics to the whole of physics, electrodynamics included.

Going further

Texts. Taylor & Wheeler, Spacetime Physics, 2nd ed., Ch. 1 · Rindler, Introduction to Special Relativity, 2nd ed., §§1–7 · Morin, Introduction to Classical Mechanics, Ch. 11 · Galilei, Dialogue Concerning the Two Chief World Systems (1632), Second Day · Einstein, "On the Electrodynamics of Moving Bodies" (1905), §1.

Provenance. The principle and the Galilean boost follow Rindler §§1–7 and Taylor & Wheeler Ch. 1; the covariance argument follows Morin Ch. 11 and Landau & Lifshitz Mechanics §3. A verifier should reproduce: $u' = u - v$ and $a' = a$ from $x' = x - vt,\ t' = t$ ; $\mathbf F' = \mathbf F$ for forces with relative arguments, hence form-invariance of $m\ddot{\mathbf r} = \mathbf F$ , with $\partial_v(\text{eom}) = 0$ ; the ball-drop numbers $t = 0.495\ \mathrm{s}$ , $\Delta x' = 0$ , $\Delta x = 4.95\ \mathrm{m}$ , landing speeds $4.85$ and $11.12\ \mathrm{m/s}$ ( $h = 1.2\ \mathrm{m}$ , $v = 10\ \mathrm{m/s}$ , $g = 9.81\ \mathrm{m/s^2}$ ); and the elastic collision giving $1$ and $4\ \mathrm{m/s}$ , conserving $p$ and $T$ in the lab ( $6\ \mathrm{kg\,m/s}$ , $9\ \mathrm{J}$ ) and under a $5\ \mathrm{m/s}$ boost ( $-9\ \mathrm{kg\,m/s}$ , $16.5\ \mathrm{J}$ ).

Builds on. Inertial Frames of Reference · Galilean Transformations. Leads to. Galilean Relativity · Electromagnetism and the Failure of Galilean Invariance · Einstein's Two Postulates.

If all inertial frames are equivalent, is it meaningless to say the train is the one that is moving?

Between two inertial frames, yes — strictly meaningless. "The train moves at 10 m/s and the platform is at rest" and "the platform moves at 10 m/s and the train is at rest" are the same physical statement written two ways, and no experiment inside either can break the tie. What is not symmetric is acceleration: the train had to accelerate to reach cruising speed, and everyone aboard felt it. So when we call the train the moving one we are remembering its history, not reporting a measurable present fact.

The passenger and the signaller see the ball follow different paths. Isn't one of them an illusion?

No — both trajectories are real, and each is correct in its own frame. A trajectory is not a property of the ball alone but of the ball and the frame you plot it in, much as a shadow is a property of an object and a light source. The ball's path is a straight vertical line with respect to the carriage and a parabola with respect to the track; no third, "true" path hides behind them. What the observers must agree on are the physical events — the ball leaves the hand, the ball strikes the floor at the dropper's feet 0.495 s later — and the law connecting them. They do, exactly.

Why is acceleration invariant under a boost but velocity is not? It feels arbitrary

It is forced, not arbitrary, and it comes from one line of calculus. Two inertial frames differ by a constant offset in velocity. Adding a constant changes a velocity ( $u' = u - v$ ) but leaves its rate of change alone ( $a' = a$ ). The disagreement is real but frozen: it shifts the first derivative, never the second — and Newton's law is written in the second derivative, so it is blind to the offset. Had nature written its dynamics in velocity instead — a first-order law like $F = ku$ , essentially Aristotle's physics — absolute velocity would be measurable and Galileo's cabin argument would fail. That mechanics is second-order in time is the mathematical reason relativity exists.

Does the principle apply to accelerating or rotating laboratories?

No, and that boundary is what gives the principle its teeth. Every step of the derivation used $\dot{\mathbf v} = \mathbf 0$ . Let the frame accelerate at $\mathbf A$ and you get $\mathbf a\,' = \mathbf a - \mathbf A$ : an untouched, free body starts to accelerate, and the observer inside must invent a fictitious force $-m\mathbf A$ to balance the books. That is a measurement — you can read your frame's acceleration off an accelerometer with your eyes shut. Uniform velocity is undetectable while acceleration is detectable, and that asymmetry is the subject of Inertial Frames of Reference.

Galileo only argued about mechanics. How do we know the principle covers electricity and optics?

We do not derive it — we postulate it, then test the consequences to death. What is proved here is precisely the mechanical half: forces built from relative positions and velocities give equations blind to a boost. Whether Maxwell's electrodynamics and atomic spectra are also blind to a boost is an empirical question, and in 1905 it was wide open, because Maxwell's equations hand light a definite speed $c$ that the Galilean transformation does not leave alone. Einstein's move was to assume the principle universally and pay whatever price it demanded — and the price was the Galilean transformation itself (see Einstein's Two Postulates).

Is the principle of relativity the same thing as the constancy of the speed of light?

No, and conflating them is the commonest error in a first SR course. The principle of relativity is a statement about the form of all laws under a change of inertial frame, shared by Newtonian and Einsteinian physics alike — Galileo had it, Newton had it, and nothing in this article needed light at all. The constancy of $c$ is a separate, far more specific empirical claim about one quantity taking the same value in every frame, and it is the input that decides which transformation implements the principle: a universal time gives the Galilean map, an invariant speed gives something else. Both are taken up in Einstein's Two Postulates.