University Physics IV · Special Relativity I · 2.1

Inertial frames and the Galilean group

Before you can say what Galileo got wrong, you have to write down exactly what he assumed. Here the boost becomes a group element you can compose and invert, Newton's laws are shown to be blind to it, and the one silent postulate — that every frame shares a clock — is dragged into the open.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An inertial frame is not a place but a verdict: release a body with no identified interaction acting on it and check that its worldline stays straight. Any frame that passes generates infinitely many more, because the map x′ = x − vt, t′ = t sends straight worldlines to straight worldlines. Those maps compose — boost by v, then by w, and you have boosted by v + w — invert by reversing v, and include the identity at v = 0, so the boosts along one axis form a one-parameter group.

Let the boost point in any direction and that is three parameters; add three rotations, three spatial displacements and one shift of clock zero and the count reaches ten, the largest group under which Newtonian mechanics is covariant. Positions and velocities change under it, but the second derivative does not, so a force built from separations gives back F = ma unaltered, which is why no mechanical experiment inside a sealed cabin can measure the cabin's velocity. The cost is buried in the second line.

Writing t′ = t declares that one clock rate and one ordering of events serve every frame at once, so simultaneity belongs to the pair of events rather than to whoever judges them, and the additive velocity law leaves no speed invariant. Nothing in mechanics tests that declaration — which is exactly why it went unexamined for two centuries. Maxwell's equations do test it, and it fails.

Simple definition: An inertial frame is one in which a body free of net force keeps a constant velocity, and the Galilean group is the ten-parameter set of coordinate changes — boosts, rotations, spatial displacements and clock shifts — that carries one such frame to another.
Example: A carriage coasting at 30 m s⁻¹ is inertial: a ball released inside lands directly below its start. Track coordinates follow from x = x′ + 30t with t = t′, so during the 0.40 s of a 0.78 m fall the ball also travels 12 m along the track.

The boost in standard configurationx′ = x − vt, y′ = y, z′ = z, t′ = t

Converts the S-coordinates of an event into the S′-coordinates of that same event.

v is the constant velocity of S′ along +x in m s⁻¹; origins coincide at t = 0

Composition, inverse, identityB(w) ∘ B(v) = B(v + w), B(v)⁻¹ = B(−v), B(0) = I

The group axioms in one line: boosts along an axis form a one-parameter group isomorphic to (ℝ, +).

v, w collinear boosts in m s⁻¹; ∘ means "then", read right to left

Velocity and acceleration under a boostu′ = u − v, a′ = a

Acceleration is boost-invariant, and that is the whole reason F = ma keeps its form.

u, v in m s⁻¹ and a in m s⁻²; differentiate with respect to t, which is legitimate because t′ = t

The full ten-parameter transformationr′ = R r − v t − d, t′ = t − s

Ten knobs, and Noether turns each one into a conservation law of Newtonian mechanics.

R rotation (3), v boost in m s⁻¹ (3), d displacement in m (3), s clock offset in s (1)

Covariance of the second lawF(rᵢ − rⱼ) = m aᵢ ⟹ F(rᵢ′ − rⱼ′) = m aᵢ′

The same equation with the same constants in every inertial frame, so no mechanical test finds v.

Separations and relative velocities are boost-invariant; the mass m in kg is invariant too

Absolute time and absolute simultaneityΔt′ = Δt, so Δt = 0 ⟹ Δt′ = 0

The assumption mechanics cannot test — and the one special relativity discards.

Δt in s between any two events, for any pair of inertial frames related by the group

An inertial frame is a test, not a place

Nothing marks a frame as inertial except how free bodies behave in it. The test: take a body far from other matter, or with its identified interactions balanced, and check that its velocity stays constant — that its worldline is straight. This looks circular, since "no net force" seems to be defined by "no acceleration". It escapes the circle by being applied to many bodies at once. In a genuinely inertial frame a body accelerates only when you can name the partner doing it, and the two accelerations satisfy m₁a₁ = −m₂a₂. In a rotating frame, by contrast, every free body accelerates at once, in a pattern that depends on its own position and velocity rather than on any partner. That is what convicts a laboratory on Earth, though only one of the usual two terms does the convicting. The spin is detectable: at the equator it removes a centripetal ω²R = (7.29 × 10⁻⁵ s⁻¹)² × 6.37 × 10⁶ m = 0.034 m s⁻² from the measured g, about 0.35% of it, and it supplies the Coriolis term that turns a Foucault pendulum. The orbit is not detectable the same way. The bench falls freely with the whole planet, so the 5.9 × 10⁻³ m s⁻² of orbital acceleration is cancelled by the solar gravity producing it, and what remains locally is only the tidal difference across an Earth radius, about 5 × 10⁻⁷ m s⁻². The bench is inertial to whatever precision your experiment fails to resolve, and no better.

The boost as a linear map on events

Write the boost as an operation on events, an event being a pair (t, x) — a time and a place. If S′ moves along +x at constant v and the origins coincide at t = 0, then B(v): (t, x) ↦ (t, x − vt), with y and z untouched. Two features matter. It is linear, so a straight worldline in S is a straight worldline in S′ and the first law survives the change of frame. And acting on the column (t, x) it is the matrix [[1, 0], [−v, 1]], a shear of determinant 1 whose entire top row is the statement t′ = t. Draw it and the meaning is immediate: lines of constant t are horizontal, and a shear slides points along those lines without tilting them, so the t′ axis (the line x = vt) leans over while the x′ axis (the line t = 0) lies exactly on top of the x axis. Take v = 30 m s⁻¹ and an event at t = 2.0 s, x = 100 m: S′ assigns it x′ = 100 − 60 = 40 m at the same t′ = 2.0 s.

Closure, inverse, identity: why the boosts form a group

Boost from S to S′ at v, then from S′ to S″ at w. The second map gives x″ = x′ − wt′, and because t′ = t you may substitute straight in: x″ = (x − vt) − wt = x − (v + w)t. So B(w) ∘ B(v) = B(v + w) — composing two boosts produces a boost, never anything new. Setting w = −v returns the identity, so every boost has an inverse B(v)⁻¹ = B(−v), and B(0) leaves everything alone; matrix multiplication supplies associativity. Those four facts are the definition of a group, and the boosts along one axis form a one-parameter group isomorphic to the real numbers under addition. Two consequences are worth naming now. "Inertial frame" is an equivalence class, not a list: pass the test once and the whole orbit of the group passes it. And the composition law is plain addition of an unbounded parameter, so no speed survives every boost unchanged. Relativity keeps that group structure intact and replaces only the addition law, with one that has a ceiling.

Ten parameters, and what each one conserves

The boosts along one axis are a single slice of the full symmetry. A general Galilean transformation is r′ = R r − v t − d, t′ = t − s: a rotation R (three parameters), a boost v (three), a spatial displacement d (three), and a shift of clock zero s (one). Ten in all, and this is the largest group under which Newtonian mechanics is covariant. Noether attaches a conserved quantity to each continuous symmetry, so the count is not idle bookkeeping. Time translation gives energy; the three spatial translations give the three components of momentum; the three rotations give angular momentum; and the three boosts give the less familiar charge G = M R_cm − P t, whose constancy is exactly the statement that an isolated system's centre of mass moves in a straight line at constant speed. Ten symmetries, ten conserved quantities — one energy, three momenta, three angular momenta, three centre-of-mass components.

Covariance is not the same as invariance

Differentiate x′ = x − vt with respect to t, which is the same operation as differentiating with respect to t′ precisely because t′ = t: u′ = u − v, and then a′ = a. Acceleration is untouched. If the force depends only on separations rᵢ − rⱼ and relative velocities — as gravity, springs and the Coulomb force do — it is untouched too, so F = ma reads identically in every inertial frame. That is covariance: same equation, same constants, new coordinates. It is not the claim that every quantity is invariant, and a single free body makes the difference plain. A 4.0 kg glider drifting at 5.0 m s⁻¹ carries p = 20 kg m s⁻¹ and K = 50 J on the platform; ride alongside it at 5.0 m s⁻¹ and both readings are zero, while the glider itself has done nothing. What the boost preserves is not the numbers but the statements about them — that p stays constant, that K stays constant while no work is done. The worked example below runs the same distinction through a collision, where the energy dissipated survives the boost even though neither p nor K does.

The line that costs the most: t′ = t

Everything above rests on the second line, and mechanics alone cannot test it. Look hard at what that line asserts. Two events given Δt = 0 in S are given Δt′ = 0 in S′ for every v, so which of two distant events came first is settled by the events themselves and by nothing else; a signal of unlimited speed sits comfortably inside the scheme; and boosting again and again walks the parameter v + w + … past any bound you name. Why did two centuries not catch it? Because catching it needs a law that names a speed and attaches no frame to it, and mechanics has none — forces set accelerations, accelerations survive the boost unchanged, and so every mechanical prediction comes out the same in every inertial frame by construction. Electromagnetism is not like that. Measure μ₀ and ε₀ on a bench with currents and capacitors, and Maxwell's equations hand back a wave speed c = 1/√(μ₀ε₀) = 3.00 × 10⁸ m s⁻¹ with no frame attached to it. Under B(v) that wave would have to run at c − v for anyone chasing it, and Maxwell's equations have no such solution. One of the two has to give way, and the next topic works out which.

Change one variable at a time

Make the relationship visible.

Interactive model

Boost v of S′ relative to S12 m s⁻¹

Further boost w of S″ vs S′8 m s⁻¹

Time t of event E3.0 s

Drag v and w and only the two solid worldlines tilt; the dashed same-time line stays horizontal, and that is t′ = t drawn. Drag te to slide it up the diagram. At v = 12 and w = 8 the arrows read x′ = 18t and x″ = 10t, so S″ depends only on the sum v + w = 20 m s⁻¹, which is closure.

x OF EVENT E IN S90 m

x′ OF EVENT E IN S′54 m

x″ OF EVENT E IN S″30 m

COMPOSED BOOST v + w20 m s⁻¹

Live interpretationx OF EVENT E IN S: 90 m. x′ OF EVENT E IN S′: 54 m. x″ OF EVENT E IN S″: 30 m. COMPOSED BOOST v + w: 20 m s⁻¹

Catch the common trap

Explain before calculating.

Misconception check

The common error is reading t′ = t as bookkeeping rather than physics — a line you write because there seems to be nothing else to write. It is a postulate with content, and the one the structure cannot survive without. Every unfamiliar result of relativity enters at exactly the point where that line is relaxed: clocks that disagree about the duration between two events, rods whose length depends on when the observer marks the ends, and a ceiling on the composition of velocities. All three have been measured, and all three are invisible while t′ = t stands. The second half of the error is believing that one null aether experiment is what killed the group. Michelson and Morley sharpened the case, but the conflict was already legible on paper: two bench constants fix a wave speed, and a Galilean boost has nowhere to put the frame that speed would need.

Which single feature of the Galilean group is responsible both for the unbounded velocity-addition law u′ = u − v and for the claim that two events simultaneous in one inertial frame are simultaneous in every inertial frame?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFrame S′ moves along +x of frame S at v = 25 m s⁻¹, the origins coinciding at t = 0. Lamp A flashes at t = 4.0 s, x = 180 m; lamp B flashes at t = 4.0 s, x = 60 m. Give both events in S′, then compare the time interval and the spatial separation the two frames assign.

Apply x′ = x − vt to A: x′_A = 180 m − (25 m s⁻¹)(4.0 s) = 180 − 100 = 80 m, and t′_A = t_A = 4.0 s.
The same map on B: x′_B = 60 − 100 = −40 m, at t′_B = 4.0 s. B lies 40 m behind the origin of S′, because that origin has already run 100 m down the track.
Time interval: Δt = 0 in S, and Δt′ = 4.0 − 4.0 = 0 in S′. The flashes are simultaneous in both frames — that is t′ = t doing the work, not a coincidence.
Separation: Δx = 180 − 60 = 120 m and Δx′ = 80 − (−40) = 120 m. The quantity −vt subtracted from each event is the same number at equal times, so it cancels in the difference: lengths are Galilean-invariant.

AnswerIn S′, A = (4.0 s, 80 m) and B = (4.0 s, −40 m). Both frames record Δt = 0 and Δx = 120 m: the boost moves the events but changes neither their separation nor their simultaneity.

MediumA 2.0 kg cart moving at 3.0 m s⁻¹ strikes a stationary 1.0 kg cart and the two stick together. Analyse the collision in the platform frame S, then again in a frame S′ moving at 3.0 m s⁻¹ in the same direction, and say which of the momentum, the kinetic energy, and the energy dissipated are frame-independent.

In S: p = (2.0)(3.0) + 0 = 6.0 kg m s⁻¹, so the joined carts leave at v_f = 6.0/3.0 = 2.0 m s⁻¹. Kinetic energy goes from ½(2.0)(3.0)² = 9.0 J to ½(3.0)(2.0)² = 6.0 J, a loss of 3.0 J.
Boost to S′ with u′ = u − v at v = 3.0 m s⁻¹: the heavy cart now starts at rest, the light cart approaches at −3.0 m s⁻¹, and the joined pair must leave at 2.0 − 3.0 = −1.0 m s⁻¹.
Check momentum in S′: before, (2.0)(0) + (1.0)(−3.0) = −3.0 kg m s⁻¹; after, (3.0)(−1.0) = −3.0 kg m s⁻¹. Conserved — and equal to the S value shifted by −Mv = −(3.0 kg)(3.0 m s⁻¹) = −9.0 kg m s⁻¹, since 6.0 − 9.0 = −3.0.
Kinetic energy in S′: before, ½(1.0)(3.0)² = 4.5 J; after, ½(3.0)(1.0)² = 1.5 J. A loss of 3.0 J again, even though both totals differ from their S values.
So p and KE are frame-dependent readings, while "momentum is conserved" and "the collision dissipated 3.0 J" hold in both. The law is covariant; the numbers are not invariant.

AnswerS: v_f = 2.0 m s⁻¹, p = 6.0 kg m s⁻¹, 9.0 J → 6.0 J. S′: v_f′ = −1.0 m s⁻¹, p = −3.0 kg m s⁻¹, 4.5 J → 1.5 J. Both frames conserve p and dissipate the same 3.0 J.

HardTake the boost B(v): (t, x) ↦ (t, x − vt) and the time translation T(s): (t, x) ↦ (t + s, x), both members of the Galilean group. Show that they do not commute, identify the group element by which the two orders differ, evaluate it for v = 8.0 m s⁻¹ and s = 3.0 s, and name the quantity Noether assigns to the boost symmetry.

Apply T first, then B. T(s) sends (t, x) to (t + s, x). Feeding that pair into B, which maps (τ, ξ) to (τ, ξ − vτ), gives (t + s, x − v(t + s)) = (t + s, x − vt − vs).
Now apply B first, then T. B(v) sends (t, x) to (t, x − vt), and T then advances the clock only, giving (t + s, x − vt).
Both orders land at the same time t + s but at positions differing by vs. Writing ∘ right to left, B ∘ T = D(−vs) ∘ T ∘ B, where D(d) is the spatial translation x ↦ x + d.
The two orders therefore differ by a spatial translation, not by the identity: with v = 8.0 m s⁻¹ and s = 3.0 s they differ by (8.0)(3.0) = 24 m. The ten parameters are not ten independent knobs — the group is a semidirect product, and boosting before or after resetting the clock is a different operation.
Noether's charge for the boost symmetry is G = M R_cm − P t. For an isolated system P is constant, so dG/dt = M(dR_cm/dt) − P − t(dP/dt) = P − P − 0 = 0, and G = constant rearranges to R_cm = G/M + (P/M)t.

AnswerThe two orders differ by a spatial translation of vs = 24 m, so B and T do not commute and the Galilean group is a semidirect product. The boost's Noether charge is G = M R_cm − P t, and dG/dt = 0 is the centre-of-mass theorem.