Inertial frames and the Galilean group

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.

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.

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.

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.

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.

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.