Galilean relativity and the invariance problem

An inertial frame is one in which a body free of forces keeps a constant velocity. Once you have one you have infinitely many: any frame moving at constant velocity relative to it is inertial too, and any frame accelerating relative to it is not. No experiment picks one of them out as at rest; rest is a choice of origin. The Earth's surface fails the test only slightly. The equatorial spin gives a centripetal acceleration ω²R = (7.29 × 10⁻⁵ s⁻¹)² × 6.38 × 10⁶ m = 0.034 m s⁻², and the orbit adds 5.9 × 10⁻³ m s⁻², so for an experiment lasting seconds the laboratory is inertial to better than half a percent of g. Galileo's version of the principle was a ship: below decks with the ports shut, no dropped weight, dripping bottle or flying insect tells you whether the ship is moored or under way. This section is about how far that principle can be pushed once the experiment involves light.

Take two inertial frames whose origins coincide at t = 0, with S′ moving along +x at speed v as seen from S. An event — a flash, a collision, a tick — has coordinates (x, t) in S and (x′, t′) in S′, and the Galilean boost relates them: x′ = x − vt, y′ = y, z′ = z, t′ = t. The inverse is the same map with v replaced by −v, x = x′ + vt′, as the symmetry of the two frames demands. A flare fired at x = 1200 m and t = 4.0 s in the ground frame sits at x′ = 1200 − 25 × 4.0 = 1100 m in the frame of a train doing 25 m s⁻¹, and at t′ = 4.0 s. That last equation is the one to stare at. It says a clock in S′ reads exactly what a clock in S reads, that the time interval between two events is the same in every frame, and that two events with the same t are simultaneous for everyone. None of that was measured. It was assumed, because no clock of the era could have detected the alternative and because it makes the boost look like a mere change of origin.

Because t′ = t, differentiating x′ = x − vt with respect to the shared time is legal and gives u′ = u − v: velocities subtract, and a ball thrown forward at 15 m s⁻¹ inside a train at 40 m s⁻¹ has u = 55 m s⁻¹ on the platform. Differentiate again and, because v is constant, a′ = a. Newton's second law keeps its form provided the forces depend only on relative positions and relative velocities, which gravity, springs, contact forces and drag through a medium all do, since the medium's own velocity transforms along with everything else. Conservation laws inherit the invariance. If Σmᵢuᵢ is conserved in S then Σmᵢ(uᵢ − v) = Σmᵢuᵢ − vΣmᵢ is conserved in S′ whenever total mass is, and for the whole system K′ = K − vp + ½Mv², so the kinetic energy lost in a collision, though the energies themselves differ between frames, comes out the same because Δp = 0. Mechanics offers no handle on v at all.

Sound obeys ∂²p/∂x² − (1/cₛ²) ∂²p/∂t² = 0 with cₛ = 343 m s⁻¹ in the frame of still air. Push that equation through the boost with the chain rule. Since x = x′ + vt′ and t = t′, ∂/∂x = ∂/∂x′ and ∂/∂t = ∂/∂t′ − v ∂/∂x′: the space derivative is untouched but the time derivative, squared, brings in cross terms, and in S′ the equation reads (1 − v²/cₛ²) ∂²p/∂x′² + (2v/cₛ²) ∂²p/∂x′∂t′ − (1/cₛ²) ∂²p/∂t′² = 0. That is not a wave equation of the standard form, and its plane-wave solutions have phase speeds cₛ − v forward and cₛ + v backward, exactly u′ = u − v applied to the wave. For a car at 30 m s⁻¹ the two sound speeds are 313 and 373 m s⁻¹, and nobody is troubled, because the air is a real thing and its rest frame is a real frame. Waves do not break the Galilean boost; they merely report which frame their medium is at rest in.

In vacuum Maxwell's equations combine into ∂²E/∂x² − μ₀ε₀ ∂²E/∂t² = 0, a wave equation with speed c = 1/√(μ₀ε₀). Evaluate it: μ₀ = 4π × 10⁻⁷ H m⁻¹ and ε₀ = 8.854 × 10⁻¹² F m⁻¹ give μ₀ε₀ = 1.113 × 10⁻¹⁷ s² m⁻², whose inverse square root is 2.998 × 10⁸ m s⁻¹, the measured speed of light. Both constants come from bench measurements — a capacitor's charge per volt, the force between steady currents — that involve no motion relative to anything. So the equations assert one speed c with no medium in sight, and the calculation of the previous section applies with cₛ replaced by c: in any frame moving at v relative to the one where Maxwell's equations hold, the wave equation acquires the (2v/c²) ∂²E/∂x′∂t′ cross term and light would travel at c − v one way and c + v the other. Two options remain. Either there is a preferred frame, the luminiferous ether, in which Maxwell's equations are exact and every other frame is detectably moving; or the boost itself is wrong and Maxwell's equations hold in every inertial frame.

If an ether exists the Earth moves through it at something like its orbital speed, v ≈ 3.0 × 10⁴ m s⁻¹, so β = v/c ≈ 1.0 × 10⁻⁴. One-way effects of order β cannot be measured directly, because two separated clocks would have to agree to 10⁻⁴ of a light transit time. Round trips remove the first-order term: light sent along the motion and back takes L/(c − v) + L/(c + v) = (2L/c)/(1 − β²) ≈ (2L/c)(1 + β²), so the ether's signature survives only as a fraction β² ≈ 1.0 × 10⁻⁸ of the transit time. That is small, but it is a definite prediction of the Galilean boost at 30 km s⁻¹, nowhere near c. Michelson and Morley built an interferometer sensitive enough to see it and in 1887 found no ether wind; how their two-arm comparison works, and what its null result forces, is where the next topic begins. This topic can already say where to look. Newton's laws are exactly invariant under the boost, so no mechanical experiment could have flagged a problem, and the line of the boost that nothing had ever tested is the universal clock, t′ = t.