Rapidity and the composition of boosts

In standard configuration the boost reads ct′ = γ(ct − βx) and x′ = γ(x − βct), so the matrix carries just two numbers, γ and γβ. They are not independent: γ = 1/√(1 − β²) forces γ² − (γβ)² = γ²(1 − β²) = 1. A pair of reals whose squares differ by one, the first of them positive, is (cosh φ, sinh φ) for exactly one real φ: sinh runs monotonically through every real value once, and cosh φ = √(1 + sinh²φ) is then fixed. Define φ by γβ = sinh φ; then γ = cosh φ and β = tanh φ. The boost becomes ct′ = ct cosh φ − x sinh φ and x′ = x cosh φ − ct sinh φ — a hyperbolic rotation, with cosh and sinh where cos and sin sit in a rotation matrix and the sign pattern that keeps (ct)² − x² fixed instead of x² + y². Nothing here is a choice of ours: the invariant interval selected the functions, and φ is simply the parameter along the hyperbola γ² − (γβ)² = 1 on which the entries of every boost must lie.

Multiply Λ(φ₂)Λ(φ₁). The diagonal entry is cosh φ₁ cosh φ₂ + sinh φ₁ sinh φ₂ = cosh(φ₁ + φ₂) and the off-diagonal is −sinh(φ₁ + φ₂), so Λ(φ₂)Λ(φ₁) = Λ(φ₁ + φ₂): two boosts along one axis are one boost, and their rapidities add exactly. Everything a group needs is now visible — the identity is φ = 0, the inverse of Λ(φ) is Λ(−φ), composition is closed and commutative, and the collinear boosts are a copy of the real line under addition. Take tanh of both sides and Einstein's law falls out: β = (β₁ + β₂)/(1 + β₁β₂) is only tanh(φ₁ + φ₂) written in terms of tanh φ₁ and tanh φ₂. Two boosts of 0.600c: φ = ln 2 each, total ln 4, and β = (16 − 1)/(16 + 1) = 15/17 = 0.882c.

Rapidity has no upper bound and tanh has one, so no finite chain of boosts arrives at c. Watch the gap close. A 0.500c boost carries φ₀ = ½ ln 3 = 0.5493; apply ten of them and φ = 5.4931, where e^φ = 3⁵ = 243, so β = (3¹⁰ − 1)/(3¹⁰ + 1) = 59048/59050 = 0.9999661. The revealing form is 1 − β = 2/(e^(2φ) + 1): after N such boosts the gap to c is 2/(3^N + 1), so each further identical boost cuts what remains by very nearly a factor of 3, and ten of them leave 2/59050 = 3.4 × 10⁻⁵. Meanwhile γ = cosh φ ≈ ½e^φ runs the other way, here to 121.5, and since a mass m carries energy γmc², each further unit of rapidity multiplies the energy the launch frame must supply by about e. Nothing stops the ship; the arithmetic simply never closes.

Ride with the ship. In its instantaneous rest frame the engine delivers proper acceleration α, and over a proper-time interval dτ that frame is boosted by dβ = α dτ/c — an infinitesimal boost, so dφ = α dτ/c as well. Because rapidities add along the line of motion, these increments accumulate with no correction term: φ = ατ/c for constant proper acceleration. That is the relativistic replacement for v = at, and it is why rapidity, not velocity, is the natural output of a motor. Take α = 9.81 m s⁻² for τ = 1.00 yr = 3.156 × 10⁷ s: φ = 1.033, so β = tanh 1.033 = 0.775 and γ = cosh 1.033 = 1.58. Burn a second year and the rapidity doubles to 2.065, yet the speed only creeps to 0.968c — while γ climbs to 4.01.

The longitudinal Doppler factor k = √[(1 + β)/(1 − β)] is precisely e^φ, because φ = ½ ln[(1 + β)/(1 − β)]. Additive rapidity therefore shows up as multiplicative Doppler: relay a light signal through a chain of frames and the k factors multiply, which is the whole engine of Bondi's k-calculus. A frame at 0.600c has k = 2, two such steps give k = 4, and β = (4² − 1)/(4² + 1) = 15/17 — the same answer the matrices gave. The same logarithm returns in particle physics as y = ½ ln[(E + p_z c)/(E − p_z c)]: a boost along the beam adds a constant to every particle's y, so a dN/dy distribution slides rigidly rather than deforming, and rapidity differences are frame-independent.

Boosts along one axis commute; boosts along different axes do not, and their product is not a pure boost at all. Take β₁ = 0.600 along x, then β₂ = 0.600 along y′, so γ₁ = γ₂ = 1.25. The composite frame moves in the original frame with components (0.600, 0.480) — the transverse part is divided by γ₁ — so |β| = 0.768 and γ = γ₁γ₂ = 1.5625. Its rapidity is arcosh 1.5625 = 1.016, short of φ₁ + φ₂ = 2 ln 2 = 1.386, and the shortfall is exact: cosh(φ₁ + φ₂) = cosh φ₁ cosh φ₂ + sinh φ₁ sinh φ₂ exceeds γ₁γ₂ by sinh φ₁ sinh φ₂ = (γ₁β₁)(γ₂β₂) = 0.5625, and the rapidities would add only if that product vanished. What is left over is a spatial rotation: Λ₂Λ₁ = R(Ω)Λ(1.016) with tan Ω = γ₁γ₂β₁β₂/(γ₁ + γ₂) = 0.5625/2.50 = 0.225, so Ω = 12.7°, against the low-speed estimate β₁β₂/2 = 0.180 rad = 10.3°. Accumulated along a curved orbit these rotations are Thomas precession, the factor of ½ that corrects the naive spin-orbit coupling in atoms.