University Physics IV · Special Relativity I · 2.6

Rapidity and the composition of boosts

Speeds refuse to add and pile up against c, which makes velocity the wrong bookkeeping for stacking boosts. Trade it for the hyperbolic angle that does add, and a chain of boosts becomes arithmetic — with the ceiling at c, the Doppler factor and the rocket's engine all falling out of the same parameter.

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Two Lorentz boosts along one line are again a Lorentz boost, but the speed of the result is not the sum of the speeds, and the usual velocity-addition law hides how simple the composition really is. The boost matrix carries only two entries, γ and γβ, locked together by γ² − (γβ)² = 1 — the same relation that binds cosh φ to sinh φ. Naming each boost by that φ, its rapidity, turns the Lorentz transformation into a hyperbolic rotation, and matrix multiplication turns into addition of angles: Λ(φ₁)Λ(φ₂) = Λ(φ₁ + φ₂) with no correction term.

The velocity law becomes the identity for tanh of a sum, and the unreachability of c is the statement that tanh never reaches 1 while φ has no ceiling. Everything an engine does — each burst of proper acceleration, each stage of a rocket, each hop in a relay of Doppler-shifted signals — is naturally counted in rapidity, which is why the speed, the Lorentz factor and the Doppler factor are one number read through tanh, cosh and exp. The limits matter as much as the result.

Additivity holds for boosts along a fixed direction, a one-parameter abelian subgroup; boosts along different directions neither commute nor add, and their product is a boost accompanied by a spatial rotation, the Wigner rotation, whose accumulation along a curved worldline is Thomas precession.

Simple definition: Rapidity is the hyperbolic angle φ of a Lorentz boost, fixed by β = v/c = tanh φ; unlike velocity it simply adds when boosts along one axis follow one another.
Example: A frame at 0.800c has φ = ½ ln(1.800/0.200) = 1.0986. Boost by the same amount again and the rapidity is exactly 2.1972 — but the speed is tanh 2.1972 = 0.9756c, not 1.600c.

Rapidity from speedφ = artanh β = ½ ln[(1 + β)/(1 − β)]

Turns a speed capped at c into a coordinate with no ceiling.

β = v/c and φ are both dimensionless; φ runs over all real values

The hyperbolic tripleβ = tanh φγ = cosh φβγ = sinh φ

Any one of β, γ and φ fixes the other two, with no square roots.

cosh²φ − sinh²φ = 1 is the identity γ² − (βγ)² = 1

Boost as a hyperbolic rotationct′ = ct cosh φ − x sinh φx′ = x cosh φ − ct sinh φ

A rotation matrix with cosh and sinh in place of cos and sin, and a minus sign in both rows.

standard configuration, boost along +x; det Λ = 1 and Λ(φ)⁻¹ = Λ(−φ)

Collinear boosts add rapidityΛ(φ₁)Λ(φ₂) = Λ(φ₁ + φ₂) ⟹ β = (β₁ + β₂)/(1 + β₁β₂)

Einstein's velocity law is the addition formula for tanh.

one axis only; the composition is closed, associative and commutative

Doppler factork = e^φ = √[(1 + β)/(1 − β)]

Rapidities add, so Doppler factors multiply along a relay chain.

k = f_received/f_emitted for head-on approach, dimensionless

Wigner rotation, perpendicular booststan Ω = γ₁γ₂β₁β₂/(γ₁ + γ₂)

Sizes the leftover rotation; Ω → β₁β₂/2 at low speed.

boost β₁ along x, then β₂ along y′; Ω is an ordinary rotation angle

Why the boost matrix has to be cosh and sinh

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.

Boosts compose by adding rapidities

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.

Why c is a horizon rather than a wall

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.

Rapidity is what the engine actually adds

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 Doppler factor is e^φ, so Doppler factors multiply

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.

Where additivity stops: boosts off the axis

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Rapidity of boost 1, φ₁0.70

Rapidity of boost 2, φ₂0.70

Both sliders start at 0.70, where the dashed classical arrow has already run through the c wall. Drag them to 2.00: the two rapidity bands stay equal, while on the velocity axis the second boost's arrow shrinks to about 4% of the first's — same rapidity, almost no speed.

RAPIDITY SUM φ₁ + φ₂1.400

COMBINED β = tanh(φ₁ + φ₂)0.8854 c

CLASSICAL β₁ + β₂1.2087 c

COMBINED γ = cosh(φ₁ + φ₂)2.151

Live interpretationRAPIDITY SUM φ₁ + φ₂: 1.400. COMBINED β = tanh(φ₁ + φ₂): 0.8854 c. CLASSICAL β₁ + β₂: 1.2087 c. COMBINED γ = cosh(φ₁ + φ₂): 2.151

Catch the common trap

Explain before calculating.

A probe leaves Earth at 0.600c. It launches a second probe at 0.600c relative to itself, in the same direction, and that one launches a third at 0.600c relative to itself. How fast is the third probe moving relative to Earth?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA muon beam travels at β = 0.800. Find its rapidity, obtain its Lorentz factor from that rapidity, and give the Doppler factor k = e^φ for light the beam emits straight ahead.

Rapidity: φ = ½ ln[(1 + β)/(1 − β)] = ½ ln(1.800/0.200) = ½ ln 9 = ln 3 = 1.0986.
That makes the exponentials exact: e^φ = 3 and e^(−φ) = 1/3 = 0.3333.
Lorentz factor: γ = cosh φ = (e^φ + e^(−φ))/2 = (3 + 0.3333)/2 = 1.6667. Cross-check the usual way: 1/√(1 − 0.640) = 1/0.600 = 1.667, which agrees.
Doppler factor: k = e^φ = 3.00, so a 500 nm line emitted forwards and received head-on arrives at 500/3 = 167 nm.

Answerφ = 1.0986, γ = 5/3 = 1.667, k = 3.00. The three are one number in three costumes: β = tanh φ, γ = cosh φ, k = e^φ.

MediumA staged rocket gains 0.300c relative to the stage it has just left, every stage identical and along the same line. How many stages does it need to pass 0.990c in the launch frame, and what are β and γ after that many? Compare with the classical count.

One stage in rapidity: φ₀ = ½ ln(1.300/0.700) = ½ ln 1.85714 = 0.30952. The stages are collinear, so N of them give exactly Nφ₀.
Target: φ_target = artanh 0.990 = ½ ln(1.990/0.010) = ½ ln 199 = 2.64665.
N ≥ 2.64665/0.30952 = 8.551, so N = 9. Eight stages reach only tanh 2.47616 = 0.9860c, short of the target.
After nine: φ = 9 × 0.30952 = 2.78568, so β = tanh φ = 0.99242 and γ = cosh φ = 8.136.
Classically 0.990/0.300 = 3.3 would call for four stages, and nine stages would have claimed 2.70c. Rapidity turns the question into one division; velocity cannot answer it without iterating the composition law nine times.

AnswerNine stages, giving β = 0.9924 and γ = 8.14. Four stages is the classical answer, and it is wrong by more than a factor of two in stage count.

HardFrame S′ moves at β₁ = 0.600 along the x axis of S. Frame S″ moves at β₂ = 0.800 along the y′ axis of S′. Find the velocity of S″ in S, its rapidity, compare that with φ₁ + φ₂, and find the leftover Wigner rotation.

Velocity components in S. The origin of S″ moves at (0, 0.800c) in S′, so uₓ = v₁ = 0.600c and u_y = u′_y/[γ₁(1 + u′ₓ v₁/c²)] = 0.800c/1.25 = 0.640c — the transverse component is divided by γ₁ = 1.25.
Speed and Lorentz factor: |β| = √(0.600² + 0.640²) = √0.7696 = 0.8773, at atan(0.640/0.600) = 46.8° to the x axis, and γ = 1/√(1 − 0.7696) = 1/0.480 = 2.0833 = γ₁γ₂ with γ₂ = 5/3.
Composite rapidity: φ = arcosh 2.0833 = ln(2.0833 + √(2.0833² − 1)) = ln 3.9110 = 1.3638.
The parts are φ₁ = ½ ln 4 = 0.6931 and φ₂ = ½ ln 9 = 1.0986, summing to 1.7918. The composite is 24% smaller: adding rapidities would have predicted tanh 1.7918 = 0.946c instead of the true 0.877c.
The missing piece is a rotation. tan Ω = γ₁γ₂β₁β₂/(γ₁ + γ₂) = (2.0833 × 0.480)/2.9167 = 1.000/2.9167 = 0.3429, so Ω = 18.9°: the axes of S″ are turned by that much relative to S.

Answerβ = 0.877 at 46.8° to the x axis, γ = 25/12 = 2.083, φ = 1.364 against a naive sum of 1.792, with a residual Wigner rotation of 18.9°.