Rapidity and the Additivity of Boosts

Derivation

Rapidity and the Additivity of Boosts

D-096 Home PU-105 Threads symmetry Depends on Lorentz Transformation from the Two Postulates, Relativistic Velocity Addition

Statement

A Lorentz boost of velocity v along a fixed axis can be written as a hyperbolic rotation of the (ct, x) plane through a real angle φ, the rapidity, defined by tanh φ = β = v/c (equivalently γ = cosh φ, γβ = sinh φ). For two collinear boosts the rapidities add: φ = φ₁ + φ₂, and taking tanh of both sides reproduces the relativistic velocity-addition law β = (β₁ + β₂)/(1 + β₁β₂).

Why it matters

Velocity does not add under boosts — it composes through a nonlinear fractional law — which makes chains of boosts awkward and obscures the group structure of special relativity. Rapidity is the coordinate in which the boost group becomes trivially additive: successive collinear boosts are just successive translations of a single real parameter, exactly as ordinary rotation angles add in the Euclidean plane.

This turns hard kinematic problems into arithmetic. Multi-stage rockets, particle-cascade energies, and constant-proper-acceleration motion (where φ grows linearly with proper time) all become linear in rapidity. Rapidity is also the natural longitudinal variable in high-energy physics, because differences of rapidity are boost-invariant along the beam axis.

Assumptions

Flat Minkowski spacetime with the standard metric signature.In curved spacetime there is no global boost and no single global rapidity; the hyperbolic-rotation picture holds only in the local tangent space. All boosts are collinear (same spatial axis).Non-collinear boosts do not commute and their composition carries a Wigner rotation; rapidities then add as non-commuting generators, not as scalars, and the simple sum fails. The two frames share a common event as origin and axes are related by a pure boost (no spatial rotation, no reflection).A residual rotation or parity flip mixes into the transformation, so the 2×2 block is no longer a pure hyperbolic rotation and cannot be labelled by a single φ. The Lorentz transformation from the postulates (a prior result) is taken as given, together with |β| < 1.Without |β| < 1 the map β = tanh φ is not onto the physical range and γ = cosh φ becomes imaginary; the real-rapidity parameterisation breaks down at and beyond c.

Derivation

ct′ = γ(ct − βx), x′ = γ(x − βct), γ = 1/√(1 − β²)

Start from the Lorentz transformation for a boost of speed v = βc along x, taken as a prior result. Writing the time coordinate as ct puts both axes in the same units of length. A

γ² − (γβ)² = γ²(1 − β²) = 1

Algebraic identity following directly from the definition of γ. It has the exact form of the hyperbolic identity cosh² φ − sinh² φ = 1, so (γ, γβ) lies on the unit hyperbola. B

γ ≡ cosh φ, γβ ≡ sinh φ ⇒ tanh φ = β

For every β ∈ (−1, 1) there is a unique real φ with tanh φ = β, and Step 2 guarantees these two definitions are consistent (cosh φ > 0 selects the future-pointing branch). This defines the rapidity. B

ct′ = ct cosh φ − x sinh φ, x′ = −ct sinh φ + x cosh φ

Substitute Step 3 into Step 1 (γ = cosh φ, γβ = sinh φ). The boost is now a hyperbolic rotation of the (ct, x) plane — identical in form to a Euclidean rotation but with cos → cosh, sin → sinh and a relative sign. B

Λ(φ) = ⎡ cosh φ −sinh φ ⎤ , ⎛ct′⎞ = Λ(φ)⎛ct⎞
⎣ −sinh φ cosh φ ⎦ ⎝x′⎠ ⎝x⎠

Repackage Step 4 as a matrix acting on the column (ct, x). Legal because Step 4 is linear in the coordinates. A

Λ(φ₂) Λ(φ₁) = ⎡ cosh φ₁cosh φ₂ + sinh φ₁sinh φ₂ ⋅ ⎤
⎣ ⋅ ⎦

Apply two collinear boosts in succession: first φ₁ (frame S → S′), then φ₂ (S′ → S″). Composition of linear maps is matrix multiplication. Both factors are hyperbolic rotations, so the product is again of that form. B

Λ(φ₂) Λ(φ₁) = Λ(φ₁ + φ₂), φ = φ₁ + φ₂

Use the hyperbolic addition identities cosh(φ₁+φ₂) = cosh φ₁cosh φ₂ + sinh φ₁sinh φ₂ and sinh(φ₁+φ₂) = sinh φ₁cosh φ₂ + cosh φ₁sinh φ₂ on every entry of Step 6. Collinear boosts therefore add their rapidities. A

β = tanh(φ₁ + φ₂) = (tanh φ₁ + tanh φ₂) / (1 + tanh φ₁ tanh φ₂)

Take tanh of Step 7 and expand with the tanh addition formula. Legal because tanh is a well-defined function of the combined rapidity. A

Result

tanh φ = β, φ = φ₁ + φ₂ ⇒ β = (β₁ + β₂) / (1 + β₁β₂)

Reading. The boost is a rotation through an imaginary-angle-like parameter φ in the time–space plane. Because rotations compose by adding their angles, collinear boosts compose by adding rapidities — a linear, group-additive law. Substituting β = tanh φ and using the tanh addition identity recovers exactly the fractional velocity-addition rule; the apparent nonlinearity of velocities is just the nonlinearity of tanh.

Units check. β = v/c is dimensionless, so φ = artanh β is dimensionless (an angle-like number in radians). cosh φ = γ and sinh φ = γβ are both dimensionless, consistent with mapping the dimensionless column (ct, x)/L. The final velocity law returns a dimensionless β ∈ (−1, 1).

Limiting cases

Small speeds β ≪ 1: φ ≈ β (since artanh β ≈ β + β³/3), and rapidity addition reduces to Galilean velocity addition v ≈ v₁ + v₂.
Ultra-relativistic β → 1: φ → +∞ logarithmically, φ ≈ ½ ln[(1+β)/(1−β)]. No finite chain of boosts reaches c, mirroring that tanh φ < 1 for all finite φ.
Equal boosts φ₁ = φ₂ = φ₀: total φ = 2φ₀, so β = tanh 2φ₀ = 2β₀/(1 + β₀²) — the relativistic "doubling" formula.
Constant proper acceleration a: rapidity accumulates linearly with proper time, φ(τ) = aτ/c, so β(τ) = tanh(aτ/c) — the hyperbolic-motion worldline.
Opposite boosts φ₂ = −φ₁: total φ = 0, returning to rest, as required for an inverse boost.

Breaks when

Boosts are not collinear. Two boosts along different axes do not commute; their product is a boost composed with a spatial (Wigner) rotation. Rapidities no longer add as scalars, and there is no single φ describing the result. This underlies Thomas precession.
Superluminal or lightlike relative motion. For |β| ≥ 1 the equation tanh φ = β has no real solution; φ → ∞ for a light signal and is complex beyond c. The real-rapidity picture cannot parameterise such "boosts."
Curved spacetime / accelerated frames globally. The additive law is a statement about the global Lorentz group of flat spacetime. In curved spacetime boosts only exist locally, and comparing rapidities at separated events requires parallel transport, which is path-dependent.

Failure modes

Adding velocities as if they were rapidities. Writing v = v₁ + v₂ for large speeds; it is φ that adds, not β.
Using φ = β instead of φ = artanh β. Valid only to first order; at β = 0.6 the error is already ~15%.
Confusing the hyperbolic rotation with a Euclidean one. Dropping the sign so cosh²+sinh² is used, or writing cos/sin, which violates cosh²−sinh² = 1.
Assuming rapidities add for non-collinear boosts. Forgetting the Wigner rotation and the non-commutativity of the boost generators.
Sign/branch slips. Taking cosh φ = −γ (wrong time-orientation), or mishandling the sign of φ for a boost in the −x direction.
Treating rapidity as a component of a 3-vector. It is an additive scalar only along one axis; combining transverse rapidities naively is meaningless.

Discussion

Rapidity exposes the Lorentz boosts as a one-parameter subgroup isomorphic to the additive real line (ℝ, +). This is the deep reason velocity addition has the structure it does: the boosts form a group, and the "natural" additive coordinate on that group is φ, not v. The nonlinear velocity law is simply what you get after pushing an additive parameter through the nonlinear chart v = c tanh φ. In this light, the speed of light as an unreachable ceiling is the geometric statement that tanh is bounded: an infinite rapidity maps to β = 1.

The analogy with Euclidean rotation is exact and useful. A rotation by angle θ mixes x and y via cos θ, sin θ and preserves x²+y²; a boost by rapidity φ mixes ct and x via cosh φ, sinh φ and preserves (ct)²−x², the invariant interval. Formally, setting φ = iθ turns one into the other — the boost is a "rotation through an imaginary angle." The minus sign in the metric is exactly what turns circular functions into hyperbolic ones.

Rapidity is also the practical variable of collider physics. Under a boost along the beam axis, longitudinal rapidity shifts by a constant, so differences of rapidity — and hence the shapes of particle distributions in rapidity — are boost-invariant. For massless or ultra-relativistic particles one uses the closely related pseudorapidity η = −ln tan(θ/2). Meanwhile, for a body under constant proper acceleration a, the equation of motion integrates trivially in rapidity: dφ/dτ = a/c gives φ = aτ/c, and the whole hyperbolic worldline follows from β = tanh φ without ever solving a nonlinear ODE in v.

At the level of Lie theory, the boost generator K satisfies Λ(φ) = exp(−φK), and the additivity Λ(φ₁)Λ(φ₂) = Λ(φ₁+φ₂) is just the statement that a one-parameter subgroup generated by a single element is abelian. For non-collinear boosts the generators K_x, K_y fail to commute — [K_x, K_y] = −J_z, a rotation generator — which is the algebraic origin of both the breakdown of rapidity additivity and of Thomas–Wigner rotation. Rapidity additivity is therefore not a universal fact about boosts but a special feature of a single boost direction.

Common misconceptions. Rapidity is not a velocity and is not bounded by 1 — it runs over all of ℝ. It is dimensionless, not measured in m/s. And "boosts add" is true only for the rapidity and only along a common axis; it is emphatically false for velocities and for boosts in different directions.

Worked examples

Two collinear boosts, β₁ = β₂ = 0.60. Find the combined β.

Convert each velocity to a rapidity, add, convert back. Symbols first, then numbers. A

φ_i = artanh β_i = ½ ln[(1+β_i)/(1−β_i)]

Definition of rapidity. φ₁ = φ₂ = ½ ln(1.6/0.4) = ½ ln 4 = 0.6931. A

φ = φ₁ + φ₂ = 2(0.6931) = 1.3863

Rapidities add (Step 7 of the derivation). A

β = tanh φ = tanh(1.3863) = 0.8824

Convert back to velocity. Cross-check with the velocity-addition law: (0.6+0.6)/(1+0.36) = 1.2/1.36 = 0.8824. A

β = 0.882, v = 0.882c ≈ 2.65 × 10⁸ m/s

Reading. Two 0.6c boosts do not give 1.2c; adding rapidities (0.693 + 0.693 = 1.386) and taking tanh lands safely below c.

Units check. φ dimensionless; β dimensionless; v = βc in m/s. Good.

A rocket accelerates at constant proper acceleration a = 9.8 m/s² for proper time τ = 1.0 yr. Find its final β.

Under constant proper acceleration the rapidity grows linearly with proper time. Symbols first. B

dφ/dτ = a/c ⇒ φ(τ) = aτ/c

Proper acceleration is the rate of change of rapidity per unit proper time (a standard property of hyperbolic motion; φ(0)=0). B

φ = aτ/c = (9.8)(3.156 × 10⁷) / (3.00 × 10⁸)

Insert numbers, with τ = 1.0 yr = 3.156 × 10⁷ s. Numerator = 3.093 × 10⁸ m/s. A

φ = 1.031, β = tanh φ = tanh(1.031) = 0.776

Divide, then convert rapidity to velocity. A

φ ≈ 1.03, β ≈ 0.78, v ≈ 0.78c, γ = cosh φ ≈ 1.59

Reading. One year at 1g (proper) builds a rapidity of about 1.03, i.e. v ≈ 0.78c — the classic near-relativistic result. Because φ is additive in τ, a second identical year simply doubles φ to 2.06, giving β = tanh(2.06) ≈ 0.968.

Units check. aτ has units m/s, divided by c (m/s) gives dimensionless φ. β, γ dimensionless. Good.

Problems

(A) Convert β = 0.80 to a rapidity, and convert φ = 1.50 back to a velocity.
Solution
φ = artanh(0.80) = ½ ln(1.8/0.2) = ½ ln 9 = ½(2.197) = 1.099. And β = tanh(1.50) = 0.905, so v = 0.905c ≈ 2.71 × 10⁸ m/s.
(A) Three collinear rockets each move at 0.50c relative to the previous. Use rapidities to find the speed of the third relative to the ground.
Solution
Each rapidity φ₀ = artanh(0.5) = ½ ln(1.5/0.5) = ½ ln 3 = 0.5493. Total φ = 3φ₀ = 1.6479. Then β = tanh(1.6479) = 0.9286, i.e. v ≈ 0.929c. (Check by iterating velocity addition: 0.5⊕0.5 = 1.0/1.25 = 0.80; 0.80⊕0.5 = 1.3/1.4 = 0.9286. ✓)
(B) Show that combining a boost φ with its inverse −φ returns the identity, and interpret physically.
Solution
Λ(φ)Λ(−φ) = Λ(φ + (−φ)) = Λ(0). Since cosh 0 = 1, sinh 0 = 0, Λ(0) = I. Physically, boosting into a frame and then back out by the equal and opposite velocity restores the original frame; equivalently tanh(0) = 0 means zero relative velocity. The inverse of a boost is the boost of opposite rapidity, confirming the boosts form a group under rapidity addition.
(B) A particle has total energy E and momentum p along x. Show that E = mc² cosh φ and pc = mc² sinh φ, and identify tanh φ.
Solution
For a massive particle E = γmc² and p = γmv = γmβc. With γ = cosh φ and γβ = sinh φ: E = mc² cosh φ, pc = mc² sinh φ. Dividing, pc/E = tanh φ = β = v/c. The energy–momentum 2-vector (E, pc) is thus a hyperbolic rotation of the rest value (mc², 0) through the particle's rapidity — and E² − (pc)² = (mc²)²(cosh²φ − sinh²φ) = (mc²)² recovers the mass-shell relation.
(C) A muon at rapidity φ = 3.0 in the lab is viewed from a frame boosted by rapidity Δφ = 1.2 along the same axis (in the direction of the muon's motion). Find the muon's rapidity and speed in the new frame, and comment on why rapidity differences are the useful quantity.
Solution
Boosting the observer toward the muon by Δφ subtracts from the muon's rapidity: φ′ = φ − Δφ = 3.0 − 1.2 = 1.8. Speed: β′ = tanh(1.8) = 0.947, so v′ ≈ 0.947c (compared with the lab value tanh(3.0) = 0.995c). Because a longitudinal boost merely shifts every rapidity by the same constant Δφ, the difference in rapidity between any two particles is unchanged by such a boost — which is precisely why rapidity (and pseudorapidity) is the coordinate of choice for describing longitudinal distributions at colliders.