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 along a fixed axis with speed \(v=\beta c\) is a hyperbolic rotation in the \((ct,x)\) plane through a hyperbolic angle \(\varphi\), the rapidity, defined by \(\tanh\varphi=\beta\) (equivalently \(\gamma=\cosh\varphi\), \(\gamma\beta=\sinh\varphi\)). Two collinear boosts of rapidities \(\varphi_1\) and \(\varphi_2\) compose into a single boost of rapidity \(\varphi=\varphi_1+\varphi_2\); taking \(\tanh\) of both sides reproduces the relativistic velocity-addition law \(\beta=(\beta_1+\beta_2)/(1+\beta_1\beta_2)\).

Why it matters

Velocities do not add in special relativity, and the nonlinear rule \(\beta=(\beta_1+\beta_2)/(1+\beta_1\beta_2)\) makes chains of boosts awkward and obscures the group structure. Rapidity is the coordinate in which boosts become linear and additive, exactly as ordinary angles are additive for rotations in the plane. This turns the Lorentz boost into a one-parameter subgroup isomorphic to the additive line \((\mathbb{R},+)\).

The payoff is both conceptual and practical: multi-stage boosts, the Doppler factor \(e^{\varphi}\), constant-proper-acceleration motion (where \(\varphi=a\tau/c\) grows linearly with proper time), and collider kinematics (where rapidity differences are boost-invariant along the beam) all become linear in \(\varphi\). The unreachability of \(c\) is simply the boundedness of \(\tanh\).

Assumptions

Boosts are collinear (all along the same spatial axis).If the boost directions differ, the composition is a boost plus a spatial rotation (the Wigner rotation, giving Thomas precession), and the scalar rapidities no longer add.

Both frames are inertial, related by a proper orthochronous Lorentz transformation, with \(|\beta|<1\).Dropping this admits parity/time reversal or superluminal "boosts", for which \(\tanh\varphi=\beta\) has no real solution and \(\gamma=\cosh\varphi\) becomes ill-defined.

The transverse coordinates \((y,z)\) are unchanged and decouple from the boost.If transverse mixing were allowed the \((ct,x)\) block would not close on itself and the \(2\times2\) hyperbolic-rotation reduction would fail.

Spacetime is flat with fixed Minkowski signature, so the interval \((ct)^2-x^2\) is globally invariant.In curved spacetime there is no global boost and no single global rapidity; the picture holds only in the local tangent space, and comparing rapidities at separated events needs path-dependent parallel transport.

Derivation

\[ ct'=\gamma\,(ct-\beta x),\qquad x'=\gamma\,(x-\beta\,ct),\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}} \]

Start from the Lorentz boost along \(x\) (assumed prior result), writing the time coordinate as \(ct\) so both axes carry length dimension. A

\[ \begin{pmatrix} ct'\\ x' \end{pmatrix}=\begin{pmatrix} \gamma & -\gamma\beta\\ -\gamma\beta & \gamma \end{pmatrix}\begin{pmatrix} ct\\ x \end{pmatrix}\equiv B\begin{pmatrix} ct\\ x \end{pmatrix} \]

Repackage the two scalar equations as one linear map; legal because Step 1 is linear in the coordinates. A

\[ \gamma^2-(\gamma\beta)^2=\gamma^2\,(1-\beta^2)=\frac{1-\beta^2}{1-\beta^2}=1 \]

The two independent entries of \(B\) obey \(X^2-Y^2=1\), the exact form of \(\cosh^2\varphi-\sinh^2\varphi=1\), so \((\gamma,\gamma\beta)\) lies on the unit hyperbola and one parameter suffices. B

\[ \gamma\equiv\cosh\varphi,\qquad \gamma\beta\equiv\sinh\varphi \quad\Longrightarrow\quad \tanh\varphi=\frac{\sinh\varphi}{\cosh\varphi}=\frac{\gamma\beta}{\gamma}=\beta \]

For every \(\beta\in(-1,1)\) there is a unique real \(\varphi\) with \(\tanh\varphi=\beta\); Step 3 makes the two definitions consistent, and \(\cosh\varphi>0\) selects the future-pointing branch. This defines the rapidity \(\varphi=\operatorname{artanh}\beta\). B

\[ B(\varphi)=\begin{pmatrix} \cosh\varphi & -\sinh\varphi\\ -\sinh\varphi & \cosh\varphi \end{pmatrix},\qquad (ct')^2-(x')^2=(ct)^2-x^2 \]

Substituting Step 4 into Step 2 recasts the boost as a hyperbolic rotation of the \((ct,x)\) plane — a Euclidean rotation with \(\cos\to\cosh\), \(\sin\to\sinh\) and a relative sign. Direct expansion using \(\cosh^2-\sinh^2=1\) confirms the interval is preserved for every \(\varphi\). B

\[ B(\varphi_2)\,B(\varphi_1)=\begin{pmatrix} \cosh\varphi_2 & -\sinh\varphi_2\\ -\sinh\varphi_2 & \cosh\varphi_2 \end{pmatrix}\begin{pmatrix} \cosh\varphi_1 & -\sinh\varphi_1\\ -\sinh\varphi_1 & \cosh\varphi_1 \end{pmatrix} \]

Apply two collinear boosts in succession (frame \(S\to S'\) by \(\varphi_1\), then \(S'\to S''\) by \(\varphi_2\)); composition of linear maps is matrix multiplication. Common direction means no rotation is generated, so the product stays in the same family. B

\[ B(\varphi_2)B(\varphi_1)=\begin{pmatrix} \cosh\varphi_1\cosh\varphi_2+\sinh\varphi_1\sinh\varphi_2 & -(\sinh\varphi_1\cosh\varphi_2+\cosh\varphi_1\sinh\varphi_2)\\ -(\sinh\varphi_1\cosh\varphi_2+\cosh\varphi_1\sinh\varphi_2) & \cosh\varphi_1\cosh\varphi_2+\sinh\varphi_1\sinh\varphi_2 \end{pmatrix} \]

Carry out the matrix product entry by entry. A

\[ B(\varphi_2)\,B(\varphi_1)=\begin{pmatrix} \cosh(\varphi_1+\varphi_2) & -\sinh(\varphi_1+\varphi_2)\\ -\sinh(\varphi_1+\varphi_2) & \cosh(\varphi_1+\varphi_2) \end{pmatrix}=B(\varphi_1+\varphi_2) \]

Apply the hyperbolic angle-sum identities \(\cosh(a+b)=\cosh a\cosh b+\sinh a\sinh b\) and \(\sinh(a+b)=\sinh a\cosh b+\cosh a\sinh b\) to every entry. The composite is a single boost of rapidity \(\varphi_1+\varphi_2\): rapidities add. B

\[ \beta=\tanh(\varphi_1+\varphi_2)=\frac{\tanh\varphi_1+\tanh\varphi_2}{1+\tanh\varphi_1\tanh\varphi_2}=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2} \]

Read the combined velocity off Step 8 via \(\beta=\tanh\varphi\) and expand the tangent of a sum. This reproduces the relativistic velocity-addition rule (assumed prior result) as a corollary of additivity. B

Result

\[ \boxed{\;\varphi\equiv\operatorname{artanh}\beta,\qquad B(\varphi_2)B(\varphi_1)=B(\varphi_1+\varphi_2),\qquad \beta=\tanh(\varphi_1+\varphi_2)=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}\;} \]

Reading. The rapidity is the hyperbolic angle of a boost. Successive collinear boosts stack by simple addition of their rapidities, just as successive rotations stack by adding angles. The measurable speed is the hyperbolic tangent of the total rapidity, which automatically enforces \(|\beta|<1\) and regenerates the fractional velocity-addition law — the apparent nonlinearity of velocities is just the nonlinearity of \(\tanh\).

Units check. \(\beta=v/c\) is dimensionless, so \(\varphi=\operatorname{artanh}\beta\) is dimensionless (an angle-like number). Both \(ct\) and \(x\) carry length, so \(B(\varphi)\) is a dimensionless \(2\times2\) matrix mapping lengths to lengths; \(\cosh\varphi=\gamma\) and \(\sinh\varphi=\gamma\beta\) are dimensionless, and the interval \((ct)^2-x^2\) has units of length\(^2\) and is preserved.

Limiting cases

Non-relativistic (\(\beta\ll1\)): \(\varphi=\operatorname{artanh}\beta\approx\beta+\tfrac13\beta^3+\cdots\approx\beta\), so \(\varphi_1+\varphi_2\approx\beta_1+\beta_2\) — rapidity addition degenerates to Galilean velocity addition.
Ultra-relativistic (\(\beta\to1\)): \(\varphi\to+\infty\) logarithmically, \(\varphi\approx\tfrac12\ln\!\frac{1+\beta}{1-\beta}\); no finite chain of boosts reaches \(c\), mirroring \(\tanh\varphi<1\) for all finite \(\varphi\).
Equal boosts (\(\varphi_1=\varphi_2=\varphi_0\)): total \(\varphi=2\varphi_0\), giving the doubling formula \(\beta=\tanh 2\varphi_0=\dfrac{2\beta_0}{1+\beta_0^2}\).
Opposite boosts (\(\varphi_2=-\varphi_1\)): total \(\varphi=0\), \(B(0)=\mathbb{1}\), returning to rest — the inverse boost is \(B(-\varphi)=B(\varphi)^{-1}\).
Doppler / acceleration: \(e^{\varphi}=\sqrt{\tfrac{1+\beta}{1-\beta}}\) is the longitudinal Doppler factor (adding rapidities multiplies Doppler factors); constant proper acceleration gives \(\varphi(\tau)=a\tau/c\), the hyperbolic worldline.

Breaks when

Non-collinear boosts. If \(\vec\beta_1\) and \(\vec\beta_2\) are not parallel, \(B(\vec\varphi_2)B(\vec\varphi_1)=R\,B(\vec\varphi_{\rm tot})\) with a nontrivial spatial rotation \(R\) (Wigner rotation, giving Thomas precession). The scalar sum \(\varphi_1+\varphi_2\) is then wrong, and rapidities behave as noncommuting generators, not additive numbers.
Superluminal or lightlike relative motion. For \(|\beta|\ge1\) the equation \(\tanh\varphi=\beta\) has no real solution: \(\varphi\to\infty\) for a light signal and is complex beyond \(c\). The real-rapidity parameterization cannot describe such "boosts".
Curved spacetime / global acceleration. Additivity is a statement about the global Lorentz group of flat spacetime. In curved spacetime boosts exist only locally, and there is no single interval-preserving \(B(\varphi)\) over finite regions.

Failure modes

Adding velocities as if they were rapidities. Writing \(\beta=\beta_1+\beta_2\) — velocities do not add; only rapidities do.
Using \(\varphi=\beta\) instead of \(\varphi=\operatorname{artanh}\beta\). Valid only to first order; at \(\beta=0.6\) the error is already about \(15\%\).
Identifying rapidity with the wrong function of \(\beta\). Setting \(\varphi=\sinh^{-1}\beta\) or \(\varphi=\gamma\); the correct definition is \(\varphi=\operatorname{artanh}\beta\), with \(\gamma=\cosh\varphi\ge1\) while \(\varphi\in(-\infty,\infty)\).
Confusing hyperbolic with circular rotation. Writing \(\cos\varphi,\sin\varphi\), or dropping the sign so \(\cosh^2+\sinh^2\) is used, violating \(\cosh^2\varphi-\sinh^2\varphi=1\).
Assuming additivity for perpendicular or skew boosts. Ignoring the Wigner rotation and the non-commutativity of the boost generators.
Sign/branch slips. Taking \(\cosh\varphi=-\gamma\) (wrong time-orientation), or mishandling the sign of \(\varphi\) for a boost in the \(-x\) direction, where \(B(\varphi)^{-1}=B(-\varphi)\).

Discussion

The deep content is that boosts along one axis form a one-parameter abelian group isomorphic to \((\mathbb{R},+)\). Rapidity is the canonical coordinate on that group — the parameter in which the group law is literally addition. The nonlinear velocity law is simply what you get after pushing an additive parameter through the nonlinear chart \(v=c\tanh\varphi\). In this light, the speed of light as an unreachable ceiling is the geometric statement that \(\tanh\) is bounded: an infinite rapidity maps to \(\beta=1\).

The analogy with Euclidean rotation is exact. A rotation by angle \(\theta\) mixes \(x\) and \(y\) via \(\cos\theta,\sin\theta\) and preserves \(x^2+y^2\); a boost by rapidity \(\varphi\) mixes \(ct\) and \(x\) via \(\cosh\varphi,\sinh\varphi\) and preserves \((ct)^2-x^2\). Formally, setting \(\varphi=i\theta\) turns one into the other — a boost is a "rotation through an imaginary angle," and the minus sign in the metric is exactly what converts circular functions into hyperbolic ones.

Rapidity is also the working variable of collider physics. Under a boost along the beam axis every longitudinal rapidity shifts by the same constant, so differences of rapidity — and the shapes of particle distributions in rapidity — are boost-invariant; for ultra-relativistic particles one uses the related pseudorapidity \(\eta=-\ln\tan(\theta/2)\). For a body under constant proper acceleration \(a\), the equation of motion integrates trivially, \(d\varphi/d\tau=a/c\Rightarrow\varphi=a\tau/c\), and the whole hyperbolic worldline follows from \(\beta=\tanh\varphi\) without solving any nonlinear ODE in \(v\).

At the Lie-algebra level, the boost generator \(K=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\) satisfies \(K^2=\mathbb{1}\), so \(B(\varphi)=e^{\varphi K}=\mathbb{1}\cosh\varphi+K\sinh\varphi\) — the matrix analogue of Euler's formula with hyperbolic functions. Additivity \(e^{\varphi_2 K}e^{\varphi_1 K}=e^{(\varphi_1+\varphi_2)K}\) is automatic because a single generator commutes with itself (the abelian case of Baker-Campbell-Hausdorff). For boosts along different axes the generators fail to commute, \([K_x,K_y]=-J_z\), a rotation generator; this nonzero commutator is the algebraic origin of both the breakdown of additivity and of Thomas-Wigner rotation.

Common misconceptions. Rapidity is not a velocity and not a Lorentz factor; it is a dimensionless hyperbolic angle running over all of \(\mathbb{R}\), unbounded even though \(\tanh\varphi\) never reaches \(1\). And "boosts add" is true only for the rapidity and only along a common axis — it is false for velocities and for boosts in different directions.

Worked examples

Two collinear boosts, each \(\beta_1=\beta_2=0.600\). Find the combined \(\beta\) via rapidity.

Convert each velocity to a rapidity, add, convert back. Symbols first: \(\varphi=\operatorname{artanh}\beta_1+\operatorname{artanh}\beta_2\), then \(\beta=\tanh\varphi\). A

\[ \varphi_i=\operatorname{artanh}\beta_i=\tfrac12\ln\frac{1+\beta_i}{1-\beta_i}=\tfrac12\ln\frac{1.600}{0.400}=\tfrac12\ln4=\ln2=0.6931 \]

Definition of rapidity; identical for \(\varphi_1\) and \(\varphi_2\). A

\[ \varphi=\varphi_1+\varphi_2=2\ln2=\ln4=1.3863 \]

Rapidities add (Step 8 of the derivation). A

\[ \beta=\tanh(\ln4)=\frac{4-\tfrac14}{4+\tfrac14}=\frac{3.75}{4.25}=0.8824 \]

Convert back; \(\tanh(\ln4)=\frac{e^{\ln4}-e^{-\ln4}}{e^{\ln4}+e^{-\ln4}}\). Cross-check via velocity addition: \((0.6+0.6)/(1+0.36)=1.2/1.36=0.8824\). B

\[ \beta=0.882,\qquad v=0.882\,c\approx2.65\times10^{8}\ \mathrm{m\,s^{-1}} \]

Reading. Two \(0.6c\) boosts do not give \(1.2c\); rapidities \(0.693+0.693=1.386\) give \(\tanh=0.882\), safely below \(c\).

Units check. \(\varphi\) dimensionless; \(\beta\) dimensionless and \(<1\); \(v=\beta c\) in \(\mathrm{m\,s^{-1}}\). Good.

A rocket holds constant proper acceleration \(a=9.80\ \mathrm{m\,s^{-2}}\) for proper time \(\tau=1.00\) year. Find its final speed relative to the launch frame.

Under constant proper acceleration the rapidity grows linearly with proper time; symbols before numbers. B

\[ \frac{d\varphi}{d\tau}=\frac{a}{c}\quad\Longrightarrow\quad \varphi(\tau)=\frac{a\,\tau}{c} \]

Proper acceleration is the rate of change of rapidity per unit proper time (hyperbolic motion, with \(\varphi(0)=0\)). B

\[ \varphi=\frac{a\,\tau}{c}=\frac{(9.80)(3.156\times10^{7})}{2.998\times10^{8}}=1.031 \]

Insert \(\tau=1\ \mathrm{yr}=3.156\times10^{7}\ \mathrm{s}\) and \(c=2.998\times10^{8}\ \mathrm{m\,s^{-1}}\); units \(\frac{(\mathrm{m\,s^{-2}})(\mathrm{s})}{\mathrm{m\,s^{-1}}}\) are dimensionless. A

\[ \beta=\tanh\varphi=\tanh(1.031)=0.776,\qquad \gamma=\cosh(1.031)=1.59 \]

Convert accumulated rapidity to velocity and Lorentz factor. A

\[ \varphi\approx1.03,\qquad \beta\approx0.78,\qquad v\approx0.78\,c,\qquad \gamma\approx1.59 \]

Reading. One year at \(1g\) proper acceleration builds a rapidity near \(1.03\), i.e. \(v\approx0.78c\) — not the Newtonian \(a\tau\approx1.03c\). Because \(\varphi\) is additive in \(\tau\), a second identical year simply doubles \(\varphi\) to \(2.06\), giving \(\beta=\tanh(2.06)\approx0.968\).

Units check. \(a\tau/c\) dimensionless; \(\beta,\gamma\) dimensionless. Good.

Problems

(A) Convert \(\beta=0.80\) to a rapidity, and convert \(\varphi=1.50\) back to a velocity.
Solution
\(\varphi=\operatorname{artanh}(0.80)=\tfrac12\ln\frac{1.8}{0.2}=\tfrac12\ln9=\ln3=1.099\). And \(\beta=\tanh(1.50)=0.905\), so \(v=0.905c\approx2.71\times10^{8}\ \mathrm{m\,s^{-1}}\).
(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 \(\varphi_0=\operatorname{artanh}(0.5)=\tfrac12\ln\frac{1.5}{0.5}=\tfrac12\ln3=0.5493\). Total \(\varphi=3\varphi_0=1.6479\), so \(\beta=\tanh(1.6479)=0.9286\), i.e. \(v\approx0.929c\). Check by iterating velocity addition: \(0.5\oplus0.5=1.0/1.25=0.80\); \(0.80\oplus0.5=1.3/1.4=0.9286\).
(B) Show directly that \(\gamma=\cosh\varphi\) follows from \(\tanh\varphi=\beta\), and hence that a boost and its inverse compose to the identity.
Solution
From \(\tanh\varphi=\beta\), \(\operatorname{sech}^2\varphi=1-\tanh^2\varphi=1-\beta^2\), so \(\cosh^2\varphi=\frac{1}{1-\beta^2}=\gamma^2\); since \(\cosh\varphi>0\), \(\cosh\varphi=\gamma\), and \(\sinh\varphi=\cosh\varphi\tanh\varphi=\gamma\beta\). Then \(B(\varphi)B(-\varphi)=B(\varphi-\varphi)=B(0)=\mathbb{1}\) because \(\cosh0=1,\ \sinh0=0\): boosting by \(v\) then \(-v\) restores the original frame.
(B) A particle of mass \(m\) has energy \(E\) and momentum \(p\) along \(x\). Show \(E=mc^2\cosh\varphi\) and \(pc=mc^2\sinh\varphi\), identify \(\tanh\varphi\), and recover the mass shell.
Solution
For a massive particle \(E=\gamma mc^2\) and \(p=\gamma m v=\gamma m\beta c\). With \(\gamma=\cosh\varphi\), \(\gamma\beta=\sinh\varphi\): \(E=mc^2\cosh\varphi\), \(pc=mc^2\sinh\varphi\). Dividing, \(pc/E=\tanh\varphi=\beta=v/c\). The 2-vector \((E,pc)\) is a hyperbolic rotation of the rest value \((mc^2,0)\) through the particle's rapidity, and \(E^2-(pc)^2=(mc^2)^2(\cosh^2\varphi-\sinh^2\varphi)=(mc^2)^2\) recovers the mass-shell relation.
(C) A muon at rapidity \(\varphi=3.0\) in the lab is viewed from a frame boosted by \(\Delta\varphi=1.2\) along the same axis, in the muon's direction of motion. Find the muon's rapidity and speed in the new frame, and explain why rapidity differences are the useful quantity.
Solution
Boosting the observer toward the muon subtracts from its rapidity: \(\varphi'=\varphi-\Delta\varphi=3.0-1.2=1.8\). Speed \(\beta'=\tanh(1.8)=0.947\), so \(v'\approx0.947c\) (versus the lab value \(\tanh(3.0)=0.995c\)). Because a longitudinal boost shifts every rapidity by the same constant \(\Delta\varphi\), the difference in rapidity between any two particles is invariant under such a boost — which is exactly why rapidity (and pseudorapidity) is the coordinate of choice for longitudinal distributions at colliders.