Lorentz Transformations, Rapidity, and Velocity Addition

The Lorentz transformation tells two inertial observers how to assign coordinates to the same events. Its mixing of space and time explains relativity of simultaneity, time dilation, and length contraction. A parameter called rapidity then makes successive boosts along one direction as easy to combine as ordinary rotation angles: their rapidities add, although their velocities do not.

Required background. Calculus and Linear Algebra provides linear maps, matrix composition, and hyperbolic functions; Variational Mechanics and Symmetries supplies the distinction between a trajectory and its description after a regular change of coordinates.

Lorentz boosts between inertial frames

Put the inertial frames in standard configuration: their spatial axes are parallel with no extra rotation, their origins coincide at $t=t'=0$, and $S'$ moves with velocity $+v\,\mathbf e_x$ relative to $S$. We transform the coordinates of the same event from $S$ to $S'$. The worldline of the $S'$ origin is $x=vt$, so the sign convention must make it satisfy $x'=0$:

$$t'=\gamma(t-vx), \qquad x'=\gamma(x-vt), \qquad y'=y,\quad z'=z,$$

with $\gamma=(1-v^2)^{-1/2}$ and $|v|\lt1$. The inverse has the same form with $v\mapsto-v$.

Spacetime homogeneity and the inertial-chart structure make the transformation affine. The coincident origin event removes its constant term, leaving a linear map. In standard configuration, the condition $x'=0$ along $x=vt$ first gives $x'=A(x-vt)$. Requiring both null lines $x=\pm t$ to remain null fixes $t'=A(t-vx)$. Substitution into

$$-t'^2+x'^2=-t^2+x^2$$

then yields $A^2(1-v^2)=1$, and continuity with the identity selects $A=\gamma\gt0$. This derives both the coefficients and their signs. Tong gives the corresponding construction in Tong 2013, §7.1, printed pp. 108–111 (PDF).

What clocks and rulers measure

Each familiar effect uses a different pair of events. Stating that pair prevents most apparent contradictions.

Simultaneity. Two separated flashes simultaneous in $S$ have $\Delta t=0$. Their time separation in $S'$ is $\Delta t'=-\gamma v\Delta x$, which generally is not zero. This is a comparison of synchronized clock readings, after accounting for light travel time; it is not merely a delay in seeing a distant flash.

Time dilation. Let two ticks occur on a clock at rest in $S'$. They obey $\Delta x'=0$, so the inverse boost gives

$$\Delta t=\gamma\Delta t'=\gamma\Delta\tau.$$

The moving clock records the proper time $\Delta\tau$ between its own ticks. The $S$ lattice records a longer coordinate interval. Exchanging $S$ and $S'$ gives the reciprocal result for a different clock at rest in $S$; it does not compare the same pair of worldlines between a departure and a reunion.

Length contraction. A rod at rest along $x'$ has rest length $L_0$. To measure its length $L$ in $S$, record the positions of its two ends at the same $S$ time, so $\Delta t=0$. The spatial boost then gives

$$L_0=\Delta x'=\gamma\Delta x=\gamma L, \qquad L=\frac{L_0}{\gamma}.$$

The two endpoint events are not simultaneous in $S'$, but its rod ends stay at fixed $x'$ and their separation remains $L_0$. For $v=3/5$, $\gamma=5/4$: a one-second tick interval becomes $1.25$ seconds of laboratory time, and a five-metre rest length becomes four metres when measured simultaneously in the laboratory. These are measurement protocols, not a claim about a photograph, whose photons can leave the rod at different times. The derivations and their event choices are developed in Tong 2013, §§7.2.1–7.2.4, pp. 115–124, PDF.

Hyperbolic angle

Define the rapidity $\chi$ by

$$v=\tanh\chi, \qquad \gamma=\cosh\chi, \qquad \gamma v=\sinh\chi.$$

The identity $\cosh^2\chi-\sinh^2\chi=1$ mirrors the invariant timelike norm. Unlike $v$, which is restricted to $(-1,1)$, $\chi$ can be any real number; no finite rapidity reaches light speed.

On the $(t,x)$ plane, the boost is

$$B_x(\chi)= \begin{pmatrix} \cosh\chi&-\sinh\chi\\ -\sinh\chi&\cosh\chi \end{pmatrix}.$$

Hyperbolic addition identities immediately give

$$B_x(\chi_2)B_x(\chi_1)=B_x(\chi_1+\chi_2).$$

Therefore two successive collinear frame velocities $v_1$ and $v_2$ produce

$$v_{\rm total} =\tanh(\chi_1+\chi_2) =\frac{v_1+v_2}{1+v_1v_2}.$$

This is why rapidity, rather than velocity, is the natural group parameter; see Tong 2013, §7.3.2, printed pp. 128–131 (PDF).

For an object whose velocity in $S$ is $\mathbf u$, differentiating the boost gives

$$u'_x=\frac{u_x-v}{1-vu_x}, \qquad \mathbf u'_\perp= \frac{\mathbf u_\perp}{\gamma(1-vu_x)}.$$

The minus sign here answers a different question from the plus sign in $v_{\rm total}$: it transforms an object’s velocity into a frame already moving at $+v$. Solving the first equation for the velocity of a second moving frame recovers the plus-sign composition law. The longitudinal and transverse formulas can be checked independently against Hughes 2024, Lecture 6, §6.1, printed pp. 46–47 (PDF).

The noncollinear boundary

Rapidity adds as a scalar only for boosts along the same oriented line. Boosts in different directions do not commute:

$$B_y(\chi_2)B_x(\chi_1)\ne B_x(\chi_1)B_y(\chi_2), \qquad \chi_1\chi_2\ne0.$$

Their product is generally a boost followed or preceded, depending on convention, by a spatial Wigner rotation. Thus “the rapidities add” must never be applied componentwise to noncollinear velocities. Ferraro and Thibeault 1999, Introduction and §II, pp. 1–3 (arXiv v1 PDF) derives this decomposition and fixes the Wigner-rotation terminology.

Collinear boosts form an Abelian one-parameter subgroup. Pure boosts in arbitrary directions do not form a subgroup at all, because their products need not be pure boosts. Closure is restored only after spatial rotations are included in the full Lorentz group.

Transfer check: two identical boosts

Frame $S'$ moves at $v_1=3/5$ relative to $S$, and $S''$ moves at $v_2=3/5$ relative to $S'$, in the same $+x$ direction. Find the velocity and Lorentz factor of $S''$ relative to $S$.

Solution

Since

$$\chi_1=\chi_2=\operatorname{artanh}\frac35 =\frac12\ln\frac{1+3/5}{1-3/5} =\ln2,$$

the total rapidity is $\chi=\ln4$. Hence

$$v_{\rm total}=\tanh(\ln4) =\frac{4-1/4}{4+1/4} =\frac{15}{17}\approx0.882.$$

Direct velocity addition gives the same result:

$$\frac{3/5+3/5}{1+9/25}=\frac{15}{17}.$$

The total Lorentz factor is

$$\gamma_{\rm total}=\cosh(\ln4) =\frac12\left(4+\frac14\right)=\frac{17}{8}.$$

As a cross-check, each individual $\gamma$ is $5/4$, and

$$\gamma_{\rm total} =\gamma_1\gamma_2(1+v_1v_2) =\frac{25}{16}\frac{34}{25} =\frac{17}{8}.$$

The result stays below unity even though the Newtonian sum is $6/5$. For small velocities the denominator tends to one, recovering ordinary addition.

The derivation assumes inertial frames and a constant relative velocity. It does not describe an accelerated coordinate system, and $|v|=1$ is a singular limit rather than a valid rest frame. The next page uses these boosts only as optional coordinate tools; the elapsed time itself is obtained invariantly from the worldline.

Proper Time, Worldlines, and the Clock Effect compares clocks without assigning causal power to acceleration.

References

• Ferraro, Rafael, and Marc Thibeault. “Generic Composition of Boosts: An Elementary Derivation of the Wigner Rotation.” European Journal of Physics 20 (1999): 143–151. Introduction and §II, pp. 1–3 (arXiv:physics/0211022v1 PDF).
• Hughes, Scott A. Introduction to Relativity and Spacetime Physics. MIT 8.033 Lecture 6, especially §6.1, printed pp. 46–47 (PDF), Fall 2024.
• Tong, David. Dynamics and Relativity. §7.1, pp. 108–111; §§7.2.1–7.2.5, pp. 115–125; §7.3.2, pp. 128–131, PDF, University of Cambridge, Lent Term 2013. Signature translated from $(+---)$ to $(-+++)$.