# How to add together non-parallel rapidities?

How to add together non-parallel rapidities?

The Lorentz transformation is essentially a hyperbolic rotation, which rotation can be described by a hyperbolic angle, which is called the rapidity. I found that this hyperbolic angle nicely and simply describe many quantities in natural units:

• Lorentz factor: $\mathrm{cosh}\,\phi$
• Coordinate velocity: $\mathrm{tanh}\,\phi$
• Proper velocity: $\mathrm{sinh}\,\phi$
• Total energy: $m\,\mathrm{cosh}\,\phi$
• Momentum: $m\,\mathrm{sinh}\,\phi$
• Proper acceleration: $d\phi / d\tau$ (so local accelerometers measure the change of rapidity)

Also other nice features:

• Velocity addition formula simplifies to adding rapidities together (if they are parallel).
• For low speeds the rapidity is the classical velocity in natural units.

I think for more than one dimensions, the rapidity can be seen as a vector quantity.

In that case my questions are:

• What's the general rapidity addition formula?
• And optionally: Given these nice properties why don't rapidity used more often? Does it have some bad properties that make it less useful?

In above Figure-01 an inertial system $$\:\mathrm S'\:$$ is translated with respect to the inertial system $$\:\mathrm S\:$$ with constant velocity
\begin{align} \!\!\!\!\!\!\!\!\!\mathbf{u}_1 \boldsymbol{=} \left(u_{1x},u_{1y},u_{1z}\right) & \boldsymbol{=}\left(u_1 n_{1x},u_1 n_{1y},u_1 n_{1z}\right) \boldsymbol{=} u_1 \mathbf{n}_1\,, \qquad u_1 \in \left(-c,0\right)\cup\left(0,c\right) \tag{01a}\label{01a}\\ \Vert \mathbf{n}_1 \Vert^2 & \boldsymbol{=} n^2_{1x}\boldsymbol{+}n^2_{1y} \boldsymbol{+} n^2_{1z} \boldsymbol{=}1 \tag{01b}\label{01b} \end{align} The Lorentz transformation $$\:\mathrm S \longrightarrow \mathrm S'\:$$ is
\begin{align} \mathrm d\mathbf{r'} & \boldsymbol{=} \mathrm d\mathbf{r}\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathrm d\mathbf{r})\mathbf{n}_1\boldsymbol{-}\gamma_1 \mathbf{u}_1\mathrm d t \tag{02a}\label{02a}\\ \mathrm d t' & \boldsymbol{=} \gamma_1\left(\mathrm d t\boldsymbol{-}\dfrac{\mathbf{u}_1\boldsymbol{\cdot} \mathrm d\mathbf{r}}{c^{2}}\right) \tag{02b}\label{02b}\\ \gamma_1 & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2_1}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{02c}\label{02c} \end{align} while its inverse $$\:\mathrm S' \boldsymbol{\longrightarrow} \mathrm S\:$$ is \begin{align} \mathrm d\mathbf{r} & \boldsymbol{=} \mathrm d\mathbf{r'}\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathrm d\mathbf{r'})\mathbf{n}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1\mathrm d t' \tag{03a}\label{03a}\\ \mathrm d t & \boldsymbol{=} \gamma_1\left(\mathrm d t'\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot} \mathrm d\mathbf{r'}}{c^{2}}\right) \tag{03b}\label{03b} \end{align}

Now, let a point particle $$\:\mathrm P\:$$ moving with velocity $$\:\mathbf{u}_2\:$$ with respect to system $$\:\mathrm S'\:$$ where \begin{align} \!\!\!\!\!\!\!\!\!\mathbf{u}_2 \boldsymbol{=}\dfrac{\mathrm d\mathbf{r'}}{\mathrm d t'} \boldsymbol{=}\left(u_{2x'},u_{2y'},u_{2z'}\right) & \boldsymbol{=} \left(u_2 n_{2x'},u_2 n_{2y'},u_2 n_{1z'}\right) \boldsymbol{=} u_2 \mathbf{n}_2\,, \quad u_2 \in \left(-c,c\right) \tag{04a}\label{04a}\\ \Vert \mathbf{n}_2 \Vert^2 & = n^2_{2x'}+n^2_{2y'} + n^2_{2z'} = 1 \tag{04b}\label{04b}\\ \gamma_2 & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2_2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{04c}\label{04c} \end{align} In order to find its velocity $$\:\mathbf{u}\:$$ with respect to system $$\:\mathrm S\:$$ where \begin{align} \mathbf{u} \boldsymbol{=}\dfrac{\mathrm d\mathbf{r}}{\mathrm d t} \boldsymbol{=}\left(u_{x},u_{y},u_{z}\right) & \boldsymbol{=} \left(u n_{x},u n_{y},u n_{z}\right) \boldsymbol{=} u \mathbf{n}\,, \qquad u \in \left(-c,c\right) \tag{05a}\label{05a}\\ \Vert \mathbf{n} \Vert^2 & = n^2_{x}+n^2_{y} + n^2_{z} = 1 \tag{05b}\label{05b}\\ \gamma & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{05c}\label{05c} \end{align} we divide equations \eqref{03a}, \eqref{03b} side by side and have $$$$\mathbf{u} \boldsymbol{=}\dfrac{\mathbf{u}_2\boldsymbol{+}(\gamma_1\boldsymbol{-}1)(\mathbf{n}_1\boldsymbol{\cdot} \mathbf{u}_2)\mathbf{n}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1}{ \gamma_1\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^{2}}\right)} \tag{06}\label{06}$$$$ Replacing $$\:\mathbf{n}_1\boldsymbol{\longrightarrow} \mathbf{u}_1/u_1\:$$ $$$$\boxed{\:\:\mathbf{u} \boldsymbol{=}\dfrac{\mathbf{u}_2\boldsymbol{+}\dfrac{\gamma^2_{1}\left(\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2\right)}{c^2 \left(\gamma_{1}\boldsymbol{+}1\right)}\mathbf{u}_1\boldsymbol{+}\gamma_1 \mathbf{u}_1}{ \gamma_1\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^{2}}\right)}\:\:} \tag{07}\label{07}$$$$ Above equation, beyond to be the transformation law for 3-velocities, is the law of relativistic addition of 3-velocities, more exactly it's the relativistic sum of $$\:\mathbf{u}_1,\mathbf{u}_2$$.

Now, between the $$\gamma-$$factors $$\gamma,\gamma_{1},\gamma_{2}\:$$ the following equation is valid

$$$$\boxed{\:\: \gamma \boldsymbol{=}\gamma_{1}\gamma_{2}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{08}\label{08}$$$$ This relation is proved as follows :

Let $$\:\mathrm S^{\mathrm P}\:$$ the rest system of the particle $$\:\mathrm P$$. In this system $$\:\mathrm S^{\mathrm P}\:$$ the time is the proper one $$\:\tau$$. The rest system $$\:\mathrm S^{\mathrm P}\:$$ is moving with velocity $$\:\mathbf{u}_2\:$$ with respect to system $$\:\mathrm S'\:$$ so according to the Lorentz transformation between these systems we have $$$$\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\gamma_2 \tag{09}\label{09}$$$$ On the same step, since the rest system $$\:\mathrm S^{\mathrm P}\:$$ is moving with velocity $$\:\mathbf u\:$$ with respect to system $$\:\mathrm S\:$$ we have $$$$\dfrac{\mathrm dt\hphantom{'}}{\mathrm d\tau}\boldsymbol{=}\gamma \tag{10}\label{10}$$$$ On the other hand from the Lorentz transformation between the systems $$\:\mathrm S\:$$ and $$\:\mathrm S'\:$$ we have, see \eqref{03b} $$$$\dfrac{\mathrm dt\hphantom{'}}{\mathrm dt'}\boldsymbol{=}\gamma_{1}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right) \tag{11}\label{11}$$$$ From equations \eqref{09},\eqref{10} and \eqref{11} the relation \eqref{08} is proved, that is $$$$\gamma\boldsymbol{=}\dfrac{\mathrm dt\hphantom{'}}{\mathrm d\tau}\boldsymbol{=}\dfrac{\mathrm dt\hphantom{'}}{\mathrm dt'}\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\gamma_{1}\gamma_{2}\left(1\boldsymbol{+}\dfrac{\mathbf{u}_1\boldsymbol{\cdot}\mathbf{u}_2}{c^2}\right) \tag{12}\label{12}$$$$ In terms of the rapidities $$\:\zeta_1,\zeta_2,\zeta\:$$ where $$$$\tanh\zeta_1\boldsymbol{=}\dfrac{u_1}{c}\,,\quad \tanh\zeta_2\boldsymbol{=}\dfrac{u_2}{c}\,,\quad\tanh\zeta\boldsymbol{=}\dfrac{u}{c} \tag{13}\label{13}$$$$ equation \eqref{08} is rewritten as $$$$\boxed{\:\:\cosh\zeta\boldsymbol{=}\cosh\zeta_1\cosh\zeta_2\boldsymbol{+}\underbrace{\left(\mathbf{n}_1\boldsymbol{\cdot}\mathbf{n}_2\right)}_{\cos\omega}\sinh\zeta_1\sinh\zeta_2\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{14}\label{14}$$$$ where $$\:\omega \in [0,\pi]\:$$ the angle between the unit vectors $$\:\mathbf{n}_1\:$$ and $$\:\mathbf{n}_2$$, see Figure-01.

In case of parallel $$\:\mathbf{n}_1,\mathbf{n}_2\:$$ we have $$$$\zeta\boldsymbol{=} \left. \begin{cases} \zeta_1\boldsymbol{+}\zeta_2 & \text{if}\:\:\: \omega=0\\ \zeta_1\boldsymbol{-}\zeta_2 & \text{if}\:\:\: \omega=\pi \end{cases}\right\} \tag{15}\label{15}$$$$

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$$

Now, we'll make a correlation with the composition of two rotations in space.

As we'll see in the following (Figure-04), Figure-02 is used for the geometric representation (construction) of the composition of two rotations in space. In this Figure two planes $$\:\rm p_1, p_2\:$$ at an angle $$\:\omega\:$$ intersect on line $$\:\epsilon$$. Two lines $$\:\epsilon_1,\epsilon_2\:$$ on planes $$\:\rm p_1, p_2\:$$ respectively are passing through a common point on line $$\:\epsilon\:$$ at angles $$\:\phi_1,\phi_2\:$$ with respect to this line. By elementary trigonometry we have $$$$\cos\phi \boldsymbol{=} \cos\phi_1 \cos\phi_2\boldsymbol{-}\cos\omega \sin\phi_1 \sin\phi_2 \tag{16}\label{16}$$$$ If in above equation we replace the real angles $$\:\phi_1,\phi_2,\phi\:$$ with imaginary ones $$$$\phi_1\boldsymbol{=}\mathrm{i}\,\zeta_1\,,\quad \phi_2\boldsymbol{=}\mathrm{i}\,\zeta_2\,,\quad \phi\boldsymbol{=}\mathrm{i}\,\zeta \tag{17}\label{17}$$$$ that is $$$$\cos\left(\mathrm{i}\,\zeta\right) \boldsymbol{=} \cos\left(\mathrm{i}\,\zeta_1\right) \cos\left(\mathrm{i}\,\zeta_2\right)\boldsymbol{-}\cos\omega \sin\left(\mathrm{i}\,\zeta_1\right) \sin\left(\mathrm{i}\,\zeta_2\right) \tag{18}\label{18}$$$$ then we'll meet equation \eqref{14} again $$$$\cosh\zeta\boldsymbol{=}\cosh\zeta_1\cosh\zeta_2\boldsymbol{+}\cos\omega\sinh\zeta_1\sinh\zeta_2 \tag{19}\label{19}$$$$ since $$$$\cos\left(\mathrm{i}\,\rho\right)\boldsymbol{=}\cosh\rho\,,\quad \sin\left(\mathrm{i}\,\rho\right)\boldsymbol{=}\mathrm{i}\,\sinh\rho \qquad \rho \in \mathbb{R} \tag{20}\label{20}$$$$ and formally we have Figure-03.

Angles work very well in two dimensions (the Euclidean plane or 1+1 spacetime), where you only need one. In three or more dimensions, there are singularities in the representation of direction or rotation/orientation by angles, and other representations are more elegant.

Also, in 3 or more dimensions, the space of directions (velocities) and the space of rotations (Lorentz transformations) are different. You can't conflate them any more. This means that there is no analogue of "rapidity addition" in higher dimensions. Instead, you must either apply a rotation (Lorentz transformation) to a direction (velocity), or else compose two rotations, and these are different operations.

The most elegant way to represent a direction is by a unit vector pointing in that direction. In spacetime, this is the four-velocity. It isn't precisely analogous to the rapidity but it's the generalization that will give you the fewest headaches. In terms of the four-velocity $$\mathbf v$$ you have

• Lorentz factor: $$\mathbf v \cdot \mathbf{\hat t}$$
• Coordinate velocity: $$\mathbf v / (\mathbf v \cdot \mathbf{\hat t}) - \mathbf{\hat t}$$
• Proper velocity: $$\mathbf v - (\mathbf v \cdot \mathbf{\hat t}) \mathbf{\hat t}$$
• Total energy and momentum: $$m\mathbf v$$
• Proper acceleration: $$\mathrm d\mathbf v/\mathrm dτ$$

In 3D Euclidean space, the most elegant way to represent a rotation or an orientation is by a unit quaternion. The generalization of this to arbitrary dimension and signature is called the even Clifford algebra of the space. By multiplying elements of the Clifford algebra, you can see why vectors are inadequate for this purpose. For example, the composition of boosts in the $$\mathbf v$$ and $$\mathbf u$$ directions is $$(A + B \mathbf{\hat t u})(C + D \mathbf{\hat t v}) = AC + \mathbf{\hat t}(BC\mathbf u + AD\mathbf v) - BD\mathbf{uv}$$ where $$A,B,C,D$$ are scalars that I'm omitting for brevity. If $$\mathbf u {\parallel} \mathbf v$$ then $$\mathbf{uv}$$ is a scalar and this is equivalent to a boost in the same direction. Otherwise, it isn't a boost, because there is an additional rotation in the $$\mathbf{uv}$$ plane.