# 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\,, \qquad 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.