Can we decompose a general Lorentz boost in a rotation followed by three boosts along the coordinate axes?

$$\newcommand{\betabold}{\boldsymbol{\beta}} \newcommand{\xbold}{\boldsymbol{x}} \newcommand{\ebold}{\boldsymbol{e}}$$ For $$\betabold\in \mathbb R^3$$, with $$0<|\betabold|<1$$, let us denote the Lorentz boost with velocity $$\betabold$$ by $$L^{\betabold}(t, \xbold):=\Big(\gamma t-\gamma \betabold\cdot \xbold, \xbold_\bot +\gamma \xbold_{\parallel}-\gamma\betabold t\Big), \quad \text{where }\ \gamma:=\frac{1}{\sqrt{1-\beta^2}}.$$ Here $$\xbold_\parallel:=\frac{\xbold\cdot\betabold}{\beta^2}\betabold$$ is the component of $$\xbold$$ in the direction of $$\betabold$$, and $$\xbold_\bot:=\xbold-\xbold_\parallel$$. (The speed of light is normalized to $$1$$).

Does there exist a spatial rotation $$R$$ and $$\alpha_1, \alpha_2, \alpha_3\in(-1, 1)$$ such that $$L^\betabold = L^{\alpha_1\ebold_1}L^{\alpha_2\ebold_2}L^{\alpha_3\ebold_3}R\quad ?$$

• A rotation can move any unit vector to coincide with any axis. So there exists a rotation which rotates $\mathbf \beta$ to the $\mathbf e_3$ axis. So yes; just look for such a rotation and then choose $\alpha_3 =|\mathbf \beta|$ and $\alpha_2=\alpha_1 = 0$. – Dwagg Oct 6 '18 at 14:41
• @Dwagg: Don't you need a second rotation after the boost? – Giuseppe Negro Oct 6 '18 at 15:15
• The commutator of two boost generators is a rotation generator. And you have $L^{\vec\varphi}=\exp(\mathrm{i}\vec\varphi\cdot \vec B)$, where $\vec B=(B_x,B_y,B_z)$ is a 3-tuple of boost generators. Your question is then if $L^{\vec\varphi}=L^{\alpha_1e_1}L^{\alpha_2e_2}L^{\alpha_3e_3}R$ for some appropriate $\alpha_i$ and $R$. The answer is yes because of the Baker-Campbell-Hausdorff formula. – user178876 Oct 6 '18 at 17:15
• Related : General matrix Lorentz transformation. By intuition there are infinitely many solutions. – Frobenius Oct 6 '18 at 18:39
• Elaborating an answer to your question first results show that my intuition about infinitely many solutions is false. I could prove that for given $\:\boldsymbol{\beta}\:$ there exists one and only one triad $\:\left(\alpha_1,\alpha_2,\alpha_3\right)\:$ and consequently one rotation R satisfying your equation. – Frobenius Oct 10 '18 at 9:35

$$$$\mathrm{L}\left(\boldsymbol{\beta}\right) =\mathrm{R}^{\boldsymbol{-1}}\cdot\mathrm{L}_3\left(\alpha_3\right)\cdot\mathrm{L}_2\left(\alpha_2\right)\cdot\mathrm{L}_1\left(\alpha_1\right) \tag{a}\label{a}$$$$ $$$$\alpha_1 =\beta_1\,, \quad \alpha_2 =\dfrac{\beta_2}{\sqrt{1-\beta^2_1}}\,, \quad \alpha_3 =\dfrac{\beta_3}{\sqrt{1-\left(\beta^2_1+\beta^2_2\right)}} \tag{b}\label{b}$$$$ $$$$\mathrm{R}=\text{space rotation} \tag{c}\label{c}$$$$ $$==================================================$$

Figure-01 3D

From Figure 01 :

Lorentz Transformation from $$\:\mathrm{S}\boldsymbol{\equiv} \{xyz\omega, \omega\boldsymbol{=}ct\}\:$$ to $$\:\mathrm{S_1}\boldsymbol{\equiv} \{x_1y_1z_1\omega_1, \omega_1\boldsymbol{=}ct_1\}\:$$ $$$$\begin{bmatrix} x_1\\ y_1\\ z_1\\ \omega_1 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \omega \end{bmatrix} \,, \quad \tanh\!\xi\boldsymbol{=}\alpha_1\boldsymbol{=}\dfrac{u_1}{c} \tag{01}\label{01}$$$$ or $$$$\mathbf{W}_1\boldsymbol{=}\mathrm{L_1}\mathbf{W}\,, \qquad \mathrm{L_1}\boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \tag{02}\label{02}$$$$

Lorentz Transformation from $$\:\mathrm{S_1}\boldsymbol{\equiv} \{x_1y_1z_1\omega_1, \omega_1\boldsymbol{=}ct_1\}\:$$ to $$\:\mathrm{S_2}\boldsymbol{\equiv} \{x_2y_2z_2\omega_2, \omega_2\boldsymbol{=}ct_2\}\:$$ $$$$\begin{bmatrix} x_2\\ y_2\\ z_2\\ \omega_2 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \begin{bmatrix} x_1\\ y_1\\ z_1\\ \omega_1 \end{bmatrix} \,, \quad \tanh\!\eta\boldsymbol{=}\alpha_2\boldsymbol{=}\dfrac{u_2}{c} \tag{03}\label{03}$$$$ or $$$$\mathbf{W}_2\boldsymbol{=}\mathrm{L_2}\mathbf{W}_1\,, \qquad \mathrm{L_2}\boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \tag{04}\label{04}$$$$

Lorentz Transformation from $$\:\mathrm{S_2}\boldsymbol{\equiv} \{x_2y_2z_2\omega_2, \omega_2\boldsymbol{=}ct_2\}\:$$ to $$\:\mathrm{S_3}\boldsymbol{\equiv}\{x_3y_3z_3\omega_3, \omega_3\boldsymbol{=}ct_3\}\:$$ $$$$\begin{bmatrix} x_3\\ y_3\\ z_3\\ \omega_3 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \begin{bmatrix} x_2\\ y_2\\ z_2\\ \omega_2 \end{bmatrix} \,, \quad \tanh\!\zeta\boldsymbol{=}\alpha_3\boldsymbol{=}\dfrac{u_3}{c} \tag{05}\label{05}$$$$ or $$$$\mathbf{W}_3\boldsymbol{=}\mathrm{L_3}\mathbf{W}_2\,, \qquad \mathrm{L_3}\boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \tag{06}\label{06}$$$$

Note that because of the Standard Configurations the matrices $$\:\mathrm{L_1}, \mathrm{L_2},\mathrm{L_3}\:$$ are real symmetric.

From equations \eqref{02},\eqref{04} and \eqref{06} we have $$$$\mathbf{W}_3\boldsymbol{=}\mathrm{L}_3\mathbf{W}_2\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathbf{W}_1\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\mathbf{W} \quad \boldsymbol{\Longrightarrow} \quad \mathbf{W}_3\boldsymbol{=}\Lambda\,\mathbf{W} \tag{07}\label{07}$$$$ where $$\:\Lambda\:$$ the composition of the three Lorentz Transformations $$\:\mathrm{L_1}, \mathrm{L_2},\mathrm{L_3}\:$$ \begin{align} &\Lambda\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \tag{08}\label{08} \end{align} that is $$$$\Lambda\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi\\ \hphantom{-}\sinh\!\eta\sinh\!\xi & \hphantom{-}\cosh\!\eta & \hphantom{-}0 & -\sinh\!\eta\cosh\!\xi\\ \hphantom{-}\sinh\!\zeta\cosh\!\eta\sinh\!\xi & \hphantom{-}\sinh\!\zeta\sinh\!\eta &\hphantom{-} \cosh\!\zeta & -\sinh\!\zeta\cosh\!\eta\cosh\!\xi\\ -\cosh\!\zeta\cosh\!\eta\sinh\!\xi & -\cosh\!\zeta\sinh\!\eta & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta\cosh\!\eta\cosh\!\xi \end{bmatrix} \tag{09}\label{09}$$$$ The Lorentz Transformation matrix $$\:\Lambda\:$$ is not symmetric, so the systems $$\:\mathrm{S},\mathrm{S_3}\:$$ are not in the Standard configuration. But it's reasonable to suppose that $$$$\Lambda=\mathrm{R}\cdot\mathrm{L} \tag{10}\label{10}$$$$ where $$\:\mathrm{L}\:$$ is the symmetric Lorentz Transformation matrix from $$\:\mathrm{S}\:$$ to an intermediate system $$\:\mathrm{S'}_3\:$$ in Standard configuration to it and co-moving with $$\:\mathrm{S}_3\:$$, while $$\:\mathrm{R}\:$$ is a purely spatial transformation between $$\:\mathrm{S'}_3\:$$ and $$\:\mathrm{S}_3$$.

Now, our target would be to express the symmetric Lorentz Transformation matrix $$\:\mathrm{L}\:$$ in terms of the rapidities $$\:\xi,\eta,\zeta\:$$ since from \eqref{10}
$$$$\mathrm{R}\boldsymbol{=}\Lambda\cdot\mathrm{L}^{\boldsymbol{-1}} \tag{11}\label{11}$$$$

The Lorentz Transformation matrix $$\:\mathrm{L}\:$$, from $$\:\mathrm{S}\:$$ to the intermediate system $$\:\mathrm{S'_3}\:$$ in Standard Configuration to it, is : $$$$\mathrm{L}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=} \begin{bmatrix} 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{x} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{y} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{z} & \!-\dfrac{\gamma\upsilon_{x}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \left(\gamma\!-\!1\right)\!\mathrm{n}_{y}\mathrm{n}_{x} & 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{y} &\left(\gamma\!-\!1\right)\!\mathrm{n}_{y}\mathrm{n}_{z} & \!-\dfrac{\gamma\upsilon_{y}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \left(\gamma\!-\!1\right)\!\mathrm{n}_{z}\mathrm{n}_{x} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{z}\mathrm{n}_{y} & 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{z} & \!-\dfrac{\gamma\upsilon_{z}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!-\dfrac{\gamma\upsilon_{x}}{c} & \!-\dfrac{\gamma\upsilon_{y}}{c} &\!-\dfrac{\gamma\upsilon_{z}}{c} & \gamma \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{12}\label{12}$$$$ In \eqref{12} \begin{align} \dfrac{\boldsymbol{\upsilon}}{c} & \boldsymbol{=} \left(\dfrac{\upsilon_{x}}{c},\dfrac{\upsilon_{y}}{c},\dfrac{\upsilon_{z}}{c}\right)\boldsymbol{=}\left(\tanh\!\xi,\dfrac{\tanh\!\eta}{\cosh\!\xi},\dfrac{\tanh\!\zeta}{\cosh\!\xi\cosh\!\eta}\right)\equiv \boldsymbol{\beta}\boldsymbol{=} \left(\beta_1,\beta_2,\beta_3,\right) \tag{13.1}\label{13.1}\\ \left(\dfrac{\upsilon}{c}\right)^{2} & \boldsymbol{=} \left(\dfrac{\upsilon_{x}}{c}\right)^{2}+\left(\dfrac{\upsilon_{y}}{c}\right)^{2}+\left(\dfrac{\upsilon_{z}}{c}\right)^{2} \boldsymbol{=}1 \boldsymbol{-}\left(\dfrac{1}{\cosh\!\xi\cosh\!\eta\cosh\!\zeta}\right)^{2}=\dfrac{\gamma^{2}\!-\!1}{\gamma^{2}} \tag{13.2}\label{13.2}\\ \gamma & \boldsymbol{=} \left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{-\frac12}\boldsymbol{=} \cosh\!\xi\cosh\!\eta\cosh\!\zeta \boldsymbol{=}\gamma_1\gamma_2\gamma_3 \tag{13.3}\label{13.3}\\ \mathbf{n} & = \left(\mathrm{n}_{x},\mathrm{n}_{y},\mathrm{n}_{z}\right) \boldsymbol{=}\dfrac{\boldsymbol{\upsilon}/c}{\upsilon/c}\boldsymbol{=}\dfrac{\left(\sinh\!\xi\cosh\!\eta\cosh\!\zeta ,\sinh\!\eta\cosh\!\zeta,\sinh\!\zeta\right)}{\sqrt{\cosh^2\!\xi\cosh^2\!\eta\cosh^2\!\zeta-1}} \tag{13.4}\label{13.4} \end{align} where $$\:\boldsymbol{\upsilon}\:$$ is the velocity vector of the origin $$\:\mathrm{O'}_{\!\!3}\left(\equiv \mathrm{O}_{3}\right)\:$$ with respect to $$\:\mathrm{S}$$(1), $$\:\mathbf{n}\:$$ the unit vector along $$\:\boldsymbol{\upsilon}\:$$ and $$\:\gamma\:$$ the corresponding $$\:\gamma-$$factor.

So the matrix $$\:\mathrm{L}\left(\boldsymbol{\upsilon} \right)\:$$ of equation \eqref{12} as function of the rapidities $$\:\xi,\eta,\zeta\:$$ is(2) \begin{align} &\mathrm{L}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\xi,\eta,\zeta \right)\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \!\boldsymbol{-} \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \!\boldsymbol{-} \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\boldsymbol{-}\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\boldsymbol{-}\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\boldsymbol{-}\sinh\!\eta\cosh\!\zeta &\!\boldsymbol{-}\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{14}\label{14} \end{align} while \begin{align} &\mathrm{L}^{\boldsymbol{-1}}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\boldsymbol{-}\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\boldsymbol{-}\xi,\boldsymbol{-}\eta,\boldsymbol{-}\zeta \right)\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\sinh\!\eta\cosh\!\zeta &\!\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{15}\label{15} \end{align}

From equations \eqref{09},\eqref{11} and \eqref{15} \begin{align} &\mathrm{R}\boldsymbol{=}\Lambda\cdot\mathrm{L}^{\boldsymbol{-1}}\boldsymbol{=} \nonumber\\ &\begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\sinh\!\eta\sinh\!\xi & \hphantom{-}\cosh\!\eta & \hphantom{-}0 & -\sinh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\sinh\!\zeta\cosh\!\eta\sinh\!\xi & \hphantom{-}\sinh\!\zeta\sinh\!\eta &\hphantom{-} \cosh\!\zeta & -\sinh\!\zeta\cosh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}}\\ -\cosh\!\zeta\cosh\!\eta\sinh\!\xi & -\cosh\!\zeta\sinh\!\eta & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta\cosh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\sinh\!\eta\cosh\!\zeta &\!\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix}\boldsymbol{=} \nonumber\\ &\begin{bmatrix} \dfrac{\cosh\!\xi\!+\!\cosh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\sinh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \hphantom{\boldsymbol{-}}\dfrac{\cosh\!\eta\!+\!\cosh\!\zeta\cosh\!\xi}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\cosh\!\xi\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi\cosh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ 0 & 0 & 0 & 1 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{16}\label{16} \end{align} $$$$\mathcal R= \begin{bmatrix} \dfrac{\cosh\!\xi\!+\!\cosh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\sinh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \hphantom{\boldsymbol{-}}\dfrac{\cosh\!\eta\!+\!\cosh\!\zeta\cosh\!\xi}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\cosh\!\xi\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi\cosh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{17}\label{17}$$$$ $$$$\mathcal R= \begin{bmatrix} \cos\theta+(1-\cos\theta)\mathrm{m}_{x}^2 & (1-\cos\theta)\mathrm{m}_{x}\mathrm{m}_{y}+\sin\theta\mathrm{m}_{z} & (1-\cos\theta)\mathrm{m}_{x}\mathrm{m}_{z}-\sin\theta\mathrm{m}_{y}\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ (1-\cos\theta)\mathrm{m}_{y}\mathrm{m}_{x}-\sin\theta\mathrm{m}_{z}& \cos\theta+(1-\cos\theta)\mathrm{m}_{y}^2 & (1-\cos\theta)\mathrm{m}_{y}\mathrm{m}_{z}+\sin\theta\mathrm{m}_{x}\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ (1-\cos\theta)\mathrm{m}_{z}\mathrm{m}_{x}+\sin\theta\mathrm{m}_{y} & (1-\cos\theta)\mathrm{m}_{z}\mathrm{m}_{y}-\sin\theta\mathrm{m}_{x} & \cos\theta+(1-\cos\theta)\mathrm{m}_{z}^2\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{18}\label{18}$$$$ \begin{align} & 2\cos\theta+1=\mathrm{trace}(\mathcal R)= \nonumber\\ &\dfrac{\left(\cosh\!\xi\!+\!\cosh\!\eta\!+\!\cosh\!\zeta\right)\!+\!\left(\cosh\!\xi\cosh\!\eta\!+\!\cosh\!\eta\!\cosh\!\zeta\!+\!\cosh\!\zeta\cosh\!\xi\right)}{\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \nonumber\\ &\dfrac{\left(1\!+\!\cosh\!\xi\right)\left(1\!+\!\cosh\!\eta\right)\left(1\!+\!\cosh\!\zeta\right)-\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)}{\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{19}\label{19} \end{align} $$$$\cos\theta=\dfrac{\left(1\!+\!\cosh\!\xi\right)\left(1\!+\!\cosh\!\eta\right)\left(1\!+\!\cosh\!\zeta\right)-2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{20}\label{20}$$$$ $$$$\cos\theta=\dfrac{\left(1\!+\!\gamma_1\right)\left(1\!+\!\gamma_2\right)\left(1\!+\!\gamma_3\right)-2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)}{2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)}\,,\quad \gamma_{\jmath}=\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\:\: \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}} \tag{21}\label{21}$$$$ \begin{align} \sin\theta\,\mathrm{m}_{x} & = \boldsymbol{-}\dfrac{\left(1\!+\!\cosh\!\xi\right)\sinh\!\eta\sinh\!\zeta}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.1}\label{22.1}\\ \sin\theta\,\mathrm{m}_{y} & = \boldsymbol{+}\dfrac{\left(1\!+\!\cosh\!\eta\right)\sinh\!\zeta\sinh\!\xi}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.2}\label{22.2}\\ \sin\theta\,\mathrm{m}_{z} & = \boldsymbol{-}\dfrac{\left(1\!+\!\cosh\!\zeta\right)\sinh\!\eta\sinh\!\xi}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.3}\label{22.3} \end{align} \begin{align} \sin\theta & =\dfrac{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}}{2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{23}\label{23} \end{align} \begin{align} \tan\theta & =\dfrac{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}}{\left(1\!+\!\gamma_1\right)\left(1\!+\!\gamma_2\right)\left(1\!+\!\gamma_3\right)-2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{24}\label{24} \end{align} \begin{align} \mathbf{m} & =\dfrac{\biggl[\left(1\!+\!\gamma_1\right)\left(\gamma_2^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_3^2\!-\!1\right)^{\boldsymbol{\frac12}}\:\boldsymbol{,}\:\left(1\!+\!\gamma_2\right)\left(\gamma_3^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_1^2\!-\!1\right)^{\boldsymbol{\frac12}}\:\boldsymbol{,}\:\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_2^2\!-\!1\right)^{\boldsymbol{\frac12}}\biggr]}{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{25}\label{25} \end{align}

$$==================================================$$

(1) see APPENDIX C - Relativistic addition of velocities

(2) see APPENDIX B - The matrix L

(3) Constructing $$\:\boldsymbol{\alpha}\:$$ from $$\:\boldsymbol{\beta}\:$$

• Impressive. That's lots of work. It will take a while for me to digest it, but thank you very much for your time. – Giuseppe Negro Oct 12 '18 at 16:14
• @Frobenius Nice! One little question: What software/package are you using for the graph? – J C Oct 13 '18 at 0:25
• @J C : GeoGebra. One of its advantages is inserting equations in $\LaTeX$. – Frobenius Oct 13 '18 at 5:37