Rapidity parametrization of collinear boosts

For every real $x$, set $a_x:=\exp (x)$. By [L2] with $y=-x$, $a_x{a}_{-x}=\exp (0)$; applying [L2] with $x=y=0$ gives $\exp (0)=\exp (0{)}^2$, and $\exp (0)\mathrm{\ne }0$ because the codomain of $\exp$ is $(0,\mathrm{\infty })$ by [L1], so $\exp (0)=1$ and $a_{-x}=a_x^{-1}$. Also $a_x=(a_{x\mathrm{/}2}{)}^2>0$ because $a_{x\mathrm{/}2}>0$ by [L1]. Consequently ${\cosh }^2(r)-{\sinh }^2(r)=\frac{(a_r+a_{-r}{)}^2-(a_r-a_{-r}{)}^2}{4}=a_r{a}_{-r}=1$ for every $r$, and $\cosh (r)=\frac{a_r+a_r^{-1}}{2}>0$.

Multiplying numerator and denominator by $a_r$ gives $\tanh (r)=\frac{a_r-a_r^{-1}}{a_r+a_r^{-1}}=\frac{a_r^2-1}{a_r^2+1}=\frac{e^{2r}-1}{e^{2r}+1}$ by [L2], since $e^{2r}=a_r^2$. Since $e^{2r}>0$, the numerator $e^{2r}-1$ is strictly smaller in absolute value than the denominator $e^{2r}+1$, so $-1<\tanh (r)<1$; thus $\tanh$ maps $\mathbb{R}$ into $(-1,1)$. For $0<u<v$ one computes $\frac{v-1}{v+1}-\frac{u-1}{u+1}=\frac{(v-1)(u+1)-(u-1)(v+1)}{(v+1)(u+1)}=\frac{2(v-u)}{(v+1)(u+1)}>0$, so $u\mapsto \frac{u-1}{u+1}$ is strictly increasing on $(0,\mathrm{\infty })$; since $r\mapsto e^{2r}=\exp (2r)$ is strictly increasing by [L1], the composition $\tanh$ is strictly increasing on $\mathbb{R}$.

Let $\beta \in (-1,1)$ and put $u:=\frac{1+\beta }{1-\beta }>0$, the denominator being positive. By [L1] there is a real $t$ with $e^t=u$; put $s:=t\mathrm{/}2$, so $e^{2s}=(e^s{)}^2=e^t=u$ by [L2]. Then step 1.2 gives $\tanh (s)=\frac{u-1}{u+1}=\frac{(1+\beta )-(1-\beta )}{(1+\beta )+(1-\beta )}=\frac{2\beta }{2}=\beta$. Hence $\tanh :\mathbb{R}\to (-1,1)$ is surjective, and with step 1.2 it is a strictly increasing bijection; its inverse ${\tanh }^{-1}:(-1,1)\to \mathbb{R}$ is strictly increasing and onto.

For $r,s\in \mathbb{R}$ put ${\beta }_1:=\tanh (r)$ and ${\beta }_2:=\tanh (s)$. Both lie in $(-1,1)$, so $1+{\beta }_1{\beta }_2>0$. Writing $u:=e^{2r}$, $v:=e^{2s}$, step 1.2 gives ${\beta }_1=(u-1)\mathrm{/}(u+1)$ and ${\beta }_2=(v-1)\mathrm{/}(v+1)$, while [L2] gives $e^{2(r+s)}=uv$. Hence $\frac{{\beta }_1+{\beta }_2}{1+{\beta }_1{\beta }_2}=\frac{(u-1)(v+1)+(v-1)(u+1)}{(u+1)(v+1)+(u-1)(v-1)}=\frac{2uv-2}{2uv+2}=\tanh (r+s)$. This checks the parameter identity of the collinear composition law; the map equality will follow from the boost plane matrices.

For $\beta \in (-1,1)$, put $q:=\frac{1+\beta }{1-\beta }>0$ and $r:=r(\beta )=\frac{1}{2}\log q$, so $2r=\log q$. By [L3], $e^{2r}=\exp (\log q)=q$. Then step 1.2 gives $\tanh (r(\beta ))=\frac{q-1}{q+1}=\frac{(1+\beta )-(1-\beta )}{(1+\beta )+(1-\beta )}=\beta$. Thus $\tanh \circ r={\mathrm{id}}_{(-1,1)}$; composing with the bijection $\tanh$ of step 2.1 gives $r={\tanh }^{-1}$, so $r$ is a strictly increasing bijection of $(-1,1)$ onto $\mathbb{R}$, and in particular $\beta =\tanh (r(\beta ))$ for every $\beta$.

Let $r\in \mathbb{R}$ and put $\beta :=\tanh (r)\in (-1,1)$, so that $r=r(\beta )$ by step 3.1. By step 1.1, $1-\tanh (r{)}^2=\frac{{\cosh }^{2}r-{\sinh }^{2}r}{{\cosh }^{2}r}=\cosh (r{)}^{-2}$; since $\cosh (r)>0$ by step 1.1, the rational power laws [L4] give $(1-\tanh (r{)}^2{)}^{-1\mathrm{/}2}=(\cosh (r{)}^{-2}{)}^{-1\mathrm{/}2}=\cosh (r)$. Multiplying the definition $\tanh r=\sinh r\mathrm{/}\cosh r$ by $\cosh r$ gives $\sinh (r)=\tanh (r)\cosh (r)=\beta \gamma (\beta )$, because $\gamma (\beta )=(1-{\beta }^2{)}^{-1\mathrm{/}2}=(1-\tanh (r{)}^2{)}^{-1\mathrm{/}2}=\cosh (r)$ by [L5].

Fix $e$ and its column $n$, and define $J(a):={\sum }_{i=1}^3{a}_i{e}_i$ for $a\in {\mathbb{R}}^3$. By [L8], $J(n)=e$ and $\eta (J(a),J(d))=a\cdot d$. Write $x=x^0{e}_0+se+y$ with $s:=n\cdot x_{\mathrm{v}\mathrm{e}\mathrm{c}}$, $y_{\mathrm{v}\mathrm{e}\mathrm{c}}:=x_{\mathrm{v}\mathrm{e}\mathrm{c}}-sn$, and $y:=J(y_{\mathrm{v}\mathrm{e}\mathrm{c}})$; then $n\cdot y_{\mathrm{v}\mathrm{e}\mathrm{c}}=\eta (e,y)=0$. Set $\beta :=\tanh (r)$ and $\mathit{\beta }:=\beta n\in {\mathbb{R}}^3$, so $\mathrm{\mid }\mathit{\beta }\mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }<1$ and $\mathit{\beta }\cdot x_{\mathrm{v}\mathrm{e}\mathrm{c}}=\beta s$. If $\beta =0$, step 4.1 gives $\cosh r=1$, $\sinh r=0$, and [L5] gives $B(0)=I$. If $\beta \mathrm{\ne }0$, [L5] gives $(B(\beta n)x{)}^0=\gamma (x^0-\beta s)$ and $(B(\beta n)x{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}=sn+y_{\mathrm{v}\mathrm{e}\mathrm{c}}+(\gamma -1)sn-\gamma \beta x^{0}n=(\cosh rs-\sinh r{x}^0)n+y_{\mathrm{v}\mathrm{e}\mathrm{c}}$ by step 4.1. Thus in both cases the action in directions $(e_0,e)$ is $\left(\begin{array}{cc}\cosh r& -\sinh r\\ -\sinh r& \cosh r\end{array}\right)$, with the identity on the Lorentz-orthogonal complement of their plane, namely $\{J(a):n\cdot a=0\}$.

By [L2] and the exponential definitions, expansion gives $\cosh (r+s)=\cosh r\cosh s+\sinh r\sinh s$ and $\sinh (r+s)=\sinh r\cosh s+\cosh r\sinh s$. The product of the two plane matrices from step 5.1 therefore has diagonal entries $\cosh (r+s)$ and off-diagonal entries $-\sinh (r+s)$; each map is the identity on the same transverse space. The direct-sum decomposition in step 5.1 then gives $B(\tanh (r)e)B(\tanh (s)e)=B(\tanh (r+s)e)$ on all of $V$, including zero rapidities. Together with step 2.2 this proves the collinear composition law stated in [L6], with additive rapidity.

Step 3.1 proves that $r$ is a strictly increasing bijection of $(-1,1)$ onto $\mathbb{R}$ with $\beta =\tanh (r(\beta ))$; step 1.1 gives ${\cosh }^2-{\sinh }^2=1$; step 4.1 gives $(1-{\tanh }^{2}r{)}^{-1\mathrm{/}2}=\cosh r$ and $\sinh r=\tanh r\cosh r=\beta \gamma (\beta )$; step 5.1 gives the matrix display and the identity on the perpendicular complement; and step 6.1 gives the addition of rapidities. This proves every clause of the statement; the characterisation $B(\tanh re{)}^{-1}=B(-\tanh re)$ used implicitly is [L7].