Boost transformation of a future null wave covector and of its frame data

Clause (i). Regard $k$ as the constant covector field on $A$; by [L3] its primed components are $k_a^{\mathrm{\prime }}=(B(-\beta ){)}^b{}_a{k}_b$. For $\beta \mathrm{\ne }0$ the matrix of $B(-\beta )$ has entries $(B(-\beta ){)}^0{}_0=\gamma$, $(B(-\beta ){)}^0{}_i=\gamma {\beta }_i$, $(B(-\beta ){)}^i{}_0=\gamma {\beta }_i$ and $(B(-\beta ){)}^i{}_j={\delta }_j^i+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}{\beta }_i{\beta }_j$ by [L1]; therefore $k_0^{\mathrm{\prime }}=\gamma k_0+\gamma {\sum }_i{\beta }_i{k}_i=\gamma (k_0+\beta \cdot k_{\mathrm{v}\mathrm{e}\mathrm{c}})$ and $k_i^{\mathrm{\prime }}=\gamma {\beta }_i{k}_0+k_i+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}{\beta }_i(\beta \cdot k_{\mathrm{v}\mathrm{e}\mathrm{c}})$, which is the displayed vector identity. For $\beta =0$ one has $B(0)=\mathrm{id}$ by [L1], so $k^{\mathrm{\prime }}=k$.

Clause (ii). By step 1.1 and [L5], $k_0^{\mathrm{\prime }}=\gamma (k_0+\beta \cdot k_{\mathrm{v}\mathrm{e}\mathrm{c}})=\gamma (-\varpi +\beta \cdot \varpi s)=-\gamma (1-\beta \cdot s)\varpi$, hence ${\varpi }^{\mathrm{\prime }}=-k_0^{\mathrm{\prime }}=\gamma (1-\beta \cdot s)\varpi$, and this is positive because $\gamma \ge 1$, $\varpi >0$ and $1-\beta \cdot s>0$ by [L6]; this also agrees with ${\varpi }^{\mathrm{\prime }}>0$ asserted by [L4]. Dividing the vector component of step 1.1 by ${\varpi }^{\mathrm{\prime }}$ gives $s_i^{\mathrm{\prime }}=k_i^{\mathrm{\prime }}\mathrm{/}{\varpi }^{\mathrm{\prime }}=[s_i+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}(\beta \cdot s){\beta }_i-\gamma {\beta }_i]\mathrm{/}[\gamma (1-\beta \cdot s)]$ for $\beta \mathrm{\ne }0$, which is the displayed $s^{\mathrm{\prime }}$. For $\beta =0$, $k^{\mathrm{\prime }}=k$ gives ${\varpi }^{\mathrm{\prime }}=\varpi$ and $s^{\mathrm{\prime }}=s$. The normalization $\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }=1$ and the component form $k_a^{\mathrm{\prime }}=(-{\varpi }^{\mathrm{\prime }},{\varpi }^{\mathrm{\prime }}{s}^{\mathrm{\prime }})$ are clauses (a)-(d) of [L4] applied in the frame of $x^{\mathrm{\prime }}$.

Clause (iii). Let $\beta =b{e}_1$ with $b\in (-1,1)$ and $\gamma =(1-b^2{)}^{-1\mathrm{/}2}$. If $b\mathrm{\ne }0$, substituting ${\beta }_1=b$, ${\beta }_2={\beta }_3=0$ and $\beta \cdot s=b{s}_1$ into step 2.1 gives $s_1^{\mathrm{\prime }}=[s_1+\frac{\gamma -1}{b^2}(b{s}_1)b-\gamma b]\mathrm{/}[\gamma (1-b{s}_1)]=(s_1-b)\mathrm{/}(1-b{s}_1)$, because $s_1+(\gamma -1)s_1-\gamma b=\gamma (s_1-b)$, and $s_2^{\mathrm{\prime }}=s_2\mathrm{/}[\gamma (1-b{s}_1)]$, $s_3^{\mathrm{\prime }}=s_3\mathrm{/}[\gamma (1-b{s}_1)]$. If $b=0$ then $\gamma =1$ and $s^{\mathrm{\prime }}=s$ by step 2.1, which is the same as the displayed formulas at $b=0$.

Clause (iv). For $\beta \mathrm{\ne }0$, step 2.1 gives $\beta \cdot s^{\mathrm{\prime }}=[\beta \cdot s+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}(\beta \cdot s)(\beta \cdot \beta )-\gamma \mathrm{\mid }\beta {\mathrm{\mid }}^2]\mathrm{/}[\gamma (1-\beta \cdot s)]=[\beta \cdot s+(\gamma -1)(\beta \cdot s)-\gamma \mathrm{\mid }\beta {\mathrm{\mid }}^2]\mathrm{/}[\gamma (1-\beta \cdot s)]=[\gamma (\beta \cdot s)-\gamma \mathrm{\mid }\beta {\mathrm{\mid }}^2]\mathrm{/}[\gamma (1-\beta \cdot s)]=(\beta \cdot s-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\mathrm{/}(1-\beta \cdot s)$; for $\beta =0$ both sides are $0$, so the identity holds always. Since $1-\beta \cdot s>0$ by [L6], $\beta \cdot s^{\mathrm{\prime }}=0$ holds exactly when $\beta \cdot s=\mathrm{\mid }\beta {\mathrm{\mid }}^2$, and $\beta \cdot s^{\mathrm{\prime }}=-\mathrm{\mid }\beta {\mathrm{\mid }}^2$ holds exactly when $\beta \cdot s-\mathrm{\mid }\beta {\mathrm{\mid }}^2=-\mathrm{\mid }\beta {\mathrm{\mid }}^2(1-\beta \cdot s)$, that is $\beta \cdot s(1-\mathrm{\mid }\beta {\mathrm{\mid }}^2)=0$, which by $\mathrm{\mid }\beta \mathrm{\mid }<1$ is exactly $\beta \cdot s=0$.

Clause (v). The transition $x^{\mathrm{\prime }}=B(\beta )x+b$ inverts to $x=B(-\beta )x^{\mathrm{\prime }}+b^{\mathrm{\prime }}$ with $b^{\mathrm{\prime }}:=-B(-\beta )b$, whose linear part $B(-\beta )$ is a proper orthochronous Lorentz map with $\mathrm{\mid }-\beta \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }<1$ by [L2]; applying step 2.1 to the chart pair $(x^{\mathrm{\prime }},x)$ with parameter $-\beta$ computes the frame-$x$ data of $k$ from its frame-$x^{\mathrm{\prime }}$ data, and that frame-$x$ data is $(\varpi ,s)$ by hypothesis. Directly, ${\varpi }^{\mathrm{\prime }\mathrm{\prime }}=\gamma (1-(-\beta )\cdot s^{\mathrm{\prime }}){\varpi }^{\mathrm{\prime }}=\gamma (1+\beta \cdot s^{\mathrm{\prime }}){\varpi }^{\mathrm{\prime }}$ and step 3.2 gives $1+\beta \cdot s^{\mathrm{\prime }}=(1-\beta \cdot s+\beta \cdot s-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\mathrm{/}(1-\beta \cdot s)=(1-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\mathrm{/}(1-\beta \cdot s)$, so ${\varpi }^{\mathrm{\prime }\mathrm{\prime }}=\gamma \cdot \frac{1-\mathrm{\mid }\beta {\mathrm{\mid }}^2}{1-\beta \cdot s}\cdot \gamma (1-\beta \cdot s)\varpi ={\gamma }^2(1-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\varpi =\varpi$; the vector part of the same instance returns a unit vector $s^{\mathrm{\prime }\mathrm{\prime }}$ with $k$'s frame-$x$ data equal to $({\varpi }^{\mathrm{\prime }\mathrm{\prime }},s^{\mathrm{\prime }\mathrm{\prime }})$, and by the uniqueness of frame data in [L4] applied to the frame of $x$, $({\varpi }^{\mathrm{\prime }\mathrm{\prime }},s^{\mathrm{\prime }\mathrm{\prime }})=(\varpi ,s)$, so $s^{\mathrm{\prime }\mathrm{\prime }}=s$. This proves (i)-(v); the argument is pure algebra of $\eta$ and the boost, and no choice principle was used.