Worked change of adapted chart with a rotation, a boost and a spatial shift

Take $V={\mathbb{R}}^4$ with coordinates $(t,x_1,x_2,x_3)$, $\tau (t,x_1,x_2,x_3)=t$, so that $K=\{0\}\times {\mathbb{R}}^3$ with the standard inner product $g$, and take $A=V$ as an affine space over itself. Let the first adapted chart be the standard one, $o=0$, $u=e_0$, $(e_1,e_2,e_3)$ the standard basis of $K$. Let the second adapted chart be

$o^{\mathrm{\prime }}=e_2,u^{\mathrm{\prime }}=e_0+e_1,e_1^{\mathrm{\prime }}=e_2,e_2^{\mathrm{\prime }}=-e_1,e_3^{\mathrm{\prime }}=e_3,$

which is adapted: $\tau (u^{\mathrm{\prime }})=1$, and $(e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is a positively oriented $g$-orthonormal basis of $K$.

For a point with first coordinates $(t,x_1,x_2,x_3)$, equating the two expansions gives $t^{\mathrm{\prime }}=t$, $x_2=x_1^{\mathrm{\prime }}+1$ and $x_1=t^{\mathrm{\prime }}-x_2^{\mathrm{\prime }}$, that is

$x_1^{\mathrm{\prime }}=x_2-1,x_2^{\mathrm{\prime }}=t-x_1,x_3^{\mathrm{\prime }}=x_3,$

which is the transformation $x^{\mathrm{\prime }}=Rx+vt+a$ of Transformation between adapted charts and its kinematic consequences with $b=0$,

$R=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right),v=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),a=\left(\begin{array}{c}-1\\ 0\\ 0\end{array}\right)\mathrm{.}$

The matrix $R$ satisfies $R^{\mathsf{T}}R=I_3$ and $\det R=1$, so $R\in SO(3)$.

• The shift $a=(-1,0,0)$ is the coordinate vector of $o-o^{\mathrm{\prime }}=-e_2$ in the primed spatial basis. The change from $u$ to $u^{\mathrm{\prime }}=u+e_1$ supplies the boost $v=(0,1,0)$; since $b=0$, it adds no constant shift.

• The matrix $R$ is the change-of-basis matrix from $(e_1,e_2,e_3)$ to $(e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$, and its determinant is $1$ because both bases are positively oriented. Had one of the two bases been negatively oriented, the same computation would give $\det R=-1$ and $R\in O(3)\setminus SO(3)$.