Screw motion

Calculation

A screw motion ${\displaystyle S}$ is the product of four non-collinear reflections, and thus ${\displaystyle S=abcd}$. But according to the Mozzi-Chasles' theorem a screw motion can be decomposed into a commuting translation $${\displaystyle T=e^{\alpha B_{1}}=1+\alpha B_{1}}$$where ${\displaystyle B_{1}}$ is the axis of translation satisfying ${\displaystyle B_{1}^{2}=0}$, and rotation$${\displaystyle R=e^{\beta B_{2}}=\cos(\beta )+B_{2}\sin(\beta )}$$where ${\displaystyle B_{2}}$ is the axis of rotation satisfying ${\displaystyle B_{2}^{2}=-1}$. The two bivector lines ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$ are orthogonal and commuting. To find ${\displaystyle T}$ and ${\displaystyle R}$ from ${\displaystyle S}$, we simply write out ${\displaystyle S}$ and consider the result grade-by-grade:$${\displaystyle {\begin{aligned}S&=TR\\&=e^{\alpha B_{1}}e^{\beta B_{2}}\\&=\underbrace {\cos \beta } _{\text{scalar}}+\underbrace {\sin \beta B_{2}+\alpha \cos \beta B_{1}} _{\text{bivector}}+\underbrace {\alpha \sin \beta B_{1}B_{2}} _{\text{quadrivector}}\end{aligned}}}$$Because the quadrivector part ${\displaystyle \langle S\rangle _{4}=\langle T\rangle _{2}\langle R\rangle _{2}}$ and ${\displaystyle B_{1}^{2}=0}$, ${\displaystyle T}$ is directly found to be$${\displaystyle T=1+{\frac {\langle S\rangle _{4}}{\langle S\rangle _{2}}}}$$and thus$${\displaystyle R=ST^{-1}=T^{-1}S={\frac {S}{T}}}$$Thus, for a given screw motion ${\displaystyle S}$ the commuting translation and rotation can be found using the two formulae above, after which the lines ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$ are found to be proportional to ${\displaystyle \langle T\rangle _{2}}$ and ${\displaystyle \langle R\rangle _{2}}$ respectively.