Decomposition of Transformations into Boost and Rotation

A HepLorentzRotation $T$ may be decomposed either into the form $B R$ or the form $R B$. Here we will refer to the first form as $T = \acute{B}(T) \grave{R}(T)$ and the second as $T = \acute{R}(T) \grave{B}(T)$.

When decomposing into the product $B R$, the boost will have the same last column as the HepLorentzRotation $T$. Using the value of $\gamma$ read off $T_{tt}$, one can apply equation () to find the matrix $\acute{B}(T)$ for that boost. Then

$$\grave{R}(T) = \left[ \acute{B}(T) \right] ^{-1} T$$ (176)

When decomposing into the product $R B$, the boost will have the same last row as the HepLorentzRotation $T$. Again one can apply equation () to find the matrix (this time $\grave{B}(T)$) for that boost. Then

$$\acute{R}(T) = T \left[ \grave{B}(T) \right] ^{-1}$$ (177)

Naively applying the above equations leads to non-neglible round-off errors in the components of the Rotation--of order $3 \cdot 10^{-14}$ if the boost has $\beta = .95$. Since a Rotation representation which is non-orthogonal on that scale would lead to errors in distance measures of more than one part in $10^{-7}$, the decompose() method rectifies the Rotation before returning.