RATTLE gives an exact step and a cubic phase defect for a free rigid bond

The initial data are consistent because $\mathrm{\parallel }{q}_0{\mathrm{\parallel }}_2=\mathrm{\ell }$ and $q_0^T{p}_0=0$. The position multiplier equation gives $q_1=q_0+\frac{h}{m}(p_0-h\lambda q_0)$; choosing $\lambda =m(1-\sqrt{1-h^2{\omega }^2})\mathrm{/}{h}^2=m{\omega }^2\mathrm{/}(1+\sqrt{1-h^2{\omega }^2})$ yields the stated $q_1$, whose squared norm is ${\mathrm{\ell }}^2$. This multiplier tends to $m{\omega }^2\mathrm{/}2$, the position branch center.

With $\mu =\lambda$, the final RATTLE equation gives the stated $p_1$; $\mu$ has the same limit $m{\omega }^2\mathrm{/}2$, so the unique local branch selects this root. Direct multiplication gives $q_1^T{p}_1=0$ and $\mathrm{\parallel }{p}_1{\mathrm{\parallel }}_2=m\omega \mathrm{\ell }$, so both exact constraints hold.

The continuous multiplier is ${\lambda }_c=m{\omega }^2$, hence the exact flow is $q(t)=\mathrm{\ell }(\cos \omega t,\sin \omega t)$ and $p(t)=m\omega \mathrm{\ell }(-\sin \omega t,\cos \omega t)$. This solves [F3] and agrees with the initial data. The RATTLE position angle is ${\theta }_h=\arcsin (h\omega )$. Taylor's theorem [F4] on $0\le h\omega \le 1\mathrm{/}2$ gives $\arcsin (h\omega )-h\omega =O((h\omega {)}^3)$. Since $\omega$ is fixed, the angle defect is $O(h^3)$ as stated.