\[\left\{z,\zeta\right\}=-2\dot{z}^{\ifrac{1}{2}}\frac{{\mathrm{d}}^{2}}{{%
\mathrm{d}\zeta}^{2}}(\dot{z}^{-\ifrac{1}{2}})=\frac{\dddot{z}}{\dot{z}}-\frac%
{3}{2}\left(\frac{\ddot{z}}{\dot{z}}\right)^{2}.\]