Taking first $\alpha = 2$, the surviving terms are \[ \frac{d^{2}x_{2}}{ds^{2}} + \{12, 2\}\, \frac{dx_{1}}{ds}\, \frac{dx_{2}}{ds} + \{21, 2\}\, \frac{dx_{2}}{ds}\, \frac{dx_{1}}{ds} + \{33, 2\}\, \frac{dx_{3}}{ds}\, \frac{dx_{3}}{ds} = 0, \] or using~\Eq{(38.5)} \[ \frac{d^{2}\theta}{ds^{2}} + \frac{2}{r}\, \frac{dr}{ds}\, \frac{d\theta}{ds} - \cos\theta \sin\theta \left(\frac{d\phi}{ds}\right)^{2} = 0. \Tag{(39.2)} \] Choose coordinates so that the particle moves initially in the plane $\theta = \frac{1}{2}\pi$. Then $d\theta/ds = 0$ and $\cos\theta = 0$ initially, so that $d^{2}\theta/ds^{2} = 0$. The particle therefore continues to move in this plane, and we may simplify the remaining equations by putting $\theta = \frac{1}{2}\pi$ throughout. The equations for $\alpha = 1$, $3$,~$4$ are found in like manner, viz.\ \begin{align*} \frac{d^{2}r}{ds^{2}} + \tfrac{1}{2}\lambda' \left(\frac{dr}{ds}\right)^{2} - re^{-\lambda} \left(\frac{d\phi}{ds}\right)^{2} + \tfrac{1}{2} e^{\nu-\lambda} \nu' \left(\frac{dt}{ds}\right)^{2} &= 0, \Tag{(39.31)}\displaybreak[0] \\ %\PageSep{86} \frac{d^{2}\phi}{ds^{2}} + \frac{2}{r}\, \frac{dr}{ds}\, \frac{d\phi}{ds} &= 0, \Tag{(39.32)}\displaybreak[0] \\ \frac{d^{2}t}{ds^{2}} + \nu'\, \frac{dr}{ds}\, \frac{dt}{ds} &= 0. \Tag{(39.33)} \end{align*} The last two equations may be integrated immediately, giving \begin{align*} r^{2}\, \frac{d\phi}{ds} &= h, \Tag{(39.41)}\displaybreak[0] \\ \frac{dt}{ds} &= ce^{-\nu} = c/\gamma, \Tag{(39.42)} \end{align*} where $h$ and $c$ are constants of integration. Instead of troubling to integrate~\Eq{(39.31)} we can use in place of it \Eq{(38.8)} which plays here the part of an integral of energy. It gives \[ \gamma^{-1} \left(\frac{dr}{ds}\right)^{2} + r^{2} \left(\frac{d\phi}{ds}\right)^{2} - \gamma \left(\frac{dt}{ds}\right)^{2} = -1. \Tag{(39.43)} \] Eliminating $dt$ and~$ds$ by means of \Eq{(39.41)} and~\Eq{(39.42)} \[ \frac{1}{\gamma} \left(\frac{h}{r^{2}}\, \frac{dr}{d\phi}\right)^{2} + \frac{h^{2}}{r^{2}} - \frac{c^{2}}{\gamma} = -1, \Tag{(39.44)} \] whence, multiplying through by~$\gamma$ or~$(1 - 2m/r)$, \[ \left(\frac{h}{r^{2}}\, \frac{dr}{d\phi}\right)^{2} + \frac{h^{2}}{r^{2}} = c^{2} - 1 + \frac{2m}{r} + \frac{2m}{r} · \frac{h^{2}}{r^{2}}, \] or writing $1/r = u$, \[ \left(\frac{du}{d\phi}\right)^{2} + u^{2} = \frac{c^{2} - 1}{h^{2}} + \frac{2m}{h^{2}} u + 2mu^{3}. \Tag{(39.5)} \] Differentiating with respect to~$\phi$, and removing the factor~$\dfrac{du}{d\phi}$, \[ \frac{d^{2}u}{d\phi^{2}} + u = \frac{m}{h^{2}} + 3mu^{2}, \Tag{(39.61)} \] with \[ r^{2}\, \frac{d\phi}{ds} = h. \Tag{(39.62)} \] Compare these with the equations of a Newtonian orbit \[ \frac{d^{2}u}{d\phi^{2}} + u = \frac{m}{h^{2}} \Tag{(39.71)} \] with \[ r^{2}\, \frac{d\phi}{dt} = h. \Tag{(39.72)} \]