The two alternative tensors are excessively complicated expressions; but when applied to determine the field of an isolated particle, they become not unmanageable. The field, being symmetrical, must be of the general form~\Eq{(38.2)}, so that we have only to determine the disposable coefficients $\lambda$ and~$\nu$ both of which must be functions of $r$~only. $K'$~can be calculated in terms of $\lambda$ and~$\nu$ without difficulty from equations~\Eq{(38.6)}; but the expression for~$K''$ turns out to be rather simpler and I shall deal with it. By the method of \SecRef{38}, we find %[** TN: Re-breaking] \begin{multline*} \mf{K}'' = K'' \sqrt{-g} = 2e^{\frac{1}{2}(\lambda + \nu)} \sin\theta \bigl\{e^{-2\lambda} (\lambda'^{2} + \nu'^{2}) \\ + 2r^{2} e^{-2\lambda} (\tfrac{1}{4}\lambda'\nu' - \tfrac{1}{4}\nu'^{2} - \tfrac{1}{2}\nu'')^{2} + 2(1 - e^{-\lambda})^{2}/r^{2}\bigr\}. \Tag{(62.3)} \end{multline*} It is clear that the integral of~$\mf{K}''$ will be stationary for variations from the symmetrical condition, so that we need only consider variations of $\lambda$ and $\nu$ and their derivatives with respect to~$r$. Thus the gravitational equations $\Ham\mf{K}''/\Ham g_{\mu\nu} = 0$ are equivalent to \[ \frac{\Ham K''}{\Ham \lambda} = 0,\quad \frac{\Ham K''}{\Ham \nu} = 0. \Tag{(62.4)} \] Now for a variation of~$\lambda$ \begin{align*} \delta\! \int\!\! \mf{K}\, d\tau &=\!\! \int\!\! \biggl(\frac{\dd\mf{K}}{\dd\lambda}\, \delta\lambda + \frac{\dd\mf{K}}{\dd\lambda'}\, \delta\lambda' + \frac{\dd\mf{K}}{\dd\lambda''}\, \delta\lambda''\biggr) d\tau\displaybreak[0] \\ &=\!\! \int\!\! \left\{\frac{\dd\mf{K}}{\dd\lambda} - \frac{\dd}{\dd r} \biggl(\frac{\dd\mf{K}}{\dd\lambda'}\biggr) + \frac{\dd^{2}}{\dd r^{2}} \biggl(\frac{\dd\mf{K}}{\dd\lambda''}\biggr)\!\!\right\} \delta\lambda\, d\tau + \text{surface-integrals.} \end{align*} Hence our equations~\Eq{(62.4)} take the Lagrangian form \[ \left. \begin{alignedat}{4} \frac{\Ham K''}{\Ham\lambda} &= \frac{\dd\mf{K}''}{\dd\lambda} &&-\frac{\dd}{\dd r} \frac{\dd\mf{K}''}{\dd\lambda'} &&+ \frac{\dd^{2}}{\dd r^{2}} \frac{\dd\mf{K}''}{\dd\lambda''} &&= 0\Add{,} \\ \frac{\Ham K''}{\Ham\nu} &= \frac{\dd\mf{K}''}{\dd\nu} &&-\frac{\dd}{\dd r} \frac{\dd\mf{K}''}{\dd\nu'} &&+ \frac{\dd^{2}}{\dd r^{2}} \frac{\dd\mf{K}''}{\dd\nu''} &&= 0. \end{alignedat} \right\} \Tag{(62.5)} \] From these $\lambda$ and $\nu$ are to be determined.