In three dimensions the vanishing of the divergence is the condition of continuity of flux, e.g.\ in hydrodynamics $\dd u/\dd x + \dd v/\dd y + \dd w/\dd z = 0$. Adding a time-coordinate, this becomes the condition of \emph{conservation} or \emph{permanence}, as \index{Permanence}% will be shown in detail later. \emph{It will be realised how important for a theory of the material world is the discovery of a world-tensor which is inherently permanent.} I think it should be possible to prove~\Eq{(52)} by geometrical reasoning in continuation of the ideas of \SecRef{33}. But I have not been able to construct a geometrical proof and must content myself with a clumsy analytical verification. By the rules of covariant differentiation \[ (g_{\mu}^{\nu} G)_{\nu} = g_{\mu}^{\nu}\, \Chg{\dd G/\dd x_{\nu}}{\frac{\dd G}{\dd x_{\nu}}} = \Chg{\dd G/\dd x_{\mu}}{\frac{\dd G}{\dd x_{\mu}}}. \] Thus the theorem reduces to \[ G_{\mu\nu}^{\nu} = \frac{1}{2}\, \frac{\dd G}{\dd x_{\mu}}. \Tag{(52.1)} \] For $\mu = 1$, $2$, $3$, $4$, these are the four identities referred to in \SecRef{37}. By~\Eq{(51.32)} \[ G_{\mu\nu}^{\nu} = \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (G_{\mu}^{\nu} \sqrt{-g}) + \tfrac{1}{2} G_{\alpha\beta}\, \frac{\dd g^{\alpha\beta}}{\dd x_{\mu}}, \] and since $G = g^{\alpha\beta} G_{\alpha\beta}$ \[ \frac{1}{2}\, \frac{\dd G}{\dd x_{\mu}} = \tfrac{1}{2} g^{\alpha\beta}\, \frac{\dd G_{\alpha\beta}}{\dd x_{\mu}} + \tfrac{1}{2} G_{\alpha\beta}\, \frac{\dd g^{\alpha\beta}}{\dd x_{\mu}}. \] Hence, subtracting, we have to prove that \[ \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (G_{\mu}^{\nu} \sqrt{-g}) = \tfrac{1}{2} g^{\alpha\beta}\, \frac{\dd G_{\alpha\beta}}{\dd x_{\mu}}. \Tag{(52.2)} \] Since \Eq{(52)}~is a tensor relation it is sufficient to show that it holds for a special coordinate-system; only we must be careful that our special choice of coordinate-system does not limit the kind of space-time and so spoil the generality of the proof. It has been shown in \SecRef{36} that in any kind of space-time, coordinates can be chosen so that all the first derivatives~$\dd g_{\mu\nu}/\dd x_{\sigma}$ vanish at a particular point; we shall therefore lighten the algebra by taking coordinates such that at the point considered \[ \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} = 0. \Tag{(52.3)} \] \PageSep{116} This condition can, of course, only be applied after all differentiations have been performed. Then \[ \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (G_{\mu}^{\nu} \sqrt{-g}) = \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (g^{\nu\tau} g^{\sigma\rho} \sqrt{-g}· B_{\mu\tau\sigma\rho}). \] Owing to~\Eq{(52.3)} $g^{\nu\tau} g^{\sigma\rho} \sqrt{-g}$ can be taken outside the differential operator, giving \[ g^{\nu\tau} g^{\sigma\rho}\, \frac{\dd}{\dd x_{\nu}} B_{\mu\tau\sigma\rho}, \] which by~\Eq{(34.5)} is equal to \[ \tfrac{1}{2} g^{\nu\tau} g^{\sigma\rho} \left( \frac{\dd^{2} g_{\rho\sigma}}{\dd x_{\mu}\, \dd x_{\tau}}