It will be seen that $(\mf{G} + \mf{L})$ has the form of a \emph{divergence}~\Eq{(51.12)}; but the quantity of which it is the divergence is not a vector-density, nor is $\mf{L}$~a scalar-density. We shall derive another formula which will be needed in \SecRef{59}, \[ d(g^{\mu\nu} \sqrt{-g}) = \sqrt{-g} (dg^{\mu\nu} + g^{\mu\nu} · \tfrac{1}{2} g^{\alpha\beta}\, dg_{\alpha\beta}) \quad\text{by~\Eq{(35.3)}.} \] Hence, using~\Eq{(35.2)}, \begin{align*} G_{\mu\nu}\, d(g^{\mu\nu} \sqrt{-g}) &= \sqrt{-g} (-G^{\mu\nu}\, dg_{\mu\nu} + \tfrac{1}{2} Gg^{\alpha\beta}\, dg_{\alpha\beta}) \\ &= -(G^{\mu\nu} - \tfrac{1}{2}g^{\mu\nu}G) \sqrt{-g} · dg_{\mu\nu} \\ &= 8\pi\, \mf{T}^{\mu\nu}\, dg_{\mu\nu}. \Tag{(58.91)} \end{align*} \PageSep{134} Accordingly \begin{align*} 8\pi\, \mf{T}^{\mu\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}} &= G_{\mu\nu} \mf{g}_{\alpha}^{\mu\nu} \\ &= \mf{g}_{\alpha}^{\mu\nu} \biggl( \frac{\dd}{\dd x_{\beta}}\, \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}} - \frac{\dd\mf{L}}{\dd\mf{g}^{\mu\nu}}\biggr) \\ &= \frac{\dd}{\dd x_{\beta}} \biggl(\mf{g}_{\alpha}^{\mu\nu}\, \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}}\biggr) - \frac{\dd}{\dd x_{\beta}}\, \mf{g}_{\alpha}^{\mu\nu} · \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}} - \mf{g}_{\alpha}^{\mu\nu}\, \frac{\dd\mf{L}}{\dd\mf{g}^{\mu\nu}}. \Tag{(58.92)} \end{align*} Now \[ \frac{\dd\mf{L}}{\dd x_{\alpha}} = \frac{\dd\mf{L}}{\dd\mf{g}^{\mu\nu}}\, \frac{\dd\mf{g}^{\mu\nu}}{\dd x_{\alpha}} + \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}}\, \frac{\dd\mf{g}_{\beta}^{\mu\nu}}{\dd x_{\alpha}}, \] and since \[ \frac{\dd\mf{g}_{\beta}^{\mu\nu}}{\dd x_{\alpha}} = \frac{\dd^{2}\mf{g}^{\mu\nu}}{\dd x_{\alpha}\, \dd x_{\beta}} = \frac{\dd\mf{g}_{\alpha}^{\mu\nu}}{\dd x_{\beta}}, \] we see that \Eq{(58.92)}~reduces to \begin{align*} 8\pi\, \mf{T}^{\mu\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}} &= \frac{\dd}{\dd x_{\beta}} \biggl(\mf{g}_{\alpha}^{\mu\nu}\, \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}}\biggr) - \frac{\dd\mf{L}}{\dd x_{\alpha}} \\ &= \frac{\dd}{\dd x_{\beta}} \biggl\{\mf{g}_{\alpha}^{\mu\nu}\, \frac{\dd\mf{L}}{\dd\mf{g}_{\beta}^{\mu\nu}} - g_{\alpha}^{\beta}\mf{L}\biggr\}. \Tag{(58.93)} \end{align*} \Section{59.}{Pseudo-energy-tensor of the gravitational field}