Comparing $g_{\mu\nu}$ and $g_{\mu\nu} + \delta g_{\mu\nu}$ by~\Eq{(23.22)}, since they correspond to a transformation of coordinates, \begin{align*} g_{\mu\nu} &= (g_{\alpha\beta} + \delta g_{\alpha\beta}) \frac{\dd(x_{\alpha} + \delta x_{\alpha})}{\dd x_{\mu}} · \frac{\dd(x_{\beta} + \delta x_{\beta})}{\dd x_{\nu}} \\ &= (g_{\alpha\beta} + \delta g_{\alpha\beta})\, \frac{\dd x_{\alpha}}{\dd x_{\mu}}\, \frac{\dd x_{\beta}}{\dd x_{\nu}} + g_{\alpha\beta}\, \frac{\dd x_{\alpha}}{\dd x_{\mu}}\, \frac{\dd (\delta x_{\beta})}{\dd x_{\nu}} + g_{\alpha\beta}\, \frac{\dd x_{\beta}}{\dd x_{\nu}}\, \frac{\dd (\delta x_{\alpha})}{\dd x_{\mu}}. \end{align*} But \[ \frac{\dd x_{\alpha}}{\dd x_{\mu}} = g_{\mu}^{\alpha},\quad \frac{\dd x_{\beta}}{\dd x_{\nu}} = g_{\nu}^{\beta}\quad\text{by~\Eq{(22.3)}.} \] Hence \[ g_{\mu\nu} = g_{\mu\nu} + \delta g_{\mu\nu} + g_{\mu\beta}\, \frac{\dd(\delta x_{\beta})}{\dd x_{\nu}} + g_{\alpha\nu}\, \frac{\dd(\delta x_{\alpha})}{\dd x_{\mu}}. \] This is a comparison of the fundamental tensor at $x_{\alpha} + \delta x_{\alpha}$ in the new coordinate-system with the value at~$x_{\alpha}$ in the old system. There would be no objection to using this value of~$\delta g_{\mu\nu}$ provided that we took account of the corresponding~$\delta(d\tau)$. We prefer, however, to keep $d\tau$~fixed in the comparison, \PageSep{141} and must compare the values at~$x_{\alpha}$ in both systems. It is therefore necessary to subtract the change $\delta x_{\alpha} · \dd g_{\mu\nu}/\dd x_{\alpha}$ of~$g_{\mu\nu}$ in the distance~$\delta x_{\alpha}$; hence \[ -\delta g_{\mu\nu} = g_{\mu\beta}\, \frac{\dd(\delta x_{\beta})}{\dd x_{\nu}} + g_{\alpha\nu}\, \frac{\dd(\delta x_{\alpha})}{\dd x_{\mu}} + \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}}\, \delta x_{\alpha}. \Tag{(61.3)} \] Hence \Eq{(61.1)}~becomes %[** TN: Not broken in the original] \begin{multline*} \delta \int K \sqrt{-g}\, d\tau \\ = -\int P^{\mu\nu} \sqrt{-g} \left( g_{\mu\alpha}\, \frac{\dd}{\dd x_{\nu}} (\delta x_{\alpha}) + g_{\nu\alpha}\, \frac{\dd}{\dd x_{\mu}} (\delta x_{\alpha}) + \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}}\, \delta x_{\alpha} \right) d\tau \end{multline*} which, by partial integration, \begin{align*} &= \int\! \left\{\!\frac{\dd}{\dd x_{\nu}} (g_{\mu\alpha} P^{\mu\nu} \sqrt{-g}) + \frac{\dd}{\dd x_{\mu}} (g_{\nu\alpha} P^{\mu\nu} \sqrt{-g}) - P^{\mu\nu} \sqrt{-g}\, \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}}\!\right\} \delta x_{\alpha}\, d\tau \\ &= 2\int \left\{\frac{\dd}{\dd x_{\nu}} \mf{P}_{\mu}^{\nu} - \tfrac{1}{2} \mf{P}^{\mu\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\alpha}}\right\} \delta x_{\alpha}\, d\tau \\ &= 2\int P_{\alpha\nu}^{\nu}\, \delta x_{\alpha}\, \sqrt{-g}\, d\tau \quad\text{by~\Eq{(51.51)}.} \Tag{(61.4)} \end{align*} This has to vanish for all arbitrary variations~$\delta x_{\alpha}$---deformations of the mesh-system---and accordingly \[ (P_{\alpha}^{\nu})_{\nu} = 0. \Tag{(61.5)} \]