By \Eq{(29.4)} \begin{align*} (A^{\mu})_{\mu} &= \frac{\dd A^{\mu}}{\dd x_{\mu}} + \{\epsilon\mu, \mu\} A^{\epsilon} \\ &= \frac{\dd A^{\mu}}{\dd x_{\mu}} + A^{\epsilon} · \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\epsilon}} \sqrt{-g} \quad\text{by~\Eq{(35.4)}} \\ &= \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\mu}} (A^{\mu} \sqrt{-g}), \Tag{(51.11)} \end{align*} since $\epsilon$~may be replaced by~$\mu$. In terms of tensor-density this may be written \[ A_{\mu}^{\mu} \sqrt{-g} = \mf{A}_{\mu}^{\mu} = \frac{\dd}{\dd x_{\mu}} \mf{A}^{\mu}. \Tag{(51.12)} \] The divergence of~$A_{\mu}^{\nu}$ is by~\Eq{(30.2)} \begin{align*} (A_{\mu}^{\nu})_{\nu} &= \frac{\dd}{\dd x_{\nu}} A_{\mu}^{\nu} + \{\alpha\nu,\nu\} A_{\mu}^{\alpha} - \{\mu\nu, \alpha\} A_{\alpha}^{\nu} \\ &= \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (A_{\mu}^{\nu} \sqrt{-g}) - \{\mu\nu, \alpha\} A_{\alpha}^{\nu}, \Tag{(51.2)} \end{align*} by the same reduction as before. The last term gives \[ -\frac{1}{2} \left(\frac{\dd g_{\mu\beta}}{\dd x_{\nu}} + \frac{\dd g_{\nu\beta}}{\dd x_{\mu}} - \frac{\dd g_{\mu\nu}}{\dd x_{\beta}}\right) A^{\beta\nu}. \] When $A^{\beta\nu}$~is a symmetrical tensor, two of the terms in the bracket cancel by interchange of $\beta$ and~$\nu$, and we are left with $-\dfrac{1}{2}\, \dfrac{\dd g_{\beta\nu}}{\dd x_{\mu}}\, A^{\beta\nu}$. Hence for \emph{symmetrical tensors} \[ (A_{\mu}^{\nu})_{\nu} = \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (A_{\mu}^{\nu} \sqrt{-g}) - \frac{1}{2}\, \frac{\dd g_{\alpha\beta}}{\dd x_{\mu}}\, A^{\alpha\beta}, \Tag{(51.31)} \] or, by~\Eq{(35.2)}, \[ (A_{\mu}^{\nu})_{\nu} = \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (A_{\mu}^{\nu} \sqrt{-g}) + \frac{1}{2}\, \frac{\dd g^{\alpha\beta}}{\dd x_{\mu}}\, A_{\alpha\beta}. \Tag{(51.32)} \] For \emph{antisymmetrical tensors}, it is easier to use the contravariant associate, \[ (A^{\mu\nu})_{\nu} = \frac{\dd}{\dd x_{\nu}} A^{\mu\nu} + \{\alpha\nu,\nu\} A^{\mu\alpha} + \{\alpha\nu, \mu\} A^{\alpha\nu}. \Tag{(51.41)} \] The last term vanishes owing to the antisymmetry. Hence \[ (A^{\mu\nu})_{\nu} = \frac{1}{\sqrt{-g}}\, \frac{\dd}{\dd x_{\nu}} (A^{\mu\nu} \sqrt{-g}). \Tag{(51.42)} \] \PageSep{115} Introducing tensor-densities our results become \begin{align*} \mf{A}_{\mu\nu}^{\nu} &= \frac{\dd}{\dd x_{\nu}} \mf{A}_{\mu}^{\nu} - \tfrac{1}{2} \mf{A}^{\alpha\beta}\, \frac{\dd g_{\alpha\beta}}{\dd x_{\mu}} & \text{(symmetrical tensors),}& \Tag{(51.51)} \\ \mf{A}_{\nu}^{\mu\nu} &= \frac{\dd}{\dd x_{\nu}} \mf{A}^{\mu\nu} & \text{(antisymmetrical tensors).} & \Tag{(51.52)} \end{align*} \Section{52.}{The four identities} \index{Identities satisfied by $G_{\mu\nu}$}% We shall now prove the fundamental theorem of mechanics--- \index{Fundamental theorem of mechanics}% \[ \text{\emph{The divergence of $G_{\mu}^{\nu} - \tfrac{1}{2}g_{\mu}^{\nu} G$ is identically zero.}} \index{Divergence!energy@of energy-tensor}% \Tag{(52)} \]