# Coordinate and tensor divergences

Using our previous results regarding the divergence, we can derive many useful coordinate dependent relations. Adopting the common abbreviation

$$\displaystyle \sqrt{g}\equiv\sqrt{\left|\det\left(g\right)\right|}$$

and including torsion for completeness, we expand both sides of the coordinate divergence expression to get

$$\displaystyle \partial_{\lambda}\sqrt{g}=\sqrt{g}\left(\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}\right).$$

From $${\det\left(\exp\left(g\right)\right)=\exp\left(\mathrm{tr}\left(g\right)\right)\Rightarrow\ln\left(\det\left(g\right)\right)=\mathrm{tr}\left(\ln\left(g\right)\right)}$$, we can take the derivative of the components to get

\begin{aligned}\frac{1}{\det\left(g\right)}\partial_{\lambda}\left(\det\left(g\right)\right) & =\mathrm{tr}\left(g^{-1}\partial_{\lambda}g\right)\\
& =g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\partial_{\lambda}\sqrt{g} & =\frac{1}{2}\sqrt{g}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda} & =\frac{1}{2}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}.
\end{aligned}

The first line above applies to any variation; applying the above to the Lie derivative and using its coordinate expression then gives us

\begin{aligned}L_{u}\sqrt{g} & =\sqrt{g}\mathrm{div}\left(u\right)\\
\Rightarrow\mathrm{div}\left(u\right) & =\frac{1}{2}g^{\mu\nu}L_{u}g_{\mu\nu}.
\end{aligned}

If we consider an anti-symmetric tensor $${F^{\mu\nu}}$$ and a symmetric tensor $${G^{\mu\nu}}$$, it is not hard to see that

\begin{aligned}\nabla_{\nu}F^{\mu\nu}-T^{\lambda}{}_{\lambda\nu}F^{\mu\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}F^{\mu\nu}\right)-\frac{1}{2}T^{\mu}{}_{\lambda\nu}F^{\lambda\nu},\\
\nabla_{\nu}G^{\mu\nu}-T^{\lambda}{}_{\lambda\nu}G^{\mu\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G^{\mu\nu}\right)+\Gamma^{\mu}{}_{\lambda\nu}G^{\lambda\nu},\\
\nabla_{\nu}G_{\mu}{}^{\nu}-T^{\lambda}{}_{\lambda\nu}G_{\mu}{}^{\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G_{\mu}{}^{\nu}\right)-\Gamma^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}\\
& =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}G_{\mu}{}^{\nu}\right)-\frac{1}{2}\partial_{\mu}g_{\lambda\nu}G^{\lambda\nu}+T^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}.
\end{aligned}

The left side of these equations can be used to extend the divergence to general tensors by operating on the last index, but other conventions are also in use. All of these expressions are more commonly presented with zero torsion.