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I'm trying to understand the alternative form of the harmonic coordinate condition, and I have two questions:

Why $(g^{{\mu \nu }}{\sqrt {-g}})_{{;\rho }}=0$, and why do we obtain the term $-g^{{\mu \nu }}\Gamma _{{\sigma \rho }}^{{\sigma }}{\sqrt {-g}}$?

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It is assumed that the metric satisfies the metric postulate, $\nabla_{\mu} g_{\nu \sigma} = 0$, or, in other words, the connection is metric compatible. This also implies $\nabla_{\mu} \sqrt{-g }= 0 = \nabla_{\mu} g^{\nu \sigma}$.

Remember that $\sqrt{-g }$ is actually a tensor density. We must be careful when expanding its (vanishing) covariant derivative. Let's try to not repeat the argument in the link you provided, instead let's use the old useful argument $\delta (\det M)= (\det M )\, \mathrm{Tr}[M^{-1}\delta M]$, for a matrix $M$. Extrapolating this for the case of a covariant derivative and considering the metric we have $\nabla_{\sigma} g = g \, g^{\mu \nu} \nabla_{\sigma} g_{\nu \mu} = - g \, g_{\mu \nu} \nabla_{\sigma} g^{\nu \mu}$, where we used $\nabla_{\sigma} (g_{\mu \nu} g^{\nu \mu}) = 0$. We have then \begin{eqnarray} 0 &=& \nabla_{\sigma} \sqrt{-g} = - \frac{1}{2} \sqrt{-g} \, g_{\mu \nu} \nabla_{\sigma} g^{\nu \mu} = - \frac{1}{2} \sqrt{-g} \, g_{\mu \nu} (\partial_{\sigma} g^{\nu \mu}+\Gamma^{\nu}_{\sigma \alpha} g^{\alpha \mu}+\Gamma^{\mu}_{\sigma \alpha} g^{\nu \alpha}) \\ &=& - \frac{1}{2} \sqrt{-g} \, (g_{\mu \nu} \partial_{\sigma} g^{\nu \mu}+2\Gamma^{\mu}_{\mu \sigma}) = - \frac{1}{2} \sqrt{-g} \,g_{\mu \nu} \partial_{\sigma} g^{\nu \mu} - \sqrt{-g} \, \Gamma^{\mu}_{\mu \sigma} \\ &=& \partial_{\sigma}\sqrt{-g} - \sqrt{-g} \, \Gamma^{\mu}_{\mu \sigma} \, . \end{eqnarray} In general we have then $0 = \nabla_{\sigma} \sqrt{-g}^W = \partial_{\sigma}\sqrt{-g}^W - W \sqrt{-g}^W \, \Gamma^{\mu}_{\mu \sigma}$ .

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