Covariant divergence of vector; physical meaning with contracted Tuv

[Tertius]

TL;DR Summary

The covariant divergence of a vector has a simplified form. I am discussing this in relation to a contraction of the SEM tensor and its meaning.

I'm studying Carroll's section on covariant derivatives, which shows that the covariant divergence of a vector $V^\mu$ is given by $$\nabla_\mu V^\mu = \partial_\mu V^\mu + \Gamma^\mu_{\mu\lambda}V^\lambda$$. Because $\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{g}}\partial_\lambda \sqrt{g}$ we can write $$\nabla_\mu V^\mu = \frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)$$. If we say $V^\mu = U_\nu T^{\mu\nu}$, then the covariant divergence looks like $$\frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g} U_\nu T^{\mu\nu})$$. The rank-1 tensor $U_\nu T^{\mu\nu}$ should represent the energy and momentum densities in each of the 4 coordinate directions. The covariant divergence of this rank-1 tensor should then be the sum of the changes of the energy and momentum densities along each coordinate. The simplified form $$\nabla_\mu V^\mu = \frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)$$ looks to be volume independent in that the partial derivative is taken of the volume element multiplied by the vector, and then divided again by $\sqrt{g}$ after the change is computed. From this perspective, it seems like $\frac{1}{\sqrt{g}} \partial_\mu(\sqrt{g}V^\mu)$ should represent changes to the components of $V^\mu$ independent of volumetric changes. Is this a correct interpretation?
For a flat FRW universe of only dust and a timelike observer, it would look like $$\frac{1}{a^3}\partial_t(-\rho a^3)$$

 

[Tertius]

I realized the problem was quite simple. The covariant derivative is "correcting" for changes to the metric. The changes in the FRW metric look volumetric because $\sqrt{g}=a^3$.

 

[vanhees71]

It's, of course, $\sqrt{-g}$ everywhere, since $g=\mathrm{det} g<0$. Otherwise it's correct.