I Covariant divergence of vector; physical meaning with contracted Tuv

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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 $$\frac{1}{a^3}\partial_t(-\rho a^3)$$
 
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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##.
 
It's, of course, ##\sqrt{-g}## everywhere, since ##g=\mathrm{det} g<0##. Otherwise it's correct.
 
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