Covariant derivative of a vector field

[physicus]

Homework Statement

Show that \nabla_a(\sqrt{-det\;h}S^a)=\partial_a(\sqrt{-det\;h}S^a)
where h is the metric and S^a a vector.

Homework Equations

\nabla_a V^b = \partial_a V^b+\Gamma^b_{ac}V^c
\Gamma^a_{ab} = \frac{1}{2det\;h}\partial_b\sqrt{det\;h}
\nabla_a\sqrt{-det\;h} (is that right??) since \nabla_a h

The Attempt at a Solution

I can't quite figure out how to get the result:
\nabla_a(\sqrt{-det\;h}S^a)
=\nabla_a(\sqrt{-det\;h})S^a+\sqrt{-det\;h}\nabla_a(S^a)
=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\Gamma^a_{ab}S^b
=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\frac{1}{2det\;h}\partial_b\sqrt{det\;h}S^b
=\sqrt{-det\;h}\;\partial_a S^a+\frac{1}{2\sqrt{-det\;h}}\partial_b\sqrt{det\;h}S^b
=\ldots

 

[Muphrid]

I'd be interested to find out if someone figures out what's going on here, because to me, it does seem unusual that the connection term must drop out for this to work.

 

[BigD]

Are you sure you have the right equations? I'm not entirely sure, but I don't think there should be square root in the second relevant equation. Also I think the problem would make more sense if one of the square roots was outside the derivative on one of the sides of the original equation.

 

[physicus]

Thank you very much for you help! You are right. There is a mistake in my second relevant equation. It should be: \Gamma^a_{ab} = \frac{1}{\sqrt{det\;h}}\partial_b\sqrt{det\;h}

Then I get:
\nabla_a(\sqrt{-det\;h}S^a)
=\nabla_a(\sqrt{-det\;h})S^a+\sqrt{-det\;h}\;\nabla_a(S^a)
=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\;\Gamma^a_{ab}S^b
= \sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\frac{1}{\sqrt{det\;h}}\partial_b \sqrt{det\;h}\;S^b
=\sqrt{-det\;h}\;\partial_a S^a+\partial_b\sqrt{-det\;h}\;S^b
=\partial_a (\sqrt{-det\;h}\;S^a)