# Homework Help: Covariant derivative of a vector field

1. Jun 10, 2012

### physicus

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $\nabla_a h$

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

2. Jun 10, 2012

### 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.

3. Jun 10, 2012

### 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.

4. Jun 11, 2012

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