- #1
physicus
- 55
- 3
Homework Statement
Show that [itex]\nabla_a(\sqrt{-det\;h}S^a)=\partial_a(\sqrt{-det\;h}S^a)[/itex]
where [itex]h[/itex] is the metric and [itex]S^a[/itex] a vector.
Homework Equations
[itex]\nabla_a V^b = \partial_a V^b+\Gamma^b_{ac}V^c[/itex]
[itex]\Gamma^a_{ab} = \frac{1}{2det\;h}\partial_b\sqrt{det\;h}[/itex]
[itex]\nabla_a\sqrt{-det\;h}[/itex] (is that right??) since [itex]\nabla_a h[/itex]
The Attempt at a Solution
I can't quite figure out how to get the result:
[itex]\nabla_a(\sqrt{-det\;h}S^a)[/itex]
[itex]=\nabla_a(\sqrt{-det\;h})S^a+\sqrt{-det\;h}\nabla_a(S^a)[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\Gamma^a_{ab}S^b[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\frac{1}{2det\;h}\partial_b\sqrt{det\;h}S^b[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\frac{1}{2\sqrt{-det\;h}}\partial_b\sqrt{det\;h}S^b[/itex]
[itex]=\ldots[/itex]