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Covariant derivative in coordinate basis
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[QUOTE="LCSphysicist, post: 6632200, member: 675151"] [B]Homework Statement:[/B] . [B]Relevant Equations:[/B] . [ATTACH type="full"]301494[/ATTACH] I need to evaluate ##\nabla_{\mu} A^{\mu}## at coordinate basis. Indeed, i should prove that ##\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt(|g|)}\partial_{\mu}(|g|^{1/2} A^{\mu})##. So, $$\nabla_{\mu} A^{\mu} = \partial_{\mu} A^{\mu} + A^{\beta} \Gamma^{\mu}_{\beta \mu}$$ The first and third terms of Christoffel will cancel, so $$ = \partial_{\mu}A^{\mu} + A^{\beta} \frac{g^{\mu x}}{2}(\partial_{\beta}g_{x \mu})$$ Now, using the fact that ##\delta g = g g^{\mu v} \delta g_{\mu v}##, we can easily find that $$\frac{g^{\mu x}}{2}(\partial_{\beta}g_{x \mu}) = \frac{\partial_{\beta}(|g|^{1/2})}{|g|^{1/2}}$$ After substitute this at our main expression, we can recover ##\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt(|g|)}\partial_{\mu}(|g|^{1/2} A^{\mu})##. The problem is, i have no idea what assumption i have made so that my result applies only to coordinate basis! That is, the problem ask for prove it at [B]coordinate basis [/B], so i guess it should be true only at these type of basis. But i haven't assumed nothing, just manipulate the terms and got the result. What am i missing? Is this expression really general like i have found? [/QUOTE]
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Covariant derivative in coordinate basis
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