Calculating Covariant Derivate of Tensor T^u_v

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if i have a tensorT^{uv}...i need to calculate the covariant derivate T^u_{v;a}

The logical thing is to do T^u_v and next to calculate T^u_{v;a}

is also correct to first calculateT^{uv}_{;a} and next T^u_{v;a}=T^{ui}_{;a}g_{iv}?
 
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Yes, provided we choose a connection which is "compatible" with the metric. There is a unique such connection, and it is the one we choose to use in GR.
 
What Ben Niehoff means by metric compatibility is that you must be careful in taking the derrivative of the metric.

Usually, in general relativity, the connection is chosen such that

\nabla_{\sigma} \ g_{\mu\nu} = 0 \ .[/itex]<br /> <br /> It simplifies things.
 

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