question tensors

[alejandrito29]

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}???

[Ben Niehoff]

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.

[Phrak]

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

[tex]\nabla_{\sigma} \ g_{\mu\nu} = 0 \ .[/itex]

It simplifies things.