Calculating Covariant Derivate of Tensor T^u_v

[alejandrito29]

if i have a tensor$$T^{uv}$$...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 calculate$$T^{uv}_{;a}$$ 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]<br /> <br /> It simplifies things.[/tex]