# Question tensors

1. Nov 15, 2009

### 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 $$T^u_{v;a}=T^{ui}_{;a}g_{iv}$$???

2. Nov 15, 2009

### 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.

3. Nov 15, 2009

### 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.

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