Calculating Covariant Derivate of Tensor T^u_v

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The discussion focuses on calculating the covariant derivative of a tensor \( T^{u}_{v} \) in the context of general relativity. It confirms that one can either compute \( T^{u}_{v} \) first and then derive \( T^{u}_{v;a} \), or calculate \( T^{uv}_{;a} \) first, followed by \( T^{u}_{v;a} = T^{ui}_{;a} g_{iv} \), provided a metric-compatible connection is used. The unique connection in general relativity ensures that the derivative of the metric satisfies \( \nabla_{\sigma} g_{\mu\nu} = 0 \), which simplifies calculations.

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if i have a tensor[tex]T^{uv}[/tex]...i need to calculate the covariant derivate [tex]T^u_{v;a}[/tex]

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

is also correct to first calculate[tex]T^{uv}_{;a}[/tex] and next [tex]T^u_{v;a}=T^{ui}_{;a}g_{iv}[/tex]?
 
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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

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

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