Covariant Derivative: What Is $\nabla^0 A_{\alpha}$?

[S.P.P]

just a quick query, I know that,

\nabla_0 A_{\alpha}= \partial_0 A_{\alpha} - \Gamma^{\beta}_{0 \alpha} A_{\beta}

But what does
\nabla^0 A_{\alpha} equal?

 

[shoehorn]

Since \nabla^0A_\alpha = g^{0\beta}\nabla_\beta A_\alpha you have

\nabla^0A_\alpha = g^{0\beta}\partial_\beta A_\alpha - g^{0\beta}\Gamma^{\gamma}_{\phantom{\gamma}\alpha\beta}A_\gamma

 

[shoehorn]

You're welcome.

 

[StatusX]

You're welcome.

Don't take it personally, it's rare to get thanked for help here.