PJPu17 wrote:
...Is there any geometric argument to prove the statement without doing any operation? Using only the compatibility of the connection with the metric, and the orthogonality of both fields in cartesians?
Maybe you're right! But I have no answer about this. My suggestion is: write $X(t)=\mathop{\sum}\limits_{k=1}^3v^k(t)\,\partial_k\big|_{\gamma(t)}$ and then apply the formula \[\frac{dX}{dt}(t)=\mathop{\sum}\limits_{k=1}^3\bigg(\frac{dv^i}{dt}+\mathop{\sum}\limits_{ij=1}^3\frac{d\gamma^j}{dt}\,\Gamma_{ji}^{k}\,v^i\bigg)\,\partial_k\big|_{\gamma(t)}\,.\]
Do the same for $Y(t)$. And don't forget the restrictions for $\alpha$ and $\beta$.