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Consider the timelike unit vector $u^{\beta}$ which satisfies $u^{\beta}u_{\beta}=-1$. Its covariant derivative vanishes: $ \nabla_{\alpha}(u^{\beta}u_{\beta})=0 $ which means we can write this as $ g^{\mu\beta}\nabla_{\alpha}(u_{\beta}u_{\alpha})=0 $. Since the metric is non-degenerate, $ \mid\nabla_{\alpha}(u_{\beta}u_{\alpha})\mid=0 $ and it follows that the only solution is $ \nabla_{\alpha}(u_{\beta}u_{\alpha})=0 $. Does that follow in every case? Is there anything that prevents this result from holding?

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  • $\begingroup$ You used the metric-compatibility of the connection to move the $g^{\mu\beta}$ outside the parentheses. $\endgroup$ – Ted Shifrin Mar 27 '20 at 22:28
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Your way of rewriting the equstion is not correct, since you have to $\alpha$'s as lower indices and one $\alpha$ as an upper index. It should read like $\nabla_{\alpha}(u^\beta u_\beta)=g^{\beta\gamma}\nabla_\alpha (u_\beta u_\gamma)$. (As mentioned in the comment by @TedShifrin, this uses metric compatibility.) The follwing arguement suffers from two mistakes: On the one hand, the expression can not be interpreted as a norm. It is the map, which associates to a vector the inner product of the derivative of $u$ in direction of that vector with $U$ itselft. To use non-degeneracy you would need to show that the inner product of a fixed object with all other objects vanishes to conclude that the first object is zero. (From vanishing of a norm you could only conclude that your object is light-like in a Lorentzian setting.)

One reasonable interpretation of the correct identity is that the trace of $\nabla_\alpha (u_\beta u_\gamma)$ over the last two indices vanishes. Alternatively, you can rewrite $ \nabla_\alpha (u_\beta u_\gamma)=u_\beta \nabla_\alpha u_\gamma+ u_\gamma \nabla_\alpha u_\beta$ and then contract with $g^{\beta\gamma}$ to conclude that $u^\beta\nabla_\alpha u_\beta$. This is the analog of the fact that constant norm of a vector field implies that the derivative is perpendicular to the field.

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  • $\begingroup$ Yes, of course the indices were wrong as I copied over the wrong expression without realizing that. Sorry for the confusion. As Andreas points out, the contraction is equivalent to the sum of two identical orthogonal terms which add to zero. We can postulate $ \nabla_{\alpha}u_{\beta}u_{\gamma}=0 $ as the trivial solution. It is consistent with the vanishing of the $u^{\beta}\nabla_{\alpha}u_{\beta}=0 $ terms and no issues should arise because of the metric compatibility of the connection. $\endgroup$ – Kolten Mar 28 '20 at 15:25

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