Integral curves of vector field are geodesics

Question

Say we have a Riemannian manifold $(M, g)$ with vector field $X$ obeying the following:

1. $g(X, X) = 1$; and
2. the $1$-form $\varphi(Y) = g(Y, X)$ is $d$-closed, $d\varphi = 0$.

Does it necessarily following that the integral curves of $X$ are geodesics: $D_X X = 0$?

Answer 1

Indeed. Let $X^a X_a =1$, $0\equiv dX = 2\nabla_{[a} X_{b]}$. We have $X^a \nabla_b X_a=0$ and $\nabla_a X_b = \nabla_b X_a$, therefore $X^a \nabla_a X^b = X^a \nabla^b X_a = 0$.