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Suppose $(M,g)$ is a Riemannian manifold. Let us assume that $X$ denotes a vector field in this manifold and consider the integral curves of this vector field. Does there exist a conformal factor $c$ such that locally these integral curves will be geodesics with respect to $ \hat{g} = c g$ ?

Thanks

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In "A foliation of geodesics is characterized by having no “tangent homologies”, J. of Pure and Applied Algebra Volume 13, Issue 1, 1978, Pages 101–104 D. Sullivan explains a necessary and sufficient condition for a flow to be "taut", i.e all leafs are geodesics. The example of the so-called Reeb flow on the annulus (Gluck example Figure 2.A in Sullivan's paper) is not. In dimension $2$, a vector field $X$ is taut iff there exists a closed $1$-form $w$ such that the product $\langle X , w \rangle$ is everywhere strictly positive.

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