I had already asked this question in MO and I present it here at physicsoverflow
Let $X$ be a geodesible non vanishing vector field on a manifold $M$. Namely there is a Riemannian structure $(M,g)$ such that all integral curves of $M$ are unparametrized geodesics of the metric $g$. [This is a beautiful non example](https://mathoverflow.net/a/274981/36688)
> Is the index of an orbit or a closed orbit (i.e the index of a geodesic or a closed geodesics ) encoded in the vector field $X$? Namely can we compute the number of conjugate points on a (closed) orbits of $X$ with information just from the vector field and nothing else?
This question could play a crucial role in investigation of the following post about a [negatively curved structure on the punctured plane for which the solution curves of the Van der Pol equation would be geodesics](https://mathoverflow.net/q/160945/36688).The post search for a negative curvature metric on the punctured plane for which the Van der Pol foliation is a foliation by geodesics.
Because if the answer to this current question is affirmative(or there are some modified way to compute the index of the closed orbits of the underling vector field as closed geodesics of the corresponding compatible metric) and we get a non zero index, then this would implies that **there is no any metric with negative curvature on the punctured plane making all solutions of the Van der Pol vector field into geodesics since negative curvature implies non existence of conjugate points**.
>The next question: Regardless of the sign of the curvature, is there a Riemannian metric on the punctured plane such that all solution curves of the Vander pol equation are geodesics and there is no any conjugate points at all?