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Let $(M, g)$ be a Riemannian manifold. Lately, I've grown interested in what you may call a "modified geodesic" problem. Given some smooth, non-negative scalar field $V$ on $M$ (aptly called the potential), what are the critical points of $E_V(\gamma) = \frac12\int_{a}^{b} \left[\left|\left|\dot{\gamma}(t)\right|\right|_g^2 + V(\gamma(t))\right]dt$ and how do they behave? The necessary conditions are easy to derive as $\frac{D}{\partial t} \dot{\gamma}(t) + \text{grad}V(\gamma(t)) = 0$ (using the Levi-Civita connection), and from here you can simply try to work through the standard results of geodesics, Jacobi Fields, etc. with this formalism.

To make things a little less tedious, my hope was to define a metric $\text{dist}_V: M \times M \to \mathbb{R}_+$ on $M$ by $$\text{dist}_V(p, q) := \inf\left\{\int_a^b \left[||\dot{\gamma}(t)||_g + V(\gamma(t))\right]dt: \gamma: [a,b] \to Q \ \text{ an admissible curve with } \ \gamma(a) = p, \ \gamma(b) = q\right\}$$ and recover from this metric a new Riemannian metric $g_V$ on $M$ such that the distance function induced by $g_V$ (in the usual way via the length of curves connecting two points) aligns with $\text{dist}_V$. In that sense, the modified geodesic problem could be understood purely from the theory of geodesics. We need only move to a new space where the geometry has been appropriately modified by the potential.

I've already seen from the discussion in this post that we in general cannot expect a given metric to be induced by a Riemannian structure, so I suppose that this may not be a fruitful method. Nevertheless, I'm curious if there is some class of potentials for which this can be done. For instance, disregarding the smoothness condition for a moment, it (perhaps naively) appeals to my intuition that with a potential like $V(q) = \text{dist}(q, p)$ for some fixed $p \in M$, the manifold would stretch/curve in a way that the geodesics get 'pulled' towards $p$.

However, in that same post I linked above, it is mentioned that it is necessary for $\text{dist}_V$ to be a "path-metric." If true, that could be a major obstruction here, as the "length" function $L_V(\gamma) = L(\gamma) + \int_{a}^{b} V(\gamma(t))dt$ is dependent upon the parameterization of $\gamma$ unless $V$ is a constant map, and so I'm not sure if this could be a well-defined notion of length.

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    $\begingroup$ If you start off not moving, then you stay not moving, following geodesic flow. But for a potential, you will start moving, from an initial zero velocity, unless the potential is constant. $\endgroup$
    – Ben McKay
    Commented May 17, 2021 at 15:34
  • $\begingroup$ @BenMcKay That makes sense. Well, tedious route it is. Thanks! $\endgroup$ Commented May 17, 2021 at 15:38
  • $\begingroup$ Even for a constant $V$, your length function will depend on the parametrization. In fact your distance will default to the standard geodesic distance, as you can traverse the same curve faster and faster to get a smaller and smaller value. You might try to only allow curves with $\|\dot{\gamma}\|_g =1$, which would get you $(1+V(x)) g(x)$ as a new metric, but I am not sure that this will help you in any way. Fundamentally, potentials should not depend on added constants, but any length involving $V$ directly will. $\endgroup$
    – mlk
    Commented May 17, 2021 at 16:30

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