Finding a Surface that admits a given geodesic?

Question

Imagine that we are given a parametrization of a geodesic $\gamma (t)$. I am curious as to whether it is possible to find a given surface or class of surfaces that admit $\gamma (t)$ as its geodesic.

As some motivation, I am curious about a variant on Abel's mechanical problem and I want to see if any curve satisfying the isochrone or tautochrone property under a given potential field will also satisfy the brachistochrone property. Knowing that the cycloid is the brachistochrone of $R^2$ under a potential field pulling in one direction with constant magnitude, it is in some way a geodesic. I was wondering that perhaps modelling this situation as a geodesic on some surface imbedded in a higher dimension would perhaps help me find a way to generalize this problem as a whole.

Also, if this is clearly ill-posed but is related to some branch of research, please let me know! Thanks

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Nml}{\mathbf{N}}\newcommand{\Bnml}{\mathbf{B}}\newcommand{\eps}{\varepsilon}$For definiteness, let's assume $\gamma:(-\eps, \eps) \to \Reals^{3}$ is a simple unit-speed curve of class $C^{2}$.

1. If $\gamma''$ is non-vanishing and $\{\gamma', \Nml, \Bnml\}$ denotes the (continuous) Frenet trihedron, then the (ruled) surface parametrized by $$ \Phi(s, t) = \gamma(t) + s\, \Bnml(t) \tag{1} $$ has $\gamma$ as a geodesic by construction (the acceleration $\gamma''(t)$ is normal to the surface, i.e., proportional to $\Nml(t)$).

2. An obvious modification of 1. gives an infinite-dimensional family of examples: Replace the line segments in (1) by an arbitrary family of plane curves $\nu(t)$ with $\nu'(t)$ proportional to $\Bnml(t)$ at each $t$.

3. Conversely, if $S$ is a regular surface containing $\gamma$ as a geodesic, then locally $S$ can be partitioned into plane sections as in 2: For each $t$, take the intersection of $S$ with the plane orthogonal to $\gamma'(t)$. (This construction does not require $\gamma''$ to be non-vanishing.)

4. The (unit-speed reparametrization of the) path $$ \gamma(t) = \begin{cases} (t, t^{3}, 0) & t < 0, \\ (t, 0, t^{3}) & t \geq 0, \\ \end{cases} $$ shows you can't generally allow $\gamma''$ to have zeros: No regular surface contains $\gamma$ as a geodesic, since $\Nml$ is discontinuous. (This example can easily be made $C^{\infty}$.)

Answer 2

Perhaps I am not understanding your question correctly, but could you not just rotate your given curve about an axis, giving a surface of revolution for which your curve is a profile curve, hence a geodesic?