Let's consider a non-relativistic particle - its position is $x(t)\in \mathbb{R}^n$ - in an external potential $\phi$, with Lagrangian $$L=\dot{x}^i \eta_{ij}\dot{x}^j/2 - \phi(x),\tag{1}$$ where $\eta_{ij}$ is the flat spacetime metric tensor. How can we (if we can) map this problem into the one for a particle in a "curved" space with spatial metric $g_{ij}$? Namely, $$ L \, \propto \, \dot{x}^i g_{ij}\dot{x}^j $$ with $g = g(\phi)$. In other words, I'd like to understand if it's possible to perform the same "trick" of General relativity (i.e.: Newtonian potential ~ metric tensor) by using only "space": how to obtain $g=g(\phi)$, if possible?
More precisely:
We know the Newtonian limit(small velocities and small curvature of spacetime) of General relativity: we recover the non-relativistic mechanics of a particle in a scalar potential $\phi$. If we are talking about gravity, then $\phi$ is the Newton potential defined in terms of the Poisson equation: I see no difficulties in applying the same construction for any generic (external) potential $\phi$ as well. However, if we follow the GR construction we have to encode the potential $\phi$ into the $g_{tt}$ part of the metric tensor. I am asking if it is possible do the same without invoking the spacetime structure and using $g_{ij}$, the metric of the "space manifold". If this proves to be impossible, does it imply that the notion of space-time is a necessity in order to "geometrize" any kind of force?
If this is possible, does it mean that it is always possible to map a problem of many particles in an external potential $\phi$ into the problem of particles that are constrained to move on a curved surface myth metric $g(\phi)$?
PS: this question is very interesting and somewhat related, but mine is probably much simpler: I am not interested in the "dynamics" of $g$ since $\phi$ is just an external potential.
