Particle in "external potential" VS particle on "curved surface": equivalence?

Question

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.

Answer 1

If I interpret the question correctly, the OP asks whether given some potential $\phi(x)$, we can find a Riemannian metric $g_{ij}$ such that the dynamics of a system described by the Lagrangian

$$ L = \frac{1}{2}|\dot x|^2 - \phi(x)$$

can be equivalently described the dynamics of a free particle moving in a curved space (Riemannian manifold), namely by the Lagrangian

$$ L = \frac{1}{2}g_{ij} \dot x^i \dot x^j$$

with a suitable choice of $g_{ij}$.

The equations of motion in the first case are given by Newton's law

$$ \ddot x^i = -\partial_i \phi(x) $$

and in the second case by the geodesic equations

$$ \ddot x^i = -\Gamma^i_{ab} \dot x^a \dot x^b $$

where $\Gamma^i_{ab}$ are the Christoffel symbols of the metric.

Apart from the immediate observation that the two equations have different forms, we can see they cannot generally have the same solutions: in the geodesic equation, $\dot x=0$ implies $\ddot x=0$. In the case of Newton's Law, the solution to the initial value problem $\dot x(0)=0, x(0)=x_0$ will violate this condition if $\nabla \phi |_{x_0} \ne 0$. Therefore unless $\phi(x)$ is a constant, such a solution of the Newtonian equation cannot be a solution of a geodesic equation$^\dagger$.

Note that the above also implies that $x(t)=x_0=const.$ is always a solution of the geodesic equation for any $x_0$ (as expected: since the particle is not subject to an external force, if it is at rest at any point it will remain at rest). This means that there cannot exist a one-to-one mapping between solutions of the two cases (that does not itself requires knowledge of $\phi(x)$): if $x(t)=x_0$ is mapped into a solution of the Newtonian equation with some $\phi(x)$, it will not be a solution of the Newtonian equation with a different potential.

Of course, there is a trivial way in which we can define such a one-to-one mapping: the initial conditions $\dot x(0)=v_0, x(0)=x_0$ define a unique solution in both cases, so we can map those to each other (for any solution $x(t)$ of the geodesic equation, find the unique solution $\tilde x(t)$ of Newton's law with $\dot{{\tilde x}}(0) = \dot x(0)$ and $\tilde x(0) = x(0)$). But such a transformation depend on $\phi(x)$, i.e. it should be formally written as $x(t),\phi(x) \mapsto \tilde x(t)$. If we require that all information about $\phi(x)$ is encoded into $g_{ij}(x)$, then we should be able to construct $x(t) \mapsto \tilde x(t)$ regardless of $\phi(x)$. But by the argument in the previous paragraph, such a transformation is impossible to construct.

$^\dagger$NOTE in relation to LPZ's answer: indeed as shown in the reference, solutions to the Newtonian equation of motion with total energy $E$ are geodesics of the given metric - in that case the Christoffel symbols $\Gamma^i_{ab}$ diverge at the point where $E=\phi(x)$ or equivalently $\dot x=0$, thus the geodesic equation is not (strictly speaking) defined at that point. Note however that arbitrary solutions to that geodesic equation are not generally solutions of the Newtonian system - it is only those that have $\dot x=0$ at the singular points, that correspond to the trajectories with a given $E$.

Answer 2

One approach is described in Arnold's Mathematical Methods of Classical Mechanics (section 9-45, p. 247 2d edition). Starting from the Lagrangian: $$ L = \frac12\dot x^2-U(x), $$ restrict to a given energy: $$ E = \frac12\dot x^2+U(x) $$ and define the metric: $$ ds^2 = (E-U)dx^2 $$ which is related to the cartesian one by a conformal transformation. As expected, the new metric is singular at the turning points $U=E$.

Hope this helps.