Can we reverse the geodesic equation to find a metric for the theory?

Question

The geodesic equation describes the motion of a particle moving in a straight line embedded in a curved geometry.

$$\frac{d^2 x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0$$ Solving for the proper acceleration we find that $$\frac{d^2x^\mu}{d\tau^2}=-\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$ So if we also find the proper acelleration with the equations of motion for a theory (e.g Electromagnetism) which we will denote as $\Omega^\mu$ we can find the relation that $$\Omega^\mu=-\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$ Expanding out the christoffel symbol out in its full form $$\Omega^\mu=-\frac{g^{\mu\lambda}}{2}(\partial_\alpha g_{\lambda\beta}+\partial_\alpha g_{\alpha\lambda}-\partial_\lambda g_{\alpha\beta}) \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$

For electromagnetism the acelleration is: $$\frac{d^2 x^\mu}{d\tau^2} = \frac{q}{m} F^{\mu} {}_{\nu} \frac{d x^\nu}{d\tau}$$ This implies that: $$\frac{q}{m} F^{\mu} {}_{\beta} \frac{d x^\beta}{d\tau}=-\frac{g^{\mu\lambda}}{2}(\partial_\alpha g_{\lambda\beta}+\partial_\alpha g_{\alpha\lambda}-\partial_\lambda g_{\alpha\beta}) \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$ $$\frac{q}{m} F^{\mu} {}_{\beta} =-\frac{g^{\mu\lambda}}{2}(\partial_\alpha g_{\lambda\beta}+\partial_\alpha g_{\alpha\lambda}-\partial_\lambda g_{\alpha\beta}) \frac{dx^\alpha}{d\tau}$$

Is there any way of solving for a metric for this theory (in general), that gives rise to the equations of motion when we use the metric in the geodesic equation? Would there be any way of "deriving" a metric to describe a particles motion. This is because from the particles frame of reference they would perceive to be stationary.

The curvature is not universal to all particles. There may be other factors influencing the metric such as the mass of the particles. However they follow geodesics from their "perspective".

But can all forces be defined to have a certain geodesic satisfied by a metric defining the curvature of the space

Thanks

Answer 1

Solving for the proper acceleration we find that $$\frac{d^2x^\mu}{d\tau^2}=-\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$

This isn't correct. The components of the proper acceleration are given by

$$a^\mu = \frac{d^2 x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$

The Lorentz force law then says

$$a^\mu = \frac{q}{m} F^{\mu} {}_{\nu} \frac{d x^\nu}{d\tau}$$

Given the $\Gamma$'s and $F$, this provides a differential equation which allows us to determine the coordinates $x^\mu$ of a charged particle moving in an electromagentic field. It does not provide a relationship between the $\Gamma$'s and $F$, just the differential equation

$$\ddot x + f(t) \dot x^2 = g(t) \dot x$$ does not provide a relationship between $f(t)$ and $g(t)$.

Answer 2

I think you would have to define clearly what you mean by "solving for the metric of the theory". The geodesic equation describes the motion of particles in a given spacetime, and this is a purely geometric property. And the spacetime is characterized by its symmetries. If you are in a flat spacetime, then the metric is the Minkowski $\eta_{\mu \nu}$ metric. If you are in a spherically symmetric spacetime, then the metric is Schwarzschild. The way you compute the metric coefficients is to solve Einstein's equations (you need to also specify if there is any matter source in your spacetime which would have an energy momentum tensor). When you have other fields (electromagnetic, say), you would first need to derive the equation of motion (by varying the action) for such fields (Maxwell's equations in curved spacetime). Then, you need to compute the force on a test charged particle (for Maxwell's theory in curved spacetime) in that curved spacetime with the field specified (note that if your field couples to the geometry, the metric can dynamically change as the field evolves so you need to start with a background metric). Note that these particles are no longer moving under the influence of curvature of spacetime alone (which is what they would if there is no additional fields) so they are not moving on 'geodesics'. Also their reference frames are not inertial ones (they are not in "free fall"). So when you want to describe their motion from "their perspective" you are going to be in a non-inertial frame. Locally, in a small region of spacetime, you can use the weak-equivalence principle and their motion will reduce to what it would be in flat-spacetime, subject to the laws of special relativity. But if you are interested in the global properties of their trajectory, you need to use their equation of motion subject to curvature and external fields (non-gravity).