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

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

• It's worth noting that the relativistic Lorentz acceleration can't be expressed in the form you want; it is$$\frac{d^2 x^\mu}{d\tau^2} = \frac{q}{m} F^{\mu} {}_{\nu} \frac{d x^\nu}{d\tau},$$and so the acceleration depends on the four-velocity linearly rather than quadratically. Commented Jun 21, 2020 at 13:49
• @MichaelSeifert Is that for electromagnetism? Commented Jun 21, 2020 at 13:52
• Some terminology issues: $-\Gamma^\mu{}_{\nu\lambda} \dot{x}^\nu \dot{x}^\lambda$ is the coordinate acceleration, not the proper acceleration. Proper acceleration is $a^\mu a_\mu$ with $a^\mu = Du^\mu/D\tau$, and it is zero for a geodesic. Commented Jun 21, 2020 at 14:11
• @Javier That's how the geodesic equation is derived, the Covariant derivative set to 0. The derivatives are with respect to the proper time so it is the proper velocity and acceleration. Not coordinate Commented Jun 21, 2020 at 15:04
• I think I haven't explained my point very well. My point was that we shouldn't even try to write the electromagnetic force law in the form you proposed, since it doesn't have the correct form for the force law to be expressible as some sort of geodesic equation. Your attempt to do so in the edited question shows why this doesn't make a lot of sense: the left-hand side of your last equation is a function of location in spacetime only, while the right-hand side depends on the velocity of a particle at that location as well. Commented Jun 21, 2020 at 20:52

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 $$g(t)$$.

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).

• Thank you for your answer. What I mean by solving for a metric, is that the electromagnetic field can be interpreted as the result a geodesic in curved space-time, and the curvature of an electromagnetic field would depend on both the attractor's charge but also the charge of the particle moving with the geodesic. This would mean that the curvature of space-time would be relative to each observer, and depend on its charge. Solving the geodesic equation for the metric would result in a metric theory that would give the particle the geodesic as the force field that we interpret them as. Commented Aug 16, 2020 at 2:46