As far as I know, the geodesic equation of motion can be directly derived from the equivalent principle. For instance, as shown by Steven Weinberg, the geodesic equation can be obtained by transforming a zero acceleration on the free-falling coordinate system to an arbitrary coordinate system.

Meanwhile, it seems that such a free-falling coordinate system can only be constructed under a very strong constraint. That is, the free-falling coordinate system can be defined and maintained, only when an observer and a target particle are at rest with respect to each other initially or at any time during their free-falling process. This is because an initial difference in velocity between the free-falling observer and particle will be increased over time in a gravitational field.

With regard to this issue, I would like to emphasize that the geodesic equation is used as an equation of motion itself, which is used to describe a motion of an object with an arbitrary initial condition, not as an auxiliary equation for determining a physical state of a system. To address this issue, is it necessary to add something else in the known geodesic equation?

  • $\begingroup$ it is valid for every possible initial condition. $\endgroup$
    – Yukterez
    Oct 30, 2018 at 2:24
  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/437494/2451 $\endgroup$
    – Qmechanic
    Oct 30, 2018 at 3:04
  • $\begingroup$ @Qmechanic. This question is related to that. However, this question is about whether it is possible to freely apply the geodetic equations itself, which are made with the constraint of free-fall observer, to an arbitrary motion. It seems to me that this is related to, but independent of, determining the A and B factors, mentioned in the previous post. $\endgroup$
    – SOQEH
    Oct 30, 2018 at 3:56
  • $\begingroup$ Another question is whether or not there exists a geodesic on M that passes through any (or every) point for any given choice of velocity at that point. This is an ODE solution question modulo special issues of completeness on manifolds. But the description of your question is incongruent with your title. $\endgroup$
    – user196418
    Nov 1, 2018 at 21:40

2 Answers 2


For any point $p$ in a (semi-)Riemanian manifold and any tangent 4-vector $u$ at that point, there exists a unique geodesic (i.e. solution to the geodesic equation) $\gamma^\mu(\lambda)$ such that $\gamma(0) = p$ and $\tfrac{d\gamma}{d\lambda}(0)$ = u$.

That being said. This does not guarantee that $\gamma$ can be interpreted as the worldline of a freely falling observer. This requires the additional condition that $u$ is a timelike 4-vector.


We do not in general have anything like a "free-falling coordinate system." It is a common misconception that a coordinate system in GR has something to do with an observer or a frame of reference. It doesn't.

Neither does the geodesic equation have anything to do with observers.

The question asks under what conditions the geodesic equation is "valid," without clearly stating what is meant by "valid." If the question is under what conditions test particles have geodesic motion, then there are no conditions on the coordinate system, because a geodesic that satisfies the equation in one coordinate system will also satisfy it in any other coordinate system related to the original system by a smooth change of coordinates.

We do need some conditions on the test particle, basically that it needs to be a test particle (small and low in mass), and also an energy condition. See Ehlers and Geroch, http://arxiv.org/abs/gr-qc/0309074v1 for a technical treatment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.