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I apologize if this question is very elementary. Somewhere I've found the following:

Freely falling observers (resp. photon) move on timelike (resp. null) geodesics.

Please note that by definition, any curve $\gamma$ in any spacetime $M$ is a geodesic if

$$\nabla_{ \dot{\gamma}} \dot{\gamma}=0$$ where $\dot{\gamma}$ represent the tangent vector of the curve $\gamma.$ In local coordinates, this is equivant to

$$\frac{d^2x^\mu}{ds^2}+{\Gamma^\mu}_{\nu\lambda}\frac{dx^\nu}{ds}\frac{dx^\lambda}{ds}=0$$ for all indices $\mu.$

However, by Free fall, I understand (classically)

$$\frac{d^2x^\mu}{ds^2}=0.$$ Clearly, these two definitions are not equivalent unless all the $\Gamma$'s are identically zero. Now the question is:

Do people say "Freely falling observers (resp. photon) move on timelike (resp. null) geodesics" a bit informally, or do they mean it literally/precisely? I would be very happy if someone could explain this without too many physics jargons.

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  • $\begingroup$ Where does your "classical" definition come from? $\endgroup$ Commented May 18 at 23:10

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Do people say "Freely falling observers (resp. photon) move on timelike (resp. null) geodesics" a bit informally, or do they mean it literally/precisely? I would be very happy if someone could explain this without too many physics jargons.

Literally and precisely. The condition you wrote down is the zero acceleration condition in a curved spacetime. To make this obvious, start from your definition of free fall in Cartesian coordinates. Then work out what do they imply in a different coordinate system, such as spherical coordinates. You will notice the Christoffel symbols appear naturally.

In summary, your definition is coordinate-dependent. The geodesic definition is a coordinate-independent (and thus objective) way of stating the zero-acceleration condition.

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