What justification do you have for setting $a^m=0$?

@QuarkSoup. I don't set $a^m=0.$ I combine the first two equations in the original post and get $a^m=0$ on a geodesic.

You do set it to zero in the first equation.$$0 = \ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$What justification do you have for this? Is there some rule of Differential Geometry that prevents the acceleration tensor field from taking on any value? Why do you assume that objects on an arbitrary manifold don't accelerate?

@QuarkSoup. No, that is just a rewrite of the first equation in the original post.

And that was the mistake that the OP made: assuming the acceleration tensor field was zero. He, like you, assumed that objects at rest remain at rest and that's why the two equations didn't reconcile. Remove that assumption and everything works.

@QuarkSoup. The OP didn't assume anything other than that two equations should coincide. The first equation describes a geodesic, the second gives an expression for some acceleration (the proper acceleration, but that is not obvious) in general coordinates. By using the first equation in the second one, I derive that the proper acceleration vanishes along a geodesic in spacetime. That is what should be expected since the geodesic describes the motion of a freely falling object.

"The first equation describes a geodesic" It describes a geodesic, it doesn't describe all geodesics. That's the point. The second equation makes no assumptions about the global acceleration of the manifold. The first equation does. That's why the OP's equations don't match:$$$$Geodesic Equation #1: assumption: $$0=\ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$ Geodesic Equation #2: No assumption $$a^m=\ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$

@QuarkSoup. What do you mean with "global acceleration of the manifold"?

@QuarkSoup. Acceleration tensor field? The $a^m$ is a tensor, but not a field.

@QuarkSoup. Where have you seen velocity being a tensor field?

@QuarkSoup. That has nothing to do with this question.

@QuarkSoup. I meant that it has nothing to do with the original question. What acceleration should be a tensor field? In the original question $a^m$ is the acceleration of a point particle; that's not a field. Talking about "the same non-zero four-acceleration" is meaningless in a curved space; we cannot compare vectors at different points.

@QuarkSoup. Vectors at very close points can (kind of) be compared so that we can define parallel transport. But in a curved space, parallel transport of one vector along different paths leading to the same end point often results in vectors pointing in different directions: en.wikipedia.org/wiki/Parallel_transport#/media/…

In physics, according to general relativity, mass (or rather mass/energy density, pressure and tension) makes space curved. en.wikipedia.org/wiki/Einstein_field_equations

This discussion has went far beyond the scope of the question. Therefore I will end it here.

@QuarkSoup. Stop trolling! I haven't made any assumptions. You seem not to have understood the question in the OP.

I just realized, you don't really understand geometry, do you? You believe that there could be a curvature tensor field everywhere, but you don't believe there could be an accompanying acceleration tensor field everywhere. Happy to chat with you more when you can connect the dots.

@QuarkSoup. There can certainly be an acceleration tensor field everywhere, but that is now what the question in the OP is about. The question is about the path of a single particle; $a^m$ is the proper acceleration of that particle.