You are asking a physics question, which you should ask in the physics site. Here the answer is that there is a zero because that is exactly what a geodesic is: no intrinsic curvature. One can consider the curves that satisfy the equation with a nonzero LHS, but its solutions are, by definition, no geodesics.

Has anyone derived for you the geodesic equation and explained what all the terms mean? I don't see how you could just hide a $1$ in with the Christoffel symbols, those symbols come from places and you can't just change what they mean.

@A.ThomasYerger If my manifold has an acceleration tensor field of $\{1, 1, 1, 1\}$ at every point, then which equation above will give me the correct answer when I try to calculate the shortest distance between two points on this manifold?

I don't even know what the acceleration tensor field is, and I do Riemannian geometry... I think you need to separate your physics intuition from the mathematical content of the question.

You do know there is a physics.stackexchange.com, right?

@A.ThomasYerger An acceleration tensor field is a vector field made up of accelerations. Please explain to me why that's difficult to imagine. Is there some rule of Differential Geometry that prevents me from imagining such a field on a manifold?

I don't think I meant to say it's difficult to imagine. What I was saying was that it's not a standard piece of mathematical terminology for geometers, at least not one I have encountered. Geodesics are supposed to describe the paths things take when NOT acted on by any other force. The standard example in physics is that objects in free-fall follow geodesics in spacetime. If you equipped your spacetime with some field that exerts a force, then the paths things follow will deviate from the shortest one because of the force.

"If you equipped your spacetime with some field that exerts a force" This has nothing to do with force at all, so I'll repeat my question: Why can't I imagine a manifold with an acceleration field of $\{1, 1, 1, 1\}$ at every point. Why must we drag physics back into the conversation?

@QuarkSoup -- Since a smooth manifold is a topological space first, consisting of open sets, which in turn consist of points, in this case spacetime points but it doesn't matter. Perhaps a line of thought would be whether you want the points themselves to shift as a function of time or some other parameter... this is not how Riemannian / pseudo Riemannian or any manifolds to my knowledge are constructed... in particular spacetime is just "always there" :) Like a stage at a play... there is movement on the stage but the stage is static...

@QuarkSoup There is no such thing as a "constant tensor field" in a general (semi-)Riemannian manifold: $\{1,1,1,1\}$ here depends on the coordinates system (this can be done however if the manifold is parallelisable). The closest you can define in full generality is a parallel tensor field.

I don't know what an 'acceleration' is outside of physics. From my point of view, you are the one making the conversation about physics.

@A.ThomasYerger 'acceleration' is the second derivative of position with respect to time. Are these concepts really beyond the realm of mathematics? Let me rephrase, it appears to me that 'force' keeps on being dragged back into the conversation. I just want to talk about the mathematical properties of an arbitrary manifold with a particular vector field.

@QuarkSoup How do you define such a thing on the whole manifold? It highly depends on the path you are following right?

@Didier -- I guess he wants it to be isotropic...

@Didier Sorry, but doesn't the path depend on the tensors on the manifold? All I'm asking is "why can't I imagine a ubiquitous, non-zero acceleration field" and, if I did, what would the Geodesic Equation look like for such a manifold.

So then what is unsatisfactory about the answer you were given? If you put a vector field on a manifold, this changes absolutely nothing about what the geodesics are. The left side of the geodesic equation is something like an acceleration. Geodesics do not have "acceleration" because they are straight. If you change it, you're asking now for curves whose "acceleration" is that vector field everywhere.

@QuarkSoup A tensor field is a map $T$ that assigns to any point $p$ of your manifold $M$ a tensor $T_p$ consistently (i.e always a vector, or always a covector, etc.). It only depends on $p$. Not on how you arrived at $p$, nor on how you're leaving $p$.

@QuarkSoup Also, acceleration isn't something that is defined on the whole manifold. This is a notion that is attached to paths. There is no "the acceleration of the manifold $M$", but only "the acceleration of the curve $\gamma\colon I \to M$". And it is defined to be $\frac{D}{dt}(\gamma')$, where $\frac{D}{dt}$ is the covariant derivative along $\gamma$.

@Didier "acceleration isn't something that is defined on the whole manifold" This is useful. Can a manifold posses constant curvature? If so, how is the constant curvature not going to translate into constant (ubiquitous) acceleration?

@QuarkSoup I mean, even physically, you can't talk about the acceleration of a location in space, you can only talk about the acceleration of a moving object in space right? This is the same idea. // Yes, there are manifolds of constant curvature, but I can't see how it is related in anyway to the question. I'll advise you to read either do Carmo, Lee or Gallot-Hulin-Lafontaine (I prefer the last two).

@QuarkSoup Maybe you are thinking that the curvature isn't present in the (true, homogeneous, usual) geodesic equation. But it really is: either it is hidden in the covariant derivative, or equivalently in the Christoffel symbols. These things contain information on the curvature.

@Didier - "But it really is: either it is hidden in the covariant derivative, or equivalently in the Christoffel symbols" YES! Exactly the point I was trying to make! If that is the case, then why is 0 (flat, Minkowski) favored over 1 (for example) for the RHS?

who says $0$ means Minkowski???

@QuarkSoup You misinterpreted my comment. I said that the equation $\nabla_{\gamma'}\gamma' = 0$ already contains all the information on the curvature you need: it is in the object $\nabla$, the Levi-Civita connection. This object is the central object of (semi-)Riemannian geometry.