Relationship between relativistic factor $\gamma$ and Christoffel symbols In special relativity, an object moving in a straight line can be described by the statement of zero proper acceleration,
$$\frac{d}{dt}\left(\gamma \frac{dx}{dt}\right)=0$$.
In contrast, in a curved space a point mass following a geodesic (with zero proper acceleration) satisfies
$$\frac{d^2x^a}{dt^2}+\Gamma^a_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}=0$$
So clearly the two gammas are related in some ways, but are generally discussed as quite different phenomena - in one sentence each, perhaps "$\gamma$ comes from transforming to different Lorentz frames, whereas $\Gamma$ comes from requiring that various quantities of interest are tensors in all frames".
I'm wondering if one can look at these two expressions and derive a more direct connection between the two gammas. For example, one could guess
$$\frac{1}{\gamma}\frac{d\gamma}{dx^b}\sim\Gamma_b$$
where perhaps $\Gamma_b$ is some contraction of velocities with tetrads or something. Is there a direct way to find this relationship?
 A: In the following, I set $c=1$. The statement of zero proper acceleration is
$$\mathbf A := \frac{d}{d\tau} \mathbf U = 0$$
where $U^\mu = \frac{dx^\mu}{d\tau}$ is the 4-velocity and $\tau$ is the proper time along the particle's worldline.  This can be written in component form as
$$A^\mu = \frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha \beta}  \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = 0$$
where the Christoffel symbols are defined to be
$$\Gamma^\mu_{\alpha\beta}:= \frac{1}{2}g^{\mu\rho}\left(\frac{\partial g_{\rho \beta}}{\partial x^\alpha} + \frac{\partial g_{\alpha \rho}}{\partial x^\beta} - \frac{\partial g_{\alpha\beta}}{\partial x^\rho} \right)$$
Adopting the $(+---)$ sign convention, the (differential) proper time is defined by
$$\mathrm d\tau^2 = g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu \equiv  \mathrm dt^2 /\gamma^2 \qquad \gamma^2 :=\frac{1}{g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}}$$
where we've singled out the coordinate time $x^0 \equiv t$.  As a result, we can re-express the above statement as
$$A^\mu = \gamma \frac{d}{dt}\left(\gamma \frac{dx^\mu}{dt}\right) + \gamma^2 \Gamma^\mu_{\alpha \beta} \frac{dx^\alpha}{dt}\frac{dx^\beta}{dt} = 0$$
This is as far as we can go without additional information.  As you can see, there is no functional relationship between the $\Gamma$'s and $\gamma$ -they both depend on the metric, but that's about all we can say. For instance, the $\Gamma$'s are functions on spacetime, while $\gamma$ also depends on the state of motion of the particle in question.
If we descend to the realm of special relativity then we may choose Cartesian coordinates, in which the (Minkowski) metric takes the form
$$g_{\mu\nu} = \pmatrix{1 &0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1}$$
In this case, we find (a) that all of the $\Gamma$'s vanish, and (b) $\gamma = 1/\sqrt{1 - v^2}$, so our expression reduces finally to
$$A^\mu = 0 \implies \frac{d}{dt}\left(\frac{1}{\sqrt{1-v^2}}\frac{dx^\mu}{dt}\right) = 0$$
