= Solution
At any event on a massive observer's <worldline>, choose an <orthonormal tetrad> with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})=\eta_{ab}=\operatorname{diag}(1,-1,-1,-1)$, so locally $ds^2=\eta_{ab}\,dX^{(a)}dX^{(b)}$: the <Minkowski metric>. A <local inertial frame> can also remove the <Christoffel symbols> at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.
In general coordinates, $X^\mu(\tau)$ labels the trajectory, while the <four-velocity> and <four-acceleration> are
$$
U^\mu=\frac{dX^\mu}{d\tau},\qquad
A^\mu=\frac{DU^\mu}{d\tau}=\frac{dU^\mu}{d\tau}+\Gamma^\mu_{\nu\rho}U^\nu U^\rho.
$$
Coordinate values $X^\mu$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the <four-momentum> is $P^\mu=mU^\mu$ and the <four-force> is $f^\mu=DP^\mu/d\tau=mA^\mu$. With the $+---$ <metric signature>,
$$
\boxed{U^\mu U_\mu=c^2,\qquad U^\mu A_\mu=0,\qquad A^\mu A_\mu=-a^2\le0,}
$$
where $a$ is the <proper acceleration>. The middle identity follows by differentiating the constant velocity norm; in the instantaneous <rest frame>, $U=(c,\mathbf0)$ and $A=(0,\mathbf a)$, proving the last identity.
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