= Solution
Nearby freely falling particles can accelerate relative to one another even though each follows a <geodesic>. A local <inertial frame> removes the common gravitational acceleration at one event; it cannot remove the variation across an extended laboratory. That variation is a <tidal force>, measured covariantly by <geodesic deviation>.
Let $u^\mu$ be the reference <four-velocity> and $\xi^\mu$ an infinitesimal separation to a neighboring <geodesic>, with matching <affine parameters>. With the convention $[\nabla_\beta,\nabla_\gamma]V^\mu=R^\mu{}_{\nu\beta\gamma}V^\nu$,
$$
\boxed{\frac{D^2\xi^\mu}{D\tau^2}=R^\mu{}_{\alpha\beta\gamma}u^\alpha u^\beta\xi^\gamma.}
$$
This relates measured relative <acceleration> to the <Riemann curvature tensor>. Reversing the curvature-tensor convention reverses the displayed curvature sign, without changing the physical relative motion.
In a small freely falling frame outside a spherical mass, the weak-field <tidal tensor> has radial and transverse components proportional to $2,-1,-1$. Thus
$$
\boxed{\ddot\xi_r=\frac{2GM}{r^3}\xi_r,\qquad
\ddot\xi_{\perp}=-\frac{GM}{r^3}\xi_{\perp}.}
$$
An initially spherical cloud stretches radially and compresses in both transverse directions. Two separated test masses, with their changing separation measured by light, form the basic idea of a gravity gradiometer; the common falling motion drops out. A uniform apparent acceleration in an accelerated laboratory can occur even in flat <Minkowski spacetime>, so it is the separation-dependent tidal response that diagnoses curvature.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2017-tidal-deviation.png]
{title=Radial stretching and transverse compression from <geodesic deviation>}
{description=Original local schematic of the weak-field <tidal tensor> around a freely falling reference particle. Arrows show relative <acceleration>, with radial direction horizontal; the cloud deformation is exaggerated and is not a trajectory calculation.}
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