= Solution
The differential <gravitational acceleration> responsible for a <tidal force> is the difference between accelerations at separated points of a body. Relative to a freely falling reference point $\mathbf R$, a small offset $\boldsymbol\xi$ experiences
$$
\Delta\mathbf g=\mathbf g(\mathbf R+\boldsymbol\xi)-\mathbf g(\mathbf R)
\simeq\mathsf T\boldsymbol\xi,\qquad
T_{ij}=\frac{\partial g_i}{\partial R_j}.
$$
The <tidal tensor> has units $\mathrm{s^{-2}}$; the vector called $\mathbf T(\boldsymbol\xi)$ in the source is the resulting differential acceleration, in $\mathrm{m\,s^{-2}}$. An actual differential force is this vector multiplied by the test mass. Measurable effects include relative motion of separated freely falling masses, radial stretching, transverse squeezing, deformation and, when self-gravity or strength is insufficient, disruption. A spatially uniform acceleration alone produces no tidal deformation.
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