= Point-mass tidal stretching and squeezing
{title2=$T=\mu(3\mathbf n\mathbf n^{\mathsf T}-I)/R^3$}
For a source <point mass> $M$ at the origin, let $\mathbf R=R\mathbf n$, where $\mathbf n$ is an outward <unit vector>, and let $\mu=GM$ be the source's <standard gravitational parameter>. Differentiating $g_i=-\mu R_i/R^3$ gives the <tidal tensor>
$$
T_{ij}=\frac{\mu}{R^3}(3n_i n_j-\delta_{ij}),\qquad T=\frac{\mu}{R^3}(3\mathbf n\mathbf n^{\mathsf T}-I),
$$
where $\delta_{ij}$ is the <Kronecker delta> and $I$ is the <identity matrix>. In an <orthonormal basis> with first axis radial, the <tidal tensor> is
$$
T=\frac{\mu}{R^3}\begin{pmatrix}2&0&0\\0&-1&0\\0&0&-1\end{pmatrix}.
$$
The radial <eigenvalue> $2\mu/R^3$ separates nearby freely falling <test particles>; the two transverse <eigenvalues> $-\mu/R^3$ bring them together. The farther radial <test particle> falls inward less rapidly, so its <acceleration> relative to the nearer one points outward. Transverse <gravitational accelerations> converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative <accelerations>; the subsequent shape also depends on initial <velocities> and the changing <tidal tensor> along the cloud's <free fall>.
Outside the source, the <matrix trace> is $2\mu/R^3-\mu/R^3-\mu/R^3=0$. This matches $\operatorname{tr}T=-\nabla^2\Phi=0$ in a region with zero <mass density>, by the <Poisson equation for Newtonian gravity>. Zero <matrix trace> means no leading change of a small initially resting cloud's <volume> at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\operatorname{tr}T=-4\pi G\rho$ instead.
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