Silly doubt about tidal effects and Einstein Field Equations

The general question seems to be why gravitational tidal effects are often described as "nonlocal", even though they are supposed to be described by the curvature tensor, which depends on only local information.

The point is that it is harder to measure a tidal effect than a static gravitational field if you do experiments in a small region, because tidal effects correspond to differences in gravitational field. In other words, tidal effects depend on the second derivative of $\phi$ (as does the curvature tensor) while static fields depend only on the first derivative. Higher-derivative terms are always "nonlocal" in the sense that it is harder to measure them with a small apparatus, but always local in the sense that derivatives are defined at points. There is no contradiction between these statements.

To slightly extend knzhou's answer. What people mean by 'nonlocal' (or conversely, by 'local') is that, however precisely you can measure curvature, there is some scale which on which spacetime is flat as well you can measure. This scale is 'local' and on that scale you don't observe tidal effects. Critically this scale is finite: it's not just a point.

This in fact does not depend on GR: any $C^\infty$ metric field $\mathbf{g}$ on a manifold $M$ is locally flat, which means that for any point $P\in M$ you can pick a coordinate system on the basis of which the metric components $g_{ij}$ satisfy:

• $g_{ij} = \pm \delta_{ij}$ (metric is orthonormal at $P$);
• $\left.\frac{\partial g_{ij}}{\partial x^k}\right|_P = 0$ (orthonormal form of the metric is decent approximation at $P$);
• Not all of $\left.\frac{\partial^2 g_{ij}}{\partial x^k\partial x^l}\right|_P = 0$ in general (the coordinates stop being orthornomal as you move away from $P$).