Tides in a Falling Laboratory: The Part of Gravity No Observer Can Remove — Epoche C1
Since Einstein set out the general theory in 1916, textbooks have repeated his claim that gravity is not a force but the curvature of spacetime. A sceptical reader may reasonably ask what this changes: the apple still accelerates downwards at $9.8\ \mathrm{m\,s^{-2}}$, and renaming the cause looks like new vocabulary rather than new physics. The doubt sharpens under the equivalence principle — the fact, established by Galileo and confirmed since to parts in $10^{13}$ by the torsion-balance experiments Will surveys in his 2014 review of the theory's experimental record, that all bodies fall with the same acceleration, so that an observer in free fall feels no gravity at all. If gravity vanishes for anyone who simply lets go, what part of it is objectively there? This essay identifies that part. Stating it needs no tensors, and the phenomenon has been on the beach all along: the tides. The quantity a change of frame destroys Begin by making the sceptic's difficulty precise, because the answer is a response to its exact shape. Release a laboratory from a tower and everything inside floats. In coordinates fixed to the ground, each object obeys $\ddot{x}^{i} = -\partial\Phi/\partial x^{i}$, where $\Phi$ is the gravitational potential, the potential energy per unit mass. Switching to coordinates that fall with acceleration $g$ subtracts the same constant from every object's acceleration and removes the uniform field from the equations of motion entirely, in the way that stepping off a carousel removes the centrifugal term. The universality of free fall is exactly what makes this possible: because the acceleration is the same for a feather and a lead brick, one substitution disposes of it for both at once. Had different materials fallen differently, no single change of coordinates could have cleared the equations, and $g$ would have been an observer-independent quantity after all. So the equivalence principle is not a curiosity alongside the problem; it is the problem. A quantity that a change of viewpoint sets to zero cannot be gravity's invariant content — the content on which all observers agree. If there is such content, some other quantity carries it. Two free-fallers, and the subtraction that survives One falling body reveals nothing about gravity; two reveal everything, and the reason is a subtraction. Drop two ball bearings side by side. Each falls towards the centre of the Earth; those two directions are not parallel; the bearings drift together. Drop them one above the other instead: the lower one sits in a slightly stronger field, pulls ahead, and the pair drifts apart. No falling frame removes this relative motion, because a change of frame adds the same acceleration to both bearings and therefore cannot alter their difference. That is the whole argument, and it is worth pausing on: a frame could refute it only by changing a difference of accelerations, and none can. The quantitative version is one Taylor expansion. Let the first bearing sit at $x^{i}$ and the second at $x^{i} + \xi^{i}$, with $\xi$ small. Each obeys Newton's second law in the potential, so $$\ddot{x}^{i} = -\frac{\partial\Phi}{\partial x^{i}}(x), \qquad \ddot{x}^{i} + \ddot{\xi}^{i} = -\frac{\partial\Phi}{\partial x^{i}}(x + \xi) = -\frac{\partial\Phi}{\partial x^{i}}(x) - \sum_j \frac{\partial^{2}\Phi}{\partial x^{i}\partial x^{j}}\,\xi^{j} + O(\xi^{2}).$$ Subtracting the first from the second cancels the field itself and leaves $$\ddot{\xi}^{\,i} \;=\; -\sum_{j}\frac{\partial^{2}\Phi}{\partial x^{i}\,\partial x^{j}}\,\xi^{j}.$$ The first derivatives of $\Phi$ — the field — have disappeared; only the second derivatives remain. That matrix of second derivatives is the tidal tensor : the table of how the field differs between neighbouring points, and the thing no free-falling observer can remove, because a change to a freely falling frame changes $\Phi$ by a linear function of position, whose second derivatives are zero. What the Earth's tidal tensor actually is Working the derivatives out for a spherical body shows the characteristic pattern, and the pattern in turn explains why the tides come twice a day. For the Earth, $\Phi = -GM/r$, so $\partial_i\Phi = GMx_i/r^{3}$ and $$\frac{\partial^{2}\Phi}{\partial x^{i}\partial x^{j}} = GM\left(\frac{\delta_{ij}}{r^{3}} - \frac{3x_ix_j}{r^{5}}\right),$$ with $\delta_{ij}$ the identity matrix. Along the radial direction, where $x_i = x_j = r$, the bracket is $1/r^{3} - 3/r^{3} = -2/r^{3}$, so the relative acceleration is $\ddot{\xi} = +\,(2GM/r^{3})\,\xi$: a stretch. Along either horizontal direction, where $x_i$ vanishes, only the first term survives and $\ddot{\xi} = -\,(GM/r^{3})\,\xi$: a squeeze. The three coefficients, $-2$, $+1$, $+1$ in units of $GM/r^{3}$, sum to zero — which is Laplace's equation $\nabla^{2}\Phi = 0$, holding in empty space. Where matter is present, Poisson's equation $\nabla^{2}\Phi = 4\pi G\rho$ makes the trace of the tidal tensor proportional to the local mass density; the trace measures the matter you are standing in, and the trace-free remainder is the tidal field of everything else. This stretch-and-squeeze is what the Moon applies to the oceans, and it accounts directly for the fact that puzzles most people about tides: there are two bulges, one facing the Moon and one directly away from it. In a frame falling with the centre of the Earth, the near side is pulled towards the Moon more strongly than the centre and the far side less strongly, so relative to the centre both move outward along the Earth–Moon line, while the flanks are squeezed inward. The two bulges are the two ends of a single stretch, and the Earth rotating beneath them delivers two high tides a day. The numbers, and a comparison that settles the point The magnitudes are small, measurable, and instructive. With $GM = 3.99\times10^{14}\ \mathrm{m^{3}\,s^{-2}}$ and $r = 6.37\times10^{6}\ \mathrm{m}$, so $r^{3} = 2.59\times10^{20}\ \mathrm{m^{3}}$, the vertical coefficient at t