# Silly doubt about tidal effects and Einstein Field Equations

Suppose I give you the following line element:

$$ds^{2} = -(1+2\phi)dt^{2}+(1-2\phi)(dx^{2}+dy^{2}+dz^{2}) \tag{1}$$

Without saying anything you might think that this is just another line element like a line element in spherical coordinates; you might think that I just perform another coordinate transformation to a system who are endowned with a particular chart in the manifold $$(M,g)$$.

But, If I say to you that $$(1)$$ line element describes the Newtonian Gravity, and moreover, that the symbols $$\phi$$ are precisely the Newtonian potential of gravity, then you might ask: "why?" I would give you the following explanation:

The motion of a Freely Falling particle in a curved spacetime is then given by the following calculation of the absolute or intrinsic derivative: $$p^{b}\nabla_{b}p^{a} = p^{b}(\partial_{b}p^{a} + \Gamma ^{a}_{cb}p^{c})$$ If you consider the line element of $$(1)$$ and non-relativistic speeds then the equation of motion is reduced to (spatial components): $$p^{b}\nabla_{b}p^{a} = m\frac{d}{d\tau}p^{i} + \Gamma ^{i}_{00}(p^{0})^{2}$$ Which, after calculations is reduced to: $$\frac{dp^{i}}{d\tau} = -m \delta^{ij}\partial _{j}\phi = -m\partial _{i}\phi$$ Now, considering that we are in a regime of low velocities, then the Newtonian Equation is valid: $$F^{i} = -m(\nabla \phi_{g})^{i}$$ Then, since is also valid, by special relativity, that: $$F^{i} = \frac{dp^{i}}{d\tau}$$ Then, $$-m(\nabla \phi_{g})^{i} = \frac{dp^{i}}{d\tau} = -m\partial _{i}\phi$$ Implies that $$\phi = \phi_{g}$$ And the line element describes the Newtonian gravity, indeed.

Now, given a reasonable explanation that $$\phi = \phi_{g}$$, you could say,also, that by equivalence principle $$(1)$$ also describes, in a local region, a relativistic form of gravity.

But here comes my doubt, I know that the geodesic deviation gives you the Riemann tensor, and for the metric $$(1)$$ you can conclude that the general quantity that encodes the notion of "gravity as curvature" is highly suggested by the Ricci tensor ,because the following expression gives you a way to talk about tidal effects:

$$a^{\mu}=\frac{D^{2} \delta x^{\mu}}{D\tau^{2}} = R^{\mu}_{\nu \gamma \beta} u^{\nu}u^{\gamma} \delta x^{\beta} \implies a^{i} = -c^{2}R^{i}_{0j0}\delta x^{j}$$

Where $$a^{\mu}=\frac{D^{2} \delta x^{\mu}}{D\tau^{2}}$$ is called acceleration between geodesics, given by geodesic deviation.

And now, the doubt: Why tidal effects are non-local? Since we can describe tidal effects with the line element of $$(1)$$ by

$$a^{i} = -c^{2}R^{i}_{0j0}\delta x^{j} = -\frac{1}{c^{2}}\partial^{k} \partial_{k}\phi_{g}$$

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.

• "is harder to measure a tidal effect" but, what about the sea tides? I mean, you can see that is occuring a gravitational influence. – M.N.Raia Jun 19 at 11:49
• @M.N.Raia It is harder to measure a tidal effect when you're sealed in a tiny box. Of course, if you have an apparatus as big as the Earth... – knzhou Jun 19 at 11:52
• So, you're saying that Earth is the "big einstein elevator how can feel the tidal effects" ? – M.N.Raia Jun 19 at 11:55
• @M.N.Raia Yeah. – knzhou Jun 19 at 11:55

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 $$P$$).