Newtonian limit of GR, geometrically!

Question

In this article, Wikipedia says that Ricci curvature represents deviation of the volume of a thin geodesic cone from the Euclidean value: $$d\mu_g = [1 - \frac16R_{jk}x^jx^k+O(|x|^3)]\,\,d\mu_{Euclidean}$$ Say I want to get an approximation for the path of a ball thrown straight up from the Earth. Take the $x$ vector above to be the Earth's time axis, and the path of the ball to be on the boundary of that "thin geodesic cone". So I should be able to restate the above equation as: $$V\approx{V}_{Euc}(1-kt^2)$$ I'm just calling the geodesic volume $V$ instead. $k$ will be the time-time component of the Ricci tensor, from the field equation - it will essentially be the mass of the Earth, so it's constant over $t$. Now, with 3 spatial dimensions, the "cone" will actually be a sphere expanding with time, so its volume will be: $$V=\int{\frac{4}{3}{\pi}r^3dt}$$ And the Euclidean volume is a sphere expanding at constant speed $v$: $$V_{Euc}=\int{\frac{4}{3}{\pi}(vt)^3dt}=\frac{4}{3}{\pi}v^3\frac{t^4}{4}$$ Combining the above 3 equations gives: $$\int{r^3dt}=\frac14v^3t^4(1-kt^2)$$ And taking the time derivative: $$r^3=v^3(t^3-\frac32kt^5)\\r=vt(1-\frac32kt^2)^{\frac13}$$ That's the height of the ball as a function of time. The problem is, if you expand it, it doesn't have a $t^2$ term, just goes right to $t^3$. Whereas we should actually get $r\approx{vt}-\frac12gt^2$ to match Newtonian gravity. So where is the error in this reasoning?? Did I misinterpret the Wikipedia article? Is my math wrong?

Answer 1

Thanks to @JohnRennie's comment I was able to understand what was wrong with my analysis. Basically I was missing the whole point that GR is a local theory, so you can't just extrapolate the value of a tensor at one point to determine a macroscopic trajectory like the arc of a ball. That equation is valid, but only in the limit as you approach the point (in this case, the Earth at the moment the ball was released), and only for objects whose velocity is infinitely close to that of the Earth (the "thin cone"). So it only determines the very beginning of the ball's trajectory, if the ball's moving really slow. Once the ball moves away from the Earth, it is at a new point in spacetime, so to find the next infinitesimal piece of it's path, you'd have to apply the Ricci tensor again, but now there is no mass locally, so the tensor is zero, so a local geodesic cone will maintain the Euclidean volume. It's almost like a zipper effect: as the Earth moves forward in time, it bends local geodesics toward it, and geodesics a little further out have to bend inward too to maintain local Euclidean volume, and so on as you move out. A geodesic appreciably far from the Earth will feel the indirect effects of all the little geodesic pieces that were directly bent by the Earth at earlier times, and I guess that adds up to an acceleration of about $GM/r^2$, but it seems you can't see that result without diving into the formal analysis. But seeing the conceptual picture is still really cool.

One more thing about the "thin cone" (low speed) approximation: I think it makes more sense when you're thinking in terms of mass density rather than a finite point mass. ie. the Ricci tensor actually means something akin to "curvature density per unit of geodesic speed and per unit of time". So each infinitesimal piece of the planet will only appreciably affect geodesics moving at infinitesimally small speeds away from it, but since those little geodesic slices border on each other, when you integrate them up, their net effect is to induce finite curvature even in higher-speed geodesics. But still, per the above paragraph, each geodesic is only affected for infinitesimal time, so you also have to integrate the curvature along the Earth's time axis and then propagate its effects outward in space.