I understand that spacetime is curved in the presence of a stress-energy tensor. I also understand that, for the example of Earth, most of that curvature is in the time-time direction. At the surface of Earth, the 4-acceleration, in coordinates of $\{t,r,\theta,\phi\}$, can be represented as:$$\boldsymbol{A}=\{\frac{9.8}{c},0,0,0\}\space m\space s^{-2}$$What is less clear to me is how the acceleration in the $t$ dimension is turned into an acceleration in the $r$ dimension. That is, how does an acceleration in time get translated into falling towards the center of gravity? I've seen videos on time-dragging, but I've not seen any mathematical formalism.
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$\begingroup$ Not sure if I understand the question because we are staying on the surface of earth and don't fall towards the center of gravity. If we did there would be a small three acceleration that would exactly offset the term $9.8/c$. In free fall the four acceleration has zero Lorentz length. $\endgroup$– Kurt G.Commented Feb 3, 2022 at 15:21
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$\begingroup$ I never said we are 'staying' on the surface, I said 'at the surface'. Imagine you're Wiley Coyote and you just stepped off a ledge. What happens to your velocity in the $r$ dimension? Why does it increase when the component of $r$ in the 4-acceleration is zero? $\endgroup$– Quark SoupCommented Feb 3, 2022 at 15:26
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$\begingroup$ Does this answer your question? Why would spacetime curvature cause gravity? $\endgroup$– John RennieCommented Feb 3, 2022 at 16:34
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2 Answers
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On the surface of the earth the four-acceleration is in the radial direction, not in the time direction. So in SI units it would be $(0,9.8,0,0)$ or in units of light years and years it is about $(0,1,0,0)$. For an object with constant invariant mass the proper acceleration is always spacelike (or 0).
To see this, recall the definition of the four acceleration: $$A^\mu = \frac{d}{d\tau}U^\mu + \Gamma^{\mu}{}_{\nu \lambda} U^\nu U^\lambda$$ so for an object at rest on the surface we have $U=(1,0,0,0)$ and thus only the $\Gamma^{\mu}{}_{tt}$ Christoffel symbols are relevant.
In Schwarzschild coordinates the only tt Christoffel symbol (p. 18) is $$\Gamma^r {}_{tt}=\frac{c^2 R(r-R)}{2 r^3}$$ so the four acceleration is purely in the r direction
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Acceleration is not a property of points in spacetime, it is a property of trajectories in space time.
The 4-acceleration you wrote is that of an observer in a trajectory that is staying at the same point on the surface of the earth. On the other hand, you are asking about the acceleration of an object that is free falling - namely, following a geodesic line. However the 4-acceleration of a geodesic line is by definition zero - this is a manifestation of the famous equivalence principle.
The quantity you are asking about - the coordinate acceleration in the $r$ direction of a free falling observer, namely $d^2 r/dt^2$ - is a different thing, that you can find by solving the geodesic equation. For a radially free falling observer in the Schwarzschild metric this turns out to be just $\ddot r = -M/r^2 = -g$ (just like in the Newtonian case, except here $\dot r = dr/d\tau$)
