Well, well -- this agrees with that Felber/Farber/whatever-his-name's "repulsive" solution that he made such a big deal about.

For a straight line free-fall, no L, no tangential component of velocity, the above Schwarzschild coordinate acceleration formula reduces to. If you're following through, be careful about the signs of the dot product and the velocity vector.

a = -GM/r^2 * [1 + 2R/r - 3 (v/c)^2 ].

You can see the v/c term reduces the acceleration, and sort of looks like a velocity dependent "drag" term. And the question is at what velocity will the term in brackets become zero. This is the turning point at which the free-falling test particle will appear to start slowing down. And that is just:

(v/c)^2 = 1/3 *(1 + 2R/r).

At r = infinity, that's just v/c = Sqrt(1/3) ~ 58%. Which means a test particle launched with this velocity or greater will only appear to slow down, not accelerate, as it approaches the source. Note this velocity increases for "launching points" deeper in the well, and at r = R, v/c becomes 1. The turnaround velocity is lower the farther away you launch. If I can still cut the differential equation mustard, I'm going to look at r(t) solutions for the above and see how this plays out for the maximum and final near-horizon velocities.

This agrees with Felber, although in his frame, he was thinking about a moving gravitating mass approaching a stationary one and seeing a repulsive "force".

I'm confident in this value now. But I wonder what is the flow in my "gamma and mc^2" reasoning in this thread, where yielded
v/c = sqrt(2/3). Note that in the above, we'd reach this value for r = 2R.

Relativistic free fall sanity check.

If I had to hazard a guess, I'd say that reasoning didn't account for the "curvature".

-Richard