Jeff and I have a little disagreement about what a stationary observer sees of a freefaller into a black hole. I maintain that the velocity must reach a peak and come to zero, which means the (coordinate) acceleration must go to zero and change sign at some point. That is, if we drop an object, it will appear to accelerate as expected, but at some point must appear to start slowing down before it reaches the horizon. Jeff doesn't agree with this.

I found the following expression for the (coordinate) acceleration field in the Schwarzschild metric for a stationary observer far away in polar coordinates. This is the modified gravitational "force law" that will give you the precession of the perhelion of Mercury and all that. It doesn' include gravitomagnetic effects or anything like that, as the Schwarzchild metric is for a stationary, non-rotating source mass.

a = d^r/dt^2 =

-GM/r^2 * [ [1 + 2R/r + v^2/c^2] *(r_n) - 4*v^2/c^2 *(v_n dot r_n)*(v_n) ]

That expression is a bit cumbersome as I was too lazy to use bold and subscripts. R is the Schwarzschild radius, and r_n is a unit radial vector while v_n is a unit vector along the particle's velocity vector.

I'm going to play with that and compare to my "rest energy to kinetic energy" logic with the slowing coordinate speed of light reasoning and see how it compares.

Note one gets an inverse cube term plus a v^2/c^2 increase in the radial force. But the second term (with a factor of 4 on it which I wonder is related to where the factor of 4 on the gravitomagnetic permeability comes from) is directed backwards along the velocity vector, multiplied by the dot product of r and v. When v is perpendicular to r, as would be for a circular orbit, that term is zero. However, for a radial velocity, that term is maximum.

And it appears that this term is what will turn the acceleration around and slow the velocity. I'll have to play with it on paper to see.

-Richard