The reason the particle slows to a stop at the horizon is not due to the acceleration of the particle but the "change in the shape of space" above the horizon, which depends on the Ricci tensor in the coordinates of the distant observer.
Also, the statement that the particle accelerates to c at the horizon is not valid, a particle can cross the horizon (in its own coodrinates) with an arbitrarily small velocity. It seems to me that a massive particle starting from infinite distance will reach a velocity which depends on the mass of the black hole. If a particle from infinitely far reaches c at the horizon, it has an infinite kinetic energy there, so the amount of work done by the black hole is infinite. However, the integral of the force acting on the particle over the path from the point at infinity to the horizon (the work) is finite since the force achieves a maximum at the horizon and goes to zero at infinity.
Where P is the path the particle travels, r(s) = s (since the path is a line) is a parameterization of P, M = Gm
BHm
p, and R
s is the Swarzchild radius.
Notice that if we substitute 0 for R
s, then the value of the integral
is infinity. With this substitution the calculation is for a path all the way to the singularity.