Imagine a rod which falls radially towards a central mass. If particles are released along the rod at an instant of time they should move downwards in the the lower and upwards in the upper part of the rod relative to it. In between one particle should not move relative to the rod for a moment. Lets call the position of this particle the force free point FFP on the rod.
Where is the FFP located? At the center of mass of the rod (i) or somewhere in the lower part (ii)? If (ii) is correct, does the FFP move along the rod as it is approaching the event horizon?
EDIT It might be challenging to rigorously do the math in order to answer my question. I would still be thankful if someone could shed light on it heuristically.
Clearly, if the rod starts the free fall from rest at infinity then the force free point coincides with the center of mass. I wonder if the question whether the FFP "moves" relative to the rod during the free fall depends on certain symmetry properties. A shell observer will measure $-\sqrt{2M/r}^{1/2}$ for the radial velocity of the FFP, thereby assuming that the trajectory of this point describes a geodesic. The radial velocity shell observer are measuring at the bottom and at the top of the rod will deviate from this value. Presumably being closer to the source of gravity the deviation will be larger at the lower end. But it seems vague to conclude from this asymmetry that the FFP should be in the lower part of the rod. Any input is welcome.
Choosing $M$ large enough we don't have to care that the rod breaks due to tidal forces.
