Clocks falling into gravity wells

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As I understand it, clocks run relatively slower the closer they are to the center of a "gravity well" and also each other clock for a pair of observers moving relatively to each other appears to be running slower, relative to each frame of reference.

So, start with a pair of highly accurate instrumented clocks at rest relative to each other within a gravity well, one higher and one lower. The lower one is on the ground, the higher one is suspended from a helicopter. At this point, the clock on the ground runs slower than the one in the helicopter. (Both clocks agree on this.) Now let the suspended clock drop from the helicopter (and ignore the friction effects from the air.)

Just before the falling clock smashes into the ground, it still reads the clock on the ground as running slower, now because of relative velocity rather than any difference in the gravity well potential.

The question is this: What variation, if any, does the falling clock observe in the ground clock? It always runs slower, but how does it change as the clock falls? Does the variation depend on the mass of the underlying ground?
 
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The lower one is on the ground, the higher one is suspended from a helicopter. At this point, the clock on the ground runs slower than the one in the helicopter. (Both clocks agree on this.)
Yes.

Just before the falling clock smashes into the ground, it still reads the clock on the ground as running slower, now because of relative velocity rather than any difference in the gravity well potential.
But now both clocks do not agree which one is running slower: the clock on the ground now reads the falling clock as running slower (whereas when both clocks were at rest the clock on the ground agreed that the clock hanging from the helicopter was running faster).

What has changed is that the two clocks are now not at rest relative to each other, so they no longer agree on simultaneity--they no longer agree on which ticks of the two clocks happen "at the same time".

What variation, if any, does the falling clock observe in the ground clock?
The simple intuitive answer is: none. Consider: the kinetic energy of the ground clock, relative to the falling clock (i.e., in the falling clock's rest frame) at any point during the fall, must be the same as the potential energy change in the falling clock from the start of its fall to that point. (This must be true by conservation of energy plus symmetry of relative motion between the clocks.) Therefore the relativistic ##\gamma## factor of the ground clock, relative to the falling clock, must be equal to the change in the gravitational redshift factor between the two clocks from the start to any given point in the fall. (This is because the ##\gamma## factor determines the kinetic energy and the redshift factor determines the potential energy.) So the falling clock will see no change in the rate of the ground clock. (The ground clock, however, will see a significant change in the rate of the falling clock, because of the lack of agreement between the clocks as discussed above.)
 

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