Can Bob be in the same reference frame as Alice while freefalling in gravitational fields of different strengths 
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*Alice and bob, while in a common inertial reference frame, synchronize watches so that they tick at the same rate.

*Alice moves to an altitude X above a large point mass and maintains her altitude by accelerating upwards at the appropriate rate.

*Bob moves to a position directly above Alice at altitude Y where Y > X and maintains his altitude by accelerating upwards at the appropriate rate.
Sanity check: Am i correct that at this point Bob should observe Alice's watch ticking slower than his own as a result of the difference in their accelerating reference frames?


*Bob adjusts his own watch's calibration such that it appears to be synchronized in both rate and value with the image of Alice's watch as it arrives at Bob's position.

*At precisely 12PM as read by their respective watches both Alice and Bob cease accelerating and enter free-fall towards the point mass.
What relative rate will Bob observe Alice's watch ticking at now that they are both in free-fall? Will the watches remain synchronized? Will Bob see Alice's watch ticking at a faster rate than his own as Alice has 'shed' more acceleration than Bob?
 A: 
Sanity check: Am i correct that at this point Bob should observe Alice's watch ticking slower than his own as a result of the difference in their accelerating reference frames?

Yes.

What relative rate will Bob observe Alice's watch ticking at now that they are both in free-fall?

$$
\Delta\tau_B = \frac{\sqrt{\frac1{r_s} - \frac1{r_{B0}}} - \sqrt{\frac1{r_B} - \frac1{r_{B0}}}}{\sqrt{\frac1{r_s} - \frac1{r_{A0}}} - \sqrt{\frac1{r_A} - \frac1{r_{A0}}}} \Delta\tau_A
$$
where indices $A,B$ denote Alice and Bob and index $0$ the starting positions. Here, $\Delta\tau_B$ is the time as measured by Bob that elapsed between the reception of two light signals emitted by Alice with a delay (from her perspective) of $\Delta\tau_A$. It does not include your calibration step 4, and has to be understood infinitesimally as $r_A, r_B$ are changing as Alice and Bob fall towards the gravitating mass.

Will the watches remain synchronized?

Probably not. I did not solve the equations of motion explicitly. I don't see why the dilation factor above should be constant, but neither did I prove that it's not....

Will Bob see Alice's watch ticking at a faster rate than his own as Alice has 'shed' more acceleration than Bob?

As mentioned, from Bob's perspective, Alice's clock (or rather it's image) starts out ticking more slowly than his own. I would expect this to continue to be the case, but note that I don't quite have a proper handle on this. Bob's motion will introduce a blue shift that will counteract the red shift introduced by Alice's motion and the gravitational field to some degree. I'm not a 100% sure that under the right conditions, this effect couldn't (temporarily?) dominate.
Why do I talk about red shift and blue shift? Because that's what determines the apparent clock rate. For example, consider a moving source in flat spacetime. In addition to the Lorentz factor accounting for time dilation, the motion of the source means that the second signal has to travel a longer distance, yielding the Doppler factor in combination.
A: 
Can Bob be in the same reference frame as Alice while freefalling in gravitational fields of different strengths

No. General relativity doesn't have global frames of reference.
See How do frames of reference work in general relativity, and are they described by coordinate systems?
A: The answer depends on what is meant by "frame of reference".  The two falling bodies, in slightly different gravities and at different speeds, are slightly (minutely) out of reference.  However, the two bodies under normal gravity could be carrying on a radio conversation together without misunderstanding, even if their watches are minutely different.  However, if one body is falling towards a black hole at close to light speed, and the other body is not, then the "frame of reference" would differ widely enough to separate the couple irrevocably.
