# Can Bob be in the same reference frame as Alice while freefalling in gravitational fields of different strengths

1. Alice and bob, while in a common inertial reference frame, synchronize watches so that they tick at the same rate.

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

3. 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?

1. 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.

2. 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?

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.

• Granted, but note that the question itself is fine otherwise. Rephrased: Will the apparent rate of a clock that is freely falling towards a gravitating mass as seen by an observer that is freely falling as well (but from a higher elevation, and with the initial conditions as specified) remain constant? Commented Jan 22, 2020 at 20:08

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.

• I had considered the distinction between the effect of the red shift and that of the Lorentz factor but my understanding of that subtlety is not as rigorous as i would like. As such i thought it better to leave it open to comment in the hopes i might learn something. Supposing it is reasonable to draw a hard line between the two phenomenon i would say i am more interested in what the Lorentz component has to say about this. That is, what would happen If Bob's watch was also calibrated such that it slowed to compensated for the red shift induced by Alice receding towards the mass? Commented Jan 23, 2020 at 3:39
• If that seems pedantic i would add that i am particularly interested in the effect of the local curvature and how this interacts with the "free-fall" dynamic and as such this scenario was constructed with the intention of eliminating as many relative-motion derived effects as possible. Commented Jan 23, 2020 at 3:56
• If we have two free faller with different r-coordinate the upper sees the lower blueshifted and the lower the upper redshifted.
– timm
Commented Jan 23, 2020 at 17:33
• Surely you mean the other way around? Commented Jan 24, 2020 at 8:57
• No, remember the extreme case, Alice at infinity, Bob just crossing the event horizon: Bob observes Alice redshifted with z = 1. To calculate the frequency shift between 2 objects in free fall one has to take gravitational shift and Doppler shift into account.
– timm
Commented Jan 24, 2020 at 9:21

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.

• Please don't use bold face for the hole answer. It is only useful, if you highlight specific things. Commented Jan 21, 2020 at 22:54