Would the one accelerating please stand up?

[Dale]

If you want a mainstream reference, look in any GR textbook; my personal favorite is Misner, Thorne, and Wheeler

And mine is Carroll's lecture notes

http://preposterousuniverse.com/grnotes/

This material is the beginning of chapter 4

 

[PeterDonis]

Using the term “relative inertia” is not a “theory” of mine, just English words trying to explain my point of view of the equivalence principle that are either common knowledge or at least things that I thought were common knowledge, not something new.

And just to clarify my request for a mainstream reference, the key thing that needs a reference is not the term "relative inertia" by itself, but whatever it is that you are describing by those words that you think is common knowledge. From your previous posts where you describe what you mean by that term, it doesn't look to me like any mainstream concept that I'm aware of. But if you have a mainstream reference and can provide it, then we can get a better idea of what you are trying to describe.

 

[MikeGomez]

The body above the Earth experiences a gravitational field.

No, it doesn't. It's in free fall; it feels no force, no acceleration, and no "field".

The body in free fall in the elevator, just like the one above the earth, is in free fall, feeling no force, no acceleration, and no "field".

A gravitational field exists in all four of these cases…

1: A gravitational field exists for the body in freefall above the earth.

2: A gravitational field exists for the body at the surface of the earth.

3: A gravitational field exists for the body in freefall above the floor in the accelerating elevator scenario.

4: A gravitational field exists for the body at the surface of the floor in the accelerating elevator scenario.

"At a fundamental level, the reason that gravitational mass and inertial mass seem to be that same is that they are the same, not because nature conspires to make them appear the same.

I agree that this is what General Relativity says. Whether this is "the correct explanation" is still, strictly speaking, an open question, since General Relativity is not a theory of everything.

Sure, everyone knows that GR is not a theory of everything, but an intellect such as Einstein makes one of the most profound statements in the history of science, and even a century later no one (and I am sure many have tried) has been able to disprove it. Yet you say that this idea in such doubt by mainstream science today that it is invalid to use in a modern day discussion of the equivalence principle?

 

[MikeGomez]

And just to clarify my request for a mainstream reference, the key thing that needs a reference is not the term "relative inertia" by itself, but whatever it is that you are describing by those words that you think is common knowledge. From your previous posts where you describe what you mean by that term, it doesn't look to me like any mainstream concept that I'm aware of. But if you have a mainstream reference and can provide it, then we can get a better idea of what you are trying to describe.

What I am trying to describe is the relative relationships between.the body in freefall and the body at the surface for both scenarios of the the planet and accelerating elevator.

Relative relationships exist for position, velocity, momentum (not inertia), gravitational attraction, and all physical aspects of what is under consideration here. The two common scenarios of the body at the planet and the body at the elevator do not apply for every physics example regarding gravitation and acceleration, but the equivalence principle does (when understood in the correct way and when applied in the correct way).

I am trying to establish the relative relationships for the bodies for the scenarios that everyone is familiar with. That is essential because once a relational understanding is established for the scenarios that everyone is familiar with, that can be extended to create appropriate scenarios for the other cases that we are discussing such as the "body falling down a hole in the earth" scenario.

 

[Dale]

The two common scenarios of the body at the planet and the body at the elevator do not apply for every physics example regarding gravitation and acceleration, but the equivalence principle does (when understood in the correct way and when applied in the correct way)

This is not correct. The equivalence principle explicitly restricts itself to local experiments. So it explicitly does not apply to non-local examples. The hole in the Earth scenario is such an example.

 

[PeterDonis]

A gravitational field exists in all four of these cases…

1: A gravitational field exists for the body in freefall above the earth.

2: A gravitational field exists for the body at the surface of the earth.

3: A gravitational field exists for the body in freefall above the floor in the accelerating elevator scenario.

4: A gravitational field exists for the body at the surface of the floor in the accelerating elevator scenario.

This is not a standard use of the term "gravitational field". (Admittedly, there is not a single "standard" use of that term; but your usage does not match either of the standard usages that I'm aware of, as I'll show below.) So you are going to have to define specifically what you mean by "gravitational field". By "specifically", I mean "in terms of the actual math of GR".

In what is probably the most common standard usage of "gravitational field", it refers to particular connection coefficients in a chosen coordinate chart. So before you can even say whether it "exists" in any of your examples, you need to specify what coordinate chart you are using. In the most natural coordinate chart for each of your cases, there is no gravitational field in cases #1 and #3, because the most natural coordinate chart is a local inertial frame in which the free-falling bodies in those cases are at rest, and in any local inertial frame the connection coefficients are zero, so there is no "gravitational field". In cases #2 and #4, the most natural coordinate chart is a non-inertial chart in which the accelerated bodies in those cases are at rest; in that chart, the appropriate connection coefficients are nonzero, so a "gravitational field" does exist in those cases. So in this usage, you are correct for cases #2 and #4 but wrong for cases #1 and #3.

The other fairly common usage of the term "gravitational field" is as a synonym for "spacetime curvature", i.e., tidal gravity. In that usage, a gravitational field exists in cases #1 and #2, but not in cases #3 and #4.

 

[PeterDonis]

Yet you say that this idea in such doubt by mainstream science today that it is invalid to use in a modern day discussion of the equivalence principle?

I said no such thing. Read what I posted again, carefully.

 

[PeterDonis]

What I am trying to describe is the relative relationships between.the body in freefall and the body at the surface for both scenarios of the the planet and accelerating elevator.

The correct term to describe that is "coordinate acceleration", not "relative inertia". In coordinates in which either body is at rest, the other body is accelerating. The other relationships all follow from that one--at least, the relationships for position, velocity, and momentum (and energy) do. I'm not sure what you mean by a "relative relationship" with reference to "gravitational attraction". (This has been a recurrent issue in this discussion--you insist on using your own idiosyncratic terminology instead of learning the standard language in which these things are described. That makes communication difficult, since the rest of us know the standard language and are using it.)

once a relational understanding is established for the scenarios that everyone is familiar with, that can be extended to create appropriate scenarios for the other cases that we are discussing such as the "body falling down a hole in the earth" scenario.

For some relationships, yes, but not for all. There are relationships in curved spacetime that cannot be duplicated in flat spacetime. We have been over this already.

 

[PeterDonis]

the equivalence principle does (when understood in the correct way and when applied in the correct way).

We have given you two mainstream references now that describe what "the correct way" is. Before making further statements about what you think "the correct way" is, I strongly advise you to read at least one of those references (the Carroll lecture notes that DaleSpam linked to are easier because they're online and free) and take some time to think over what it is telling you.

 

[Dale]

@MikeGomez FYI, it is considered very poor form on PF to not provide a reference when asked, even if the concept seems to you to be completely standard or obvious. Such references provide valuable learning material, clarify the point being made, and serve to ensure that the content of PF remains consistent with the professional scientific community. Please take all such requests seriously.

 

[MikeGomez]

@MikeGomez FYI, it is considered very poor form on PF to not provide a reference when asked, even if the concept seems to you to be completely standard or obvious. Such references provide valuable learning material, clarify the point being made, and serve to ensure that the content of PF remains consistent with the professional scientific community. Please take all such requests seriously.

Sorry if I'm taking too long to reply guys. I am extremely hard pressed for time here.

My reference is "Relativity" by Albert Einstein, last printed in 1952.

Thank you both for the references you have provided for me. I have the link that you have provided and at first glance it looks great. I'll download the MTW also. I've seen plenty of references to that one so it so it must be good as well. Please be a little patient and I will get back.

 

[MikeGomez]

You're missing the point. The elevator's gravity can be made negligible in the flat spacetime case (it would have to be for spacetime to be flat), but the Earth's gravity cannot be made negligible in the clock in the hole case. The Earth's gravity completely changes the relationship (in comparison with the flat spacetime case) between the proper acceleration experienced by the "elevator" (the Earth in the clock in the hole case) at a given point on the clock's worldline, and the relative velocity of the clock and the "elevator" (the Earth) at the same point.

We need to produce a gravitational field in flat spacetime for the clock in the hole, which is uniform (uniform in the sense that it does not create proper acceleration). I do not agree about the worldlines and that has to with the ongoing issue we have about the meaning of the equivalence principle.

You can reproduce the profile of relative velocity in flat spacetime with a variably accelerating elevator, but you cannot, with the same variable acceleration of the elevator that you need to reproduce the profile of relative velocity, also reproduce the profile of proper acceleration of each point on the Earth as the clock passes it. And the proper acceleration of the piece of the Earth that the clock is passing at any given point is a locally measurable quantity. So if the EP applied to an "elevator" the size of the Earth in this scenario, it would be easy to tell the "elevator" with the real Earth moving in it from the elevator variably accelerating in flat spacetime, by the different proper acceleration profiles--meaning the EP would be violated.

Once again issues regarding equivalence principle.

It does not apply. There are three different equivalence principles (taken from http://en.wikipedia.org/wiki/Equivalence_principle):

Strong: The outcome of any local experiment (gravitational or not) in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

Einstein: The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

Weak: The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat spacetime, without exception.

All three refer to "local" which means that the experiment is conducted in such a small region of space and brief duration of time that tidal effects are negligible. This is simply not the case in the hole-in-the-earth scenario. The scenario involves two clocks, one at the surface of the Earth and one that falls through a hole in the Earth and back up again. The time that this requires and the distance from one side of the Earth to the other are both sufficiently large that the experiment is non-local and tidal effects are non-negligible.

Einstein EP or stong EP.

In the hole-in-the-earth scenario the following are observed:
1) the distance between the two clocks starts at 0, increases, and then decreases back to 0 (as measured by radar)
2) the relative velocity between the two clocks starts at 0 when they are co-located, increases away from each other, decreases away from each other, increases towards each other, and decreases towards each other back to 0 (as measured by Doppler radar).
3) the proper acceleration of the first clock is always 0 (as measured by an accelerometer on the first clock)
4) the proper acceleration of the second clock is always g in the direction away from the first clock (as measured by an accelerometer on the second clock)
5) the proper time accumulated on the second clock is greater than the proper time of the first clock (as measured by the two clocks themselves)

These 5 results are impossible to replicate in flat spacetime. Even if you are using only the weak equivalence principle this scenario fails to qualify. The effects of motion in the hole-in-the-earth scenario are, in fact, distinguishable from an accelerated observer in flat spacetime, therefore the scenario is not local.

Number 1 looks fine, although we shouldn't need to measure the distances by radar.

Number 3 is fine. The proper acceleration for the hole in the Earth clock must remain zero by definition of the problem.

Number 5 is the difference in time in the Earth case which must be reproduced in the flat spacetime scenario, so that is fine.

Numbers 2 & 4 are the problem, and yet again (no surprise), this is due to what we think about the meaning of the equivalence principle.

Please be a little patient, I am under serious time constraints here.

Thank you.

 

[harrylin]

A gravitational field exists in all four of these cases…

That depends... I can easily see how that phrasing resulted in some disagreement! I have the impression that - as all too often - this discussion has deteriorated onto a rather useless discussion about words. Perhaps you will agree with the following refinement of your statement, as follows.

Coordinate systems can be chosen by means of which according to Einstein's GR a gravitational field appears to exist or may be held to exist in all four of these cases:

1: The case of a body in freefall above the earth.

2: The case of a body at the surface of the earth.

3: The case of a body in freefall above the floor in the accelerating elevator scenario.

4: The case of a body at the surface of the floor in the accelerating elevator scenario.

[..]
My reference is "Relativity" by Albert Einstein, last printed in 1952 [..].

That reference may suffice indeed. Here's my attempt on specifying appropriate references for 1-4 as reformulated by myself, taken from https://en.wikisource.org/wiki/Relativity:_The_Special_and_General_Theory/Part_II:

1. Check - using any reference system fixed to the Earth:
"The action of the Earth on the stone takes place indirectly. The Earth produces in its surrounding a gravitational field, which acts on the stone and produces its motion of fall."
- OK.
2. Check - using any reference system fixed to the Earth: I did not find an appropriate reference in that discussion.
However, a body at the surface of the Earth is surely not an issue.
3. Check - using any reference system fixed to the accelerating chest:
"the body will approach the floor of the chest with an accelerated relative motion. The observer will further convince himself that the acceleration of the body towards the floor of the chest is always of the same magnitude, whatever kind of body he may happen to use for the experiment. Relying on his knowledge of the gravitational field (as it was discussed in the preceding section), the man in the chest will thus come to the conclusion that he and the chest are in a gravitational field which is constant with regard to time."
- OK.
4. Check - using any reference system fixed to the accelerating chest:
"the man in the chest will thus come to the conclusion that he and the chest are in a gravitational field which is constant with regard to time."
- OK.

 

[A.T.]

We need to produce a gravitational field in flat spacetime for the clock in the hole, which is uniform (uniform in the sense that it does not create proper acceleration).

That is not what uniform gravitational field usually means. Uniform gravitational field usually means that it produces no tidal effects. Free falling bodies (point masses) always have zero proper acceleration, regardless whether the gravitational field is uniform or not.

The proper acceleration for the hole in the Earth clock must remain zero by definition of the problem.

The proper accelerations of all involved objects must be reproduced, in order to have an equivalence of physical laws between the two cases.

 

[PeterDonis]

(uniform in the sense that it does not create proper acceleration)

A gravitational field never "creates" proper acceleration; objects moving solely under gravity are always in free fall, feeling zero proper acceleration. This is true for the Earth's gravitational field (or any field) just as much as for a "uniform" gravitational field. Any object that feels proper acceleration, feels it because some non-gravitational force is acting on it.

As A.T. said, the usual way of distinguishing a "uniform" gravitational field (i.e., one produced by acceleration in flat spacetime) from a "real" gravitational field (i.e., one produced by a massive body in curved spacetime) is by tidal effects; the latter has them, the former does not. However, as I pointed out before, in both cases the presence of the "gravitational field" in this sense depends on a particular choice of reference frame: you have to pick a frame in which the object feeling proper acceleration is at rest (i.e., the object at rest in the "elevator", the object at rest on the Earth's surface). If you pick a frame in which a freely falling object is at rest, the "gravitational field" vanishes.

I do not agree about the worldlines and that has to with the ongoing issue we have about the meaning of the equivalence principle.

We should defer further discussion of that until you've had time to look at references, since you said you were doing that.

 

[Dale]

Numbers 2 & 4 are the problem, and yet again (no surprise), this is due to what we think about the meaning of the equivalence principle.

All 5 points (including 2 and 4) are the experimental results that are predicted by GR. Any way you think about the equivalence principle MUST be consistent with them.

 

[MikeGomez]

All 5 points (including 2 and 4) are the experimental results that are predicted by GR. Any way you think about the equivalence principle MUST be consistent with them.

I don’t disagree that GR predict all those things.. It’s just that I thought that there might be a couple less restraints in the flat spacetime scenario such as the one regarding adherence to relative velocity. I guess my mistake, based on what everyone here is telling me is my lack of understand of what the equivalence principle means.

I have looked at Sean Carroll’s description of the equivalence principle and of course it seems reasonable, although there are a couple points which may come into consideration in our discussion.

When he says “…we can no longer speak with confidence about the relative velocity of far away objects…” this may come into play.

When he says “…It is the EEP which implies (or at least suggests) that we should attribute the action of gravity to the curvature of spacetime…” This might not be an issue, and it might not come up at alll, and even if it does come up it doesn’t mean that it will necessarily affect the argument one way or the other.

He says”… It is impossible to "prove" that gravity should be thought of as spacetime curvature, since scientific hypotheses can only be falsified, never verified…” That may or may not be an issue. I just feel more comfortable conceptually with the stress-energy tensor side of the EFE equations.

I’m afraid I won’t be able to keep up with a lot of that follows when I starts talking about the tensor calculus, but if it gets that far we probably will have settled the issue anyway.

 

[Dale]

When he says “…we can no longer speak with confidence about the relative velocity of far away objects…” this may come into play.

Which is why I specified the measurement process (Doppler radar). We can speak with confidence about the outcome of a Doppler measurement.

For a flat spacetime scenario to be equivalent it must reproduce all measurements, not just a chosen few. That includes Doppler shift. A flat spacetime scenario can be devised which reproduces anyone of the above measurements, but not all of them together. In fact, I think that even just 3 and 4 together are impossible.

 

[PeterDonis]

I think that even just 3 and 4 together are impossible.

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

 

[MikeGomez]

That is not what uniform gravitational field usually means. Uniform gravitational field usually means that it produces no tidal effects. Free falling bodies (point masses) always have zero proper acceleration, regardless whether the gravitational field is uniform or not.

So uniform gravitational fields mean flat spacetime, and non-uniform gravitational fields are mean curved spacetime, but what would be the terminology of another type of non-uniform gravitational field which is non-uniform in the sense that the gravitational strength of the source various with time, not distance (i.e. like if the source is a pulsar)?

The proper accelerations of all involved objects must be reproduced, in order to have an equivalence of physical laws between the two cases.

Maybe the non-zero proper acceleration clock in the experiment would be able to transition from linear acceleration to rotational acceleration while maintaining its constant proper acceleration.

 

[A.T.]

what would be the terminology of another type of non-uniform gravitational field which is non-uniform in the sense that the gravitational strength of the source various with time,

Static vs. non-static?

Maybe the non-zero proper acceleration clock in the experiment would be able to transition from linear acceleration to rotational acceleration while maintaining its constant proper acceleration.

Every possible measurement has to be reproduced to have equivalence, including what a gyroscope would show.

 

[MikeGomez]

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

I was also thinking that the direction of the proper acceleration of the second clock in #4 would not matter in order for the final clock reading to be the same. I was thinking there is a reduction in the accuracy of which one body may know about the other body when separated, but if each body from both scearios locally maintains the exact same results, and the end results from both scearios match up, then that would be an acceptable definition for reproducing the experiment.

 

[MikeGomez]

Static vs. non-static?

Every possible measurement has to be reproduced to have equivalence, including what a gyroscope would show.

Gyroscope would be difficult, but I am not sure if that would be impossible. Anyway I still think the experiment can be done for normal non-gyroscopic clocks that I thought we were talking about.

 

[A.T.]

for normal non-gyroscopic clocks that I thought we were talking about.

I thought we are talking about an equivalence in the sense of the equivalence principle: a general equivalence of all laws of physics. Some hand picked cases and measurements that happen to match are not of much interest.

 

[MikeGomez]

I thought we are talking about an equivalence in the sense of the equivalence principle: a general equivalence of all laws of physics. Some hand picked cases and measurements that happen to match are not of much interest.

I am not hand picking anything. If you take the limiting case to such extremes then the argument you are making would mean that the standard man in the elevator scenario is hand picked (and technically it is). There are approximations in the flatness of the spacetime in the elevator case, and there are approximations of flatness of spacetime on the Earth side of it (you need to neglect tidal forces).

 

[A.T.]

I am not hand picking anything.

You can't get beyond the most trivial case of a single point mass, and just a few measured quantities.

the argument you are making would mean that the standard man in the elevator scenario is hand picked

The EP states that all measurements in that elevator in the will be equivalent. That is a general local equivalence.

and there are approximations of flatness of spacetime on the Earth side of it (you need to neglect tidal forces)..

Without tidal (non-uniform) gravity the clock in the hole wouldn't even come back to the clock on the surface. How can you neglect that?

 

[PeterDonis]

the direction of the proper acceleration of the second clock in #4 would not matter in order for the final clock reading to be the same.

Even if the direction doesn't matter (I actually think it does, but I'll assume it doesn't here for the sake of argument), the magnitude certainly does. In your flat spacetime version where the "elevator" is the size of the Earth, and varies its acceleration in order to duplicate the relative motion of the Earth and the clock in the hole, the proper acceleration of every piece of the "elevator" must vary with time. That is obviously different from the actual Earth scenario, where the proper acceleration of a clock at one end of the hole is constant.

(Actually, it's even worse than that; in the Earth scenario, the proper acceleration of every single piece of the Earth is constant--the magnitude of the acceleration varies with distance from the center, but at any given distance it's constant. There's no way to match that in flat spacetime while also matching the relative motion of the clock and the "elevator". This has been pointed out before.)

if each body from both scearios locally maintains the exact same results, and the end results from both scearios match up

They won't. See above.

 

[Dale]

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

You are of course correct. I was thinking of the distance being 0 at two points, which is a condition beyond 3 and 4, but not quite as strong as 1.

 

[MikeGomez]

Ok, finally I see the source of confusion (predicable it’s my fault).
------------------------------------------------------------------------------------------
Case A: Standard Earth versus flat spacetime scenario.

All the physics in the Earth lab frame are the same as in the elevator lab frame.
-------------------------------------------------------------------------------------------
Case B: Hole in Earth, Mike’s versus

In the Earth scenario clock 1 drops down a hole in the Earth (freefall) and takes its lab frame with it on its journey. Clock 2 (proper acceleration) remains at the surface and it has a separate lab frame. When the two clocks meet up again they will show different time readings.

In the flat spacetime scenario clock 1 in its freefall lab frame must match the physics from clock 1 from the Earth scenario. Clock 2 in its proper acceleration lab frame must match the physics from the Earth clock 2 during its journey. When the clocks meet up again they show the same time difference as in the Earth scenario.

-----------------------------------------------------------------------------------------
Case C: Hole in Earth, Everybody else

In the Earth scenario, clock 1 in its freefall communicates with clock 2. In this way the size of the lab frame is the diameter of the entire earth.

The flatspace scenario here is impossible.
------------------------------------------------------------------------------------------

Based on the context of the conversation I should have realized the situation a long time ago, but I was kind of blinded because I was thinking in terms of smaller lab frames. Profuse apologies.

..In your flat spacetime version where the "elevator" is the size of the Earth,...

The giant Earth sized thing was just a random idea and I indicated after that, that I had other ideas, but I guess I was not clear enough.

You can't get beyond the most trivial case of a single point mass

I agree. I was confused about certain requirements that I thought people were making about the experiment. I hope that is cleared up now.

Without tidal (non-uniform) gravity the clock in the hole wouldn't even come back to the clock on the surface. How can you neglect that?

In my version of the flat spacetime scenario the clock in the hole stays in the same location, so it doesn't need to come back, but it does need a changing gravitational field in order to reproduce the proper time differences. The proper acceleration clock in the elevator does need to come back. That's why I was talking about transitioning from linear acceleration to rotational acceleration, so it could make the return journey. The rotational acceleration will produce small gyroscopic effects, and those need to be compensated for by manipulating its gravitational field (which is what I am also assuming we can do for the clock in freefall).

 

[PeterDonis]

finally I see the source of confusion

I still don't think you see all of it. See below.

clock 1 drops down a hole in the Earth (freefall) and takes its lab frame with it on its journey

Yes, but this "lab frame" is not a "local inertial frame" in the sense of the EP. It is localized in space but not in time. The technical term for this "lab frame" is "Riemann normal coordinates". If you look in the mainstream references we gave you, you will see that these are carefully distinguished from "local inertial coordinates". They are not the same thing.

Clock 2 (proper acceleration) remains at the surface and it has a separate lab frame.

Just as a note, this "lab frame" is non-inertial. More importantly, the caveats about it not being "local" apply here as well. In this case, since the "lab" is accelerated, the technical term for what you are calling the "lab frame" here is "Fermi normal coordinates". Again, in the mainstream references we gave you, these are carefully distinguished from "local inertial coordinates". They are even different from "accelerated coordinates in a local inertial frame". These are all important distinctions and you need to grasp them if you are going to understand what the mainstream references are saying.

In the flat spacetime scenario clock 1 in its freefall lab frame must match the physics from clock 1 from the Earth scenario. Clock 2 in its proper acceleration lab frame must match the physics from the Earth clock 2 during its journey. When the clocks meet up again they show the same time difference as in the Earth scenario.

No; this is impossible. If clock 2 in flat spacetime matches the proper acceleration of Earth clock 2, there is no way for it to meet up again with clock 1. If clock 2 in flat spacetime meets up again with clock 1, there is no way for its proper acceleration to match that of Earth clock 2. Not even if you throw in rotation. See below.

Please, instead of waving your hands about this scenario, take the time to actually do the math. It will not work out the way you apparently think it will.

In the Earth scenario, clock 1 in its freefall communicates with clock 2. In this way the size of the lab frame is the diameter of the entire earth.

The flatspace scenario here is impossible.

This is all true, but irrelevant to the actual issue with your version. See above.

I was thinking in terms of smaller lab frames

Yes, but that's not the problem. See above.

The proper acceleration clock in the elevator does need to come back. That's why I was talking about transitioning from linear acceleration to rotational acceleration, so it could make the return journey.

That won't work. See below.

The rotational acceleration will produce small gyroscopic effects, and those need to be compensated for by manipulating its gravitational field

In flat spacetime, there is no "gravitational field" except what is produced by the clock's proper acceleration. So if the clock's proper acceleration has a rotating component, the "gravitational field" will have a rotating component, i.e., it will not be uniform. There's no other "gravitational field" to adjust to compensate; spacetime is flat, so there are no other free parameters besides the magnitude and direction of the clock's proper acceleration.

which is what I am also assuming we can do for the clock in freefall

See, this is why I objected to your usage of the term "gravitational field" before. It has misled you into thinking that you can somehow "adjust the gravitational field" of a free-falling clock in flat spacetime. You can't; there's nothing to adjust. The clock is in free fall; the spacetime it is in is flat; there are no free parameters whatsoever that you can use to "adjust the gravitational field".