High School Is Inertia Tied to the Stars According to Mach's Principle?

[timmdeeg]

I see it differently. According to Mach's Principle the rising water is related to the distant stars. If the water rises without the distant stars (the case of the "the otherwise empty universe") then there is no need for Mach's principle.

 

[Dale]

Hence, one cannot claim that Mach's principle is inadequate by citing the fact that it doesn't resolve the ambiguities of an empty universe-- it is the universe itself that is inadequate in such a case

I like that suggestion. It is the obsessive discussion of empty universes that always makes the usefulness of Mach's principle questionable to me.

 

[PAllen]

But it isn't just empty universes that is an issue for Mach. One might oversimplify by saying any universe not matter/energy dominated is primarily determined by boundary conditions. Barbour, in the paper linked earlier in this thread, is critical of Einstein's abandonment of hoping GR was Machian, but Barbour's formal arguments start from an assumption of a closed universe of simplest topology, which makes them matter dominated (among other aspects of victory by definition). Einstein's doubts focused precisely on the importance of boundary conditions in GR. He hoped, initially, that there would be no freedom to choose these, or, at least they would be highly constrained. His Machian doubts precisely focused on this key question of boundary conditions.

Consider any asymptotically flat universe, alternatively any open asymptotic geometry not determined by the matter content. Then the inertial structure (the answer to bucket questions) is not primarily determined by matter. Just because these don't match our universe, is it really correct to say the are edge cases in GR? At best one might claim that our universe as modeled in GR is Machian, but not that GR per se is a Machian theory.

 

[Ken G]

I would agree with that, but I would add that Mach's principle is not intended to be a universal rule for all universes, but rather an explanation of inertia in our universe. I could easily imagine universes that were so poorly constrained by their mass/energy content that you would in effect need some kind of "external hand" (i.e, arbitrary boundary conditions) to resolve the potential ambiguities in how inertia works. So we might generalize from a version of Mach that says "matter there establishes inertia here" to the more general statement that "the history of what has gone before controls how inertia works now." When we do have plenty going on by way of matter and energy, and it obeys simplifying symmetries like the cosmological principle, only then does Mach's principle take on its familiar form that stresses distant matter rather than external boundary conditions (such as a concept of "absolute space"). For more weakly matter-controlled cases, we'd need some other consideration to stand in for the "distant stars," but the important point would still be that inertia here is related to the properties of spacetime, which have a dynamical history that must be tracked. Rather than seeing inertia as some kind of inherent property of matter, and straight lines as some kind of preferred inertial state, the whole business is unified with the very dynamics we are trying to understand that plays out within that spacetime.

So to me, that's the essence of Mach's principle-- the rejection of the Newtonian picture that motion through space is somehow a two step process where first you have the space and then you have the motion within it. Rather, motion and space are as interrelated as water and an olympic swimmer. In that rather loose analogy, the "distant stars" of Mach are like the walls of the pool and the people who poured the water into it, and they established that the water the swimmer traverses will not have a current in it. But if the pool was made differently, then it could.

 

[Ken G]

Maybe another way to say all this is that rotation is not built, in GR, to be absolute. In Newtonian physics, you can always tell if you are rotating because you experience fictitious forces, but in GR, even fictitious forces can be regarded as gravitational, stemming from the behavior of the distant stars. In other words, since motion is not absolute, there is also no way to talk in absolute terms about the motions of the distant stars, that too depends on our choice of reference frame. If we choose language for which the Earth is not spinning, then the distant stars are revolving instead, and all our observations are explained perfectly well by GR.

An analogous situation occurs for linear acceleration. It seems easy to say whether or not one is undergoing linear proper acceleration, one simply looks for a constant unexplained acceleration in everything one observes that isn't nailed down. One can translate that into an experiment via the use of an accelerometer. But the accelerometer must first be calibrated, and of course we are going to calibrate it such that it should read zero when we are not experiencing any known forces. But imagine you do that, you go way out into space where you wouldn't imagine there being any forces (and gravity isn't a force in GR), and you calibrate your accelerometer to read zero there, but then you notice something strange (and entirely hypothetical)-- what if every distant galaxy is accelerating in the same linear direction! What do you now do? You basically have two choices: if you are Machian, you assert that you miscalibrated your accelerometer, and you need to reset it to match what you are seeing in all those other galaxies, even though you can't understand why your proper acceleration is not zero. If you are not Machian, you say to heck with the distant galaxies, you don't have any force on you so the acceleration should be zero.

In terms of the above experiment, the fact that this very problem does not in fact arise is tantamount to saying that the universe is Machian. In other words, the Machian solution does not present the problem I just described: when we set our accelerometer to zero when we expect it to be zero, we find the distant stars are not linearly accelerating, and we are happy. This doesn't imply the universe must obey Mach's principle, in the sense that no other approach would also serve to explain what we see, it only means that the universe we see is consistent with Mach's principle, so if GR is going to describe the universe, it must be built to be Machian also.

Given that state of affairs, I cannot see why Einstein would have reversed his opinion. Perhaps he had some more specific version of Mach's principle in mind, something less dependent on the actual history of our own universe. But when we start talking about physical principles of universes that are not our own, I think we start to have a very significant problem in trying to demonstrate the truth in what we are saying.

 

[Torog]

Given that state of affairs, I cannot see why Einstein would have reversed his opinion. Perhaps he had some more specific version of Mach's principle in mind, something less dependent on the actual history of our own universe. But when we start talking about physical principles of universes that are not our own, I think we start to have a very significant problem in trying to demonstrate the truth in what we are saying.

Thanks Ken G, I agree that maybe we should work with Universe we find ourselves in. And thanks and to all who have contributed to this discussion.

[Moderator's note: personal speculation deleted.]

 

[timmdeeg]

So we might generalize from a version of Mach that says "matter there establishes inertia here" to the more general statement that "the history of what has gone before controls how inertia works now."

Let's consider the Milne universe which occurs in flat Minkowski spacetime. It doesn't contain matter including it's history, but is equivalent to the empty FRW universe. So following your "more general statement" and if I understood you correctly you wouldn't expect inertia in this case, right?

 

[Ken G]

Can you explain why you think the Milne universe doesn't matter including its history? It seems to me making that assumption is assuming an anti-Machian conclusion. I would expect any Machian to claim that a Milne universe occurs due to a particular history, and that history will allow there to be inertia in a Milne universe.

Perhaps I can clarify my position by repeating the (loose) analogy to Poisson's equation. We would say that a universe in which there is no charge anywhere must obey the equation that the divergence of the electric field is zero. However, we cannot say the universe has zero field everywhere, that is only one possible solution to Poisson's equation. To determine if that is the correct solution, we must know the history of that universe.

The above analogy draws out the connection between Machian thinking and the role of symmetries. I would be inclined to say that one does not need a Mach's principle if one thinks one can adopt certain symmetry assumptions as absolute truths, just as Newton didn't need it because he thought space was absolute. But it is most common for people to regard cosmological symmetry principles, such as the cosmological principle itself, as a function of the history of the universe, rather than as an absolute truth. As such, we can imagine universes that don't present that symmetry, and we can imagine that someday we might detect deviations from the cosmological principle even on the largest scales in our own universe. So if one holds that this is possible, then one does regard Mach's principle as a statement connecting the history of our particular universe with the way inertia works in our particular universe, and speculating about other universes (such as Milne universes) comes with an implicit assumption of a different history.

 

[timmdeeg]

Can you explain why you think the Milne universe doesn't matter including its history?

In the Milne universe special relativity holds globally and it obeys the cosmological principle. It is based on an explosion scenario. So as this spacetime is flat I can not think of any possible contribution of matter at any time. Also how would you argue regarding the case of the empty (the energy density is zero) FRW universe? I see no way that matter was somehow included in its history and then disappeared.

 

[PeterDonis]

My personal opinion – very strongly based on my lack of formal education

Please bear in mind the PF rules on personal speculation. I have edited your post to remove the quoted statement and what follows from it.

 

[Torog]

Thank everybody. It was fun until it was discovered that I was a genuine crackpot.

All the best,

 

[PeterDonis]

It was fun until it was discovered that I was a genuine crackpot.

Personal speculation, by itself, doesn't make you a crackpot. It just is off topic for this forum.

"Crackpot" would be failing to distinguish between your own personal speculation and mainstream physics. But you didn't do that; you quite clearly identified your personal speculation.

 

[Torog]

Personal speculation, by itself, doesn't make you a crackpot. It just is off topic for this forum.

"Crackpot" would be failing to distinguish between your own personal speculation and mainstream physics. But you didn't do that; you quite clearly identified your personal speculation.

Peter,

Having read the rules (a bit late) I now realize that I shouldn't have posted at all. I find the Big Bang abhorrent from both a technical physical and from a philosophical point of view. Also I have no degrees.

I'll send you a personal note with a question or two. I hope you will have time for a short answer.

All the best,

 

[PeterDonis]

Having read the rules (a bit late) I now realize that I shouldn't have posted at all.

Your original question in this thread was fine.

 

[andrewkirk]

Can't we explain Mach just by considering a part of the rotating object? If we have a ring rotating around its axis of rotational symmetry in an otherwise empty universe then the worldline of a particle of the ring is not a geodesic. The particle is being accelerated by the electrostatic forces applied to it by surrounding particles.

In contrast, in a non-rotating ring, all particles have geodesic worldliness.

To move from the latter to the former state, we fire rockets at antipodal points on the ring, whereby the rockets apply force to certain particles of the ring, which then transmit to all particles in the ring, until it is rotating uniformly.

Am I missing something? Is there something about the curvature of a universe containing nothing except a ring that I am failing to take account of?

 

[timmdeeg]

Can't we explain Mach just by considering a part of the rotating object? If we have a ring rotating around its axis of rotational symmetry in an otherwise empty universe then the worldline of a particle of the ring is not a geodesic. The particle is being accelerated by the electrostatic forces applied to it by surrounding particles.

In contrast, in a non-rotating ring, all particles have geodesic worldliness.

To move from the latter to the former state, we fire rockets at antipodal points on the ring, whereby the rockets apply force to certain particles of the ring, which then transmit to all particles in the ring, until it is rotating uniformly.

Am I missing something? Is there something about the curvature of a universe containing nothing except a ring that I am failing to take account of?

I think we have to be clear of how to define the "otherwise empty" universe. If we stay with the empty FRW universe (which is a solution of the Einstein equations and which is expanding) then - if I see it correctly - we can distinguish two situations.

1. The ring doesn't rotate relative to thought comoving test particles (here taking the role of the fixed stars in the non-empty universe). Then if one is very strict the points of the ring aren't describing geodesics because they aren't in free fall.

2. The ring rotates relative to thought comoving test particles. Then I can't imagine that centrifugal forces aren't created and in the case of the bucket the water wouldn't rise.

 

[andrewkirk]

Then if one is very strict the points of the ring aren't describing geodesics because they aren't in free fall.

True. I hadn't thought of that. The geodesics would lead towards the centre of mass as, under the influence of the ring's own minuscule self-gravity, free fall leads to each particle falling to there. The particles are being incrementally accelerated away from that geodesic by their electrostatic bonds to their neighbours.

Hence, in the non-rotating case, we have a person sitting on the ring experiencing a tiny gravitational force inwards. As the rotation rate increases from zero, that force reduces, becomes zero, then starts to point outwards.

I think the best coordinate system to use might be Schwarzschild rather than FLRW, since the latter is designed for homogeneous spacetimes, which this categorically is not. To keep with the spherically-symmetrical paradigm of Schwarzschild, we could change our ring to a solid, perfect sphere. IIRC, the coordinates of the Schwarzschild Metric are non-rotating by definition, so we can just talk about rotation relative to those coordinates, without having to invoke thought particles.

 

[timmdeeg]

I think the best coordinate system to use might be Schwarzschild

This spacetime is empty in the sense of the vacuum solution but curved. Are you tempting to argue pro- resp. contra Mach's principle here?

 

[PeterDonis]

If we have a ring rotating around its axis of rotational symmetry in an otherwise empty universe then the worldline of a particle of the ring is not a geodesic

Yes, but why not? That's the question. Saying that each piece of the ring has a force exerted on it by neighboring pieces if the ring is rotating, and therefore moves on a non-geodesic path, just restates the problem. The question is why certain particular paths through spacetime are geodesic and others are not, and how that is determined. In an empty universe (i.e., if we assume the ring itself has no self-gravity--see below), the only answer is really "just because"--we just declare by fiat that spacetime has a certain geometry which imposes certain inertial properties. In a universe with matter and energy in it, where we can use a dynamical equation, the Einstein Field Equation, to link the spacetime geometry, and hence inertial properties, to the distribution of matter and energy, we at least have something like an answer to the "why" question that isn't "just because". (Though even then, as has been pointed out, we have to worry about boundary conditions--unless the universe is spatially closed and there are no boundaries at all.)

If we stay with the empty FRW universe (which is a solution of the Einstein equations and which is expanding)

This "expanding" universe is just flat Minkowski spacetime in unusual coordinates. That means the ring itself cannot have any self-gravity; if it did, spacetime wouldn't be flat and the empty FRW universe would not describe it. (Neither would any other FRW solution, since all of those assume uniform density of matter/energy throughout the universe, which clearly doesn't describe a ring surrounded by empty space.) Assuming that the ring has no self-gravity is the usual assumption in these discussions.

in the non-rotating case, we have a person sitting on the ring experiencing a tiny gravitational force inwards

Only if you assume the ring has non-negligible self-gravity--which means spacetime can't be flat and you have to start all over again trying to find a solution to the Einstein Field Equation that describes it. And until you've done that, you can't say you know what the inertial properties would be.

I think the best coordinate system to use might be Schwarzschild

Not for a ring, because a ring is not spherically symmetric, and the Schwarzschild geometry assumes spherical symmetry.

If you switch to a sphere, then it still can't rotate, because rotation would break the spherical symmetry.

the coordinates of the Schwarzschild Metric are non-rotating by definition, so we can just talk about rotation relative to those coordinates

No, you can't, because coordinates aren't physical things, they're just labels we put on events. The definition of "non-rotating" in Schwarzschild spacetime is relative to the asymptotically flat spacetime geometry at infinity--in fact it's really no different from flat Minkowski spacetime in that respect. You're still assuming particular boundary conditions and the inertial properties still depend on those boundary conditions.

 

[timmdeeg]

This "expanding" universe is just flat Minkowski spacetime in unusual coordinates. That means the ring itself cannot have any self-gravity; if it did, spacetime wouldn't be flat and the empty FRW universe would not describe it.

I think the Milne universe which occurs in Minkowski spacetime and which is mathematically equivalent to the empty FRW universe is expanding too. But please correct if wrong.

But let me please come back to another question regarding the statement of Max Jammer mentioned in post 12:

"It could be shown that a particle in an otherwise empty universe can possesses inertia or that the first Machian effect is not at all a truly physical effect but can be eliminated by an appropriate choice of a coordinate system."

Would you agree that this "particle" (I think the usual term is test particle) is understood to possesses negligible mass such that it doesn't contribute to any ambient gravitational field (which is lacking here anyhow) in the "otherwise empty universe". If yes it seems legitimate to discuss this particle in the empty FRW universe.

Max Jammer doesn't mention any specific spacetime though, just "empty". He also doesn't mention how it could be shown that the particle can possesses inertia. Could you explain in ordinary language how in principle this can be shown? I had in mind that it was shown by Gödel, but couldn't find any reference unfortunately. I might be mistaken.

 

[PeterDonis]

I think the Milne universe which occurs in Minkowski spacetime and which is mathematically equivalent to the empty FRW universe is expanding too.

It's "expanding" if you pick a particular set of worldlines and call them the "comoving" ones. But those worldlines, from the viewpoint of a standard inertial frame on Minkowski spacetime, are just the worldlines of a set of inertial observers who all pass each other at a particular event (the spacetime origin) and then recede from each other at a constant speed in all possible directions.

Would you agree that this "particle" (I think the usual term is test particle) is understood to possesses negligible mass such that it doesn't contribute to any ambient gravitational field (which is lacking here anyhow) in the "otherwise empty universe".

Yes. But then the statements you and @andrewkirk were making about the points of the non-rotating ring not traveling on geodesics because they are experiencing a tiny gravitational force inward are not correct. The points of the non-rotating ring are "test particles" in the sense you have described, so they don't cause any spacetime curvature and don't cause any gravitational field. So the "empty universe" is just flat Minkowski spacetime.

Max Jammer doesn't mention any specific spacetime though, just "empty".

Yes, that's a good example of why pop science sources are not good ones for learning actual science. Any actual textbook or peer-reviewed paper would have to specify exactly which spacetime geometry they mean by "empty universe".

He also doesn't mention how it could be shown that the particle can possesses inertia. Could you explain in ordinary language how in principle this can be shown?

The principle that the spacetime geometry determines the inertial properties of worldlines (which ones are geodesics and which ones are not) is not "shown"; it's just part of what we mean by "spacetime geometry". In GR there is no other principle from which this one is derived.

 

[timmdeeg]

The principle that the spacetime geometry determines the inertial properties of worldlines (which ones are geodesics and which ones are not) is not "shown"; it's just part of what we mean by "spacetime geometry". In GR there is no other principle from which this one is derived.

Yes. What I don't get is why Einstein obviously was impressed by the argument of the particle possessing inertia in an otherwise empty universe. Isn't it obvious that any deviation from geodesics in flat Minkowski spacetime is due to inertia per definition?

 

[PeterDonis]

Isn't it obvious that any deviation from geodesics in flat Minkowski spacetime is due to inertia per definition?

Objects don't follow non-geodesic paths because of "inertia". They follow non-geodesic paths because a force is exerted on them. (In GR, gravity is not a force.) An object's "inertia" (more properly it's "inertial mass") determines the amount of force that is necessary to make it deviate from a geodesic by a particular amount, i.e., to give it a particular proper acceleration. (Note that the term "inertia" is also sometimes used to refer to the property that objects that do not have a force acting on them will follow geodesics--another good example of the vagueness of ordinary language and why it's always a good idea to precisely specify things in terms of math, or technical terms like "geodesic" and "proper acceleration" that are defined in terms of math.)

As for Einstein being "impressed", can you be more specific about what is confusing you?

 

[PAllen]

Yes. What I don't get is why Einstein obviously was impressed by the argument of the particle possessing inertia in an otherwise empty universe. Isn't it obvious that any deviation from geodesics in flat Minkowski spacetime is due to inertia per definition?

Einstein hoped he would be able to do away with or highly constrain freedom to choose boundary conditions. If you have no choice in boundary conditions, then the theory is determining the inertial structure rather than something arbitrary. He was very disappointed at how significant boundary conditions remained, and considered this to spoil the idea of GR being truly Machian. Look back at one of my earlier examples: a universe that is Minkowski except for an small region of matter, versus a universe with an eternal BH and one small isolated matter region. The inertial structures are completely different, and neither is caused by the matter configuration. The only difference is boundary conditions.

 

[andrewkirk]

Yes. But then the statements you and @andrewkirk were making about the points of the non-rotating ring not traveling on geodesics because they are experiencing a tiny gravitational force inward are not correct.

I didn't say that, or at least have no recollection of saying anything like that. What I did say was:

The particles are being incrementally accelerated away from that geodesic by their electrostatic bonds to their neighbours.

That is, it is electrostatic forces that make the atoms' worldlines deviate from geodesics. The points do experience a small gravitational force, but it would make no sense to say that makes their worldlines deviate from geodesic (since a geodesic is the path of a particle that is acted on by gravity alone).

IIRC, the coordinates of the Schwarzschild Metric are non-rotating by definition, so we can just talk about rotation relative to those coordinates, without having to invoke thought particles.

No, you can't, because coordinates aren't physical things, they're just labels we put on events.

The IIRC ('If I Remember Correctly' - sorry for the use of internet slang) was critical here. I'm going off my vague memory of Schutz's derivation of the metric. I thought there was a constraint equation in that derivation that means that no non-radial preferred direction in space can be derived from the coordinates and the spacetime. My not-fully-mapped-out-yet reasoning says that means that any free-falling particle whose four-velocity at time $t$ has no non-radial component in the coordinate system will continue to have no non-radial component throughout its future worldline until it hits the central mass. That's what I meant by 'non-rotating coordinates'.

But it's a while since I've thought about this so I'll go and refresh on the Schutz derivation and see how many holes my memory has.

 

[PeterDonis]

I didn't say that, or at least have no recollection of saying anything like that

See here:

in the non-rotating case, we have a person sitting on the ring experiencing a tiny gravitational force inwards

And you repeat it in this post:

The points do experience a small gravitational force

If the ring is made of test particles, and the universe is otherwise empty, this statement is simply false, because there is no source of gravity present.

it would make no sense to say that makes their worldlines deviate from geodesic

If we consider the case of a ring not made of test particles, i.e., with non-negligible stress-energy, and the ring is not rotating and is static, then the worldlines of its particles will not be geodesics, because geodesic worldlines would (heuristically, assuming that the gravitational effect of the ring is to curve spacetime inward towards its center) fall inward, not remain static. So if the ring is static, that means the electromagnetic forces between its atoms are keeping those atoms from moving geodesically, which of course they must in order to keep the ring from collapsing. In other words, the ring is under stress due to its self-gravity. The spacetime in this case will of course not be flat, since there is stress-energy present. (It will be asymptotically flat, if the ring is the only stress-energy in the universe.)

My understanding up to now has been that the case I just described is not the case we are supposed to be discussing in this thread--that in this thread we are supposed to be discussing the case of a ring made of test particles in an otherwise empty universe, which I have been assuming means a ring of test particles in flat Minkowski spacetime. In that case, as I said above, the points of the ring experience no gravitational force at all.

My not-fully-mapped-out-yet reasoning says that means that any free-falling particle whose four-velocity at time $t$ has no non-radial component in the coordinate system will continue to have no non-radial component throughout its future worldline until it hits the central mass.

This is correct, but the usual interpretation of that fact is not that the coordinates are "non-rotating", but that the particle's worldline is "non-rotating" (in the sense of not "rotating" about the central "mass").

There is a property of Schwarzschild coordinates that you might be intuitively trying to capture here, which is most easily observed by noting that they are diagonal everywhere. That means that the surfaces of constant coordinate time $t$ are everywhere orthogonal to the integral curves of the vector field $\partial / \partial t$. It turns out that this property actually reflects an invariant property of the Schwarzschild geometry, which is that it has a Killing vector field (which in Schwarzschild coordinates is $\partial / \partial t$) which is everywhere hypersurface orthogonal (i.e., the spacetime can be foliated by hypersurfaces which are everywhere orthogonal to this KVF). This property, hypersurface orthogonality, turns out to be equivalent to the KVF itself being irrotational (having zero vorticity), which in turn can be interpreted physically as saying that the "source of gravity" in the spacetime is not rotating. (Contrast, for example, with Kerr spacetime, usually described as containing a rotating black hole, which has a KVF that is not hypersurface orthogonal.)

You were actually involved in a thread several years ago that touched on this:

https://www.physicsforums.com/threa...-large-scale-homogeneity.706736/#post-4483918

This is getting well beyond the level of this thread, though, so we should spin off a separate one if you want to go into it further.

 

[andrewkirk]

See here:

in the non-rotating case, we have a person sitting on the ring experiencing a tiny gravitational force inwards

And you repeat it in this post:

The points do experience a small gravitational force

Yes, they experience a gravitational force since, in the case I was describing, the sphere is made of actual massive particles, not test particles. I presume we both agree on that force applying in that scenario. The scenario of the OP is of real spinning space capsules in an otherwise empty universe, not test particles.

But I am 99.5% confident that nowhere in this thread did I say what you thought I said, which is that gravitational force causes the particles' worldline to deviate from a geodesic. If I did say that and forgot it, I will be very surprised, since I do not believe it. I am usually pretty careful about the exact words I use, and am triply careful with the words 'cause' and because' because they are such minefields.

If we consider the case of a ring not made of test particles, i.e., with non-negligible stress-energy, and the ring is not rotating and is static, then the worldlines of its particles will not be geodesics, because geodesic worldlines would (heuristically, assuming that the gravitational effect of the ring is to curve spacetime inward towards its center) fall inward, not remain static

I agree. I corrected this in response to @timmdeeg 's observation about this, here:

The geodesics would lead towards the centre of mass as, under the influence of the ring's [now "sphere's"] own minuscule self-gravity, free fall leads to each particle falling to there. The particles are being incrementally accelerated away from that geodesic by their electrostatic bonds to their neighbours.

 

[PeterDonis]

The scenario of the OP is of real spinning space capsules in an otherwise empty universe, not test particles.

I don't think the OP had in mind assigning non-negligible stress-energy to the real spinning space capsules (i.e., not assuming they were made out of test particles), but that's ultimately a question only the OP can answer.

nowhere in this thread did I say what you thought I said, which is that gravitational force causes the particles' worldline to deviate from a geodesic

That wasn't the statement I was attributing to you. The statement I was attributing to you was that, if gravity (spacetime curvature) is present, due to non-negligible stress-energy of the object (ring, sphere, space capsule, whatever), and electrical forces between the atoms are present and are keeping the object static (i.e., its atoms are not moving solely under gravity, because if they were the object would not be static), then the atoms in the object will not be moving on geodesics. So what causes the atoms' worldlines to not be geodesics is non-gravitational (electrical) forces, as you say; but what causes the geodesic worldlines to be non-static (and therefore the static worldlines to be non-geodesic) is the spacetime curvature (gravity) due to the non-negligible stress-energy of the object.

 

[andrewkirk]

The statement I was attributing to you was that, if gravity (spacetime curvature) is present, due to non-negligible stress-energy of the object (ring, sphere, space capsule, whatever), and electrical forces between the atoms are present and are keeping the object static (i.e., its atoms are not moving solely under gravity, because if they were the object would not be static), then the atoms in the object will not be moving on geodesics.

I regard that as a fair paraphrase of what I wrote.

So what causes the atoms' worldlines to not be geodesics is non-gravitational (electrical) forces, as you say; but what causes the geodesic worldlines to be non-static (and therefore the static worldlines to be non-geodesic) is the spacetime curvature (gravity) due to the non-negligible stress-energy of the object.

But I don't think that follows from it.

The problem is the word 'cause' which, I agree with Bertrand Russell, is an unscientific word, and hence I try to assiduously avoid using it except when speaking very loosely. On a quick check with the text search I am relieved to find that I have not used the word 'cause' at all in this thread, but only mentioned it. That was deliberate.

So it's not so much that I think the second sentence is false, as that I find it lacking in clarity, and hence would be reluctant to either affirm or deny it, because of the presence of that word.

It seems to me that using the dreaded word 'cause' nearly always implies an interpretation of some scientific fact, and interpretations IMHO are philosophy rather than science. Interpretations are fun to discuss, but since I am not yet confident that I understand the science of Mach's principle, I need to focus on that rather than on interpretations.

Having said that, wikipedia seems to imply that the idea of what Mach's principle actually is is so vague that perhaps it is not unreasonable to interpret it as a philosophical rather than a scientific principle. It even says

'It was never made clear by Mach himself exactly what his principle was.'

 

[PeterDonis]

it's not so much that I think the second sentence is false, as that I find it lacking in clarity, and hence would be reluctant to either affirm or deny it, because of the presence of that word

I don't see why "cause" is any worse than "make", which you used:

it is electrostatic forces that make the atoms' worldlines deviate from geodesics

We need some kind of shorthand word for "when you look at the relevant particular solution of the applicable equations, it has these particular properties, which are traceable to these other particular inputs that you gave". What word would you like to use?