# Inertia and Mach's Principle

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PeterDonis
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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 possess 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 possess 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
Gold Member
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?

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PeterDonis
Mentor
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
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Gold Member
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
Mentor
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
Homework Helper
Gold Member
See here:
andrewkirk said:
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:
andrewkirk said:
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
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.

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PeterDonis
Mentor
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
Homework Helper
Gold Member
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
Mentor
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?

timmdeeg
Gold Member
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. ... 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.
I think to have a better understanding of Einstein's doubts.

To put it as simple as possible could one say the existence of any inertial structure (which requires defined geodesics) is inseparably linked with the rising water question in the bucket, so that there is no need to "show" that (here remembering Max Jammers wording "it could be shown ...)?
Then it should make no difference if we talk about the asymptotically flat Schwarzschild spacetime or about flat Minkowski spacetime. If true whether or not there is a history of matter (to mention Ken G's idea) doesn't make any difference either.

timmdeeg
Gold Member
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.
Ok thanks for clarifying.

Ken G
Gold Member
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.
Indeed, and here is where I think Einstein might have been using too specific a version of "Mach's principle." It seems to me the essence of that principle is simply "dynamical history determines the geodesics," as in, the geodesics are not handed to us by some absolute geometry (as Newton thought), but rather, they are part and parcel of the dynamics that involves the devations from them. This is what I would call the "big idea" of looking to the distant stars to understand local inertia, the idea that you need information to know the local geodesics, and that information involves everything that is within your past light cone, including boundaries (after all, what is a "boundary" if not simply a proxy for everything that has gone before that you are choosing not to explicitly include?). So when framed that way, there is no reason to wish to elevate the effects of matter to some higher status than the effects of boundary conditions, it's all about what conditions the geodesics as emergent phenomena rather than dictated by mystery or fiat. In that sense, the ambiguities in the geodesics in an empty universe that foists it all onto the boundary conditions is not a problem for Mach, it is a problem for the universe itself-- a universe in which nothing happens becomes a universe in which it is hard to talk about what will happen, that doesn't sound like a problem.

In post N.4 Chronos sent a link to a paper by D. W. Sciama which agrees nicely with my non-academic (a polite way of saying it) world view.

You are correct in that GR is not fully consistent with Mach's principle regarding inertia. Einstein himself acknowledged this fact. The seminal paper; On the Oriicely gin of Inertia, by Sciama: http://articles.adsabs.harvard.edu/...=2&data_type=GIF&type=SCREEN_VIEW&classic=YES, has a nice summar. You may also find this discussion of interest; A Look at the Abandoned Contributions to Cosmology of Dirac, Sciama and Dicke, https://arxiv.org/abs/0708.3518

From Sciama's paper:

“If the rest of the universe determines the inertial frames, it follows that inertia is not an intrinsic property of matter, but arises as a result of the interaction of matter with the rest of the matter in the universe”

and from his calculations:

“This means that the main contribution comes from distant matter – (I) shows that 99 percent of local inertia arises from matter further away than 108 light years.”

I present this as a bit of a poll to understand if any (or all) of the esteemed contributors to this discussion agree with this view.

PeterDonis
Mentor
From Sciama's paper

Sciama's paper was using a simplified model of gravity, which was known even at the time to be incorrect; he just used it to illustrate the kind of thing he was talking about. He said he was working on applying his ideas to a more realistic model of gravity, but AFAIK he never completed that or published anything based on it.

from his calculations

Which, as above, should not be taken as a claim about actual numbers, just as an illustration of a general method he was proposing, but which he never actually developed to the point where it could be used with a realistic model of gravity.

Sciama's paper was using a simplified model of gravity

The paper is about the origin of Inertia. I suppose, however, with the principle of equivalence you can interchange gravity and inertia at will.

And you are right he never came up with his promised complete theory.

Ken G
Gold Member
One way to focus the discussion is point out the empirical fact that if we enter a frame that is both distant from any known sources of forces, and where the distant galaxies have no net rotation around us, and no net acceleration in a particular direction, then this is also the frame in which geodesics are consistent with our concept of no net forces. In short, in a frame like that, inertia behaves in the way Newton expected on small scales (say, water in a bucket). This is an empirical fact, and we have two basic approaches for addressing it. The first I would call Newtonian, the second, Machian:
1) (Newtonian) Our local inertia, and the behavior of the distant galaxies, all respect the same intrinsic geometry that is handed to us by fiat. This is the flavor of Newton's first law, and it does indeed require its own law to stipulate the intrinsic character of geodesics. It has the same flavor of when the Greeks thought orbits had to be perfect circles, as that geometry was in some sense "intrinsic" to how motion through space worked.
2) (Machian) Our local inertia is coupled to the behavior of the distant galaxies, the latter (along with any boundary conditions that are proxies for the influences we are choosing not to treat in detail) unfolding dynamically in a unified and self-consistent way to how the geodesics behave. This has the flavor of a dynamical theory rather than an assertion by fiat of some kind of intrinsic character of geodesics.
So framed like that, I cannot see how GR does not take the second perspective. It seems to me the issue basically comes down to whether you put gravity into a modification of Newton's first law, or if you put it into the second law as a particular type of force. I realize that many physicists are attempting to treat gravity as the fourth force, and model gravitons and so forth, which would be fundamentally non-Machian because it would place gravity on the same footing as other forces and thus place it in the second law. But it seems to me that this is not what GR does, so GR should therefore be regarded as Machian. When people say it is technically not Machian, it must therefore be some more specific, and I would argue less useful and less generalizable, version of Mach's principle than the way I framed it above.

PeterDonis
Mentor
The paper is about the origin of Inertia.

Yes, but his model for how inertia is produced was that it is due to the interaction of a given piece of matter with all the rest of the matter in the universe, and the interaction he used was a simplified model of gravity.