How Does Mach's Principle Connect Distant Stars to Local Motion?

[CyberShot]

I was reading this example from Wikipedia:You are standing in a field looking at the stars. Your arms are resting freely at your side, and you see that the distant stars are not moving. Now start spinning. The stars are whirling around you and your arms are pulled away from your body. Why should your arms be pulled away when the stars are whirling? Why should they be dangling freely when the stars don't move?

I am experiencing some difficulty understanding why Mach's principle says that, somehow, the stars moving and your hand being pulled is not a coincidence. To me, the answer to the above question is that whirling stars imply you are spinning, and vice versa. How can there be an intimate relation between something far away out there and local events?

 

[Timothy Takemoto]

> How can there be an intimate relation between something far away out there and local events?
If there is a lot of mass out there then even if it is distant it can still exert a gravitational pull that some have termed "star suck."

I am not sure if this is relevant but interestingly, Mach is not someone to insist that the stars are "far away". Mach believed that the elements of the universe are our sensations, in particular our visual field (which he famously drew from a first person perspective) having had an experience of oneness with the heavens as a young man.

Here is the mach quote [with notes by me]

"For us, therefore, the world does not consist of mysterious entities [fundamental particles, Kantian "things in themselves", Tegmark's math], which by their interaction with another equally mysterious entity, the ego [the mind], produce sensations [the qualia], which alone are accessible. For us, colours, sounds, spaces, times, are the ultimate elements whose given connexion it is our business to investigate.

Mach's Footnote to the above "I have always felt it as a stroke of special good fortune, that early in life, at about the age of fifteen, I lighted, in the library of my father, on a copy of Kant's Prolegomena zu jeder Künftigen Metaphysik. The book made at the time a powerful and ineffaceable impression upon me, the like of which I never afterwards experienced in any of my philosophical reading. Some two or three years later the superfluous role played by "the thing in itself" abruptly dawned upon me. On a bright summer day under the open heaven, the world with my ego suddenly appeared to me as one coherent mass of sensations..." (Mach, 1897, p23. See also the brilliant diagram on p.16)

Mach, E. (1897). Contributions to the Analysis of the Sensations. (C. M. Williams, Trans.). The Open court publishing company. Retrieved from http://www.archive.org/details/contributionsto00machgoog"

The distance of the stars is "emergent" based upon our analysis of our sensations, and with the holographic principle, perhaps not a feature of the universe in itself. We see the stars as distant but this is just a way of seeing, and they can fall flat on our face as it were. For me the holographic principle if true would demonstrate that the sensationalism of Mach is very likely to be correct.

 

[PeterDonis]

To me, the answer to the above question is that whirling stars imply you are spinning, and vice versa.

The issue with this as you state it, from a relativity perspective, is that motion should be relative--at least at first blush, you spinning and the stars being at rest should be equivalent, physically, to the stars spinning and you being at rest.

The first response to this is to note that, if you are spinning relative to the stars, you feel a force, whereas the stars don't. (At least, as far as we can tell, they don't--but we can run the same experiment in, say, a spaceship out in free space, and verify that, if the ship is not spinning relative to the stars and you are, you are the one that feels the force.) So you and the stars are not equivalent in this respect; there is an observable physical difference between you.

However, that just raises the question of where this force that you feel when you spin relative to the stars comes from. See below.

How can there be an intimate relation between something far away out there and local events?

Notice that this is similar to the question people asked about gravity in Newton's time: how can (for example) the distant Sun exert a force on the Earth? It turns out that GR gives the same answer for both questions. Heuristically, the Sun curves spacetime in its vicinity; that curvature propagates to the vicinity of the Earth, and the Earth moves in response to the geometry of spacetime in its vicinity. If you were out in space without a rocket, you would similarly move in response to the spacetime geometry in your vicinity: but if you had a rocket and fired it, you would feel a force, because you were moving differently than the spacetime geometry in your vicinity is "telling" you to move.

Similarly, the distant stars (and galaxies, etc.) have an overall effect that creates a spacetime geometry throughout the entire universe--a sort of "average" geometry based on the average distribution of matter and energy. One might think this makes things a lot more complicated; but there is a nice theorem called the "shell theorem" that greatly simplifies it. According to this theorem, if the distribution of matter and energy is spherically symmetric outside of some sphere, then that matter and energy has no effect on the spacetime geometry inside the sphere--which means that, if there is no matter or energy inside the sphere, the spacetime geometry there will be flat. And if inside the sphere is mostly empty, but with some isolated pieces of matter, then the spacetime geometry there will be close to flat, with significant curvature only being observable near one of the isolated pieces of matter.

So, for example, if we think of a sphere around our solar system, the distribution of matter and energy in the rest of the universe is almost spherically symmetric outside it. So the spacetime geometry in our solar system is close to flat; we only notice significant curvature near massive objects like the Sun and the Earth. This is why you feel a force when you are spinning relative to the distant stars--which is to say relative to the spherically symmetric mass distribution outside the solar system, which is what makes the geometry inside the solar system close to flat.

 

[Elemental]

PeterDonis' post above reflects how I've always understood Mach's principle. Put another way, inertia originates in the preponderance of mass distributed throughout the universe, with most massive bodies moving relatively slowly relative to other massive bodies.

From special relativity all inertial frames (having uniform velocity) are equivalent, so if you are coasting along at high speed relative to distant stars, it is just as valid to say it is the stars that are moving instead of you. No experiment will decide which is right.

But if you are spinning, instead of moving in a straight line, the above argument doesn't work. You are in a rotating, not an inertial frame of reference, and an experiment quickly detects centrifugal force ("star suck"), thereby determining that you are rotating rather than the "fixed stars". Einstein gives a similar thought experiment in his 1915 (I think) general relativity paper where he discusses two isolated celestial bodies rotating relative to one another, one being spherical and the other ellipsoidal. Which one is really spinning? It will be found that the body which is not spinning relative to the distant stars is the spherical one.

But whoa - so why isn't acceleration just as "relative" as linear motion? In fact, it is if you equate gravity with acceleration, and then redefine that acceleration as moving along a geodesic in curved space-time. The rotating reference frame is no longer flat; it has curvature which gives rise to fictitious centrifugal force.

The real mystery here (to me) is why does the preponderance of mass in the universe cause inertia in the first place? Einstein thought that general relativity went a long way toward explaining this, and I've seen attempts to derive it based on analogous calculations with Poisson's equation in three-space; but it only works if you make special assumptions about conditions at infinity (which appear unwarranted based on astronomy). Here I get lost.

 

[Timothy Takemoto]

Dear Elemental
> it only works if you make special assumptions about conditions at infinity (which appear unwarranted based on astronomy).

I have heard and attempted to read something about "boundary conditions," here.
Khoury, J., & Parikh, M. (2006). Mach’s Holographic Principle. arXiv:hep-th/0612117. Retrieved from https://arxiv.org/pdf/hep-th/0612117v1.pdf
They do not mention Poisson's equation, so I presume this is different.

With regard to the boundary matter they write
"Remarkably the boundary matter typicallyturns out to be simply pressureless dust."

 

[PeterDonis]

it is if you equate gravity with acceleration, and then redefine that acceleration as moving along a geodesic in curved space-time

No, this is not correct. Moving along a geodesic is inertial, free-falling motion; an object moving on a geodesic feels no force and has zero proper acceleration. The correct way to say what I think you are trying to say here is that accelerating in flat spacetime is equivalent, locally, to being at rest in a gravitational field in a curved spacetime. But in both cases, you feel acceleration, and you aren't moving along a geodesic.

The rotating reference frame is no longer flat; it has curvature which gives rise to fictitious centrifugal force.

This isn't correct either. Curvature isn't a property of a reference frame; it's a property of spacetime (if we're talking about tidal gravity) or of a worldline (if we're talking about proper acceleration). The rotating frame is non-inertial, but "non-inertial" is not the same as "curved". The "fictitious" force arises because the frame is non-inertial; that's all there is to it. (The "force" of gravity is similarly a "fictitious" force, because it only appears in non-inertial frames, such as a frame at rest on the surface of the Earth, which is an accelerated frame. In an inertial frame free-falling in the Earth's vicinity, there is no "force" of gravity.)

 

[Elemental]

I am referring to the example of a rotating disk discussed in 1919 by Einstein in Relativity, Chapter 23 (Three Rivers Press, 1961). The observer on the disk sees a body, not attached to the disk, deflected from rectilinear motion in the rotating coordinate system. So it appears "accelerated" in the rotating system. This observer, believing general relativity, concludes he is in a gravitational field. He confirms this by measuring the circumference and radius of the disk, and finds their ratio greater than 2π. His space is thus not flat, and he ascribes the apparent acceleration of the body as due to its following a geodesic in curved space-time.

 

[PAllen]

I am referring to the example of a rotating disk discussed in 1919 by Einstein in Relativity, Chapter 23 (Three Rivers Press, 1961). The observer on the disk sees a body, not attached to the disk, deflected from rectilinear motion in the rotating coordinate system. So it appears "accelerated" in the rotating system. This observer, believing general relativity, concludes he is in a gravitational field. He confirms this by measuring the circumference and radius of the disk, and finds their ratio greater than 2π. His space is thus not flat, and he ascribes the apparent acceleration of the body as due to its following a geodesic in curved space-time.

In point of fact, no matter how coordinates for the 'disk at rest'' are established, they will show zero curvature. The vanishing of crvature tensor is a coordinate invariant fact, for any coordinates at all. This means, in SR, you have no curvature at all in any coordinates. Einstein's argument here was a heuristic for why one might expect curvature for gravity. However, if made rigorous for rotating disk coordinates, there would be no curvature.

 

[Dale]

He confirms this by measuring the circumference and radius of the disk, and finds their ratio greater than 2π. His space is thus not flat, and he ascribes the apparent acceleration of the body as due to its following a geodesic in curved space-time.

There is a subtlety here that was not recognized in Einstein's early days. The space that Einstein describes is curved, but the spacetime is still flat. However, there is no reason to prefer Einstein's curved foliation to any flat foliation. So it turns out that the curvature he identified is simply an artifact of his choice of simultaneity.

It is always possible to foliate any flat manifold into curved sub manifolds. So the existence of curved "space" is not particularly meaningful.

 

[PAllen]

There is a subtlety here that was not recognized in Einstein's early days. The space that Einstein describes is curved, but the spacetime is still flat. However, there is no reason to prefer Einstein's curved foliation to any flat foliation. So it turns out that the curvature he identified is simply an artifact of his choice of simultaneity.

Yes, this clarifies more than my post. You can have coordinate in SR with curvature of spatial slices, while the overall space time (4d) curvature remains zero. Only with gravity do you get 4d curvature.

 

[PeterDonis]

His space is thus not flat

Yes.

and he ascribes the apparent acceleration of the body as due to its following a geodesic in curved space-time.

No. Spacetime curvature is not the same thing as space curvature. As Dale and PAllen have posted, the particular choice of "rotating disk" coordinates in question makes space look curved, but spacetime is still flat.

Also, you are still incorrectly using the term "geodesic" to refer to the motion of a body with nonzero proper acceleration; that is not correct regardless of whether spacetime is curved or flat. A body moving on a geodesic, in either flat or curved spacetime, is in free fall, with zero proper acceleration.

 

[Elemental]

Also, you are still incorrectly using the term "geodesic" to refer to the motion of a body with nonzero proper acceleration

The body in the example is in free-fall - that's the intent of saying "not attached to the disk". Thus it follows a geodesic. Nevertheless, in the rotating frame, its trajectory is curved. To what do you attribute the curvature of this trajectory if not gravity?

 

[PeterDonis]

The body in the example is in free-fall - that's the intent of saying "not attached to the disk".

Ah, ok, I had misunderstood what body you were referring to. I agree that a body not attached to the disk is in free fall.

Nevertheless, in the rotating frame, its trajectory is curved.

It appears to be spatially curved; but it is a geodesic, therefore it is not curved in any invariant sense. "Geodesic" is the proper generalization of "straight" in this context. The proper term for what you are calling "curved" here is "coordinate acceleration"

To what do you attribute the curvature of this trajectory if not gravity?

The coordinate acceleration of this trajectory in the rotating frame can indeed be attributed to "gravity", or at any rate to some "fictitious force". But the trajectory is not curved in any invariant sense, and neither is spacetime. The trajectory is a geodesic, and a geodesic is straight; the appearance of spatial curvature is due to the distortion of the coordinates in the rotating frame, so that "space" itself is curved in this frame. But spacetime is still flat, as has already been said.

 

[Elemental]

I apologize for the confusion and agree there is no intrinsic curvature to the trajectory. Assuming that the space-time defined by the frame of reference of the falling body is flat, it still should be flat when re-expressed in a rotating coordinate system even though a space-like hyper-plane in this system can be curved.

This ought to be simple problem (being over 100 years old), but I scanned through several recent papers and still see controversy. For instance, this paper, https://www.amherst.edu/media/view/10267/original/reden05.pdf , calculates three different metric tensors for a rotating frame, all of which have non-zero components in their curvature tensors.

 

[PeterDonis]

I scanned through several recent papers and still see controversy

No, you are misunderstanding what you see. See below.

this paper

It's a Bachelor's Thesis. Those do get some review, but they are not peer-reviewed research in the usual sense.

calculates three different metric tensors for a rotating frame, all of which have non-zero components in their curvature tensors

No, it calculates a spatial metric for a rotating frame. Space is curved in this frame, but spacetime is flat. There is no controversy at all about that and hasn't been for a long time.

This particular author appears to be confused about what he has found, and to have made some incorrect derivations. See my comment above about the actual nature of this source. For a correct expression for the metric of a "rotating frame" on a disk, see here:

https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

It is simple to show that the Riemann tensor of this metric vanishes, as it should.

 

[Jonathan Scott]

Note that General Relativity actually predicts "frame-dragging" effects of rotation and relative acceleration which are very similar to those expected from Mach's Principle. MTW "Gravitation" and other sources refer to this as the "sum for inertia". The effect of a rotating or linearly accelerating gravitational source on a test object is like a tiny rate of rotation or relative acceleration. If you add up what those effects would be due to the apparent relative rotation or linear acceleration of every mass in the universe relative to a test object which would normally be considered to be rotating or accelerating itself, the predicted effects are very similar to the actual effects observed for rotational effects (centrifugal and coriolis forces) and for inertia.

Without a specific model of the shape of the universe and the distribution of masses within it, it is not possible to calculate an exact result, but the effects are qualitatively similar and quantitatively around the right order of magnitude. However, as an explanation of inertia this is frustratingly incompatible with General Relativity, as moving some nearby mass would modify the total effect and it would no longer exactly match inertia unless the gravitational constant G were modified as well, but GR requires big G to be a constant (and the experimental evidence also supports that).

 

[PeterDonis]

calculates three different metric tensors for a rotating frame

Following up further on this, in the section where the author "adds time", the first of his three frames (the one using "central time") is the Born coordinate chart given in the Wikipedia page I linked to. The author notes, correctly, that the spatial portion of this chart does not look like the "spatial metric" he derived for a "rotating frame". The reason is that the "spatial metric" he derived previously does not describe any spacelike slice of the actual spacetime, so it cannot appear as part of any "rotating frame". More on that below.

Also, the author's claim that this metric gives a nonzero Riemann curvature tensor is incorrect; as I noted in a previous post, it is easy to show that the Born chart has a vanishing Riemann curvature tensor. The author does not give explicit computations, so it's impossible to tell where he has gone wrong. I'll post a transcript of my Maxima session that computes the Riemann tensor (and other quantities of interest) separately to avoid cluttering up this post.

The second of his three frames (the one using "local time") is just the Born chart with the time coordinate rescaled by an $r$ dependent transformation. The spacelike slices of this chart are the same as the spacelike slices of the standard Born chart and have the same metric, as is evident from the metric given.

The third of his three frames (the one using "central time") is not a valid coordinate chart at all, because of the "cut" he describes, which violates continuity of coordinates (heuristically, nearby events must have nearby coordinates, but events on opposite sides of the "cut" are nearby and yet have time coordinates that are not). So the "metric tensor" he derives for this "frame" is not a valid metric tensor, at least not for the purpose he is using it, which is to investigate global properties of the spacetime.

To correctly understand what the "spatial metric" the author derives is telling us, we need to understand what "space" it is a metric of. This is explained in the Wiki article I linked to, in the section on "Radar distance in the small":

https://en.wikipedia.org/wiki/Born_coordinates#Radar_distance_in_the_small

Briefly, the "space" described by this spatial metric is a quotient space, not a spacelike slice of the spacetime. So the spacetime as a whole cannot be described as an infinite sequence of "spatial slices", each with the given spatial metric. That means there is no valid spacetime metric for this spacetime which contains the given spatial metric as its spatial part.

 

[PeterDonis]

Here is a transcript of the Maxima session I ran that computes the Riemann tensor (and other quantities of interest) for the Born chart:

Maxima 5.32.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.10 (a.k.a. GCL)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) batch("born.dem")

read and interpret file: #p/home/peter/projects/physics/born.dem
(%i2) if get('ctensor,'version) = false then load(ctensor)
(%o2)          /usr/share/maxima/5.32.1/share/tensor/ctensor.mac
(%i3) "Metric for the Born chart."
(%i4) "The dimension of the manifold"
(%i5) dim:4
(%o5)                                  4
(%i6) "The coordinate labels"
(%i7) ct_coords:[t,z,r,phi]
(%o7)                           [t, z, r, phi]
(%i8) "Rational simplification of geometrical objects"
(%i9) (ratwtlvl:false,ratfac:true,ctrgsimp:true,ratchristof:true)
(%o9)                                true
(%i10) "This is the metric matrix:"
(%i11) lg:matrix([-(1-w^2*r^2),0,0,w*r^2],[0,1,0,0],[0,0,1,0],[w*r^2,0,0,r^2])
                           [  2  2             2   ]
                           [ r  w  - 1  0  0  r  w ]
                           [                       ]
                           [     0      1  0   0   ]
(%o11)                     [                       ]
                           [     0      0  1   0   ]
                           [                       ]
                           [    2               2  ]
                           [   r  w     0  0   r   ]
(%i14) "Compute metric inverse and determine diagonality"
(%i15) cmetric()
(%o15)                               done
(%i16) "Compute and display mixed Christoffel symbols"
(%i17) christof(mcs)
                                                2
(%t17)                        mcs        = - r w
                                 1, 1, 3

                                             w
(%t18)                          mcs        = -
                                   1, 3, 4   r

(%t19)                        mcs        = - r w
                                 1, 4, 3

                                             1
(%t20)                          mcs        = -
                                   3, 4, 4   r

(%t21)                         mcs        = - r
                                  4, 4, 3

(%o21)                               done
(%i22) "Compute the (3,1) Riemann tensor"
(%i23) riemann(true)
This spacetime is flat
(%o23)                               done
(%i24) "Compute the Ricci tensor"
(%i25) ricci(true)
THIS SPACETIME IS EMPTY AND/OR FLAT
(%o25)                               done
(%i26) "Compute the scalar curvature"
(%i27) scurvature()
(%o27)                                 0
(%i28) "Compute the mixed Einstein tensor"
(%i29) einstein(true)
THIS SPACETIME IS EMPTY AND/OR FLAT
(%o29)                               done

 

[vanhees71]

Honestly, I never understood what Mach's principle actually is and why GR is assumed to realizes it. The principle stated by Mach is quite vague; it's just saying that inertia of a body is due to the presence of all other masses in the universe. I can understand that there is a vague analogy of GR with the vague statement of Mach's principle in the sense that free test particles move on geodesics, where free means that the test particle is subject to gravity as the only interaction and no other interactions (like electromagnetic fields). But in which sense explains GR that inertia is just due to "all masses in the universe"? I think GR is rather starting from the equivalence between inertial and gravitational mass, i.e., the equivalence principle, i.e., it doesn't "explain" it in any sense but postulates its validity which is indeed empirically very well confirmed. Also GR, as a relativistic theory, corrects the statement that gravity is due to masses as sources but it's the entire energy-momentum-stress tensor of matter and radiation that necessarily must universally couple to the gravitational field.

 

[vanhees71]

Following up further on this, in the section where the author "adds time", the first of his three frames (the one using "central time") is the Born coordinate chart given in the Wikipedia page I linked to. The author notes, correctly, that the spatial portion of this chart does not look like the "spatial metric" he derived for a "rotating frame". The reason is that the "spatial metric" he derived previously does not describe any spacelike slice of the actual spacetime, so it cannot appear as part of any "rotating frame". More on that below.

Also, the author's claim that this metric gives a nonzero Riemann curvature tensor is incorrect; as I noted in a previous post, it is easy to show that the Born chart has a vanishing Riemann curvature tensor. The author does not give explicit computations, so it's impossible to tell where he has gone wrong. I'll post a transcript of my Maxima session that computes the Riemann tensor (and other quantities of interest) separately to avoid cluttering up this post.

The second of his three frames (the one using "local time") is just the Born chart with the time coordinate rescaled by an $r$ dependent transformation. The spacelike slices of this chart are the same as the spacelike slices of the standard Born chart and have the same metric, as is evident from the metric given.

The third of his three frames (the one using "central time") is not a valid coordinate chart at all, because of the "cut" he describes, which violates continuity of coordinates (heuristically, nearby events must have nearby coordinates, but events on opposite sides of the "cut" are nearby and yet have time coordinates that are not). So the "metric tensor" he derives for this "frame" is not a valid metric tensor, at least not for the purpose he is using it, which is to investigate global properties of the spacetime.

To correctly understand what the "spatial metric" the author derives is telling us, we need to understand what "space" it is a metric of. This is explained in the Wiki article I linked to, in the section on "Radar distance in the small":

https://en.wikipedia.org/wiki/Born_coordinates#Radar_distance_in_the_small

Briefly, the "space" described by this spatial metric is a quotient space, not a spacelike slice of the spacetime. So the spacetime as a whole cannot be described as an infinite sequence of "spatial slices", each with the given spatial metric. That means there is no valid spacetime metric for this spacetime which contains the given spatial metric as its spatial part.

Well, I recently had a quite lively discussion with a colleague, who claimed that spacetime is curved in a rotating reference frame, but that cannot be true since Minkowski spacetime is flat, and this cannot change only because you introduce "noninertial" spacetime coordinates. The Riemann curvature tensor is (as the name says) a tensor, and since it's identically 0 in flat Minkowski space, so it must come out using any coordinates. This seems to be a common misconception of curvature in GR. What's right is that for an accelerated observer his space in the sense of a timeslice is in general non-Euclidean, but that has to be clearly distinguished from non-Euclidean (or non-pseudo-Euclidean) spacetime. Anyway, in the end, I also could convince my colleauge by explicitly calculating the Riemann curvature tensor using the Born chart you described above.

 

[Jonathan Scott]

If you place gravitational sources near a test mass and accelerate them linearly, GR says that the test mass will accelerate slightly in the same direction, although this effect is too small to detect in practice, especially in comparison with the ordinary fields created by the same masses. However, if you add up what the effect would be if somehow the whole universe were accelerated relative to the test mass, it predicts that the test mass would naturally accelerate by an amount of the same order of magnitude in the same direction, and would require a force proportional to its mass in the opposite direction to prevent such an acceleration. If we could replace "of the same order of magnitude" with "equal" then this would describe exactly the observed effect of inertia as a purely relative effect. A similar result applies for rotation.

I've heard that Mach himself said nothing of the kind, but this is rather how Einstein interpreted Mach's ideas.

 

[Elemental]

Here is a transcript of the Maxima session I ran that computes the Riemann tensor (and other quantities of interest) for the Born chart:

Thanks for the post; I concede this space-time is flat.
Back to Mach's principle, Ciufolini and Wheeler discuss it at length in Gravitation and Inertia (Princeton University Press, 1995). They provide an analogy from electrodynamics. The electromagnetic radiation from a suddenly accelerated test charge will tend to drag other charges (eventually even faraway charges) in the same direction. Now, reverse the picture and say all of the other charges accelerated. Then they would drag the test charge with it. This looks like frame-dragging, but it is only an analogy and I'm sure it doesn't reproduce general relativity's prediction of, say, the Lense-Thirring effect. It does seem to me however to be a promising way forward if the electromagnetic field were replaced with the gravitational field.

 

[Jonathan Scott]

The closest electromagnetic analogy for linear frame dragging is the E field component due to the rate of change of the vector potential with time (where the gradient of the scalar potential is equivalent to the ordinary gravitational field). Accelerating charges would produce such a component.

The electromagnetic equivalent of the Lense-Thirring (rotational) frame-dragging effect is magnetism.

I remember seeing a paper by Kenneth Nordtvedt on varieties of frame-dragging which I found in some obscure place on the internet, including a derivation of the exact expression for the induced linear frame dragging, but I'm sorry I can't remember right now what it was called or where I found it.

 

[PeterDonis]

What's right is that for an accelerated observer his space in the sense of a timeslice is in general non-Euclidean

"Space in the sense of a timeslice" has to be very carefully defined in this context. As I noted in my last post, the "space" to which the non-Euclidean metric applies in the case of the rotating disk is a quotient space; it is not a spacelike slice of the spacetime.

In some special cases, the "space" of an accelerated observer can be defined as a spacelike slice of the spacetime--for instance, in the case of Rindler observers. But the space in that case is flat (Euclidean). I'm not aware of any case of an accelerated observer's "space" in flat spacetime being both curved and definable as a spacelike slice of the spacetime. In curved spacetime of course there are such cases, for example observers following orbits of the timelike Killing vector field in Schwarzschild spacetime, who are accelerated and whose "space" is a slice of constant Schwarzschild coordinate time and is indeed non-Euclidean.

 

[vanhees71]

Yes, that refers to finite spacetime regions. What's usually defined are local "splittings" in space and time of an observer. You can define it as a set of tetrades (e.g., the Fermi-Walker transported tetrades) carried along a time-like curve (not necessarily a geodesic). For finite regions of spacetime the usual definitions of the split lead in general indeed to non-Riemannian geometries (as written in the Wikipedia article on the usual rotating coordinates you discussed above).

 

[Dale]

The principle stated by Mach is quite vague

This is my primary objection to Mach's principle. It is a gem of physics: pretty but useless.

 

[PAllen]

"Space in the sense of a timeslice" has to be very carefully defined in this context. As I noted in my last post, the "space" to which the non-Euclidean metric applies in the case of the rotating disk is a quotient space; it is not a spacelike slice of the spacetime.

In some special cases, the "space" of an accelerated observer can be defined as a spacelike slice of the spacetime--for instance, in the case of Rindler observers. But the space in that case is flat (Euclidean). I'm not aware of any case of an accelerated observer's "space" in flat spacetime being both curved and definable as a spacelike slice of the spacetime. In curved spacetime of course there are such cases, for example observers following orbits of the timelike Killing vector field in Schwarzschild spacetime, who are accelerated and whose "space" is a slice of constant Schwarzschild coordinate time and is indeed non-Euclidean.

An analog is the Milne foliation of a quadrant of Minkowski space. It is the hypersurface orthogonal foliation for the isotropic, homogenous Milne congruence. Each slice is hyperbolic. However, it is the natural foliation of the congruence, rather than corresponding to one observer (each observer of the congruence being inertial).

 

[Jonathan Scott]

This is my primary objection to Mach's principle. It is a gem of physics: pretty but useless.

It is a very attractive gem and I find it hard to believe there isn't something important in it, even through it is difficult to see how it can be consistent with the experimental evidence.

I noticed long ago that Newtonian gravity could be approximated on the scale of the universe by simply taking ratios of sums of m/r as the potential, in which case G simply became an abbreviation for the sum for the "rest of the universe", and even scales of mass and distance simply become relative (so e.g. if all masses were doubled, the physics would be unaffected). When I posted to a newsgroup about that something like 30 years ago, someone kindly referred me to Machian theories and to Sciama's beautiful paper "On the Origin of Inertia" which relates to the same relative concepts, and I followed up Machian articles from various other sources such as Julian Barbour (I met him at a local physics conference many years ago).

The usual main consequence of such ideas is that some form of the Whitrow-Randall relation would hold, which means that the sum of "Gm/rc^2" for everything in the universe as seen from any point is a constant of order 1. This also relates to Dirac's Large Numbers Hypothesis. This effectively defines G as a slightly variable quantity, and experimental evidence does not support any variation in G with time or location (although "G due to all other masses other than the central one under consideration" is still effectively constant over a local region). There are however various ways (so far none of them very satisfactory) in which a Machian theory could result in locally or temporarily constant G.

An alternative way of expressing this is that we would expect the metric to be to a good approximation defined by some function of the scalar ratio of the sum of m/r for all masses in the universe as seen at some reference point and at the local point. This isn't as simple as it seems at first sight, as the values of m and r are normally slightly dependent on the coordinate system, and what we need is something which works globally.

Brans-Dicke theory effectively attempts to create a Machian theory by adding a controlled amount of "Machian-ness" to GR, but unfortunately the best fit occurs when the controlled amount is set to zero.

I would happily dismiss all of this as being irrelevant to the real universe if it were not for the fact that GR theory reproduces the same Machian effects of rotational and linear frame dragging, and the rotational aspect has been experimentally confirmed (although unfortunately with low accuracy) by results from GP-B. When mass estimates are extrapolated to the size of the observable universe, it seems that the Whitrow-Randall relation holds at least to around the right sort of order of magnitude for this to be a physical explanation of inertia and rotational effects in terms of relative motion (as propagated through a field; I don't subscribe to the ideas of Woodward and his followers that Machian effects should be non-local). However, the fact that according to that scheme moving one local mass would effectively change the value of G means that in that simple form it is not compatible with General Relativity nor with experiments showing that G is constant.

 

[stevendaryl]

I think that to make sense of Mach's principle, you have to formulate your theory in terms of equations of motions of objects or particles with definite, localized positions at all time. This excludes fields as fundamental components of your theory. (For example, Feynman with his "absorber" theory of electrodynamics managed to eliminate the electromagnetic field as an independent entity.) If all the equations of motion are in terms of particle locations, then the equations are Machian if they only involve relative positions of the form \vec{r}_n - \vec{r}_m, where \vec{r}_n is the location of the n^th particle. General Relativity is not Machian in this sense, because the equations of motion for one particle involves the metric tensor (and it's derivatives), not the relative positions of the other particles. (As a matter of fact, there is no coordinate-independent way to make sense of a difference of two positions in curved spacetime.) For some solutions of the field equations of General Relativity, it might be possible to integrate out the metric tensor in terms of particle locations, in the same sense that Feynman's theory integrates out the electromagnetic field. So those particular solutions might be Machian. Julian Barbour discusses this in his book "The End of Time". I don't quite understand the motivation in wanting to eliminate fields in favor of particle positions, but I think that it's necessary if you want to make sense of Mach's principle.

 

[Jonathan Scott]

I think that to make sense of Mach's principle, you have to formulate your theory in terms of equations of motions of objects or particles with definite, localized positions at all time. This excludes fields as fundamental components of your theory.

There are many ways to use Mach's Principle, and I do not see any need to assume anything non-local (that is, involving "faster than light" communication or direct "action at a distance"). I have tried out various ideas with considerable success, but they usually disagree with GR and experiment at the Post-Newtonian level, and even when they don't there's usually some other problem. This isn't the place to discuss speculative theories, but I find the general Machian idea that gravity might be an entirely relative effect which also explains inertia, mass, space and time to be extremely compelling. I also have other speculative personal reasons (related to the universality of the mass scale of elementary particles and the Large Numbers Hypothesis) to expect gravity to be Machian. For me, the big question is not whether gravity is Machian but rather how it manages to hide its Machian nature from both theoretical and experimental investigations!

 

[stevendaryl]

There are many ways to use Mach's Principle, and I do not see any need to assume anything non-local (that is, involving "faster than light" communication or direct "action at a distance").

Well, at first glance (or first think), Mach's principle is saying that only relative motion is meaningful. So rotation or acceleration of a single object doesn't mean anything--the only meaningful notion is rotation or acceleration relative to some other object (the "fixed stars", for example). But if there is no action-at-a-distance, then how can the rotation or non-rotation of distant stars be relevant here? The answer given by General Relativity is that spacetime itself is an entity that provides a reference for acceleration or rotation. Rotation or acceleration is not measured relative to the distant stars, but relative to the local geodesics determined by the metric tensor. So I would say that GR fails to satisfy Mach's principle.

A way to see this is that Mach's principle would say that without the fixed stars to provide a reference for acceleration and rotation, a single rigid object cannot be said to be rotating. However, there are solutions of GR that consist of a single rotating star in an otherwise empty universe. This is a different solution than a single nonrotating star. So rotation makes a difference, even in the absence of distant stars to provide a reference for rotation.

 

[Jonathan Scott]

I totally agree that if spacetime is assumed to be able to exist without the masses that would define it according to Mach's Principle then any such model doesn't satisfy Mach's Principle, and that includes hypothetical GR solutions where the contents of the universe are anything other than the actual contents.

In my preferred model, the shape and scale of space and time is defined by the wave functions of all masses, and how those propagate is defined by the shape of space and time, so the local observation of a remote value of m/r is defined by the frequency of the wave for m and parallax of the wave fronts for r.

I don't think I can get any further into this without getting into details of speculative personal theories, which aren't allowed here.

 

[stevendaryl]

I totally agree that if spacetime is assumed to be able to exist without the masses that would define it according to Mach's Principle then any such model doesn't satisfy Mach's Principle, and that includes hypothetical GR solutions where the contents of the universe are anything other than the actual contents.

In my preferred model, the shape and scale of space and time is defined by the wave functions of all masses, and how those propagate is defined by the shape of space and time, so the local observation of a remote value of m/r is defined by the frequency of the wave for m and parallax of the wave fronts for r.

I don't think I can get any further into this without getting into details of speculative personal theories, which aren't allowed here.

Fair enough, as long as we agree that it isn't GR that you're talking about, but some possible extension or correction or replacement to GR.

 

[PeterDonis]

I totally agree that if spacetime is assumed to be able to exist without the masses that would define it according to Mach's Principle then any such model doesn't satisfy Mach's Principle, and that includes hypothetical GR solutions where the contents of the universe are anything other than the actual contents.

But even in our actual universe there are regions where there are no masses but only vacuum--yet it is not "the distant stars" that immediately determine which states of motion are inertial and which are not in such regions, but the local geometry of spacetime (at least according to GR). So for the interpretation of Mach's Principle that you appear to be using, even the GR solution that describes our actual universe does not satisfy Mach's Principle. In other words, I don't think the interpretation of Mach's Principle that you appear to be using only requires that spacetime can only exist if there are masses somewhere in it; I think the interpretation you are using requires that there is some invariant meaning to "the distance from here to distant masses" at every event, even events in the middle of a vacuum region. GR does not satisfy that requirement.

 

[Dale]

It is a very attractive gem and I find it hard to believe there isn't something important in it, even through it is difficult to see how it can be consistent with the experimental evidence.

In science experimental evidence beats attractiveness and belief.

Brans-Dicke theory effectively attempts to create a Machian theory by adding a controlled amount of "Machian-ness" to GR, but unfortunately the best fit occurs when the controlled amount is set to zero

To me this is a strong reason to reject Mach's principle, regardless of its philosophical appeal.

 

[Jonathan Scott]

But even in our actual universe there are regions where there are no masses but only vacuum--yet it is not "the distant stars" that immediately determine which states of motion are inertial and which are not in such regions, but the local geometry of spacetime (at least according to GR). So for the interpretation of Mach's Principle that you appear to be using, even the GR solution that describes our actual universe does not satisfy Mach's Principle. In other words, I don't think the interpretation of Mach's Principle that you appear to be using only requires that spacetime can only exist if there are masses somewhere in it; I think the interpretation you are using requires that there is some invariant meaning to "the distance from here to distant masses" at every event, even events in the middle of a vacuum region. GR does not satisfy that requirement.

The scale and shape of space and time is determined by the sum of m/r which is generally dominated by distant masses everywhere in the observable universe (except when very close to compact objects), regardless of how empty it is locally.

GR requires additional boundary conditions to match it to the actual universe, and it does not give a reason for the actual value of G. For the actual universe, the sum of Gm/rc^2 is of order 1, which is very Machian and suggests a relationship between G and the distribution of mass in the universe, which might for example turn out to be related to such boundary conditions via additional physical constraints on GR or something similar.

 

[vanhees71]

In GR the full energy-momentum-stress tensor is universally coupled to the gravitational field, not only mass, and gravity is not simply described by a potential which goes like $1/r$ as in Newtonian gravity theory. I don't think that the original version of Mach's principle can be accommodated with the local field-theory picture that's provided by Einstein's theory. Of course, Mach's principle is vague enough so that you can reformulate to match in some sense GR, but what's the merit of this?

 

[Jonathan Scott]

In science experimental evidence beats attractiveness and belief.

It isn't that Machian theories can't match the experiments (specifically to give the same PPN $\beta$ as GR and to give the appearance of constant $G$) but rather that none of the ways of doing so seems particularly natural at present.

GR is neat but it is only part of a theory, requiring additional constraints and parameters. Machian theories also naturally extend to the scale of the universe and explain inertia, mass and the value of G as relative effects with no additional parameters. So in many ways the Machian theories seem to give more value for less complexity. However, the simplest Machian theories are too simple in some ways theoretically and also conflict with experiment, so one needs to fill in some more detail. These problems are only at the post-Newtonian level, relating for example to non-linearity due to the gravitational effect of potential energy. Although one can arbitrarily choose parameters to match experiment, this undermines the compelling simplicity of the Machian theories.

 

[vanhees71]

How do you come to the conclusion that "GR is neat but it is only part of a theory, requiring additional constraints and parameters"? I don't know any example, where GR fails to predict observations. So where does it need additions?

 

[Jonathan Scott]

How do you come to the conclusion that "GR is neat but it is only part of a theory, requiring additional constraints and parameters"? I don't know any example, where GR fails to predict observations. So where does it need additions?

Firstly, GR solutions require assumed boundary conditions, even at the scale of the universe.
Secondly, in GR the constant G is effectively an arbitrary parameter which is matched to experimental observations.

 

[vanhees71]

Any field theory includes appropriate initial and boundary conditions, and of course $G$ is a parameter which is mathced by experimental observations. So what?

 

[Jonathan Scott]

Any field theory includes appropriate initial and boundary conditions, and of course $G$ is a parameter which is mathced by experimental observations. So what?

In basic Machian theories, $G$ is effectively an abbreviated way of representing the relative distribution of mass in the universe, rather than being an arbitrary parameter. For boundary conditions, I can't be so specific without getting into speculative details, but I feel that without additional rules for the boundary conditions GR has exotic but implausible solutions which simply don't arise in Machian theories because the boundary conditions are simpler.

The simplest Machian theories clearly diverge from GR in one important aspect; they don't quite lead to black holes, in that the metric factor for the time dilation is not of the form $\sqrt{1-2Gm/rc^2}$ but rather typically approximately $1/(1+Gm/nrc^2)^n$ where $n$ depends on the specific theory (and this expression is not necessarily in Schwarzschild coordinates).

 

[Dale]

For boundary conditions, I can't be so specific without getting into speculative details,

This seems like a good point and a good reason to close the thread.

IMO, your defense of Mach's principle sounds like other people's defense of the luminiferous aether. If there arises any compelling scientific evidence to support it then we can open a new thread at that time.