# Mach's principle

1. Jul 23, 2011

### 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?

2. Sep 13, 2016

### 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.

3. Sep 13, 2016

### Staff: Mentor

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.

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.

4. Sep 29, 2016

### 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.

5. Sep 29, 2016

### 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."

6. Sep 30, 2016

### Staff: Mentor

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.

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.)

7. Sep 30, 2016

### 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.

8. Sep 30, 2016

### PAllen

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.

9. Sep 30, 2016

### Staff: Mentor

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.

10. Sep 30, 2016

### PAllen

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.

11. Sep 30, 2016

### Staff: Mentor

Yes.

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.

12. Oct 1, 2016

### Elemental

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?

13. Oct 1, 2016

### Staff: Mentor

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.

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"

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.

14. Oct 1, 2016

### 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.

15. Oct 1, 2016

### Staff: Mentor

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

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 Those do get some review, but they are not peer-reviewed research in the usual sense.

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.

16. Oct 1, 2016

### 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).

17. Oct 1, 2016

### Staff: Mentor

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":

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.

18. Oct 1, 2016

### Staff: Mentor

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

Code (Text):

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

19. Oct 1, 2016

### 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.

20. Oct 1, 2016

### vanhees71

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.