What is the Earth truly rotating about/revolving around?

Question

Earth rotates on its axis and revolves around the sun, the sun revolves around the galaxy, the galaxy is also moving. So Earth's net rotation as observed from a fixed inertial frame consists of all these contributions (and is rather complex).

Now a Foucault pendulum on earth is supposed to tell the experimenter whether the earth is rotating or not. See a recent question on this forum Proof that the Earth rotates? Basically for a Foucault's pendulum, the plane of the oscillation of the bob rotates, as the Earth rotates. But, the Foucault's pendulum does not single out one type of rotation, it gets affected by all. Therefore the rotation of plane of oscillation can be used to measure rotation of Earth around its own axis, around the sun, everything.

But the question is, around what is the Foucault pendulum measuring Earth's rotation with respect to? That is, finally what is the Earth rotating around/revolving around? Isn't it intriguing that the effect of the entire Universe on Earth can be measured by a pendulum?

I understand that the effect on the pendulum due to say the revolution of solar system around the galaxy is going to be small, but still, it can be measured say in an experiment going on for 10 years. You can keep on increasing the accuracy by measuring for longer periods.

Answer 1

Your question will eventually lead you to Mach's Principle. It is an old, yet unsolved question, that still remains at the stage of "philosophical idea".

I understand that your question is equivalent to "What would be found if we could measure all effects on the pendulum with infinite accuracy?", what if even the tiniest contributions could be registered? (Please read the note at the end as well, regarding the effect on any pendulum of the proximity of mass, whether that pendulum is in a free-fall orbit or not. The effect of earth's orbital motion is not zero because it affects the speed rate of proper time)

Yes, some components of the acceleration on the pendulum allow to deduce that the pendulum belongs to a rotating frame. That leads to think that the pendulum and the whole Universe may eventually be found to be rotating around some point, but that idea makes no sense (what is that point then, if everything is rotating? Rotation relative to what?). Then Mach's principle comes to the rescue, telling us that inertia effects on your pendulum arise somehow from the influence of all the other objects of the Universe, from here to the most distant ones. But there is no mathematical model for such thing, not even in General Relativity.

The pendulum is blindly affected by the local conditions of space and time, which constantly change in time and from one point to another (although all effect other than those arising from the rotating frame on top of the bulk mass of the Earth are extremely tiny). Those conditions are determined by the arrangement of energy/mass and momentum around. In the newtonian model, by the mass distribution. This is useful because you can idealize a portion of the Universe in a model that allows you to predict some behaviour of the system: for instance the Schwarzschild metrics allow to accurately synchronize the clocks of the GPS satellites in their motion around the Earth, and to accurately model orbits close to the Sun. The homogeneous and isotropic Universe model allows to derive properties of the expansion in the past, etc. But there is no model for an accurate description of how the whole universe is affecting your pendulum.

In other words, the essential origin of inertia is still unknown. What is a Foucault pendulum eventually rotating around? There is no answer to that question. Moreover, it is not yet clear whether the question makes sense or not.

The most close answer to your question may be found in our motion relative to the Background radiation, found by means of the dipole anisotropy of the CBR. This is the closest thing that there is, to an "absolute reference frame" but it makes sense only for us. Other distant observers in out expanding Universe will have a completely different perception.

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EDIT:

As correctly stated by Ben Crowell, the orbital motion is a free fall, and therefore its dynamical effects on the pendulum are different from those of being on top of the rotating Earth. However, that free fall happens along places with different values of the gravitational potential (bigger in January, for instance) and therefore the speed rate of pendulums is affected. Thus, your pendulum, as any other clock-alike device, is affected by all the other masses in the Universe.

You might think about placing several synchronized pendulums at different distant points on the surface of the Earth and, by measuring (with infinite accuracy) their speed rate differences, map some properties of the gravitational potential in which you are embedded, deducing for example the direction of a center of mass. This makes an interesting question if you want to start another post.

As for Mach's principle, let me stress that it is merely a philosophical idea, that may or may not some day lead to a real theory. It is neither correct nor incorrect.

There is often a fallacy motivated by the Equivalence Principle, in which people ignore the different speed rate of proper time inside the free-falling elevator. Yes, the man inside the free-falling elevator is unable to distinguish if he is in a gravitational field (but in free fall), or if he is floating in interstellar space, far away from any mass. But in the second case, the man inside the elevator is ageing faster that the one that is in free fall (orbit) around the Sun. This is another kind of twins paradox that is often forgotten.

Answer 2

The effect on a Foucault pendulum due to the earth's orbital motion is not just small, it's zero. You can't analyze the orbital motion the same way as the rotation of the earth about its axis. The earth free falls around the sun. A Foucault pendulum isn't free-falling around the center of the earth.

One way of stating the equivalence principle is that in a local experiment in a free-falling laboratory, you can't detect any effect of gravity. "Local" means that all such effects approach zero as the size of the experiment approaches zero.

None of this requires Mach's principle (which is basically false, in the sense that the Brans-Dicke $\omega$ parameter is large, see Is Mach's Principle Wrong? ). In fact, none of it even requires general relativity.

What we can theoretically detect on earth as a result of gravitational effects from distant bodies is small tidal effects. These are not "local" in the sense defined above. In practice, I don't think any tidal effects can be detected from bodies outside the solar system.

To see any other effects from the gravity of bodies outside the solar system, you need to do completely non-local measurements, such as measuring the sun's acceleration relative to some other galaxy.

Answer 3

There's a nice subtlety that hasn't been addressed yet. First of all, a Foucault pendulum is influenced by

• The rotation of the Earth around its axis.
• The change in the orientation of the Earth's rotation axis: precession, nutation and long-term changes in the obliquity of the axis.

In principle, it is also influenced by changes in the local gravitational field of the Earth, but let's ignore those. Also, a pendulum is way too small to experience tidal effects.

The question is: in which reference frame should we describe the rotation of the Earth and its changes in axial orientation? The answer is: in the Earth's local inertial frame. As Ben Crowell pointed out, the orbit of the Earth around the Sun is a free fall, so the Earth's orbital motion is in fact a local inertial frame. In other words, to an observer on Earth, the Sun has no effect on the pendulum.

But it doesn't answer the question what the orientation of this inertial frame is. One would think that it is fixed with respect to the distant stars. This is almost correct. The International Celestial Reference Frame (ICRF) is a reference frame fixed with respect to the positions of 212 extragalactic sources (mainly quasars). However, this frame is centred at the barycenter of the Solar System. There is an equivalent reference frame, fixed to the same sources, and centred on the Earth: the so-called Geocentric Celestial Reference Frame (GCRF).

The GCRF however is not an exact inertial frame, due to subtle general-relativistic effects: the Geodetic effect and the Lense–Thirring precession. Both these effects change the orientation of an object orbiting a central mass, causing it to precess (the geodetic effect is due to the presence of the central mass, the Lense–Thirring precession is an additional frame-dragging effect if the central mass rotates about its axis). These effects have been measured by the Gravity Probe B, demonstrating that gyroscopes precess as they orbit the Earth.

There are two consequences:

• The Earth causes a geodetic effect and Lense–Thirring precession, influencing a Foucault pendulum (because the daily motion of the pendulum can be thought of as an orbit). The wiki article on the Lense–Thirring precession mentions that these cause an extra precession of a pendulum of 1 degree in 16000 years.
• The Sun causes a geodetic effect and Lense–Thirring precession on the Earth itself. Since this precession is a gravitational effect, it has to be taken into account when defining an inertial frame. In other words, the Earth's local inertial frame is precessing with respect to the distant stars. However, the effect of the Sun is the same for everything on the Earth, so it doesn't cause an extra precession of a pendulum with respect to an observer on Earth.

Answer 4

I hope that this simplistic attenpt at answering in an elementary way will nevertheless be an acceptable one, at least as first approximation. If I am wrong please tell me where and why. Several answers analyze in very fine details what influences the Foucault pendulum motion, but I am under the impression that the arguments are more complex than they should be, at least to answer the question asked, and they leave me ill at ease. In particular, I do not see what gravitation and free fall have to do with the problem (given our ignorance of the deep reasons for the working of the Foucault pendulum, and excepting of course the analysis of relativistic effects, but that was not really in the question).

A gyroscope (invented like the Foucault pendulum by Leon Foucault to study Earth rotation) is supposed to point always in the same direction (up to precession and nutation), relatively to an inertial frame.

My intuitive understanding of the gyroscope and the Foucault pendulum is that they depend exclusively on inertia. The fact that the device is in a gravitational field or not, or in free fall or not is irrelevant, except for the obvious fact that the original pendulum uses gravity as part of the pendulum mechanism. But this use of gravity is not essential as it can be replaced by springs, as in vibratory gyroscopes and "pendulums".

Our rotating Earth should qualify as a gyroscope, and thus have an axis that is always pointing in the same direction (up to the 26.000 years precession). Indeed, that is what allows people to use stars for navigation.

The freefall situation of the Earth does not seem very relevant for this analysis (outside relativistic effects), except for the fact that it may be the best frictionless gimbals ever invented for a gyroscope (tidal effects that cause precession are not a necessary condition for free fall). Actually I would think that all rotating structures in space will generally keep a stable axis (still up to precession and long term tidal effects).

If a gyroscope is used at the surface of Earth it will only have to move its orientation with respect to the (rotating frame of the) planet, to account for the rotation of the planet and the fixed position of its support on the planet surface. All other rotations are already compensated for (if need be) by the planet itself (as a gyroscope). The same should hold true regarding a Foucault pendulum.

Consider now a Foucault pendulum, or rather its spring equivalent, that is fixed to a small craft with attitude thrusters, so that the orientation of a frame attached to the craft will stay parallel to a frame attached to a point on Earth surface where another identical Foucault pendulum is similarly fixed. I would expect that, up to the relativistic effects described by Pulsar, the behavior of the (spring) Foucault pendulum in the craft is the same as that of the corresponding pendulum on Earth. The only thing that matters is the rotation of the Foucault pendulum frame with respect to an inertial frame.

The experiment could actually be tried on Earth, using small electrical motors to move the orientation of a spring Foucault pendulum in order to mimic the orientation of the local frame attached to another point of the planet surface.

Since the only rotation of (the gyroscope) Earth with respect to an inertial frame is its own rotation (with precession and nutation), this is all the rotation the Foucault pendulum will ever measure (up to relativistic effects, some of which are due to rotations not invloving Earth, such as the Sun own).

Answer 5

what is the Earth rotating around/revolving around?

Others have given the long answer. Here's the short one:
The rotation is in the frame where the directions to quasars are at rest.

Quasars are the most distant object we can observe (due to their enormous luminosity), have no measurable apparent motion (unlike stars in the milkyway), and for all purposes appear point-like (unlike galaxies).

Quasars make for the best reference we can calibrate our positional measurements against. It is the method of choice for the most delicate of measurements, such as Gravity Probe B.

Answer 6

It's all an approximation, isn't it ? And it's all about scale. Everything is affected by "Mach's" gravity, say $G$. But if we break it down to what is causing this gravity we have something like

$$ G=G( \text{due to Earth})+G( \text{due to other planets in the solar system})+G( \text{due to the Sun})+G( \text{due to other suns in our galaxy})+G( \text{due to the central structure of the galaxy})+G( \text{due to other galaxies})+ \dots $$

where we consider higher and higher length scales. The "strength" of the length scales are not comparable. Starting with Earth's gravity, each length scale creates gravity many orders of magnitude smaller than the next scale, according to the vast distances from us.

In a lab experiment on Earth the Earth's gravity is approximately uniform (which is the crudest approximation for a field) and since it affects everything in the lab we neglect it and focus on differences between stuff fixed on the Earth and things moving wrt Earth.

Moving on to larger scales, in an experiment on a satellite in orbit around the Earth this approximation is not good any more because we can distinguish between points on Earth at which the gravity vector changes by a big amount (e.g. $g \to -g$). Still gravity from the sun and all other higher length scales can be approximated as uniform, because they don't vary much through the satellite's orbit. So we ignore them and consider Earth's non-constant gravity. In fact Earth's gravity is many orders of magnitude greater than all larger length scales.

People into astrology believe that gravity from other planets affects us but if you put it down to numbers, it's many orders of magnitude less than e.g. the moon's gravity.

We can go on in this process to answer questions like what's the sun's angular speed, or even our galaxy's rotational frequency, our local cluster of galaxies and so on, who knows how many levels up...

Answer 7

Foucault's pendulum measures rotation relative to the local spacetime metric which forms the gravitational field in General Relativity. Each observer such as someone standing on the surface of the Earth follows a worldline through spacetime. In GR the gravitational field is represented by the components of a metric tensor. It is possible to use the metric to transport a vector along any world line by requiring that its covariant derivative along the direction of the worldline is always zero. This allows you to define a local non-rotating frame of reference for any observer. In practice you can determine the local non-rotating frame using gyroscopes.

As the Earth follows its worldline round the sun it rotates relative to its local non-rotational frame and that is what Foucault's pendulum shows.

In our solar system and our position in our galaxy gravitational fields are relatively weak and we find that this local non-rotating frame coincides very closely with the global non-rotating frame defined by the directions of distant stars or galaxies. However, the match is not quite perfect and it is possible to detect rotation of the local frame relative to the global frame. This was one of the purposes of the gravity probe B experiment. The observable difference can be accounted for by effects due to the Earth's rotation (and geodetic effect of the satellite's orbit). Any effect from rotation on larger scales is too small to detect.

Answer 8

The framework of space.

First, imagine a universe with only light and no mass to warp its direction of travel. Thus, each beam will travel in a straight line forever (or to the end of the universe whe[nr]ever that is), and we can thus get an idea of the geometry of the framework of space from it.

Second, imagine the earth is put into this universe. We can tell if it is rotating simply by comparing its motion to our light-grid. In other words, if the beams of light appear to repeatedly wind around an axis, the earth is rotating around that same axis. This is very similar to how we can tell that our earth is rotating relative to space close to a full cycle once per day by observing the paths of the stars across the sky. Similarly, our Foucault's pendulum's swing path will be determined by its relationship to the light-grid.

Finally, imagine the sun is put into this universe and the earth is put into orbit around it. Again, the pendulum will still stay fixed to the localized light-grid. To clarify this, let's say that we fixed the Pacific Ocean to always face the sun. In this case, a pendulum at the top of the world would rotate relative to the earth's surface a full cycle about once per year.

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Note that our light-grid is just an attempt at a visual method of gauging the correct general shape of the framework of space. Because masses warp space, an actual beam of light will not be straight across the whole of the universe in the Euclidean sense, as it will bend around stars, black holes, and other masses. Thus, in practice only the behavior of light in your (or Foucault's) localized space is really a good gauge of its orientation.

Answer 9

This is an interesting question and does not have 1 answer (just bear with me). According to Einstein's Theory of Relativity, the speed of an object is relative to the viewer; we can then expand on this idea and say that the direction of an object is also relative to the viewer. So in order to answer the question "What is the Earth truly rotating about/revolving around?", we must first answer from what perspective?

I wonder if you'll find that in a book, because I just came up with this in the spur of the moment on things that I already have knowledge about.

One final note: From your perspective, the earth is staying still. From the perspective of the sun, we are revolving around it with some spin with a moon revolving us with a slower spin. This is one of many ways to interpret this question, so to truly get an answer you are looking for please specify from what perspective.

Answer 10

If you consider the entire Universe then the Earth is rotating/revolving around the center of mass of the Universe. If the Big Bang Theory is true then this point would be the origin of the Big Bang.