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.

  • $\begingroup$ When experimenting on Earth, physicists always idealize that they aren't affected by Sun or the Andromeda (which truly happens) ;-) $\endgroup$ – Waffle's Crazy Peanut May 30 '13 at 13:36
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    $\begingroup$ But if someone carries out an experiment precisely and for long enough all effects will come into picture. $\endgroup$ – Man May 30 '13 at 13:40
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20048/2451 , physics.stackexchange.com/q/3193/2451 and links therein. $\endgroup$ – Qmechanic May 30 '13 at 13:41
  • $\begingroup$ You will not be able to disentangle them with this one measurement, even if you do wait long enough.You can only solve for one unknown if you have one measurement. $\endgroup$ – anna v May 30 '13 at 13:49
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    $\begingroup$ Not really I think. If I plot the $\phi$ of the plane of the rotation of the plane of the bob with time, I will see the discrepancies. I can separate the first order, second order deviations and associate them with different kinds of motion. $\endgroup$ – Man May 30 '13 at 14:00

10 Answers 10


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.


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.

  • $\begingroup$ The answer by Ben Crowell says that the Mach's principle is false? $\endgroup$ – Man May 30 '13 at 15:34
  • $\begingroup$ Mach's principle is neither true nor false, but merely a germinal idea that might lead to a theory, but that hasn't happened yet. What Ben Crowell states it that the motion of the Earth around the Sun is a free fall, thus very different in its dynamical effects on the pendulum, than the motion on the surface of the Earth (not a free fall). He is right about that, but I think nevertheless the motion of the pendulum is affected by that free-fall trajectory, since the proper time at the pendulum will be affected, and that depends on the gravitational potential due to the rest of the Universe. $\endgroup$ – Eduardo Guerras Valera May 30 '13 at 15:49
  • $\begingroup$ Whether you could have for instance several Foucault pendulums, say, in Australia, North America and Africa and by comparing their speed rate (with infinite precision) map the gravitational potential changes and thus derive some sort of curvature center around you are going, I don't know. Ben or any other user might probably know that better than me. That is an interesting question. $\endgroup$ – Eduardo Guerras Valera May 30 '13 at 15:53
  • $\begingroup$ @EduardoGuerrasValer I don't agree with your edit. The Earth's orbit around the Sun is a free fall, which means by definition that it doesn't feel the gravity of the Sun (apart from tidal effects), so there's no seasonal effect. The period of the pendulum is only affected by the Earth's rotation and possible local changes in the Earth's gravitational field, not by the orbit of the Earth. $\endgroup$ – Pulsar May 30 '13 at 16:29
  • $\begingroup$ @EduardoGuerrasValer Actually, I just thought of another possible effect, which does depend on the Earth's orbit. I have to think about it first before I write an answer. $\endgroup$ – Pulsar May 30 '13 at 16:39

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.

  • $\begingroup$ Are you sure its zero? I have distictly read that its not. $\endgroup$ – Man May 30 '13 at 15:32
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    $\begingroup$ A measurement over time (as in over the course of an orbit) in a freely falling reference frame is not a local measurement. A local frame is a coordinate system that is restricted to a small region of space time. For instance if you fall towards the sun, you can do no local measurement at any instant that will confirm you are in a gravitational potential well. However if you take many measurements several minutes apart you might notice the sun accelerating towards you. $\endgroup$ – ejrb May 30 '13 at 15:36
  • $\begingroup$ It is true that the orbital motion is a free fall and thus it is different of the motion around the Earth, and therefore it must be accounted for in a different way. But that elliptical free falls happens along different values of the gravitational potential. Thus, at least the period of the pendulum (as any other clock speed rate) will be affected, as seen by an external observer. There is no acceleration, but there are other effects due to that movement, it is not just zero. $\endgroup$ – Eduardo Guerras Valera May 30 '13 at 15:42
  • $\begingroup$ @ejrb: "A measurement over time (as in over the course of an orbit) in a freely falling reference frame is not a local measurement." This is certainly a natural way of thinking about it relativistically, since time is not separate from space. But in fact the extension of the experiment in time does not have any effect. "However if you take many measurements several minutes apart you might notice the sun accelerating towards you." Observing the sun is not a local measurement. $\endgroup$ – user4552 May 30 '13 at 17:19
  • $\begingroup$ @EduardoGuerrasValera: "Thus, at least the period of the pendulum (as any other clock speed rate) will be affected, as seen by an external observer." The use of the external observer makes it not a local experiment. $\endgroup$ – user4552 May 30 '13 at 17:19

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.
  • $\begingroup$ Very interesting links, thanks (upvoted +1). At the very core of the De Sitter Precession is the same time dilation I address when I think that a pendulum is affected too by the orbital motion of the Earth with respect to the Sun and other more distant masses. Time dilation affects the angular frequency of a gyroscope on board a satellite. But you can move a pendulum along the equator of the Earth, for instance, and then detect anomalies due to "something more" than the potential of the Earth, specially if you compare the time delays along several different closed pathways. $\endgroup$ – Eduardo Guerras Valera May 30 '13 at 20:13

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).


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.

  • $\begingroup$ GPB used IM Pegasi, which is a star within our own galaxy, not a quasar. In any case, this doesn't really address the OP's question, which was specifically about what effects would be observed on local experiments such as the Foucault pendulum. $\endgroup$ – user4552 May 30 '13 at 18:42
  • $\begingroup$ @BenCrowell GBP used IM Pegasi as a guide star; but the guide star's apparent motion was measured against quasar positions. I also question your judgement what my answer answers and what not. At least on the poles, the plane of the pendulum rotates as answered. $\endgroup$ – Jens May 30 '13 at 19:00

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.

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.


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.

  • $\begingroup$ The physics in the last paragraph is garbled. The largest effect detected by GPB was the geodetic effect, which is not related to the earth's rotation. The earth's rotation causes frame dragging, which is a much smaller effect and just barely detectable by GPB. The geodetic effect is also not the same for a frame at the earth's surface as it is for an orbiting satellite. The latter is much stronger, because the effect is proportional to the area enclosed by the orbit and the number of orbits per unit time. $\endgroup$ – user4552 May 30 '13 at 19:33
  • $\begingroup$ OK I should have said "Earth's rotation and orbital geodetic effect" They both contribute to the difference between local and global non-rotating frames that I described. You seem to have ignored these crucial effects altogether in your answer. You seem to have said that they are not just small, they are zero. $\endgroup$ – Philip Gibbs - inactive May 30 '13 at 20:14
  • $\begingroup$ I still don't understand how the final paragraph relates to the question. The question is about the Foucault pendulum, which is a local experiment that measures to what extent the earth's surface is noninertial. GPB was measuring the precession of an inertially moving gyroscope relative to distant objects. They're not at all analogous. $\endgroup$ – user4552 May 31 '13 at 3:57
  • $\begingroup$ @Ben Crowell. Yes it is a local effect, not global, that is exactly what I have said. It is important to state why these things are different because the question indicates confusion about it. $\endgroup$ – Philip Gibbs - inactive May 31 '13 at 9:21

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.

  • $\begingroup$ You're right that motion and velocity is relative however you also didn't really answer (or attempt to answer) the question. $\endgroup$ – Brandon Enright May 31 '13 at 3:47
  • $\begingroup$ I'm sorry you feel that way, but I indeed did answer the question. One does not truly know what an object is rotating, for it differs from person to person; so condemning me for pointing this out and obviously not reading my entire entry is so blatantly absurd; this and your comment are both quite preposterous. $\endgroup$ – Z_Doctor May 31 '13 at 4:11
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    $\begingroup$ I'm not the downvoter. One does know when an object is rotating. A rotating reference frame is not an inertial reference frame and an experiment like the Foucault pendulum demonstrates that. $\endgroup$ – Brandon Enright May 31 '13 at 4:18
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    $\begingroup$ For that I am mistaken, I have made and an assumption and have truly made a sceptical; but let me say this, I'm not one to waste words, but my intention was to answer his question "What is the Earth truly rotating about/revolving around?" It is true I did not mention the Foucault Pendulum for I did not feel it was necessary to answer his question, for though he did mention it he made quite clear that his question was what is the earth rotating around by saying "That is, finally what is the Earth rotating around/revolving around?" I hope there are no hard feelings. $\endgroup$ – Z_Doctor May 31 '13 at 4:25

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...


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.

  • $\begingroup$ Thats a really profound statement! :) $\endgroup$ – Man May 31 '13 at 0:45
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    $\begingroup$ The Big Bang didn't happen at one point. The universe doesn't have a center of mass. $\endgroup$ – user4552 May 31 '13 at 3:45
  • $\begingroup$ @BenCrowell: "Big Bang didn't happen at one point": true. "Universe has no center of mass": proof please? $\endgroup$ – Rody Oldenhuis May 31 '13 at 8:03
  • $\begingroup$ @BenCrowell: Every system of particles must have a COM $\endgroup$ – ABC May 31 '13 at 8:29

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