Newton's Bucket

This thought experiment is originally due to Sir Isaac Newton. We have a sphere of water floating freely in an opaque box in intergalactic space, held together by surface tension and not rotating with respect to the distant galaxies. Now we set the box and water to rotate about some axis and we notice that the sphere flattens into an oblate spheroid.

How does the water know it’s spinning?

NOTE: Newton thought this proved the concept of absolute rotation with respect to a preferred spatial frame of reference. Perhaps these days we can do better, or different?

  • $\begingroup$ Just to be clear this is a question about (the bucket in) General Relativity? $\endgroup$ – Roy Simpson Jan 27 '11 at 13:17
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    $\begingroup$ For complete clarity, could I request people to address the following issue? In a universe which was empty of all matter except for the sphere of water, does General Relativity allow us to distinguish between a "non-rotating" vs. "rotating" state of the sphere? If yes, why doesn't this re-introduce a concept of absolute space? $\endgroup$ – Nigel Seel Jan 27 '11 at 18:48
  • $\begingroup$ Won't the water stick to the inside of the box and freeze due to lack of pressure? $\endgroup$ – Cees Timmerman Oct 27 '12 at 14:53

Dear Nigel, Newton had to postulate an absolute space. In fact, he used his physics insights to support the idea of a "spirit" that is filling the space - a paradigm this greatest scientist and a devoted Christian was as passionate about as about physics itself. The absolute space determined geometry everywhere except that it didn't know about any preferred velocity; it only knew about preferred accelerations.

Inertial systems in classical physics

Newton's laws of physics were valid in inertial frames only. If the laws have the usual forms in one frame, one can show that they also have the same form in all frames that are moving by a constant speed in the same direction. But one can also show that the form of the laws changes if we switch to a different system that is accelerating or spinning because this system is not inertial.

The difference between inertial and non-inertial frames is surely a basic postulate of classical mechanics and it is one that is extremely well established by the experiments, too. Newton's bucket is one of the simple ways to show that rotating frames and non-rotating frames simply differ, so the hypothesis (assumed in between lines of your question) that there is a "complete democracy" between all frames, regardless of their rotation, is instantly falsified.

Special relativity

Similar "absolute structures" filling space and time survived in relativity as well, despite Einstein's original fascination with the so-called Mach's principle that de facto wanted to deny that the rotating bucket behaves differently than the non-rotating one. General relativity ultimately rejected Mach's principle even though one may see some individual effects - memories - predicted by general relativity that are similar to those discussed by Mach.

In special relativity, there exists a "metric tensor" in the whole spacetime that tells all the buckets - and all other objects - whether they're rotating (and accelerating) or not. If they're not rotating, the metric will be given by $$\eta(x,y,z,t)=\mbox{diag}(-1,+1,+1,+1)$$ I chose the sign convention randomly. However, if one transforms this metric to a frame that is inertial - it is spinning or accelerating - the metric tensor will be transformed into a different one, namely a set of 10 non-constant functions.

General relativity

The very same thing is true in general relativity where the metric tensor becomes dynamical and may be curved by the presence of heavy objects. It is still true that the metric in non-rotating frames will be given by $$ds^2 =-c^2dt^2+dx^2+dy^2+dz^2$$ which is just a different way of writing the metric $\eta$ a few lines above. However, if one transforms this metric tensor to a spinning frame, one gets a different metric tensor. The deviation from the flat space metric may be interpreted as a "gravitational field". The equivalence principle guarantees that the effect of gravitational fields is indistinguishable from the effect of inertial forces resulting from spin or acceleration.

So the extra corrections in the metric tensor will know all about the centrifugal, centripetal, and Coriolis forces that are responsible for the modified shape of the water surface, among many other effects.

To summarize, the bucket - and all other objects - know how to behave and whether they're spinning because they interact with the metric tensor that fills the whole spacetime and that allows one to distinguish straight lines (or world lines) from the curved lines (or world lines) at any point. It's important to realize that the metric tensor, while it allows to distinguish accelerating (curved) lines from the non-accelerating (straight) lines, can't distinguish "moving objects" from "objects at rest". This is the principle of relativity underlying both Einstein's famous theories but in this general form, it was true already in Newton's mechanics - and realized by Galileo himself.

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    $\begingroup$ So what you're saying is that in the spinning case, the molecules of water (apart from those on the rotation axis) are not following geodesics due the inter-molecular forces keeping the spheroid intact. And that's the difference? $\endgroup$ – Nigel Seel Jan 27 '11 at 12:33
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    $\begingroup$ Dear Nigel, if you search through my comment, you will see that there is nothing about molecules of water or intermolecular forces, so it's pretty self-evident that I am not saying what you're attributing to me, or isn't it obvious? Isn't it better to read the answers instead of trying to force the people to say completely different answers that you could a priori prefer? $\endgroup$ – Luboš Motl Jan 27 '11 at 12:47
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    $\begingroup$ I answered your question how the bucket figures out whether it's spinning or not. It determines it from the metric tensor that fills the space. If it is not spinning, the laws of physics are simple and dictate that the surface of the water is flat, to minimize the water's potential energy. If the bucket is spinning, there are extra centrifugal forces - that may be derived from the metric tensor transformed to the spinning bucket's frame. These centrifugal forces grow away from the axis and when combined with the gravitational "down" force, they lead to the curved shape of the surface. $\endgroup$ – Luboš Motl Jan 27 '11 at 12:50
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    $\begingroup$ So the molecules of water "know" they're spinning because they're not in a local inertial frame (easy to find out). In their own frame of reference they figure out a metric tensor which works and it has all those extra components you mentioned. So they figure out they're in some kind of force field (centrifugal or weird cylindrical gravitational field as mentioned by kakemonsteret). Hopefully I've now understood your reply. $\endgroup$ – Nigel Seel Jan 27 '11 at 13:41
  • $\begingroup$ Thanks a lot, Nigel, much better! By the way, the surface will be flat until the water really orbits, i.e. until it is pushed by the centrifugal force. Similarly, if you stop the bucket, you must wait until the water itself stops, and only then it recovers the flat surface. Take the latter example: the surface is concave for a while. But gravity is eventually able to make the surface flat because of forces you can understand. The situation is fully analogous when the surface is getting concave: it may be flat to start with but the rotating water sees a different, concave, "natural surface". $\endgroup$ – Luboš Motl Jan 27 '11 at 14:55

The answers that have already been posted are correct, but @kakemonsteret raises a followup question in the comments that's worth addressing:

Lets say you are spinning somewhere in outer space, can you know you are spinning, ie can you rule out that the forces you feel are not caused by a mass distribution somewhere ?

This question is getting at some ideas about Mach's principle and its relation to general relativity, which is a somewhat complex subject. But there is a well-known effect in general relativity that bears directly on this question: the Lense-Thirring effect.

Imagine a large spinning spherical massive shell. The local inertial reference frames inside the shell will be "dragged" around by the rotating mass, so that they rotate with respect to the "fixed stars" (i.e., the inertial reference frames far outside the shell). So if you lived inside this shell, and felt like you weren't rotating, you would "really" be rotating relative to the fixed stars. If you then started turning the opposite way at just the right rate, you could make it so that you weren't "really" rotating relative to the fixed stars, but you felt like you were.

I put scare quotes aroung the word "really" there for a reason: in general relativity, the most natural meaning to ascribe to the phrase "really rotating" is "rotating with respect to your local inertial frame" -- that is, if you feel like you're rotating (or if your Newton's bucket indicates you're rotating), then you are. But if you define "really rotating" to mean rotating with respect to very distant inertial objects, then yes, you can feel like you're rotating, even when you're not "really rotating," due to being surrounded by lots of spinning mass.

Needless to say (I presume), this is all very much in-principle stuff: the frame-dragging effect is very small in practice.


A synopsis of what Lubos wrote: It is possible to tell what an inertial reference frame is, locally (in an infinitesimal neighborhood of any point of spacetime), with respect to the local gravitational field (it's the one that is "freely falling"). The bucket "knows" that it is rotating because it rotates with respect to the local inertial frame, that is because it "rotates relative to the local gravitational field".

  • $\begingroup$ Can the bucket know whether it is rotating or if there is some large cylindrical mass distribution far out somewhere ? $\endgroup$ – TROLLHUNTER Jan 27 '11 at 12:49
  • $\begingroup$ Sorry, I don't understand your question. In GR we have a spacetime, it is a solution to the Einstein field equations. Any mass distribution influences this solution, therefore any mass will influence the local gravitational field and therefore will influence what the local reference frame is (unless, of course, it does not because the mass and the point we are looking at are for ever spacelike separated, for example). $\endgroup$ – Tim van Beek Jan 27 '11 at 14:08
  • $\begingroup$ Lets say you are spinning somewhere in outer space, can you know you are spinning, ie can you rule out that the forces you feel are not caused by a mass distribution somewhere ? $\endgroup$ – TROLLHUNTER Jan 27 '11 at 14:33
  • $\begingroup$ Where would the coriolis force come from if you had the cylindrical mass distribution as the field source? Perhaps that's the mechanism by which you could tell the difference? Now, I wonder what would happen if your external massive cylinder was spinning .... $\endgroup$ – Nigel Seel Jan 27 '11 at 16:54
  • $\begingroup$ (I see Ted Bunn has picked up this point in his answer). $\endgroup$ – Nigel Seel Jan 27 '11 at 17:01

If this is really about General Relativity then this is a description of a rotating body in General Relativity. As this question impinges on some issues in my General relativity Stack questions I shall make a few remarks about it.

General Relativity provides a solution (which we can think of for now as a generalised metric) given some conditions: usually matter conditions. Your basic scenario is of a bounded perfect fluid which is understood in GR. The other aspect of your condition is that it is rotating. A google search shows that for bounded matter a full rotational GR solution may not yet be known. The mechanism by which it is studied is that of a perturbation of a non-rotating solution based around the Schwarzchild solution. This is the model for a rotating star and so on.

The matter in the solution will follow the metric, which has curved spacetime inside and around the fluid. The surface tension and other things like that are meant to be incorporated in the Stress-Energy Tensor: if those features are present the fluid isnt perfect, and so another layer of approximation is used in practice.

I believe that this rotating scenario also generates (weak) gravitational waves!

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    $\begingroup$ Newton is no doubt spinning in his grave! $\endgroup$ – Nigel Seel Jan 27 '11 at 14:35

The reason that the Bucket knows that it is spinning is that the Universe has a horizon in much the same way that the Earth has a horizon. The Earth's Horizon is curved line that is there not only because of Perspective Geometry, but also because the Earth is curved. Because the Earth has a two dimensional surface the height tilts back away from the observer, so no matter how tall the height, all heights will tend to an line, which we call the Horizon. The surface of the Universe has a three dimensional surface, the height is called "time" and the curvature of the Universe causes all lengths of time to disappear and become a "red-shift" that we call the CMB. The CMB is not really the "Event Horizon", nor is it the start of the "Big Bang". It may also be those things, but that is not the importance of the CMB. The Bucket knows it is rotating, because the Universe has a definite Up and Down, or Inside and Outside (which is probably a more accurate description). This is a very ingenius solution to the problem, and proves that the current theory is wrong, and gives a reason for Quantum Mechanics as well. The other type of spin is also the direction showing that the Universe has a definite outside, but it is a second dimension of time--that dimension of time is set on a Negatively Curved surface, unlike the one we are used to, which from the argument above is set on a Positively Curved surface (which gives the CMB). The negatively curved surface gives rise to Non-Laplacian statistics, since any given point can come from an infinite number of timelines, whereas the Positively curved surface gives rise to a time that is eerily similar to classical Electrodynamics. There is probably only one surface, which has time "vectors" on both sides, one side is Positively curved the other Negatively curved. This is possible in a curved three space.

  • $\begingroup$ Have you any evidence for anything you've written here, is it all just person feelings? $\endgroup$ – Kyle Kanos Dec 16 '14 at 20:04
  • $\begingroup$ "This is a very ingenius solution to the problem" Right or wrong, I have to disagree with this point. If it's wrong, then the reason for disagreeing is obvious, if it's right, then I would call it novel, insightful, maybe even inspired. However, for a solution to be ingenious, let alone very ingenious, you would have to have a way of formulating it mathematically and it would have to generate testable predictions. As it stands, this is an unpublished personal theory that seems more speculative than scientific. Be it right or wrong, it's poor science $\endgroup$ – Jim Dec 17 '14 at 13:48

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