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Let's assume that there is nothing in the universe except Earth. If the Earth rotates on its axis as it does, then would we experience the effects of rotational motion like centrifugal force and Coriolis force?

The meaning of my question is: is Earth rotating relative to space?

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Related What if the universe is rotating as a whole? – ja72 May 21 '15 at 12:53
up vote 9 down vote accepted

The same Wikipedia article everyone else is citing is a decent reference on this. Basically, we don't know, and probably never will, because we can't put an object in an otherwise empty universe.

Suppose you could, though. So we've got a planet in an otherwise empty universe. To test the hypothesis of absolute rotation, you could do various experiments on the planet's surface to measure the fictitious forces arising from being in a rotating reference frame. For example, you could set up Foucault's pendulum at various locations on the planet and measure the precession rate at each location. (I think you'd need at least 3 locations) From those results you could determine the planet's rotation axis and rotational velocity.

The Newtonian viewpoint holds that yes, rotation is relative to space. If this view is correct, and the isolated planet were rotating relative to space, you would see your pendulums (pendula?) precessing at a nonzero rate, and you could solve for the planet's rotational velocity.

On the other hand, the Einstein/Mach viewpoint holds that rotation is not relative to space, but rather is defined relative to the matter in the universe. If this viewpoint is correct, you would never see any precession of the pendulums because the bulk of the matter in this experimental universe is the planet itself, so it basically defines the frame of zero rotation. In our universe, of course, there is a much larger distribution of matter to define a nonrotating rotational reference frame. Mathematically, this results from a phenomenon in GR known as frame dragging.

The Newtonian/absolute view has the advantage of being kind of intuitive, but it does require that space defines some sort of absolute rotational reference frame. Given that we know all linear motion is relative, it seems odd (to me, and others) that rotational motion could be absolute. In addition, if rotation could be absolute, for any nonzero rotational velocity, a large enough distribution of matter in the universe would require the outer objects to be moving at the speed of light relative to a nonrotating reference frame. This could conceivably be allowed, it would just mean that no matter could be boosted into that nonrotating reference frame, but again, it seems odd. The Einstein/Mach view has the advantage that it makes this "faster-than-light rotation" extremely unlikely as a consequence of the structure of the theory alone.

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David, I must disagree with you on this. There are numerous experiments that would confirm the Earth's rotation: if they all succeeded then we would conclude that the Earth is rotating, if they all failed, we would conclude that the Earth is not rotating. In any case we can measure the rotation, with or without stars and galaxies. If only some succeed then the laws of physics are different from what we know and the question is moot (anything can be invented if we allow physics to change). Also, the rotation is relative to an axis (no rotational velocity is measured at the poles). – Sklivvz Nov 29 '10 at 11:32
@Sklivvz: sorry, but I don't understand your objections. I mean, you're certainly right that there are numerous experiments that we can perform on the real Earth to show that it is rotating, but are you saying that those experiments would necessarily work the same if there were no other matter in the universe? – David Z Dec 2 '10 at 2:15
@David Zaslavsky: Yes, that's exactly what I am saying. Actually when we model those experiments theoretically, we actively exclude the rest of the universe as irrelevant. E.g. when we model Focault's pendulum we only consider gravitational attraction between the Earth and the pendulum, we don't include the moon, the sun etc. We work under the active assumption that the rest of the universe is irrelevant. If this assumption is false, then probably we can't even prove that the Earth is rotating with all the matter in the universe (since our theoretical model has wrong assumptions). – Sklivvz Dec 2 '10 at 8:13
@David Zaslavsky: I am not excluding frame dragging, since it is a measurable effect of the Earth's rotation. Since this effect seems to be an inherent feature of GR and not of the particular solution of our universe, I am quite unconvinced that the rest of the universe makes much of a difference in this context. – Sklivvz Dec 2 '10 at 14:55
@David Zaslavsky: Can you explain this a bit better? The only references to the Lense–Thirring effect on Wikipedia talk about a Kerr solution or a rotating shell solution - neither of these seems cosmological, so I don't understand why you need that reference frame. – Sklivvz Dec 2 '10 at 19:53

This is an old question, but it might be possible to put the old saw to rest for good.

If you have a deSitter space, it can't be rotating--- deSitter space is unique. If you have a black hole in deSitter space, it can rotate (this is the deSitter Kerr solution recently discovered), but it is only one body rotation, the cosmological horizon can't rotate independently of the black hole horizon.

This might not be surprising, except that if you make the nonrotating deSitter black hole bigger and bigger, there is a point where the black hole and Cosmological horizon are symmetrical. In this case, you have two horizons. If you rotate one horizon, the other rotates in the opposite sense, so that only their relative rotation is meanigful. The two horizons are symmetric now, so you can't differentiate between their motions.

If you add matter in-between the two horizons, you will curve the universe inbetween, and if you put a lot of static dust in, you get to an Einstein static universe with two black holes at opposite ends. In this universe, the two horizons are clearly matter. So there is no boundary between matter and cosmological horizons, and it is a fair statement to equate all matter with some sort of horizon-object, so that the electron is like a little micrscopic black hole.

This is the point of view most consistent with string theory, since the strings in string theory are dual under strong-weak coupling dualities to objects which are clearly black holes in the classical limit, namely D-branes. The point of view that matter is the same as horizon puts Mach's principle to rest--- all motion is relative to distant "matter", either matter matter or horizon matter, which is also matter.

This statement is consistent with the holographic principle, and the holographic principle can be thought of as the ultimate in Mach's principle, since it says that all motion is relative to a distant holographic screen, so that the whole thing is moving relative to a distant horizon. This principle is more precise, more quantum, and more general than Mach's principle, and it is consistent with the solutions of GR in a space bounded by a horizon. I must say, however, that the deSitter formulation of string theory is not available right now, so that the full holographic principle is not completely known.

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you have an unfinished sentence – anna v Jun 4 '12 at 5:27
You say "If you have a black hole in deSitter space, it can rotate" in the sense measurable with a pendulum? What about the lone planet that the questioner asks? – anna v Jun 4 '12 at 5:32
Erased the dangling sentence. A black hole in deSitter space rotates because it has an ergosphere and all that. A lone planet can also rotate in deSitter, but the point is that the rotation is relative to the cosmological horizon, and this can be considered a form of matter. So the rotation is relative to "matter" in an extended sense that would probably have satisfied Einstein, since all matter is horizons in string theory. – Ron Maimon Jun 4 '12 at 6:03
@RonMaimon, Thanks for your answer. I just want to understand it correctly. In the beginning you said: if you have a deSitter space, it "can't" be rotating. But in the previous comment you said: A lone planet "can" rotate in deSitter space relative to the cosmological horizon. Can you clarify this to me please. – stupidity Jun 4 '12 at 14:01
@stupidity: A patch of deSitter space is surrounded by a cosmological horizon. This horizon can't rotate--- there's no deformation corresponding to this. If you put a planet inside, you can spin up the planet, but this shows up in the horizon spinning up "the other way" where I put it in quotes because the definition of "other way" is not obvious for a horizon, it's only obvious in the symmetric black hole case. – Ron Maimon Jun 4 '12 at 18:05

thank you for your question. Indeed, you are pointing in the directions of two big unifications made in physics: Linear motion and rest by Newton and acceleration by Einstein.

If it is convenient for you, I will devide your question into two parts concerning Newtonian and Einsteinian physics which will be related to who is observing the rotation.

First of all, if you are fixed to a point on the surface of the earth which is rotating, you will be able to measure it by means of fictitious forces acting on you, also in a non-relativistic Newtonian setup. But the interesting question is what happens if you are somewhere in space.

Then, assuming the earth as axisymmetric, Newtonian physics will tell you that there will be no difference to a non-rotating earth in the gravitational field. You will have to consider general relativity to answer this question. It turns out, that rotating bodies curl spacetime and you can even assign an angular momentum to it. By measuring this effect, which might be very hard, you would be able experimentally verify earth's rotation.



PS.: For further information, see e.g. On the multipole moments of a rigidly rotating fluid body

PPS.: @David: If I understand your argumentation correctly, you state that earth's rotation would define the zero-rotation reference-frame. I must admit that this might not be correct. Think of the Kerr spacetime for a non-vanishing rotation of a black hole. This is a similar situation, you cannot find a coordinate system with vanishing angular momentum of the spacetime as a whole.

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The Earth is rotating relative to its axis. So, yes, the effects would be there - Focault's pendulum would still precede, etc.

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Sorry, but your answer is plain wrong. See other answers on this page. -and an axis is a mathematical object, there is no physical axis sticking out of the earth. – Hans-Peter E. Kristiansen Jun 4 '12 at 7:00

From a popular physics standpoint, a good source that constantly discusses the history of the debate about this problem is the popular book The Fabric of the Cosmos by Brian Greene. The answer to this problem has been different in different eras since the time of Newton, who I believe first proposed this question in the form of a person spinning in an empty universe and whether or not their arms splay outward. Greene seems to think that string theory has something to do with the answer, though I'm not so convinced he's right. A lot of pseudoscientific philosophy goes into these types of questions, which is why I think it's best addressed in a popular book rather than in a scientific context. Anyway, there are lots of different angles and approaches to this problem, and this book is a good read if you're interested in knowing what a lot of people think about this problem throughout history, among other things.

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David Z, excellent answer - You comment that "... it does require that space defines some sort of absolute rotational reference frame". I believe that the existence of the reference frame is provable -

In a nutshell - The proof is based on the relationship between angular velocity and the centripetal force it produces. It assumes that the same angular velocity will produce the same amount of centripetal force (given the same mass and radius) anywhere in the universe. Rotation is measured or calculated as angular displacement from a reference direction. For rotations with the same angular velocity to produce the same magnitude of centripetal force, their reference directions must be rotationally static relative to each other.If this were not so, rotations that are measured as being the same would produce differing amounts of centripetal force. The reference direction must in fact, be the same for all rotations with parallel axes, in any location. If this were not so, translational motion of a rotation would cause its reference direction, which must be rotationally static, to change. To provide references for rotations around three perpendicular axes, three perpendicular reference directions are necessary. A more detailed version of this "proof" with diagrams is at:

This validates the requirement for the reference framework, but to answer the original question of "Is rotation relative to space", we need to know what form this framework might take. The three-dimensional Cartesian coordinate system can be a graphical representation of the three reference directions. This set of three perpendicular reference directions is more than an abstraction. There must be a real physical means of ensuring that the reference directions are rotationally static and parallel to each other for all rotations with parallel axes. If the relation between centripetal force is equally valid anywhere in the universe, then this means of ensuring static directions must be universal in extent. If a reference direction is delineated by something with a physical existence, then there must be something physical that ensures that all reference directions for rotations with parallel axes, even those separated by the breadth of the universe, are fixed or held parallel to each other in some manner. The Cartesian coordinate system is a graphical representation of a cubic lattice of universal extent. A real physical cubic lattice framework that extends throughout the universe would provide the fixed directions required, but the existence of such a lattice cannot be detected by our senses, so what is required is a real physical lattice that is undetectable to the senses. A cubic lattice comprised of elementary electric charges, with alternating positive and negative charges at the vertices, provides such a physical manifestation of the Cartesian coordinate system, and is undetectable to our senses. So, in my opinion, rotation is relative to this lattice of elementary charges that occupies all of space.

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The statement "The Earth is rotating relative to its axis" is circular, what defines the axis is the rotation, and without anything else in universe to observe the rotation of earth by, is earth really rotating?

The assumption of a universe with only earth in it should imply that the space and time and anything else are also bounded to the surface of the earth. Now in such a universe Earth is rotating respective to what?

If space is necessary as a medium for earth to rotate in, in a universe that there is nothing except earth, is space also part of that universe? What about dark matter, dark energy and anything else that could be the cause of all the "rotational effects" in a the normal universe?

and according to the there is no reason to conclude the effects will still be there.

Whoever down voted without commenting as to why, does have any logical reason?

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Downvoted because it's not answering the question, but it should be a comment on my answer. Also, the Wikipedia article clearly states: "Rotation was thus concluded to be absolute rather than relative." – Sklivvz Nov 28 '10 at 11:26
Is that the only reason? Is final line in a wikipedia article is the absolute answer where the Machs principle clearly states that such discussions are meaningless? What about answering the questions that I raised in my post? Wikipedia artivle does not include anything as to the cause of the "rotational effects", in a universe where those causes are excluded but only earth is the only object in that universe how could one talk about absolute rotation? Absolute rotation with respect to Luminiferous aether? Also I don not see a comment link on your post to comment with respect to. – Arjang Nov 28 '10 at 11:35
As long as there's gravity, it's possible to detect Earth's rotation absolutely. The rest of the universe is completely ignored in these experiments and does not make a contribution. I sincerely don't see your point, at all. – Sklivvz Nov 28 '10 at 11:48
In a universe where Earth is the only object, i.e. Earth is the universe. What is gravity? (If gravity is a force that acts between at least 2 entities.) The observations of this universe can not be extended to a universe witch is made up only with Earth. The effects that have been observed are from a universe where Earth is not the only Entity. Now the causes of the of the effects are unknown (although the effects are known). Now my point is this: If everything besides Earth is excluded in the new universe, what would guarantee that the causes of rotational effects will still to be present? – Arjang Nov 28 '10 at 12:05
There are two bodies, the earth and your pendulum. For all we know, GR still holds. If you want to dispute the correctness of GR: a) propose a viable experiment and b) do so in another question. – Sklivvz Nov 28 '10 at 12:29

protected by Qmechanic Oct 31 '15 at 12:56

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