I am totally confused about Euler's rotation theorem.

Normally I would think that an asteroid could rotate around two axes simultaneously. But Euler's rotation theorem states that:

In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about a fixed axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation.

But asteroids do rotate around two axes. Look at the videos from this website: http://csep10.phys.utk.edu/astr161/lect/asteroids/features.html

The Spin of Asteroid Toutatis

By making a series of observations, it is possible to study the rotation of some asteroids. Most have simple rotations around a fixed axis, with periods typically between one hour and one day. For example, here is a movie (83 kB MPEG) made by the Hubble Space Telescope of the asteroid Vesta in rotation (Ref). However, the asteroid 4179 Toutatis (which crosses Earth's orbit) has been found through radio telescope observations to have an irregular shape and a complex tumbling rotation---both thought to arise from a history of violent collisions. Here is a short animation (47 kB MPEG) of the spin of Toutatis; here is a longer animation (288 kB MPEG).

Here are the movies:




You can clearly see that those rotations are not possible around one axis. But then isn't it contradicting Euler's rotation theorem?

  • $\begingroup$ Take a look at math.stackexchange.com/questions/10741/… $\endgroup$
    – JavaMan
    Jun 11, 2011 at 3:24
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    $\begingroup$ @DJC: That post is not related to this one. The asteroids zsero is talking about are real asteroids, not the game Asteroids (the topology of which is the subject of the post you linked to). $\endgroup$ Jun 11, 2011 at 3:27
  • $\begingroup$ Oh. That's embarassing and makes me laugh at the same time! Thanks for pointing out my error @Zev. $\endgroup$
    – JavaMan
    Jun 11, 2011 at 3:32
  • $\begingroup$ @Arturo: that's not the point. see my answer $\endgroup$
    – leonbloy
    Jun 11, 2011 at 3:49
  • $\begingroup$ @leonbloy: Yes, you are right. The difference here is between "composition of two motions" (a beginning and an end) as opposed to trying to reproduce a continuous motion. $\endgroup$ Jun 11, 2011 at 4:01

2 Answers 2


I think you are confusing a single rotation (as a fixed displacement) with a rotational motion.

Euler theorem says that the composition of two individual rotations (say, I rotate a body 15 degrees around a vertical axis, then I rotate it 10 degrees around some horizontal axis) is equivalent to a single rotation around some axis.

But suppose I do the same double rotation, with the same axes but, say, double angles ( 30 and 20 degrees): they would be, again, equivalent to a single rotation, but the equivalent axis would be different.

Hence, the composition of two rotation (motions) with fixed axis is not equivalent to some other rotation (motion) with another (fixed) axis. Then, to speak of the superposition of two rotational motions (that cannot be reduced to a single rotational motion) makes perfect sense.

Example: Take a long cilinder, make it rotate quicky along its longitudinal axis. Superpose to that a slow rotation along a transversal axis. If both axis passes through the center of the cylinder, there is a fixed point. However, the resulting rotational motion cannot be expressed as a single rotational motion (with a fixed axis).

  • $\begingroup$ I'm sorry, but I'm confused by your fourth paragraph. In the previous two you give two examples where the composition of two rotations is equivalent to a single rotation around some other axis. The next paragraph begins with "hence..." and asserts the composition of two rotations is not equivalent to some other rotation. $\endgroup$ Jun 11, 2011 at 3:56
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    $\begingroup$ @zsero: If you only consider the beginning and end of the motion, and there is a fixed point, then you can find a single rotation that will begin in the same place and end in the same place; however, that does not mean that the motion produced by that single rotation will equal the motion of the original action. You are only reproducing the beginning and end positions, not the continuous motion. $\endgroup$ Jun 11, 2011 at 4:03
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    $\begingroup$ @zsero: Actually the expresion 'fixed axis' makes no sense in that quote. It should simply say 'about an axis'. But the important thing is to see that it's speaking of a "displacement" (from point A to B, or from angle A1 to angle A2), it's NOT speaking of a (rotational) "motion". If you change in that quote the expressions "displacement/rotation" for "rotational motion", then it's false. See my example of the cylinder. Its motion is NOT equivalent to a rotation (as motion) about a fixed axis. $\endgroup$
    – leonbloy
    Jun 11, 2011 at 4:08
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    $\begingroup$ OK, I see it now! Thanks! So Euler's rotation theorem is only about "states", like how to get from a start "state" to a result "state". OK, and the fixed part doesn't make any sense there. Maybe we should correct the Wikipedia article? en.wikipedia.org/wiki/Euler's_rotation_theorem $\endgroup$
    – hyperknot
    Jun 11, 2011 at 4:12
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    $\begingroup$ There, now the wikipedia pages does not have that confusing 'fixed axis' expression any more :-) $\endgroup$
    – leonbloy
    Jun 11, 2011 at 4:13

Hmm well actually for any rigid body, it does rotate along an axis. (at least for a infinitesimal amount of time) and this rotation will stay around the same axis over time as long as no torque is applied. It may just be difficult to see that axis.

  • $\begingroup$ The rotation of a rigid body will not stay around the same axis over time, except in the special case that the three principal axes of rotation happen to pass through the center of mass of the rigid body. Moment of inertia is a tensor, not a vector. See en.wikipedia.org/wiki/… $\endgroup$ Jul 25, 2013 at 20:02
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    $\begingroup$ We learned in experimental physics class at the university that a rigid body always can rotate stably around 2 of its 3 axes. The longest axis and the shortest one. The rotation is totally unstable around the middle axis. You can show that by throwing a rotating book up into the air and watch how it performs. $\endgroup$ Jan 8, 2015 at 5:34

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