Euler's Rotation Theorem states that given an arbitrary motion of a sphere about its center, there exists a diameter of the sphere (the 'Euler Axis') and axial rotation about it which produces the same net displacement.

Euler's original proof [1, sections 24-28] makes use of spherical 'non-Euclidean' geometry, for example spherical triangles, and is discussed in [2] and [3].

What other methods of proof exist, which require only elementary Euclidean geometry, and are purely geometric, not requiring any algebra or matrix theory ?


[1] Euler's original proof of 1775, with English translation, http://www.17centurymaths.com/contents/euler/e478tr.pdf, retrieved 14th October 2019.

[2] Euler's Rotation Theorem, https://en.wikipedia.org/wiki/Euler's_rotation_theorem, retrieved 14th October 2019.

[3] Bob Palais, Richard Palais, and Stephen Rodi, A Disorienting Look at Euler’s Theorem on the Axis of a Rotation, American Mathematical Monthly 116:10, 25th August 2009, p892-909. https://www.researchgate.net/publication/233611890_A_Disorienting_Look_at_Euler's_Theorem_on_the_Axis_of_a_Rotation, retrieved 14th October 2019.


1 Answer 1


Two geometric proofs are given below. Both proofs start off as in Euler's proof, by considering the image $C_{2}$ of a great circle $C_{1}$ under the motion. In proof (1) this is used to construct a non-great circle which must be mapped onto itself due to a certain orientation property that is preserved - the axis of that circle is then the Euler Axis. In proof (2) it is shown the final displacement of the great circle can be achieved via a composition of two $ 180^\circ $ axial rotations which then gives the Euler Axis as the normal to the plane containing these two axes. For a non-zero motion the Euler Axis must be unique since it implies there exist exactly two fixed points, namely its endpoints - with all the other points rotated by a common angle not a multiple of $ 360^\circ $.

The following terms are used :

  • a 'motion' of a sphere means a general arbitrary motion in 3D about its center
  • an 'axial rotation' is a special case of a 'motion' which is a rotation about a fixed axis (diameter) of the sphere
  • two motions are 'equivalent' if they produce the same net displacement
  • a 'zero motion' is one with no net displacement
  • a 'fixed point' is a point whose final position equals its initial position
  • the 'axis' of a circle on the sphere (great or non-great) is the sphere diameter perpendicular to the circle plane
  • the 'poles' of a circle on the sphere are the end points of its axis
  • the 'great circle of a diameter' is the great circle perpendicular to it
  • the antipode of a point on the sphere is the diametrically opposite point
  • any 'circle' will be assumed to have a non-zero radius

The lemmas cover the simple special cases and define the notion of 'orientation' used in Proof (1).

Lemma 1

A motion of a sphere about its center $O$ which leaves a point $P$ on the sphere fixed is equivalent to an axial rotation about $OP$. Hence the antipode $P'$ of $P$ is fixed also, and if the motion is non-zero $P$ and $ P' $ are the only fixed points.


There are no possible final positions of the sphere in which $ P $ is fixed, other than axial rotations about $ OP $ from the original position, since with $ P $ fixed the situation of the sphere is constrained from every other possible motion. The antipode $ P' $ is the opposite end of this axis and hence is fixed also. For a non-zero motion the angle of axial rotation cannot be a multiple of $ 360^\circ $, thus ALL points other than $ P $ and $ P' $ must be moved.

Lemma 2

Given a motion $M_{1}$ of a sphere $ S $ about its center $ O $, then a second motion $M_{2}$ which places a circle $ C $ of $ S $ (great or non-great) identically to $M_{1}$ is equal to $M_{1}$.


No other possible final position of $ S $ than that of $M_{1}$ can have $ C $ placed completely 'correctly' because the sphere is completely constrained by this criteria - for, once the final positions of all the points of a circle on a fixed center sphere have been determined the final positions of all the other points of the sphere have been determined also. Thus $M_{2}$ must equal $M_{1}$.

Lemma 3

A motion of a sphere about its center $ O $ which overlays a circle $ C $ (great or non-great) onto itself in some manner is equivalent to an axial rotation.


(i) If $ C $ is non-great then as in Lemma 1 the sphere is constrained so no net displacement other than a rotation about the circle's axis is possible.

(ii) If $ C $ is a great circle then it must either be :

(a) overlayed the 'same way up', in which case the same argument as (i) applies, or

(b) overlayed but 'flipped over'. Consider an arbitrary point $ P $ on $ C $ and its image $ P' $ under the motion ($ P' $ may equal $ P $), as in the 'plan view' of Fig 1 :

enter image description here

A $180^\circ$ rotation $\phi$ about axis $ D $ of symmetry of $ P $ and $ P' $ places $ C $ the 'right way up' and puts $ P $ onto $ P' $. This must then place all the other points of $ C $ in the correct position, and so by Lemma 2, $\phi$ is equivalent to the original motion.

Lemma 4

Any motion of a sphere about its center $ O $ in which a diameter is flipped is equivalent to a $ 180^\circ $ axial rotation.


This causes the great circle of the diameter to be flipped over onto itself, and thus by Lemma 3 case (ii)(b), the result follows.

Lemma 5

Given two non-diametrical points $ A $ and $ B $ on a sphere $ S $ of radius $ R $, then if $ d $ is the straight-line distance $ AB $, the circles on the sphere which contain $ A, B $ are :

(i) a unique minimal radius circle of radius $ r = d/2 $,

(ii) a unique maximal radius great circle of radius $ r = R $,

(iii) for every intermediate radius $ r \in (d/2, R) $, exactly 2 circles of radius $ r $.


In the 'Hoopla Construction' in Fig 2 below, $ A $ and $ B $ are viewed at $ D $, with $ A $ in front of $ B $. The set of circles on $ S $ containing $ A, B $ corresponds to the set of planes through the axis $ AB $, as they cut $ S $, such as $ \Gamma $ and $ \Delta $. $ \theta = 0^\circ $ gives case (i), $ \theta = 90^\circ $ gives case (ii), and $ \theta \in (0, 90^\circ) $ gives case (iii), with $ r = \sqrt{ R^2 - l^2 cos^2 \theta } $ (an increasing function of $\theta$).

enter image description here


Given two non-diametrical points $ A $ and $ B $ on a circle $ C $, 'orientation of $A, B$ on $C$' is either CW or ACW according to the sense of the minor arc from $ A $ to $ B $.

With this definition : (i) 'orientation of $A, B$ on $C$' flips when we view from the other side of $C$, (ii) the orientation is undefined for diametrical points $A, B$ of $C$, and (iii) the 'orientation of $B, A$ on $C$' is opposite from the 'orientation of $A, B$ on $C$'.

Lemma 6

Given any non-great circle $C$ on a sphere $S$ and two non-diametrical points $A, B$ of $C$, the orientation of $A, B$ (as viewed from 'non-$O$' side of circle $C$, ie the 'outside' of $S$) is preserved after any motion of $S$ about $O$.


Center $O$ of sphere never crosses or touches the plane of circle $C$, so circle $C$ is always being viewed from the same side, and so as $A, B$ are fixed onto $C$, their orientation on $C$ remains the same.

Lemma 7

If two non-diametrical points $A, B$ on a sphere $S$ of radius $ R $ lie on two distinct circles of common radius $ r $ on the sphere, then $A, B$ (viewed from non-$O$ side) have opposite orientations on these respective circles.


From Lemma 5, the two distinct circles of same radius implies case (iii), so the circles are non-great.

Thus from the 'Hoopla' diagram of Fig 2, with $A, B$ seen at $ D $ with $ A $ in front of $ B $, we have $\theta \in (0, 90^\circ)$.

The circle in plane $ \Gamma $ gives $A, B$ with orientation ACW, whilst the other circle in the mirror image plane $ \Delta $ gives $A, B$ with orientation CW.

Lemma 8

Given two diameters $ L $ and $ M $ of sphere $S$, then the motion which is the composition of a $ 180^\circ $ rotation about $ L $ followed by a $ 180^\circ $ rotation about $ M $ is equivalent to a single axial rotation.


The case $ L = M $ is trivial as the composition is a zero motion.

Otherwise consider the great circle $C$ defined by the plane containing $ L $ and $ M $, and let $ P $ and $ P' $ be the poles of $C$, as in Fig 3.

enter image description here

The rotation about $ L $ flips $ P $ and $ P' $, as does the rotation about $ M $. Hence the composition leaves $ P $ fixed. By Lemma 1 the result then follows, with the axis being the normal to the plane containing $ L $ and $ M $.

Proof 1

Suppose great circle $C_{1}$ is mapped onto great circle $C_{2}$. Assume planes $C_{1}$ and $C_{2}$ do not coincide (otherwise Lemma 3 completes the proof).

Let $C_{1}$ and $C_{2}$ intersect along a diameter $ BF $ (the 'line of nodes'), as shown in Fig 4.

enter image description here

Since $ B $ lies on $C_{2}$, it must have been mapped from some point $ A $ on $C_{1}$. Assume $ A \neq F $ (otherwise the proof follows from Lemma 4), and $ A \neq B $ (otherwise the proof follows from Lemma 1). $ A $ is shown on the left of $ BF $ in Fig 4 - if it was on the right, we could rotate the diagram around $ 180^\circ $ about $ BF $ so $ A $ is on the left. The dihedral angle $ \delta \in (0, 180^\circ) $.

Let $ \Omega \in (0, 180^\circ) $ be the angle $\angle A\widehat{O}B$.

$ B $ also lies on $C_{1}$ so it maps to some point $ E $ on $C_{2}$. So from the rigid body motion of $C_{1}$, angle $\angle B \widehat{O} E = \Omega$, and $\mbox{chord } |AB| = \mbox{chord } |BE| $ (on $C_{1}$, $C_{2}$ respectively). $E$ is shown above $C_{1}$ in Fig 4, but the same argument as below applies if it is below.

Also plane $A, O, B = \mbox{plane } C_{1}$, and plane $B, O, E = \mbox{plane } C_{2}$.

$ A, E, B $ cannot be collinear because that would imply $ E $ to be in plane $C_{1}$, so the plane $C_{2}$ defined by $ B, O, E $ would then be in plane $C_{1}$ - a contradiction.

Thus $ A, E, B $ define a unique plane, containing 3 distinct points of sphere $S$. That plane cannot pass through $O$ since then all of $ A, E, B, O $ would lie in the same plane, again implying $C_{1}$, $C_{2}$ coincident - a contradiction. Let the non-great circle defined by this plane be $C$, and let its image be $ D $.

We show $ C = D $. Firstly note that although $C$ is a smaller radius circle than $C_{1,2}$, chords $AB$ and $BE$ can't be diameters of $C$ because that would imply $A = E$ - a contradiction - so the orientations below are well-defined. Consider the points $ B, E $ which lie on $C$. They must also lie on $ D $, being the image of $A, B$. But (viewing from non-$O$ side) :

orientation of $ B, E $ on $C$ = orientation of $A, B$ on $C$,

because the non-diametrical equal length chords $AB$ and $BE$ of $C$ subtend the same angle within $C$, and these chords lie to either side of point $B$ by virtue of $A \neq E$.

And secondly, considering the rigid motion taking $C$ to $ D $ :

orientation of $ B, E $ on $ D $ = orientation of $A, B$ on $C$,

by Lemma 6.

So $ B, E $ have the same orientation on circles $ C, D $. But by Lemma 7, as $ C, D $ have common radius, this means $ C = D $, from which the proof now follows from Lemma 3 case (i), the Euler Axis being the axis of circle $C$.

Proof 2

View $C_{1}$ and $C_{2}$ as shown in Fig 5. Cases $ \theta = 0^\circ $ and $ \theta = 90^\circ $ follow from Lemma 3 case (ii), so assume $ \theta \in (0, 90^\circ) $.

enter image description here

$C_{1}$ can be made to overlay $C_{2}$ by $ 180^\circ $ rotation about $ L $ or about $ M $.

The first of these places upper side of $C_{1}$ onto lower side of $C_{2}$, while the second places the upper side of $C_{1}$ onto the upper side of $C_{2}$.

Choose whichever of these results in $C_{1}$ being overlayed onto $C_{2}$ the 'wrong way up'. Then by Lemma 3 case (ii) (b) a $ 180^\circ $ rotation about some axis within $C_{2}$ places $C_{1}$ exactly in the 'correct' position of $C_{2}$.

Thus we have achieved the correct final position for $C_{1}$ by a succession of two $ 180^\circ $ axial rotations, and thus by Lemma 8 and Lemma 2 the proof follows.


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