# Elementary Geometric Proofs of Euler's Rotation Theorem

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 ?

References

[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.

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.

Proof

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 $$M_{1}$$.

Proof

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.

Proof

(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 :

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.

Proof

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 $$r$$.

Proof

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 $$\theta$$).

Definition

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 $$O$$.

Proof

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.

Proof

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.

Proof

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.

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.

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 $$\theta \in (0, 90^\circ)$$.

$$C_{1}$$ 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.