Euler's rotation theorem revisited - Elementary geometric proofs

Question

This is a very elementary topic but I thought it might be worth giving it a try here, I would be very interested in any comments - I originally posted it to Maths SE.

Euler's Rotation Theorem, proved by Euler [1] in 1775, is an important theorem in the study of general 3D motion of rigid bodies, as well as an early example of a fixed point theorem in mathematics. It 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

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.

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.

Answer 2

How is the standard proof non-elementary? You take an arbitrary point $V$ in the unit sphere and look at its image $V'$. You then compose your rotation $A$ with the reflection $R_1$ with respect to the plane orthogonal to the line $(V'V)$ and passing through the origin $O$; clearly the composition maps $V$ back to $V$. Therefore, the composition must preserve the plane through the origin orthogonal to $V$.

You now take an arbitrary point $W$ in that plane, look at its image $W'$ under the composition, and compose $R_1\circ A$ with the reflection $R_2$ w. r. t. plane orthogonal to $(W;W')$ and passing through the origin. That plane contains $V$, therefore $R_2\circ R_1\circ A$ preserves both $V$ and $W$; and hence the set of the (two) vectors in the sphere orthogonal to both.

You take one of these vectors $U$ and look at its image $U'$ and compose everything with another reflection... we get that $R_3\circ R_2\circ R_1\circ A$ preserves the mutually orthogonal points $U$, $W$ and $V$. Therefore, it preserves distances to $U$, $W$ and $V$. But these distances uniquely determine a point in the sphere, for the set of points in the sphere at a certain distance, e. g., from $U$ is a circle lying in the plane orthogonal to $OU$, and an intersection of three such planes contains just one point.

We conclude that $R_3\circ R_2\circ R_1$ undoes $A$, which means that $R_1\circ R_2\circ R_3$ is the same as $A$. But each reflection changes the orientation. This means that the construction must have failed at some point, i. e., we must have either $V=V'$, or $W=W'$, or $U=U'$. The first possibility can be disregarded: if there is no point with $V\neq V'$, then our map is the identity and there's nothing to prove. The second one can also be excluded, for if $W=W'$ for all $W$, then $A$ is just one reflection and thus changes orientation. Therefore, $U=U'$. and we did not in fact need $R_3$, so, $A=R_1\circ R_2$ and thus $A$ preserves the common line of the planes of reflections $R_1$ and $R_2$.

Arguably, the by-product - that any rotation is a composition of two reflections - is also of fundamental importance, and generalizes straightforwardly to any dimension.

Answer 3

Assume that the sphere has radius $R\neq 0$ with centre $O$. Denote the image of a point $P$ on the sphere under the motion by $P'$.

The assumptions we are making about the sphere motion is that it preserves distances between sphere points and if we take a triangle that does not lie along a great circle (i.e. it is not in a plane passing through the sphere centre) then it's orientation relative to a vector from $O$ to the triangle centroid does not change under the transformation.

We first try to find a fixed point under the motion noting that if we have one fixed point $F$ then automatically the antipodal point $F^\#$ is also fixed since $|FF^\#|=2 R$ and as the motion preserves distance between sphere points, $|F'F^{\#'}|$=$|FF^{\#'}|=2 R$ which implies that $F^{\#'}=F^\#$.

Take a point $A$ on the sphere surface with image $A'$ under the mapping. If $A=A'$ we are done. If not there exists a unique plane $A^*$ perpendicular to the line segment $AA'$ and passing through its midpoint. Take another sphere point $B$, distinct from $A$ and $A'$ lying on $A^*$ with image $B'$. If $B=B'$ again we are done, so assume $B\neq B'$. Now as $B$ lies on $A^*$ the midpoint of $BB'$ cannot line on $A^*$ unless both $B$ and $B'$ lie on $A^*$. In this case the plane $B^*$ perpendicularly bisecting $BB'$ is perpendicular to $A^*$ and hence distinct from it. If the midpoint of $BB'$ which is on $B^*$ does not lie on $A^*$ then clearly $B^*$ is distinct from $A^*$.

We can now take the intersection of the two distinct planes $A^*$ and $B^*$ which consists of a line, L, passing through the sphere centre intersecting it in 2 antipodal points, $N$ and $S$. Since the sphere motion preserves distances between points the image $N'$ of $N$ must be equidistant from $A$ and $A'$ since $N$ was and also it must be equidistant from $B$ and $B'$. This is equivalent to it lying on the planes $A^*$ and $B^*$ and hence on their intersection. Therefore either $N'=S$ (and $S'= N$) or $N'=N$ (and $S'=S$). But the two directed arcs $AB$ and $A'B'$ of equal length can be moved into coincidence by a rotation about $NS$. (This was the point of the construction and is obvious since we can drop equal length perpendiculars from $A$ and $A'$ to a common point $A_p$ on $NS$ since all points of $NS$ are equidistant from $A$ and $A'$ by construction. Similarly for $B$ and $B'$ to point $B_p$. The two planes containing $AA_pA'$ and $BB_pB'$ are perpendicular to $NS$ hence parallel. Bringing the two planes into coincidence by translating along NS we can see that $|AB|=|A'B'|$ implies that the directed angle $\angle AA_pA'$ equals the directed angle $\angle BB_pB'$ and hence a rotation by this angle about $NS$ takes the directed arc $AB$ into $A'B'$ ). This means that the triangles $NAB$ and $NA'B'$ have the same orientation but $NAB$ and $SA'B'$ do not. Hence the sphere motion will reverse orientation if $N'=S$ so this cannot be true and both $N$ and $S$ must be fixed under the motion.

Hence we can now assume that the sphere motion itself has 2 antipodal fixed points and we now show this means it must be a rotation. We can complete this with 2 different arguments:

Argument 1) If we have an additional fixed point to the 2 antipodal points then any point on the sphere can be categorised uniquely by it's distance to these three points up to a reflection in the plane $F$ through $O$ containing them. Hence a given point either stays fixed or is reflected in this plane. Take a point $M$ not on $F$. Then if $M$ stays fixed then it's reflection $M_r$ does as well. Take an arbitrary point $X$ not on $F$. Then we have $|XM|=|X'M|$ and $|XM_r|=|X'M_r|$ and hence $X'$ cannot be the reflection of $X$. Hence either all points stay fixed or they are all reflected in $F$. However a reflection is not orientation preserving so the sphere motion must be the identity.

Therefore we have only 2 antipodal fixed points. Since we have proved that any two sphere points can be mapped to any other 2 points spaced the same by a rotation about an axis passing through $O$ and there are only 2 fixed points all pairs of points can be rotated into their images about the same axis. If we take 2 arbitrary non-antipodal points $X$ and $Y$ on the sphere then the both must be rotated about this axis in exactly the same way to preserve the distance between them. Hence all points on the sphere are transformed by a common rotation about a common axis.

This proves that the sphere motion is a rotation about an axis passing through its centre.

Argument 2) For any point $X$ on the sphere $|NX|=|NX'|$ and $|SX|=|SX'|$. This implies that $X'$ lies on the plane perpendicular to $NS$ passing through $X$ which is a rotation of $X$ about $NS$. Pick any other point $Y$ distinct from $X$ on the great circle passing through $N$, $S$ and $X$. $Y'$ is the rotation of $Y$ about $NS$. The quadrilateral $XOSY$ is congruent to $X'OSY'$ since all the distances between corresponding points are the same. Therefore $Y'$ lies on the great circle passing through $N$, $X'$ and $S$. (Another way to see this is that $Y$ and $Y'$ lie on the same plane $Q$ perpendicular to $NS$ and $Y$ is the closest sphere point on $Q$ to $X$, hence $Y'$ must be the closest to $X'$ which implies it lies along the great-circle containing $X'$ and perpendicular to $Q$). This means that all points on a given great circle passing through $N$ and $S$ are rotated about $NS$ by the same angle to another such great circle.

Now take the great circle $H$ lying in the plane perpendicular to $NS$. Take distinct points $X$ and $Y$ on this circle. Since the sphere motion preserves distances between points and is orientation preserving we know that the triangles $NXY$ and $NX'Y'$ must be congruent and oriented in the same sense which implies that $Y$ is rotated about $NS$ identically to $X$ and hence this circle is mapped to itself under a rotation about $NS$.

Since each great circle passing through $N$ and $S$ shares a point with $H$ they all must rotate about $NS$ as $H$ does.

This proves that the sphere motion is a rotation about an axis passing through its centre.