A famous theorem of Leonard Euler states that given any two positions of a sphere with its center fixed, there exists a single rotation taking the sphere from one position to the other. In other words, positing that two positions of a sphere have in common their center entails that they also have in common a diameter, and the sphere can be moved from one position to the other by rotating about that diameter. [1]

The problem we explore here is how to compose two rotations. Given a sphere with its center fixed, rotating it about a diameter through an angle, then rotating it about another diameter through another angle, takes the sphere from its original position to a final position via an intermediate position.

We seek a purely geometric solution. To that end a possible analogy comes to mind. A translation of the plane can be represented by an arrow in the plane. Perhaps something similar can represent a rotation of the sphere, where great circles are analogous to straight lines. Focus on the plane for a moment. The translation arrow’s direction is the direction of the movement and its length tells how far to go.

Given two translations, the result of performing first one then the other can be obtained by a single translation. We can determine its arrow by adding the arrows of the given translations using the following geometric construction. We consider only the case when the lines of the two arrows intersect. (Parallel lines have no analogue on the sphere. All great circles intersect.) Slide the first arrow along its line so its head meets the intersection of the two lines. Slide the second arrow so its tail meets the first arrow’s head. Then the arrow from the first’s tail to the second’s head is the sum of the two given arrows.

   

The head-to-tail composition of translations suggests trying to do something similar with rotations of the sphere. We shall represent a rotation by a directed arc of a great circle, a curved arrow, on the unit sphere. Its direction is the direction to rotate, its arc length measures how far.

The plane of the curved arrow’s great circle is perpendicular to the axis of rotation. (The axis is considered as a full line. No right hand rule is necessary to determine the rotation’s direction.)

From the point of view of the sphere the curved arrow is straight: its lateral curvature is zero.

We can slide a curved arrow anywhere along its great circle, its arc length remaining constant, and it will specify the same rotation. In mathematical jargon, one curved arrow determines an “equivalence class” of curved arrows on its great circle. We call such an equivalence class a turn. At times we may just refer to a curved arrow and leave the equivalence class it represents, the turn, understood.

The  plane – translation / sphere – rotation  analogy suggests that we might be able to find the result of two rotations in space, one after the other, by adding their corresponding turns, head-to-tail, with sliding along a line changed to sliding along a great circle.

We said that a turn or curved arrow represents the act of rotating through an angle indicated by its length. The word “indicated” was left undefined intentionally. Suppose we took it to mean that the angle of rotation is the length of the curved arrow, understood as a radian measure. To see if this works, consider a simple example. Rotate 180° about an axis – a flip – then 180° about another axis perpendicular to the first. A moment’s thought shows that the end result is the same as a flip about the axis perpendicular to the given axes. Does the head-to-tail construction work? The two curved arrows corresponding to the two flips would be great-semicircles – if their arc lengths were the angle of rotation – and then their sum per the head-to-tail construction would be a curved arrow starting and ending at the same point, that is, no rotation at all – which is about as wrong as it could be.

In fact, this construction usually fails to give the correct answer. It is correct only when the two arrows lie on the same great circle. Somehow the curvature of the sphere breaks what works in the plane.

However a simple modification or reinterpretation of our preliminary definition of a curved arrow / turn will exactly compensate for the curvature of the sphere. To wit: make its arc length exactly half the rotation angle. When we come to follow an arrow’s instructions to rotate, we double its arc length to obtain how much to rotate. When adding them head-to-tail we use them as they are, with the half angle. With this modification it happens that the head-to-tail construction gives the correct answer in every situation.

This amazing and little known theorem is due to Lawrence Biedenharn & James Louck. It can be found in “The Theory of Turns Adapted from Hamilton” which is chapter 4 of their Angular Momentum in Quantum Physics (1985), a volume of Encyclopedia of Mathematics and its Applications edited by G.-C. Rota.

The unit sphere with its freight of potential curved arrows whose lengths are half the angles of the rotations they represent will be called the Turn Sphere.

Applied to our example of the two flips, the given curved arrows are now each 90° (π/2 radians) and adding them head-to-tail yields an arrow of 90° perpendicular to the given arrows – the correct result.

To repeat, when defining the curved arrow, only a rotation angle factor of one-half will make head-to-tail addition work. The fact that in all cases that particular factor exactly compensates for the curvature of the sphere is what is so amazing.

We shall now prove that the head-to-tail construction works in general, and thus turns are the rotational analog of vectors.

The theorem can be proven using the idea of reflection and two lemmas of solid geometry.

A reflection, like a rotation, takes objects in space to objects in space. It is determined by one plane and defined as follows. Given a plane, the reflection of a point is the point on the other side of the plane directly opposite, that is, the line between the given point and its reflection is perpendicular to the plane, and the two points are the same distance from the plane. A reflection changes an object into its mirror image. Points on the plane remain where they are.

Lemma #1  Given two planes  A  and  B,  intersecting in the line  L,  the result of reflecting an object first through  A  and then through  B  is the same as rotating it about  L  in the direction  A  to  B,  through twice the dihedral angle between them.

In other words, given two intersecting planes, if we first reflect an object in space “through” one plane and then the other, the resulting position of the object will be the same as if it had been rotated about the line common to the two planes, the angle of rotation being exactly twice the dihedral angle between the planes in the direction from the plane used first to the one used second.

This lemma, though about solid geometry, in essence is about plane geometry. Given an arbitrary point to be moved by the rotation, view the two reflecting planes end on, or what amounts to the same thing, consider the plane slice perpendicular to their intersection and through the given point. In that slice we have two intersecting lines and the arbitrary point, the angle between the lines being the dihedral angle. The lemma says that reflecting that point through one line and then the other in effect rotates the point by twice the angle between the lines.

The proof is based on the fact that the mirror image created by a reflection is undistorted; reflecting is a rigid transformation. The mirror image is, in every respect except “handedness” and position, just like the original. Consequently, if you expand a vector, its reflection – its mirror image – expands by the same amount. And if you add two vectors, their reflections add likewise, that is, the reflection of “their sum on the original side” is the sum of their reflections on the other side. These two properties can be written algebraically (for the moment, instead of using juxtaposition for multiplication we use a dot):      Reflect (c ·V)       =  c·  Reflect (V)      Reflect (V + W)  =  Reflect (V)  +  Reflect (W)An operation having these two properties is called a linear transformation. [3]

Thus the effect of reflecting any position vector (x,y) can be determined by its effect on just two position vectors:  (1,0) and (0,1).  First decompose the vector like this (now using juxtaposition for multiplication):      (x, y)   =   x (1, 0)  +  y (0, 1)Then use the general equations above:      Reflect (x, y)   =   x Reflect (1, 0)  +  y Reflect (0, 1)

In what follows we assume the reader knows the formula for rotating a point about the origin:      x’  =      x cos α +  y sin α
      y’  =  – x sin α  +  y cos α(x’, y’)  is   (x, y)  rotated about  (0, 0)  by the angle  α.

Now suppose we are given two lines through the origin that define the first and second reflections. Choose an  x, y  coordinate system so that the intersection of the lines is the origin and the first line is the  x-axis.  Let   θ  be the angle from the first line (the  x-axis)  to the second line. Let  (x₀, y₀)  be a given point. Then the coordinates of its reflection in the first line are      x₁  =      x₀
      y₁  =  – y₀Where does reflecting in the second line take the point  (x₁, y₁) ?  Again, all we need to compute is the action on the two basis position vectors  (1, 0)  and  (0, 1).  Call the operation of reflecting in the second line, Ref.  The diagram below shows that in effect it rotates the vector  (1, 0)  by  2θ,      Ref  (1, 0)  =  (cos 2θ, – sin 2θ)

A reflection is a discontinuous operation: it is done entire or not at all. A rotation on the other hand can be thought of as the culmination of a continuum of infinitesimal rotations. Despite that, the final result of a rotation is the same as the final result of two reflections.

Lemma #2  Given a line  L  and a plane containing it, and a second line  M  intersecting  L,  the plane can be rotated about  L  so that it contains  M.

This is obvious; rotate the plane to the plane determined by the intersecting lines.

       

       

By Lemma #1 a rotation is equivalent to two reflections about an ordered pair of planes through the rotation axis, where the angle between the planes is exactly half the angle of rotation. We say that the ordered pair of planes represents the rotation.

Call such an ordered pair a paddelrad (German for “paddlewheel”) and each plane a paddel.

Now we will use paddelrads to prove that the head-to-tail construction works for composing rotations.

Euler’s theorem says that given two orientations of an object there exists a rotation taking one to the other. Thus the change in orientation after two rotations is the same as the change made by a some single rotation. Paddelrads enable us to find that single rotation, as follows.

Consider two rotations (as usual about the same point) and let  A  and  B  be two paddelrads representing them per Lemma #1. Call their ordered planes  A_tail A_head  and  B_tail B_head  respectively, taking the names from the corresponding curved arrow. Consider the plane common to the axis of  A  and the axis of  B  and call it  P.  By Lemma #2, if we choose a paddel from  A  and one from  B,  their respective paddelrads can be rotated so that the chosen paddels coincide, specifically in the plane determined by the paddelrads’ axes,  P.  Rotate  A  on its axis so that  A_head  lies on  P  and rotate  B  on its axis so that  B_tail  lies on  P  as well.  Thus the four planes
      A_tail A_head B_tail B_head
become
      A_tail P P B_head
Now imagine the reflections being done through these planes in that their given order. The two reflections through plane  P  in the middle cancel, and we are left with the single paddelrad
      A_tail B_head

Tom Verhoeff puts it this way in “Composing Rotations via Hamilton Turns” (abridged and copyedited):

The way of adding paddelrads in the proof is identical to the head-to-tail way of adding the associated curved arrows.

By the way, the proof shows that given two rotations one followed by the other, there exists a single rotation that has the same effect, which is a special case of Euler’s theorem.

The validity of turn addition is so important it warrants more than one proof. We will try to use a pair of flips instead of reflections for such a proof. A flip is a rotation through 180° (π radians). It is simpler than a reflection in that it preserves orientation (handedness). It is more complicated in that the action of the two flips cannot be seen in a single two-dimensional slice.

First a few words about flips. A flip about a line in space is the same as a “reflection” through that line, where a point is moved as follows: take the perpendicular from the point to the line and move the point an equal distance beyond the line. As noted above, unlike with reflection in a plane orientation (handedness) is preserved. A diagram with numbers indicating positions of some points before and after the flip will make this clear. The line is viewed end on:

Lemma #1 of the paddelrad proof applied to reflections in two planes. A similar lemma holds for flips about two lines.

Lemma #3  Given two lines  L  and  M,  intersecting at the point  P,  the result of flipping an object first about  L  and then about  M  is the same as rotating it about their common perpendicular line through  P  in the direction  L  to  M,  through twice the angle between them.

If this statement is true then rotating the two lines  L  and  M  about their common perpendicular line, keeping the angle between them fixed, has no effect on the resultant rotation. We call the two lines a propeller, analogous to the paddelrad of Lemma #1, and each line a blatt (German for “blade”).  We call the plane containing the propeller the plane of the propeller. We call the line through the center and perpendicular to that plane the axis of the propeller.

We could use Lemma #1 to prove Lemma #3 fairly easily and without additional analytic geometry, however we seek an independent proof so the lemma, and hence this second proof using flips, will stand on its own legs. [4]

We will need the following fact:  Given any rotation  R  and flip  F  whose axes are perpendicular,

Thus a rotation is the result of two reflections or of two flips. In the first case the reflecting planes are through the axis of rotation, in the second the flipping lines are in the plane perpendicular to the axis of rotation and through the fixed center. In each case the angle between the lines / planes is half the angle of rotation.

Though a flip can be performed continuously, two partial flips are not equivalent to a partial rotation. This suggests that thinking of a flip about a line as reflecting through the line is the correct understanding instead of thinking of it as rotating about the line by 180°. Reflecting through a line is, like reflecting through a plane, a discontinuous operation done entire or not at all

There is another reason for believing that line reflection is the correct understanding. Line reflection is outside the class of rotations, even though the effect is the same as one, just as plane reflection is outside. This emphasizes that R, F, and H above are not to be considered as turns. In fact a flip – that is, a 180° rotation – is not, as a turn, its own inverse. If understood as a turn, the proof breaks down at the last step.

As noted earlier, because of Lemma #3, we can rotate a propeller about its axis by any amount and it will continue to represent the same rotation.

Thus a propeller can represent a curved arrow or turn, with the first line corresponding to the tail and the second to the head.

Now we use propellers to prove that head-to-tail composition works.

The analogue of Lemma #2 of the paddelrad proof for flips is:

Lemma #4  Given a plane and two intersecting lines within it, one line can be rotated in the plane about the intersection point so it coincides with the other line.

This is obvious.  Thus given two planes intersecting in a line  L,  and a line in one plane and a line in the other plane that intersect on  L,  each line can be rotated in its plane so the two lines coincide. (Rotate each line to  L.)

Now consider two rotations (as usual about the same point) and let  A  and  B  be two propellers representing them per Lemma #3. Call their ordered lines  A_tail A_head  and  B_tail B_head  respectively, taking the names from the corresponding curved arrow. Consider the intersection of the plane of  A  and the plane of  B  and call it line  L.  (If the planes coincide, let  L  be any line in the plane through the common center of the two propellers.)  Per Lemma #4,  rotate  A  on its axis so that  A_head,  rotating in the plane of A, lies on  L  and rotate  B  on its axis so that  B_tail,  rotating in the plane of B, lies on  L  also.  Thus the four lines
      A_tail A_head B_tail B_head
become
      A_tail L L B_head
Now imagine the flips being done about these lines in that their given order. The two flips about the line  L  in the middle cancel, and we are left with the single propeller
      A_tail B_head

In words:

The way of adding propellers in the proof is identical to the head-to-tail way of adding the associated curved arrows.

Thus we have two proofs, one using paddelrads the other using propellers, that Turn Sphere head-to-tail composition works.

Given a turn it is easy to construct a propeller for it, as follows. Consider the axis of the rotation the turn represents, in other words, the line through the center of the Turn Sphere perpendicular to the turn’s great circle. It intersects the Turn Sphere in two polar opposite points. Choose one of them and call it p. Then one flip is the curved arrow (directed great circle segment) from the tail of the given turn to p while the other flip is the curved arrow from p to the head of the given turn.

We have discovered how to compose rotations by adding turns so we will use the plus sign (+) to stand for the composition of turns.

Given any rotation there is an ordered pair of points (the head and tail of a curved arrow) that represents it. Indeed, viewing a turn as just an ordered pair of points instead of a curved arrow has the advantage that we needn’t specify that the length of the curved arrow lies between 0 and π.

At first glance this space of turns has four dimensions (two times two) but the fact that the pair of points can slide along a great circle (one dimension) and represent the same rotation reduces that by one. Thus the “space” of rotations has three dimensions. [6]

On the Turn Sphere any curved arrow of zero length – an overlapping pair of points, anywhere on the sphere – represents the null rotation. All such degenerate arrows are equivalent; there is only one null turn. We designate the null turn by  T₀.

Given a turn  A,  we will write its inverse (the turn with the same axis as  A  rotating by the same amount but in the opposite direction) as  –A.  Then we see graphically that for any turn  A,

The order of performing two rotations, and hence of adding two turns, usually matters. Unless the turns lie on the same great circle (when the rotations have a common axis),  A + B  ≠  B + A.  This is because, having added the turns by aligning them head-to-tail, after reversing the order each curved arrow must be slid on its great circle to again align them head-to-tail and the result lies on a different great circle than before. The angle of rotation is the same but the axis differs. Turn addition, and rotation composition, is not a commutative operation.

Turn addition is associative, that is, the grouping of several additions does not affect the result. It suffices to prove this for any three turns:

There is another special turn besides the null turn; it represents rotation through 360° (2π radians) about any axis. The turn’s (arc) length is π. At first glance there might appear to be many such turns, one for each possible axis of rotation, however consistency requires that they must all be the same turn. To prove this, let  T_π  be such a turn. Its head and tail are on opposite sides of the Turn Sphere. Using head-to-tail construction we see that  T_π  is its own inverse

T₀  and  T_π  each possess a length (0 and π respectively) but neither has a unique axis or great circle. In the head-to-tail construction, any great circle can be used for  T_π  to line it up with another turn.

There are no other special turns in the same vein. The turn for a rotation through 720° equals the null turn.

Codman’s Paradox provides an example of three turns that cannot be slid head-to-tail simultaneously. Most any three turns would do for such an example but these particular three are easy to compose and the result is the basis of a good parlor trick.

Ernest Amory Codman, a Boston surgeon (and very interesting man, look him up), described something like the following in a medical book of 1934 about the shoulder joint. [8]

Start with your right arm hanging by your side with the thumb of the hand pointing forward. With your shoulder as the pivot, lift your arm sideways straight out so the arm points to the right. Then swing your arm forward horizontally so it points ahead. Finally, let your arm drop to your side where it started. But it is not quite where it started. Your thumb is now pointing toward you, that is, to the left instead of forward as at the start. The end result is that your arm rotated 90° within itself.

The three rotations about the shoulder were 90° each, around three perpendicular axes: the first about the horizontal axis pointing backwards, the second about the vertical axis pointing upwards, the third about the horizontal axis pointing leftward. That resulted in a 90° rotation about the vertical axis pointing upwards – a rotation about the long axis of your arm despite your never seeming to have done it.

The paradox is resolved by observing that though you didn’t directly so rotate your arm, it is the composition of the three rotations that you did do.

Here are three curved arrows (turns) for the three rotations.