Euler's Rotation Theorem - Euler's Theorem (1776)

Euler's Theorem (1776)

Euler states the theorem as follows:

Theorema. Quomodocunque sphaera circa centrum suum conuertatur, semper assignari potest diameter, cuius directio in situ translato conueniat cum situ initiali.

or (in free translation):

When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position.

To prove this, Euler considers a great circle on the sphere and the great circle to which it is transported by the movement. These two circles intersect in two (opposite) points of which one, say A, is chosen. This point lies on the initial circle and thus is transported to a point a on the second circle. On the other hand, A lies also on the translated circle, and thus corresponds to a point α on the initial circle. Notice that the arc aA must be equal to the arc Aα.

Now Euler needs to construct point O in the surface of the sphere that is in the same position in reference to the arcs aA and αA. If such a point exists then:

• it is necessary, that the distances OA and Oa are equal to each other; the arcs Oa and OA must be equal,
• it is necessary that the angles OaA and OAα are equal.

Now Euler points out that the angles OAa and OaA must also be equal, since Oa and OA have the same length. Thus OAa and OAα are equal, proving O lies on the angle bisecting αAa. To provide a complete construction for O, we need only note that the arc aO may also be constructed such that AaO is the same as αAO. This completes the proof.

Euler provides a further construction that might be easier in practice. He proposes two planes:

• the symmetry plane of the angle αAa (which passes through the centre C of the sphere), and
• the symmetry plane of the arc Aa (which also passes through C).

Proposition. These two planes intersect in a diameter. This diameter is the one we are looking for.

Proof. Let's call O to any of the endpoints (there are two) of this diameter over the sphere surface. Since αA is mapped on Aa and the triangles have the same angles, it follows that the triangle OαA is transported onto the triangle OAa. Therefore the point O has to remain fixed under the movement.

Corollaries This also shows that the rotation of the sphere can be seen as two consecutive reflections about the two planes described above. Points in a mirror plane are invariant under reflection, and hence the points on their intersection (a line: the axis of rotation) are invariant under both the reflections, and hence under the rotation.

Another simple way to find the rotation axis is by considering the plane on which the points α, A, a lie. The rotation axis is obviously orthogonal to this plane, and passes through the center C of the sphere.

Given that for a rigid body any movement that leaves an axis invariant is a rotation, this also proves that any arbitrary composition of rotations is equivalent to a single rotation around a new axis.