Editing Euler's rotation theorem (section)

==Euler's theorem (1776)==

Euler states the theorem as follows:<ref>Novi Commentarii academiae scientiarum Petropolitanae 20, 1776, pp. 189–207 (E478)</ref>

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

or (in English):
<blockquote>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.</blockquote>

[[Image:Euler Rotation 1.JPG|200px|left|thumb|'''Figure 1''': Blue great circle on sphere transforms into red great circle when rotated about diameter through {{math|'''O'''}}.]]
===Proof===

Euler's original proof was made using [[spherical geometry]] and therefore whenever he speaks about triangles they must be understood as [[spherical triangle]]s.

====Previous analysis====
To arrive at a proof, Euler analyses what the situation would look like if the theorem were true. To that end, suppose the yellow line in '''Figure 1''' goes through the center of the sphere and is the axis of rotation we are looking for, and point {{math|'''O'''}} is one of the two intersection points of that axis with the sphere. Then he considers an arbitrary great circle that does not contain {{math|'''O'''}} (the blue circle), and its image after rotation (the red circle), which is another great circle not containing {{math|'''O'''}}. He labels a point on their intersection as point {{math|'''A'''}}. (If the circles coincide, then {{math|'''A'''}} can be taken as any point on either; otherwise {{math|'''A'''}} is one of the two points of intersection.) 

[[Image:Euler Rotation 2.JPG|200px|right|thumb|'''Figure 2''': Arcs connecting preimage {{math|'''α'''}} and image {{math|'''a'''}} of {{math|'''A'''}} with bisector {{math|'''AO'''}} of the angle at {{math|'''A'''}}.]]

Now {{math|'''A'''}} is on the initial circle (the blue circle), so its image will be on the transported circle (red). He labels that image as point {{math|'''a'''}}. Since {{math|'''A'''}} is also on the transported circle (red), it is the image of another point that was on the initial circle (blue) and he labels that preimage as {{math|'''α'''}} (see '''Figure 2'''). Then he considers the two arcs joining {{math|'''α'''}} and {{math|'''a'''}} to {{math|'''A'''}}. These arcs have the same length because arc {{math|'''αA'''}} is mapped onto arc {{math|'''Aa'''}}. Also, since {{math|'''O'''}} is a fixed point, triangle {{math|'''αOA'''}} is mapped onto triangle {{math|'''AOa'''}}, so these triangles are isosceles, and arc {{math|'''AO'''}} bisects angle {{math|∠'''αAa'''}}.

{{clear}}
[[Image:Euler Rotation 3.JPG|200px|left|thumb|'''Figure 3''': {{math|'''O'''}} goes to {{math|'''O′'''}}, but {{math|'''O′'''}} must coincide with {{math|'''O'''}}.]]

====Construction of the best candidate point====
Let us construct a point that could be invariant using the previous considerations. We start with the blue great circle and its image under the transformation, which is the red great circle as in the '''Figure 1'''. Let point {{math|'''A'''}} be a point of intersection of those circles. If {{math|'''A'''}}’s image under the transformation is the same point then {{math|'''A'''}} is a fixed point of the transformation, and since the center is also a fixed point, the diameter of the sphere containing {{math|'''A'''}} is the axis of rotation and the theorem is proved.

Otherwise we label {{math|'''A'''}}’s image as {{math|'''a'''}} and its preimage as {{math|'''α'''}}, and connect these two points to {{math|'''A'''}} with arcs {{math|'''αA'''}} and {{math|'''Aa'''}}. These arcs have the same length. Construct the great circle that bisects {{math|∠'''αAa'''}} and locate point {{math|'''O'''}} on that great circle so that arcs {{math|'''AO'''}} and {{math|'''aO'''}} have the same length, and call the region of the sphere containing {{math|'''O'''}} and bounded by the blue and red great circles the interior of {{math|∠'''αAa'''}}. (That is, the yellow region in '''Figure 3'''.) Then since {{math|'''αA''' {{=}} '''Aa'''}} and {{math|'''O'''}} is on the bisector of {{math|∠'''αAa'''}}, we also have {{math|'''αO''' {{=}} '''aO'''}}.

====Proof of its invariance under the transformation====
Now let us suppose that {{math|'''O′'''}} is the image of {{math|'''O'''}}. Then we know {{math|∠'''αAO''' {{=}} ∠'''AaO′'''}} and orientation is preserved,{{efn|Orientation is preserved in the sense that if {{math|'''αA'''}} is rotated about {{math|'''A'''}} counterclockwise to align with {{math|'''OA'''}}, then {{math|'''Aa'''}} must be rotated about {{math|'''a'''}} counterclockwise to align with {{math|'''O′a'''}}. Likewise if the rotations are clockwise.}} so {{math|'''O′'''}} must be interior to {{math|∠'''αAa'''}}. Now {{math|'''AO'''}} is transformed to {{math|'''aO′'''}}, so {{math|'''AO''' {{=}} '''aO′'''}}. Since {{math|'''AO'''}} is also the same length as {{math|'''aO'''}}, then {{math|'''aO''' {{=}} '''aO′'''}} and {{math|∠'''AaO''' {{=}} ∠'''aAO'''}}. But {{math|∠'''αAO''' {{=}} ∠'''aAO'''}}, so {{math|∠'''αAO''' {{=}} ∠'''AaO'''}} and {{math|∠'''AaO''' {{=}} ∠'''AaO′'''}}. Therefore {{math|'''O′'''}} is the same point as {{math|'''O'''}}. In other words, {{math|'''O'''}} is a fixed point of the transformation, and since the center is also a fixed point, the diameter of the sphere containing {{math|'''O'''}} is the axis of rotation.

{{clear}}

====Final notes about the construction====
[[Image:Eulerrotation.svg|200px|left|thumb|Euler's original drawing where ABC is the blue circle and ACc is the red circle]]

Euler also points out that {{math|'''O'''}} can be found by intersecting the perpendicular bisector of {{math|'''Aa'''}} with the angle bisector of {{math|∠'''αAa'''}}, a construction that might be easier in practice. He also proposed the intersection of two planes:

*the symmetry plane of the angle {{math|∠'''αAa'''}} (which passes through the center {{math|'''C'''}} of the sphere), and
*the symmetry plane of the arc {{math|'''Aa'''}} (which also passes through {{math|'''C'''}}).

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

:'''Proof'''. Let us call {{math|'''O'''}} either of the endpoints (there are two) of this diameter over the sphere surface. Since {{math|'''αA'''}} is mapped on {{math|'''Aa'''}} and the triangles have the same angles, it follows that the triangle {{math|'''OαA'''}} is transported onto the triangle {{math|'''OAa'''}}. Therefore the point {{math|'''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 {{math|'''α'''}}, {{math|'''A'''}}, {{math|'''a'''}} lie. The rotation axis is obviously orthogonal to this plane, and passes through the center {{math|'''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.