Proof Euler's rotation theorem

1 proof

1.1 previous analysis
1.2 construction of best candidate point
1.3 proof of invariance under transformation
1.4 final notes construction

proof

euler s original proof made using spherical geometry , therefore whenever speaks triangles must understood spherical triangles.

previous analysis

to arrive @ proof, euler analyses situation if theorem true. end, suppose yellow line in figure 1 goes through center of sphere , axis of rotation looking for, , point o 1 of 2 intersection points of axis sphere. considers arbitrary great circle not contain o (the blue circle), , image after rotation (the red circle), great circle not containing o. labels point on intersection point a. (if circles coincide, can taken point on either; otherwise 1 of 2 points of intersection.)

figure 2: arcs connecting preimage α , image of bisector ao of angle @ a.

now on initial circle (the blue circle), image on transported circle (red). labels image point a. since on transported circle (red), image of point on initial circle (blue) , labels preimage α (see figure 2). considers 2 arcs joining α , a. these arcs have same length because arc αa mapped onto arc aa. also, since o fixed point, triangle αoa mapped onto triangle aoa, these triangles isosceles, , arc ao bisects angle ∠αaa.

figure 3: o goes o′, o′ must coincide o.


construction of best candidate point

let construct point invariant using previous considerations. start blue great circle , image under transformation, red great circle in figure 1. let point a point of intersection of circles. if a’s image under transformation same point fixed point of transformation, , since center fixed point, diameter of sphere containing axis of rotation , theorem proved.

otherwise label a’s image , preimage α, , connect these 2 points arcs αa , aa. these arcs have same length. construct great circle bisects ∠αaa , locate point o on great circle arcs ao , ao have same length, , call region of sphere containing o , bounded blue , red great circles interior of ∠αaa. (that is, yellow region in figure 3.) since αa = aa , o on bisector of ∠αaa, have αo = ao.

proof of invariance under transformation

now let suppose o′ image of o. know ∠αao = ∠aao′ , orientation preserved, o′ must interior ∠αaa. ao transformed ao′, ao = ao′. since ao same length ao, ∠aao = ∠aao. ∠aao = ∠aao′, ∠aao = ∠aao′ , therefore o′ same point o. in other words, o fixed point of transformation, , since center fixed point, diameter of sphere containing o axis of rotation.

final notes construction

euler s original drawing

euler points out o can found intersecting perpendicular bisector of aa angle bisector of ∠αao, construction might easier in practice. proposed intersection of 2 planes:

the symmetry plane of angle ∠αaa (which passes through center c of sphere), and
the symmetry plane of arc aa (which passes through c).


proposition. these 2 planes intersect in diameter. diameter 1 looking for.


proof. let call o either of endpoints (there two) of diameter on sphere surface. since αa mapped on aa , triangles have same angles, follows triangle oαa transported onto triangle oaa. therefore point o has remain fixed under movement.


corollaries. shows rotation of sphere can seen 2 consecutive reflections 2 planes described above. points in mirror plane invariant under reflection, , hence points on intersection (a line: axis of rotation) invariant under both reflections, , hence under rotation.

another simple way find rotation axis considering plane on points α, a, lie. rotation axis orthogonal plane, , passes through center c of sphere.

given rigid body movement leaves axis invariant rotation, proves arbitrary composition of rotations equivalent single rotation around new axis.

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