NNoname manuscript No. (will be inserted by the editor)
An alternative proof for Euler rotation theorem
Toby Joseph
Received: date / Accepted: date
Abstract
Euler’s rotation theorem states that any reconfiguration of a rigid bodywith one of its points fixed is equivalent to a single rotation about an axis passingthrough the fixed point. The theorem forms the basis for Chasles’ theorem whichstates that it is always possible to represent the general displacement of a rigidbody by a translation and a rotation about an axis. Though there are many ways toachieve this, the direction of the rotation axis and angle of rotation are independentof the translation vector. The theorem is important in the study of rigid bodydynamics. There are various proofs available for these theorems, both geometricand algebraic. A novel geometric proof of Euler rotation theorem is presented herewhich makes use of two successive rotations about two mutually perpendicularaxis to go from one configuration of the rigid body to the other with one of itspoints fixed.
Keywords
Rigid bodies · Rotations · Euler rotation theorem · Chasles’ theorem
Study of rigid body dynamics is one of the important topics in classical mechanics.As in the case of point particle dynamics, a good understanding of the kinematicdescription of the rigid body motion is essential for studying its dynamics. One ofthe crucial results related to the kinematic description of motion of a rigid bodyis Chasles’ theorem[1,2,3]. It states that a general displacement of a rigid bodycan be described by a translation and a rotation about an axis. Further, eventhough the translation vector is not unique, the orientation of axis of rotationand the angle of rotation will be the same for going from one configuration to theother [4,6,5]. This natural split of general motion into translational and rotationalparts allows for studying the dynamics of translation and rotation separately. The former reduces to a point particle like dynamics and the latter is described using
T. JosephBITS Pilani, K. K. Birla Goa Campus, Goa, IndiaTel.: +91-832-2580145E-mail: [email protected] a r X i v : . [ m a t h . HO ] A ug Toby Joseph
Euler equations in rigid body dynamics. It turns out that one can use the freedomin choosing the translation vector so as to make its direction coincide with that ofaxis of rotation. This is the content of the Mozzi-Chasles’ theorem and is centralto the study of dynamics of rigid body using screw theory[7]. Screw theory isextensively used in the modern day study of robotics [8,9].Chasles’ theorem is an extension of an earlier theorem due to Euler, referredto as Euler’s rotation theorem. Euler’s rotation theorem states that any reconfigu-ration of a rigid body with one of its points fixed is equivalent to a single rotationabout an axis passing through the fixed point. In other words, whatever way asphere might be rotated around its center, a diameter can always be chosen whosedirection in the rotated configuration would coincide with that in the original con-figuration. The original proof given by Euler himself is a geometrical one[10,11].The proof looks at the initial and final configuration of a great circle on the sphereand gives the recipe to construct a point, which is subsequently proved to be thepoint through which the axis of rotation passes.There are various other proofs available for the rotation theorem both geomet-ric [6,5] as well as algebraic ones [11,12]. The algebraic proofs typically require afamiliarity with rotation matrices and their properties or with ideas from grouptheory. The most commonly seen analytic proof [13,14,15] uses the orthogonalityproperty of three-dimensional rotation matrices to show that they always havean eigenvector with eigenvalue equal to one. This proof makes use of the resultthat eigenvalues of an orthogonal matrix have modulus one. Another one of thegeometric proofs [6] involves looking at the displacement of a segment under therotations and constructs planes of symmetry using the end points of the originaland displaced segments. The intersection of these symmetry planes is then shownto be the axis of rotation. This is by far the most transparent of the existing proofs.The proof for Euler rotation theorem given by Pars [5] involves going from the ini-tial configuration to the final configuration using two rotations: the first one is arotation by angle π about the center of great circle connecting one of the originalpoints and its final location and a second rotation about an axis passing throughthis final location. The invariant point is then established by a construction.Aim of the current work is to give a geometric proof of Euler rotation theo-rem that is different from the existing ones. We shall first develop the key ideasinvolved in the description of rigid body and use this in setting up the proof ofthe theorem. After proving Euler’s rotation theorem, which is crux of the paper,Chasles’ theorem is derived. The plan of the paper is as follows: In section I, wederive the number of degrees of freedom of a rigid body. Further, we set up ascheme for describing the general displacement of the rigid body which we makeuse of subsequently. Euler’s rotation theorem is derived in section II and the prooffor Chasles’ theorem is given in section III. To make the scope of discussion widerand complete, we shall then look at a few consequences of the theorems provedinvolving the idea of screw axis and rigid body motion in two dimensions. Rigid body is defined as a collection of particles whose mutual distances remainsinvariant. In three dimensions, N independent particles have 3 N degrees of free-dom. But if the particles constitute a rigid body, the degrees of freedom is reduced n alternative proof for Euler rotation theorem 3 to 6 (for the case when N ≥ N − N variables dependent on the remaining 6. Let us prove this result rigorously.We first show that a rigid body configuration is completely defined once coor-dinates of any three non-collinear particles in the rigid body are specified. Assumethat positions of particles (named A, B and C ) are given. Consider now a fourthparticle, D . Since the body is rigid, distance between particles A and D (say, d ), B and D ( d ) and between C and D ( d ) are fixed. Construct a sphere of radius d centered around particle A as shown in Fig. 1. It is clear that particle D has tobe residing on the surface of this sphere. Now construct a second sphere of radius d centered around particle B . The rigidity constraint will imply that particle D has to lie on the circle (call it S ) formed by the intersection of these two spheres.Note that if the spheres do not intersect, the constraints will not be consistentwith that of a rigid body configuration. A third sphere of radius d centered at C will intersect the circle S at two points ( D and D (cid:48) in Fig. 1) implying that,once the three points are fixed a fourth particle can only be placed at two pos-sible points consistent with the rigidity constraints. The two possible points arerelated to each other by a reflection about the plane containing particles A, B and C and corresponds to the mirror images of each other. So if we put an additionalconstraint that the handedness of rigid body is preserved there is only a uniqueposition where particle D can be placed and hence does not require any furthercoordinates to be specified. But particle D is completely arbitrary and could beany of the particles in the rigid body other than the original three particles whosepositions were specified.Thus we see that the number of degrees of freedom of a rigid body in threedimensions is the same as that of a rigid body containing three non-collinearparticles. Three independent particle have 9 degrees of freedom. Since the body isrigid, there are 3 constraint equations specifying the mutual separation betweenthe 3 particles. The difference gives the degrees of freedom of a rigid body to be6. The next question we address is about how to describe the displacement ofthe rigid body from one configuration to another. There are multiple ways to dothis. We shall adopt a scheme which will make it convenient for us to construct theproof for Chasles’ theorem. Consider two configurations I and II of the rigid body.Consider three points P , P and P (which are non-collinear) in the rigid body inconfiguration I. In the final configuration let these points move over to locations P (cid:48) , P (cid:48) and P (cid:48) respectively. To go from the configuration I to configuration II , wewill carry out the following three steps:1. Translate the body by vector −−−→ P P (cid:48) . This ensures that the point P is in itsfinal position P (cid:48) .2. Let P (cid:48)(cid:48) be the location of points P after this translation. Consider the planeformed by the vectors −−−→ P (cid:48) P (cid:48)(cid:48) and −−−→ P (cid:48) P (cid:48) . This is the equatorial plane shown inFig. 2. Rotate the rigid body about an axis perpendicular to this plane and passing through the point at P (cid:48) such that the point P (cid:48)(cid:48) is in its final position P (cid:48) . Note that if P (cid:48)(cid:48) is the same as P (cid:48) then this step need not be carried out.3. Let P (cid:48)(cid:48) be the location of the original point P after the above two operations.The rigid body can now be rotated about the axis passing through P (cid:48) and P (cid:48) (see Fig. 2) such that the point P (cid:48)(cid:48) is in its final position P (cid:48) . Toby Joseph
Fig. 1
The construction for finding out the number of degrees of freedom of a rigid body in 3dimensions. If three particles A , B and C are fixed, then there are only two possible locationsfor the fourth particle whose distance from the other there are fixed by rigid body constraints.The possible locations for the fourth particle D are shown as D (red dot) and D (cid:48) (blue dot) inthe figure. They are related by a reflection about the plane containing A , B and C particles.The dashed circle is the intersection of the spheres centered around A and B (referred toas circle S in the text). Points D and D (cid:48) are the intersections of this circle with the spherecentered at C . These steps will ensure that the rigid body has been displaced to its final config-uration.
We are now in a position to prove Euler rotation theorem. In order to prove thetheorem let us look at the steps 2 and 3 in the scheme described above to go fromone configuration to another of a rigid body. Note that after step 1 the point P isin its final position and P has moved over to the point P (cid:48)(cid:48) . Steps 2 and 3 involverotations to be carried out with P (cid:48) fixed. These operations are shown in Fig. 2.We represent these rotations by R AB and R P (cid:48) P (cid:48) . R AB is rotation by an angle φ about the axis AB (the axis perpendicular to vectors −−−→ P (cid:48) P (cid:48)(cid:48) and −−−→ P (cid:48) P (cid:48) and passingthrough the point P (cid:48) ) that is involved in step-2 above. The value of φ can vary between 0 and 2 π . R P (cid:48) P (cid:48) is rotation by an angle θ about an axis connecting points P (cid:48) and P (cid:48) which corresponds to step-3 above. θ can take values from − π to π .The sphere shown in the figure has got P (cid:48) at its center and has a radius equalto the distance between points P (cid:48) and P (cid:48)(cid:48) . For convenience we have oriented thefigure such that the AB axis is vertical. n alternative proof for Euler rotation theorem 5 Fig. 2
Steps 2 and 3 of the procedure for going from one configuration to the other of therigid body. The point P (cid:48) is in its final position after step-1. Rotation about AB axis (whichis perpendicular to the vectors connecting P (cid:48) to P (cid:48)(cid:48) and P (cid:48) to P (cid:48) ) by amount φ carries P (cid:48)(cid:48) to its final position. The second rotation by angle θ about P (cid:48) P (cid:48) will bring the rigid body toits final configuration. Note that we have oriented the figure such that the AB axis is vertical.We have also not shown P (cid:48)(cid:48) , or P (cid:48) in the figure. These points in general will not lie on thesurface of the sphere shown. As the rotation R AB is carried out, the great circle arc AP (cid:48)(cid:48) B will move overinto the great circle arc AP (cid:48) B . And the entire region that lies between these twoarcs before rotation will now lie between the great circle arcs AP (cid:48) B and AP (cid:48)(cid:48)(cid:48) B (see Fig. 2). In particular, the great circle arc ADB which bisects the region AP (cid:48)(cid:48) BP (cid:48) A (see Fig. 3) will move over to great circle arc AD (cid:48) B . Consider nowan arc of latitude like LM N , where L lies on AP (cid:48)(cid:48) B , N on AP (cid:48) B and M on ADB . Note that M is the midpoint of the arc. Under rotation R AB , LM N willmove over to latitude arc
N M (cid:48) N (cid:48) . Similarly the latitude arc HQS ( Q being the midpoint) will move over to SQ (cid:48) S (cid:48) under rotation (Fig. 3). It is interesting to notewhat happens to points like M (cid:48) and Q (cid:48) under the second rotation (Step-3 above).They are candidates for points that could fall back to their original position afterthe two rotations! This is so because the great circle arc P (cid:48) M ( P (cid:48) Q ) is equal inmagnitude to the great circle arc P (cid:48) M (cid:48) ( P (cid:48) Q (cid:48) ) and hence under rotation about Toby Joseph
Fig. 3
The figure shows how arcs of latitudes shift under the rotation about AB axis. Theequatorial arc P (cid:48)(cid:48) DP (cid:48) , D being the midpoint of the arc, moves over to P (cid:48) D (cid:48) P (cid:48)(cid:48)(cid:48) . The arc LMN , M being the midpoint of the arc, moves over to NM (cid:48) N (cid:48) . Similarly, point Q , which isthe midpoint of arc of latitude HQS goes to point Q (cid:48) . Points like D (cid:48) , M (cid:48) and Q (cid:48) can moveback to their original positions under the rotation about P (cid:48) P (cid:48) (see Fig 4). an axis passing through P (cid:48) and P (cid:48) both the points will fall on the same latitudecircle with P (cid:48) as the pole.We will now argue that depending on the value of θ , there is going to be exactlyone such point that will go back to its original position (that is, the position beforeStep-2). Fig. 4 shows a few of the candidate points that can come back to theiroriginal location. It is clear from the figure that the angles shown have the followingordering: DP (cid:48) D (cid:48) = π > ..M P (cid:48) M (cid:48) .. > ..QP (cid:48) Q (cid:48) .. > ..AP (cid:48) A = 0Even though this ordering is apparent from the figure, one can establish it more rigorously in the following manner. The spherical triangle P (cid:48) DQ (see Fig. 3) hasin it the spherical triangle P (cid:48) DM included. This is because the base P (cid:48) D is com-mon for both the triangles and the great circle arcs DQ and DM are part of thesame great circle with DQ being longer than DM . This implies that that theangle N P (cid:48) M is larger than SP (cid:48) Q (both being defined as angles between the cor- n alternative proof for Euler rotation theorem 7 responding great circle arcs). But M P (cid:48) M (cid:48) = 2 M P (cid:48) N and QP (cid:48) Q (cid:48) = 2 QP (cid:48) S . Therelationship in above equation follows.Thus for any positive value of θ in the interval from 0 to π , one and only one ofthe points of the kind M (cid:48) that lie in the hemisphere containing point A will comeback to its original position. There would be a similar point in the diametricallyopposite side of the sphere. If θ were negative and lies between 0 and − π , therewould be a point in the lower region below the equatorial plane that would go backto its original position and a corresponding point in the diametrically opposite side.Thus for any given value of φ and θ there are diametrically opposite pair of pointsthat do not change their position under steps 2 and 3. This implies that the effectof both the rotations considered above should be the same as that due to a singlerotation about an axis that passes through these invariant points and the centerof the sphere ( P (cid:48) ). Since the effect of any arbitrary set of rotations with P (cid:48) fixedcan be described using steps 2 and 3 above, we see that the net effect of theserotations can be attained by a single rotation about an axis. This proves Eulerrotation theorem. If we know the value of θ , we can find the invariant point byconstruction. To find this, pick the point (say X ) on the great circle arc ADB such that the angle between the great circle arcs P (cid:48) X and P (cid:48) A is θ . We have already shown that the last two steps in our procedure to represent ageneral displacement of a rigid body corresponds to a rotation about a singleaxis. But step 1 involved a pure translation that took point P to P (cid:48) . Thus wecan conclude that a general displacement of the rigid body can be obtained bya translation and a rotation about an axis. To complete the proof of Chasles’theorem we also need to show that a different choice of point (instead of P )for translation will not alter the direction and amount of rotation to be carriedout.To prove this, imagine we had chosen a different point Q instead of P fortranslation. Pick points Q and Q such that −−−→ P P = −−−→ Q Q and −−−→ P P = −−−→ Q Q , asshown in Fig. 5. One can now repeat the arguments above for proving the Eulerrotation theorem. Since the vectors involved in steps 2 and 3 ( −−−→ Q Q and −−−→ Q Q ) inthis case are identical to the old ones (even though the new displacement vectorcould be different), we will end up with the same rotation axis and angle. Thiscompletes the proof of Chasles’ theorem. It may well be that there is no materialpoint in the rigid body at the location of Q or Q . One can nevertheless think ofimaginary points rigidly attached to the body and moving in accordance with therigidity constraints. In fact, the displacing points (like P or Q ) themselves neednot form the part of the rigid body.An important corollary of Chasles’ theorem is Mozzi-Chasles’ theorem whichstates that a general displacement of the rigid body can be obtained by a rotationabout an axis and a translation along the same axis. For completeness, we givehere a proof of this theorem. Consider a rigid body displacement described by the displacement vector F and the rotation about direction ˆ n by an amount Θ as shown in Fig. 6. F may not be parallel to ˆ n . We will assume that the rigiddisplacement affects all the points in space and not just those belonging to therigid body. The translation vector F can be expressed as a sum of vector pointingalong ˆ n ( g in the figure) and a vector lying in the plane perpendicular to ˆ n ( s in Toby Joseph
Fig. 4
The pairs like points D (cid:48) and D , M and M (cid:48) and Q and Q (cid:48) are equidistant from point P (cid:48) . This implies that these pairs of points will lie in the same latitudinal circles (red dashedcurves) with P (cid:48) as the pole. This in turn makes it possible for these points to move back totheir original position (that is, the one prior to rotation about AB ) after the rotation about P (cid:48) P (cid:48) . In fact, one can show that (see text) exactly one of these set of points will fall back onto the original position for θ lying between 0 and π . Note that the blue curves in the figureare great circle arcs connecting the points involved. the figure). Under the rotation about ˆ n , the different points in space will undergodisplacements that lie in a plane perpendicular to ˆ n . For any given value of Θ , theset of displacement vectors will contain all possible vectors in the plane. This isbecause the magnitude of the rotation vector will vary from zero to infinity as thedistance of the points from the axis of rotation increases from zero to infinity andall points lying on a circle at fixed distance from the axis of rotation will generatedisplacements in all possible directions in the plane. Thus one should be able tofind points whose displacement is − s under the rotation. If one choose one of thesepoints as the translating point, it will ensure that the displacement vector is alongˆ n itself proving the corollary. The common axis that is involved in this description,in the direction of ˆ n , is referred to as the screw axis or Mozzi axis.Mozzi-Chasles’ theorem leads to another important result concerning motion of a rigid body in two dimensions. The counterpart of Chasles’ theorem in twodimensions, sometimes referred to as the first Euler rotation theorem, states thatany displacement of a rigid body in two dimensions can be achieved by either asingle rotation or a translation. There exists a straight forward geometric proof byconstruction for this theorem [6]. We shall prove the result using Mozzi-Chasles’ n alternative proof for Euler rotation theorem 9 Fig. 5
The proof of Chasles’ theorem also involves proving that irrespective of the translatingpoint the direction of the rotation axis and the angle of rotation are the same. If one choosesthe translating point to be Q instead of P , the constancy of direction of axis of rotation andthe amount of rotation can be seen by considering points Q and Q that are related to Q as P and P are to P . theorem. For this, note that in two dimensions the axis of rotation is always per-pendicular to the plane. By Mozzi-Chasles’ theorem (since two-dimensional dis-placements are a subset of possible displacements in three dimensions), the rigiddisplacement can be achieved using a translation along an axis and rotation aboutthat axis. Since the only possible translation along rotation axis is one with zeromagnitude, there must be a point that does not change its position under rigiddisplacement in two dimensions. The other possibility is a pure translation in theplane in which case the screw axis will lie in the plane and the rotation about thescrew axis will be zero. It follows that in two dimensions a rigid displacement iseither a pure translation (screw axis lies in the plane) or a pure rotation (screwaxis is normal to the plane). We have derived Euler’s rotation theorem using a novel geometric proof. Theproof involves using a set of three steps that takes the rigid body from its initialto final state. The Euler rotation theorem is derived using the last of the twosteps in this procedure. The proof is presented in a manner that helps one in thevisualization of how the invariant points arise and will be of pedagogic interest.
Fig. 6
To prove Mozzi-Chasles’ theorem, the translation vector connecting P to P (cid:48) , F , isresolved into a part that is along the axis of rotation ( g = g ˆ n ) and another that lies theplane perpendicular to it ( s ). Under an arbitrary rotation Θ about ˆ n , one can find pointsthat will have a displacement equal to − s (for example, displacement of Q (cid:48)(cid:48) to Q (cid:48) in thefigure). Choosing one of these points as the translating point, one can arrive at a translationthat is along ˆ n . In the figure shown, picking Q as the translating point will ensure that thetranslation is along the rotation axis itself. But it should be kept in mind that the sequences by which one chooses to movefrom one configuration to the other is neither unique nor special. The first part ofthe Chasles’ theorem, which asserts that the general displacement of a rigid bodyis a combination of translation and a rotation about an axis, follows immediatelyfrom Euler theorem and the first step in the procedure for carrying out rigiddisplacement. The fact that the orientation of axis of rotation and amount ofrotation is independent of the translation vector involved is proved by a separateconstruction. For completeness, we have also presented proofs for the existence ofscrew axis for motion in three dimensions and that in two dimensions any rigiddisplacement can be achieved by a pure rotation or a translation.
Acknowledgements
The author thanks Vibhu Mishra for useful discussions on existingproofs of Euler’s rotation theorem. The author would like to acknowledge financial supportunder the DST-FIST scheme (SR/FST/PS-1/2017/21). This is a pre-print of an article pub-lished in Mathematical Intelligencer. The final authenticated version is available online at:https://doi.org/10.1007/s00283-020-09991-z