Projective Bivector Parametrization of Isometries Part I: Rodrigues' Vector
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1 Part I: Rodrigues' Vector Department of Mathematics, UACEG International Summer School on Hyprcomplex Numbers, Lie Algebras and Their Applications, Varna, June 09-12, 2017
2 Eulerian Approach The Decomposition Problem Quantum Mechanics Classical Mechanics A long time ago in a galaxy far, far away... O. Rodrigues [1840] W. Hamilton [1843] A. Cayley [1846] byul ogo
3 Recommended Readings Rodrigues O., Des lois g eom etriques qui regissent les d eplac ements d'un syst eme solide dans l'espace, et de la variation des coordonn ees provenant de ces d eplac ements consid er es ind ependamment des causes qui peuvent les produire, J. Math. Pures Appl. 5 (1840). Fedorov F., The Lorentz Group (in Russian), Science, Moscow Mladenova C., An Approach to Description of a Rigid Body Motion, C. R. Acad. Sci. Bulg. 38 (1985). Bauchau O., Trainelli L. and Bottaso C., The Vectorial Parameterization of Rotation, Nonlinear Dynamics 32 (2003). Pi na E., Rotations with Rodrigues' Vector, Eur. J. Phys. 32 (2011). Brezov D., Mladenova C. and Mladenov I., Vector Decompositions of Rotations, J. Geom. Symmetry Phys. 28 (2012).
4 Rotation Vectors and Rodrigues' Formula The axis-angle representation of rotations in R 3 yields the rotation vector s = ϕ n, (n, ϕ) S 2 S 1 which the Hodge map transforms into a so(3) generator as s g[ ] s = ϕ n so(3). Now, one obtains the group element via the exponential map s exp R(n, ϕ) = cos ϕ I + (1 cos ϕ) nn t + sin ϕ n that is the famous Rodrigues' rotation formula.
5 Pros and Cons On he positive side: one has a handy explicit formula for the matrix entries the relation to mechanics is straightforward: ṡ = Ω (fixed axis). On the other hand: composing spherical vectors is cumbersome, while working with matrices is rather inecient this representation involves transcendent functions, so one needs to work with approximations the parametrization is topologically incorrect since SO(3) = RP 3 S 2 S 1.
6 Euler's Trigonometric Substitution The famous Euler trigonometric substitution allows for writing s = ϕn c = τn, τ = tan ϕ 2 sin ϕ = 2τ 1 + τ 2, cos ϕ = 1 τ τ 2 which yields rational expressions for the rotation matrix entries R(c) = (1 c2 ) I + 2 cc t + 2 c 1 + c 2
7 Dealing with Innities As τ ϕ π one applies l'h opital's rule to obtain R(c) = (1 c2 ) I + 2 cc t + 2 c 1 + c 2 c 2 2nn t I = O(n). Half-turns are mapped on the plane at innity in RP 3. Moreover c 2, c 1 c 2 1,2 ĉ 1 ĉ 2 (ĉ 1, ĉ 2 ) where ĉ k denote the corresponding unit vectors.
8 The Vector-Parameter Some major advantages of Rodrigues' construction: compact expressions and no excessive parameters whatsoever topologically correct parametrization of SO(3) = RP 3, instead of coordinates on T 3 (e.g., Euler angles), which yield singularities allows for rational expressions for the rotation's matrix entries R(c) = (1 c2 ) I + 2 cc t + 2 c 1 + c 2 an ecient composition to replace the usual matrix multiplication c 2, c 1 = c 2 + c 1 + c 2 c 1 1 (c 2, c 1 ) R(c 2 ) R(c 1 )=R( c 2, c 1 ) numerically fast and analytically convenient representation.
9 Generalized Euler Decomposition Consider the decomposition R(c) = R(c 3 )R(c 2 )R(c 1 ) with c = τn and similarly c k = τ k ĉ k, ĉ k = 1. Denoting g ij = (ĉ i, ĉ j ), r ij = (ĉ i, R(c) ĉ j ), ω = (ĉ 1, ĉ 2 ĉ 3 ) the explicit form of some matrix entries r ij yields a system of QE's (r 32 + g 32 2g 12 r 31 ) τ1 2 2 ωτ 1 + r 32 g 32 = 0 (r 31 + g 31 2g 12 g 23 ) τ ωτ 2 + r 31 g 31 = 0 (r 21 + g 21 2g 23 r 31 ) τ3 2 2 ωτ 3 + r 21 g 21 = 0 where the two solutions for τ 2 determine the double-valued parameter ω ± = ( R(τ ± 2 ĉ2) ĉ 1, ĉ 2, ĉ 3 ), ω + ω + = 0.
10 The Solutions NSC for real solutions (Gram determinant for the moving frame): 1 g 12 r 31 = g 21 1 g 23 r 31 g Moreover, with the notation ω 1 = ( ĉ 1, ĉ 2, R t (c) ĉ 3 ), ω2 = ω, ω 3 = (R(c) ĉ 1, ĉ 2, ĉ 3 ) the solutions for τ k can be given in a closed decoupled form as τ ± k = σ k ω k ±, σk = ε ijk (g ij r ij ), i > j.
11 Specic Cases Similarly, we have NSC for decomposition in two factors and the solution is given as where we have denoted r 21 = g 21 R(c) = R(c 2 ) R(c 1 ) τ 1 = r 22 1 ω 1, τ 2 = r 11 1 ω 2 ω 1 = ( ĉ 1, ĉ 2, R t (c) ĉ 2 ), ω2 = (R(c) ĉ 1, ĉ 1, ĉ 2 ). In the critical points of the map RP 3 = S 3 /Z 2 T 3 = (RP 1 ) 3 given as ĉ 3 = ±R(c) ĉ 1 the rank drops (gimbal lock) and ϕ 1,3 become dependent, namely ϕ 2 = 2 arctan r 11 1 ω 2, ϕ 1 ± ϕ 3 = 2 arctan r 22 1 ω 1
12 Linear-Fractional Relations The composition law yields a linear-fractional dependance between each pair of parameters in the decomposition, namely Γ1 ij τ j τ i (τ j ) = Γ 2 ij τ j + Γ 3 ij With the notation c k = (c, ĉ k ) and κ ij = g ij + r ij, the matrices Γ k are given in the generic case as 1 g 32 r 32 r 32 g 32 Γ 1 = g 31 r 31 1 g 31 r 31 = (Γ 3 ) t g 21 r 21 g 21 r 21 1 Γ 2 = 0 κ 32 c 1 κ 31 c 2 κ 21 c 3 κ 32 c 1 κ 32 c 1 κ 31 c 2 0 κ 21 c 3 κ 31 c 2 κ 21 c 3 κ 32 c 1 κ 21 c 3 κ 31 c 2 0
13 Coordinate Changes in SO(3) Euler to Bryan angles: τ ± φ = cos φ sin ϑ cos ϑ ± 1 sin 2 φ sin 2 ϑ, τ ± ϑ = sin φ sin ϑ 1 ± 1 sin 2 φ sin 2 ϑ τ ± ψ = cos φ sin ψ + sin φ cos ϑ cos ψ cos φ cos ψ sin φ cos ϑ sin ψ ± 1 sin 2 φ sin 2 ϑ Bryan to Euler angles: τ ± φ = sin ϑ sin φ cos ϑ ± 1 cos 2 φ cos 2 ϑ, τ ± ϑ = ± 1 cos φ cos ϑ 1 + cos φ cos ϑ τ ± ψ = cos φ sin ϑ cos ψ + sin φ sin ψ sin φ cos ψ cos φ sin ϑ sin ψ ± 1 cos 2 φ cos 2 ϑ
14 Weakening Davenport's Condition In order to ensure the NSC 0 for all group elements, one needs ĉ 2 ĉ 1,3 (Davenport s condition). It is possible, however, to use weaker restrictions if we: parameterize only regions of SO(3) decompose into more factors. For example, in the case ĉ 3 ĉ 1 one may decompose into three factors if for the rotation angle and the one between the axes one has ϕ 2γ, γ = min (ĉ 1, ĉ 2 ) the invariant axis satises β = min (n, ĉ 1,2 ) γ. Similarly, one may always decompose into N factors with [ ] + π N 1 + (Lowenthal s formula). γ
15 Variations and Lie Derivatives We consider both left and right shifts of a given rotation Direct dierentiation yields and respectively u k (t) c = t ĉ k, c, u + k (t) c = c, t ĉ k k u ± k c = (I + cct ± c ) ĉ k k R(u ± k c) = 2 1+c 2 [ (I ± c ) ĉ k c t sym + (ĉ k ρ k c ± c ĉ k ) 2ρ k I ]. Since we also have k = k r 31, 31 = g 32 g 21 r 31 the variations of the decomposition parameters are given as k τ ± i = kσ i τ i ( k ω i ± k r 31 ) ω i ±
16 The Angular Momentum in Bryan Angles In a decomposition with respect to the coordinate axes, one has r 32 r 31 0 r 33 0 r 31 k ω j = , k σ j = 2 0 r 33 r 32 0 r 13 r 12 0 r 23 r 22 which yields the QM angular momentum in these coordinates L 1 = ϕ L 2 = tan ϑ sin ϕ ϕ + cos ϕ + sec ϑ sin ϕ ϑ ψ L 3 = tan ϑ cos ϕ ϕ sin ϕ + sec ϑ cos ϕ ϑ ψ and we easily obtain the hamiltonian (Laplace operator) as ( = L 2 = sec 2 2 ( ϑ ϕ sin ϑ 2 ϕ ψ + cos ϑ ) ) ϑ ψ 2
17 Decomposition in a Precessing Frame For any ĉ 1, such that r 11 ±1, we may decompose R = R 2 R 1 choosing ĉ 2 = (1 r 2 11) 1/2 ĉ 1 R(c) ĉ 1 and with the notation ρ k = (ĉ k, c) the decomposition angles are given as ϕ 1 = 2 arctan ρ 1, ϕ 2 = arccos r 11. Denoting κ the precession rate of the coordinate frame {K}, we have = sec 2 ϑ [ 2 2 ϕ 2 + ( cos 2 ϑ )] + csc 2 ϑ 2 ϑ 2 ϑ κ 2
18 Rigid Body Mechanics Using the kinematical expressions Ω = c 2 (I + c ) ċ, ċ = 1 ( I + cc t c ) Ω 2 one easily obtains the system of ODE's ϕ 1 = Ω 1 Ω 3 tan ϕ 2 2 ϕ 2 = Ω 2 κ = Ω 1 + Ω 3 cot ϕ 2 and in the case of rotational inertial ellipsoid the Euler equations yield Ω 1 (t) = λω cos(ωt + ϕ ), Ω 2 (t) = λω sin(ωt + ϕ ), Ω 3 = ω µ
19 Possible Generalizations What else do we want? Higher-dimensional generalizations; Pseudo-Euclidean groups; Non-homogeneous isometries. To do all this, however, we need a shift in our perspective...
20 Thank You! THANKS FOR YOUR PATIENCE!
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