From: Pertti Lounesto Subject: Re: Help X Product, Determinants Date: Tue, 24 Aug 1999 21:14:28 +0200 Newsgroups: sci.math Keywords: Lie group of rotations Kevin Foltinek wrote: > Hop David writes: > > > My > > ignorance is keeping me locked out of the garden of orbital mechanics. > > I hear there are wonderful things to be found in this garden. > > There are. The key is Lie groups; in particular, the group of 3x3 > rotations, which are those matrices R such that I=R*R' (R' is the > transpose of R), with determinant 1. You can differentiate this > defining relation (with respect to t, if you think of R as a function > of t): > d/dt(I) = d/dt(R*R') > 0 = d/dt(R)*R' + R*d/dt(R') > 0 = d/dt(R)*R' + R*d/dt(R)' > and then, at t such that R=I, > 0 = d/dt(R)*I' + I*d/dt(R)' . > Letting d/dt(R)=A, > 0 = A+A'. > Therefore, the antisymmetric matrices (A+A'=0, or A=-A') are of some > importance in rotations. You'll note that E_x, E_y, E_z, and M_v are > all antisymmetric; you'll also note that the antisymmetric 3x3 > matrices form a vector space of dimension 3, which is part of one > reason why 3-dimensional space has a cross product, but 4-dimensional > space doesn't. (The antisymmetric 4x4 matrices form a vector space of > dimension 6, not 4.) The cross product axr, where a=(a1,a2,a3) can be represented by Ar, where [0 -a3 a2 A = a3 0 -a1 -a2 a1 0]. The exponetial of A is exp(A) = I+A/|a|*sin(|a|)+A^2/|a|^2*(1-cos(|a|)). The rotation sending r to Rr, where R = exp(A), results in exp(A)r = cos(|a|)r +sin(|a|)/|a|*axr+(1-cos(|a|)/|a|^2*(a.r)a. In 4D, the corresponding formula is more involved, unless the eigenvalues of A have the same absolute value. Then it is remarkably easy. Can you work it out?