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?