In the attached paper by Francois Leyvraz, I could not get past eqs.(6-7b) for the following reason: if you differentiate eq.(5) with respect to t, you obtain R(dot) super T times R + R super T times R(dot) = 0 , or, using nomenclature from the paper, Omega sub b + R(dot) super T times R = 0. Omega sub l does not appear. Moreover, if R is antisymmetric (as it must be), that does not imply that R(dot) is also antisymmetric. The product of any matrix with an antisymmetric one does not have to be antisymmetric...
Firstly, by taking the transpose of Omegab you obtain dot(R)TR, so that the first equation gives Omegab+OmegabT=0, so Omegab is antisymmetric as claimed. The same trick can be used for Omegal after multiplying eq.(6) with R from the left and R-1 from the right.
Second, R is not antisymmetric, it is an orthogonal matrix. dot(R) is not antisymmetric either, in general, it fulfills:
dot(R)T=-R-1dot(R)R-1
I now realize that rotation matrices are not antisymmetric; if they were, they would have to have r_ii = 0. Thanks, S. (I hate to ask stupid questions.)
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