I am trying to understand the following proof form the book "Classical Mechanics" by Gregory:
Question
What I do not understand is were the formula $\frac{\partial r_{i}^{\lambda}}{\partial\lambda}=k\times r_{i}^{\lambda}$ comes from. The explanation from Goldstein on pages 58-59 is definetly plausible, however I would like it a bit more rigorous. What I think I can show is that $\frac{\partial r_{i}^{0}}{\partial\lambda}=k\times r_{i}$.
Goldstein has the following formula for rotation:
If I differentiate this w.r.t. $\Phi$ or in the case of Gregories notation $\lambda$ I do not get same answer. Why does this not work?
I would really appreciate some tips. Many thanks in advance.
$\br^λ$ is the result of a rotation of $\br$, $\br^λ=A(λ)\br$, $A:I\to SO(3)$ some path in the matrix group of orthogonal matrices. $\newcommand{\br}{{\mathbf r}}\newcommand{\bv}{{\mathbf v}}\newcommand{\bk}{{\mathbf k}}\newcommand{\bn}{{\mathbf n}}$
For such continuous groups it is most convenient to express the tangent of the curve or the full tangent space relative to its base point, so either as $A'(λ)=X(λ)A(λ)$ or $A'(λ)=A(λ)X(λ)$. The first is closer to the usual format of linear ODE systems, the second is more common in the definition of Lie-groups and their algebras. Your first source is based on the first convention, the second on the second.
Orthogonal matrices satisfy $AA^T=I=A^TA$. Taking the derivative with the product rule and inserting the tangent equation results in $X+X^T=0$. In 3D such a skew-symmetric matrix can be parametrized as $$X=\pmatrix{0&-k_3&k_2\\k_3&0&-k_1\\-k_2&k_1&0}\implies X\bv=\bk\times \bv.$$ Due to the anti-symmetry of the cross-product $\bk\times \bv=-\bv\times \bk$, so that in the second source $\bn=-\bk$ for compatibility. In the context of the question we consider rotation paths that have a constant $X$ and thus constant rotation axes $k$ and $n$. Moreover, these are unit vectors, so that $λ$ is the arc length, which is the angle in radians.
It follows that $$\frac{d}{dλ}\br^λ=A'(λ)\br=XA(λ)\br=\bk\times \br^λ=\br^λ\times \bn.$$
For the connection to the Rodrigues rotation formula, recall that $$ \bv·(\bk\times \br)=\det(\bv,\bk,\br)=\det(\br,\bv,\bk)=(\br\times \bv)·\bk $$ and $$ (\ba\times(\bb\times \bc))·\bv=\det(\ba,\bb\times \bc,\bv)=(\bb\times \bc)·(\bv\times \ba)\\=\det((\bb,\bc)^T(\bv,\ba))=(\bb·\bv)(\bc·\ba)-(\bb·\ba)(\bc·\bv)\\~\\ \implies (\ba\times(\bb\times \bc))=(\bc·\ba)\bb-(\bb·\ba)\bc $$ $\newcommand{\ba}{{\mathbf a}}\newcommand{\bb}{{\mathbf b}}\newcommand{\bc}{{\mathbf c}}$
In solving $A'=XA$ via matrix exponential, one finds $$X^2v=\bk\times(\bk\times \bv)=(\bk·\bv)\bk-(\bk·\bk)\bv=(\bk·\bv)\bk-\bv,$$ so $X$ acts as a complex unit in the plane orthogonal to $\bk$, while vectors parallel to $\bk$ get mapped to zero.
Decompose $\br=\br_\parallel+\br_\perp$ in components parallel and orthogonal to $\bk$. Then $$ \br^λ=A(λ)\br=\exp(λX)\br=\exp(λX)\br_{\parallel}+\exp(λX)\br_\perp =I\br_{\parallel}+(\cos(λ)I+\sin(λ) X) \br_\perp \\=(\bk·\br)\bk+\cos(λ)(\br-(\bk·\br) \bk)+\sin(λ)(\bk\times \br). $$