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I am trying to understand the following proof form the book "Classical Mechanics" by Gregory: enter image description here 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: enter image description here

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.

Hans
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  • You can simply verify the result. Compute $\frac{dr'}{d\Phi}$ and compare to $r'\times n$, using the double cross product formula. – Lutz Lehmann Dec 06 '22 at 22:18
  • @LutzLehmann I have been trying this too. All I get is $(1-cos(\lambda)(n* r)(r\times n)+sin(\lambda)(rn)r-sin(\lambda)(r r)n$. I used the double cross prdouct formula only once,i.e. on the last term of $r\times r'$. – Hans Dec 06 '22 at 23:18
  • There is no $r\times r'$, there is $(r\times n)\times n$ in $r'\times n$ – Lutz Lehmann Dec 06 '22 at 23:32
  • @LutzLehmann Why is there no $r\times r'$? I thought $r'$ from Goldstein is what Gregory denotes by $r_{i}^{\lambda}$. – Hans Dec 06 '22 at 23:35
  • Yes, and $n$ is $k$. But still I do not see any cross product of two versions of $r$. – Lutz Lehmann Dec 06 '22 at 23:38
  • @LutzLehmann What I did is $r\times r'=r\times(rcos(\lambda)+n(n*r)(1-cos(\lambda))+(r\times n)sin(\lambda))$ – Hans Dec 06 '22 at 23:50
  • And again I ask you, where in your cited text do they consider $r\times r'$? The cross product is always with the rotation axis. – Lutz Lehmann Dec 07 '22 at 00:06
  • @LutzLehmann Of course....yes, I dont know what went wrong there! The rotation axis in Goldstein is $n$, isn't it? But I have to use $n\times r $ instead of $r\times n$ in the rotation formula from Goldstein in order to get the signs in $n\times r'$ right. That kind of puzzles me a bit. – Hans Dec 07 '22 at 12:32
  • Yes. You need to use $r\times n=-n\times r$ which gives $n=-k$. – Lutz Lehmann Dec 07 '22 at 13:01

1 Answers1

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$\br^λ$ 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 $$ \br^λ=A(λ)\br=\exp(λX)\br=\exp(λX)\br_{\parallel}+\exp(λX)\br_\perp =I\br_{\parallel}+(\cos(λ)I+\sin(λ) X) \br_\perp \\=(\bk·\br)\bk+\cos(λ)(\br-(\bk·\br) \bk)+\sin(λ)(\bk\times \br). $$

Lutz Lehmann
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  • I am not sure I completly understand whats going on here. I am not that familiar with Lie Groups. In the beginning you write scalar product but then use a $\times$. Did you mean cross-product by any chance? – Hans Dec 07 '22 at 12:36
  • Yes, $Xv$ is the matrix form of the linear map $k\times v$. – Lutz Lehmann Dec 07 '22 at 13:00
  • In the first part are so many things I haven't seen.Why is $A'(\lambda)=X(\lambda)A(\lambda)$. To name just one question. Do you have a book recommendation by any chance? I am not sure, but is $X$ the complex unit because $X^2$ maps elements v of the plane othogonal to $k$ onto -v? This is sort of similiar to $i^2=-1$...but why do you emphasize it that it behaves as a complex unit? The part with the exponential...are you using uniquness of ODEs in some way? And to get to Rodrigues rotation formula, are you using taylor expansion of exp? I am sorry for all the questions. – Hans Dec 07 '22 at 13:54
  • This is a definition for $X$ à la Lie-Algebra (not really, one usually takes the inverse from the left, $X=A^{-1}A'$). Then insert into the derivative of $AA^T=I$ to get the skew-symmetry. // This should be contained in the chapter on matrix groups in any linear algebra book, books discussion curvature and tension of curves, on the mechanics of solid bodies,... // Yes, the defining property of something to be called "complex unit" is $X^2=-I$. Prominent are the generators of the quaternions. // ... – Lutz Lehmann Dec 07 '22 at 14:02
  • Because then $e^{λX}=\cos(λ)I+\sin(λ)X$ on the orthogonal plane, without long discussion of the powers in the exponential series. // Yes, if you want to go the long way, the power series expansion of the exponential series., using $Iv = v_∥+v_⊥$ and $X^kv=X^kv_⊥=(-1)^mX^{k-2m}v_⊥$ for $k\ge 1$, $k-2<2m\le k$. – Lutz Lehmann Dec 07 '22 at 14:06
  • I am sorry, but for me there are too many things skipped to be able to really understand what is going on. As I said, I am not familiar with Lie-Algebras. Therefore the definition of X doesn't really make much sense to me. Furthermore, why is $exp(\lambda X)$ supposed to be the rotation matrix? If so, why is that the case. I tried to google Lie-groups and lie-groups and rotations, but I couldn't really find any reference/book that looked like it covered this. But thank you anyway! – Hans Dec 07 '22 at 15:32