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I’m reading charpter 4.8 of Goldstein’s classical mechanics 3rd edition that deals with infinitesimal rotations, and the following is the part I got stuck:

(p.166~167) If $d\boldsymbol{\Omega}$ is to be a vector in the same sense as $\mathbf{r}$, it must transform under $\mathbf{B}$ in the same way. As we shall see, $d\boldsymbol{\Omega}$ passes most of this test for a vector, although in one respect it fails to make the grade. One way of examining the transformation properties of $d\boldsymbol{\Omega}$ is to find how the matrix $\boldsymbol{\epsilon}$ transforms under a coordinate transformation. The transformed matrix $\boldsymbol{\epsilon}’$ is obtained by a similarity transformation: \begin{equation} \boldsymbol{\epsilon}’=B\boldsymbol{\epsilon}B^{-1}\end{equation} As the antisymmetry property of a matrix is preserved under an orthogonal similarity transformation, $\boldsymbol{\epsilon}’$ consists of nonvanishing elements $d\Omega’_i$ such that \begin{equation}d\Omega'_i=|B|b_{ij}d\Omega_j.\end{equation}

How can I derive the last formula about $d\Omega’_i$? Also, why does the transformation law for an axial vectors which do not change sign under inversion have to be of the form of this formula? I appreciate your help.

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Let $\boldsymbol{\epsilon}$ the $3\times 3$ antisymmetric matrix in the textbook (equation 4.69) \begin{equation} \boldsymbol{\epsilon}\boldsymbol{=} \begin{bmatrix} \hphantom{-} 0 & \hphantom{\boldsymbol{-}}\mathrm d\Omega_3 & \boldsymbol{-}\mathrm d\Omega_2 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}}\\ \boldsymbol{-}\mathrm d\Omega_3 & \hphantom{-}0 & \hphantom{\boldsymbol{-}}\mathrm d\Omega_1 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}}\\ \hphantom{\boldsymbol{-}}\mathrm d\Omega_2 & \boldsymbol{-}\mathrm d\Omega_1 & \hphantom{-}0 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}} \end{bmatrix} \tag{01}\label{01} \end{equation} For the matrix $\boldsymbol{\epsilon}$ we have formally \begin{equation} \boldsymbol{\epsilon}\boldsymbol{=} \boldsymbol{-}\mathrm d\boldsymbol{\Omega}\boldsymbol{\times} \tag{02}\label{02} \end{equation} where \begin{equation} \mathrm d\boldsymbol{\Omega}\boldsymbol{\equiv}\left(\mathrm d\Omega_1,\mathrm d\Omega_2,\mathrm d\Omega_3\right) \tag{03}\label{03} \end{equation}

Now, under a general change of the coordinate system, that is under a similarity transformation by an invertible matrix $\mathrm S$, the antisymmetric matrix $\boldsymbol{\epsilon}$ in the new system is \begin{equation} \boldsymbol{\epsilon'}\boldsymbol{=}\mathrm S\,\boldsymbol{\epsilon}\,\mathrm S^{\boldsymbol{-}1} \tag{04}\label{04} \end{equation} The new matrix $\boldsymbol{\epsilon'}$ is not antisymmetric in general. But if the similarity is produced by an orthogonal matrix $\mathrm B$ with the well-known properties(1) \begin{equation} \mathrm B^{\boldsymbol{-}1}\boldsymbol{=}\mathrm B^{\boldsymbol{\top}}\,,\quad \det(\mathrm B)=\vert \mathrm B \vert \boldsymbol{=}\boldsymbol{\pm}1 \qquad (^{\boldsymbol{\top}}\boldsymbol{\equiv}\text{transpose}) \tag{05}\label{05} \end{equation} then \begin{equation} \boldsymbol{\epsilon'}\boldsymbol{=}\mathrm B\,\boldsymbol{\epsilon}\,\mathrm B^{\boldsymbol{-}1}\boldsymbol{=}\mathrm B\,\boldsymbol{\epsilon}\,\mathrm B^{\boldsymbol{\top}} \tag{06}\label{06} \end{equation} so \begin{equation} \boldsymbol{\epsilon'}^{\boldsymbol{\top}}\boldsymbol{=}\left(\mathrm B\,\boldsymbol{\epsilon}\,\mathrm B^{\boldsymbol{\top}}\right)^{\boldsymbol{\top}}=\mathrm B\,\boldsymbol{\epsilon}^{\boldsymbol{\top}}\,\mathrm B^{\boldsymbol{\top}}\stackrel{\boldsymbol{\epsilon}^{\boldsymbol{\top}}\boldsymbol{=-}\boldsymbol{\epsilon}}{\boldsymbol{=\!=\!=\!=}}\boldsymbol{-}\mathrm B\,\boldsymbol{\epsilon}\,\mathrm B^{\boldsymbol{\top}}\boldsymbol{=-}\boldsymbol{\epsilon'} \tag{07}\label{07} \end{equation} that is the matrix $\boldsymbol{\epsilon'}$ is antisymmetric and it could be represented by a vector $\mathrm d\boldsymbol{\Omega'}$ \begin{equation} \mathrm d\boldsymbol{\Omega'}\boldsymbol{\equiv}\left(\mathrm d\Omega'_1,\mathrm d\Omega'_2,\mathrm d\Omega'_3\right) \tag{08}\label{08} \end{equation} as follows \begin{equation} \boldsymbol{\epsilon'}\boldsymbol{=} \begin{bmatrix} \hphantom{-} 0 & \hphantom{\boldsymbol{-}}\mathrm d\Omega'_3 & \boldsymbol{-}\mathrm d\Omega'_2 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}}\\ \boldsymbol{-}\mathrm d\Omega'_3 & \hphantom{-}0 & \hphantom{\boldsymbol{-}}\mathrm d\Omega'_1 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}}\\ \hphantom{\boldsymbol{-}}\mathrm d\Omega'_2 & \boldsymbol{-}\mathrm d\Omega'_1 & \hphantom{-}0 \hphantom{-}\vphantom{\dfrac{a}{\frac{a}{b}}} \end{bmatrix} \boldsymbol{=} \boldsymbol{-}\mathrm d\boldsymbol{\Omega'}\boldsymbol{\times} \tag{09}\label{09} \end{equation}

Now, we'll try to find the relation between the infinitesimal vectors $\mathrm d\boldsymbol{\Omega},\mathrm d\boldsymbol{\Omega'}$.

So, let a vector $\mathbf r$ in the unprimed coordinate system. In this system the vector $\mathbf r$ is transformed by the infinitesimal transformation $\boldsymbol{\epsilon}$ to an infinitesimal vector $\mathrm d\mathbf s$
\begin{equation} \mathrm d\mathbf s \boldsymbol{=}\boldsymbol{\epsilon}\,\mathbf r \boldsymbol{=} \boldsymbol{-}\mathrm d\boldsymbol{\Omega}\boldsymbol{\times}\mathbf r \tag{10}\label{10} \end{equation}

In the primed coordinate system produced from the uprimed one via an orthogonal similarity transformation $\mathrm B$ we'll have \begin{equation} \mathrm d\mathbf s' \boldsymbol{=}\boldsymbol{\epsilon'}\,\mathbf r' \boldsymbol{=} \boldsymbol{-}\mathrm d\boldsymbol{\Omega'}\boldsymbol{\times}\mathbf r' \tag{11}\label{11} \end{equation} where \begin{equation} \mathbf r'\boldsymbol{=}\mathrm B\,\mathbf r\,, \quad\boldsymbol{\epsilon'}\boldsymbol{=} \mathrm B\,\boldsymbol{\epsilon}\,\mathrm B^{\boldsymbol{-}1} \tag{12}\label{12} \end{equation} But then we must have \begin{align} \mathrm d\mathbf s'\boldsymbol{=}\mathrm B\,\mathrm d\mathbf s \quad &\stackrel{\eqref{10},\eqref{11}} {\boldsymbol{=\!=\!=\!\Longrightarrow}} \boldsymbol{-}\mathrm d\boldsymbol{\Omega'}\boldsymbol{\times}\mathbf r' \boldsymbol{=}\mathrm B\,\left(\boldsymbol{-}\mathrm d\boldsymbol{\Omega}\boldsymbol{\times}\mathbf r\right)\quad \stackrel{\eqref{12}} {\boldsymbol{=\!=\!=\!\Longrightarrow}} \nonumber\\ & \mathrm d\boldsymbol{\Omega'}\boldsymbol{\times}\mathrm B\,\mathbf r \boldsymbol{=}\mathrm B\,\left(\mathrm d\boldsymbol{\Omega}\boldsymbol{\times}\mathbf r\right) \tag{13}\label{13} \end{align}

The following identity shows how the outer product of two vectors is transformed by an orthogonal transformation $\mathrm B$ in terms of the transforms of these two vectors \begin{equation} \boxed{\:\: \mathrm B\left(\mathbf a \boldsymbol{\times}\mathbf b\right)\boldsymbol{=}\vert \mathrm B \vert \cdot\left(\mathrm B\mathbf a \boldsymbol{\times}\mathrm B\mathbf b\right)\stackrel{\eqref{05}} {\boldsymbol{=\!=}}\boldsymbol{\pm}\left(\mathrm B\mathbf a \boldsymbol{\times}\mathrm B\mathbf b\right) \vphantom{\dfrac{a}{b}}\:\:} \tag{14}\label{14} \end{equation} For a proof of the identity \eqref{14} see the ADDENDUM(2).

Using identity \eqref{14} equation \eqref{13} yields \begin{align} \mathrm d\boldsymbol{\Omega'}\boldsymbol{\times}\mathrm B\,\mathbf r \boldsymbol{=}&\mathrm B\left(\mathrm d\boldsymbol{\Omega}\boldsymbol{\times}\mathbf r\right) \boldsymbol{=}\vert \mathrm B \vert \cdot\bigl[\mathrm B\left(\mathrm d\boldsymbol{\Omega}\right) \boldsymbol{\times}\mathrm B\mathbf r\bigr]\boldsymbol{=\!=\!=\!\Longrightarrow} \nonumber\\ & \underbrace{\bigl[\mathrm d\boldsymbol{\Omega'}\boldsymbol{-}\vert \mathrm B \vert \cdot\mathrm B\left(\mathrm d\boldsymbol{\Omega}\right)\bigr]}{}\boldsymbol{\times}\mathrm B\,\mathbf r \boldsymbol{=}\boldsymbol{0} \tag{15}\label{15} \end{align} Since \eqref{15} must be valid for any $\mathbf r$, so for any $\mathrm B\,\mathbf r$, the infinitesimal vector over the under-brace must be $\boldsymbol{0}$ that is \begin{equation} \boxed{\:\: \mathrm d\boldsymbol{\Omega'}\boldsymbol{=}\vert \mathrm B \vert \cdot\mathrm B\left(\mathrm d\boldsymbol{\Omega}\right)\boldsymbol{=\pm}\mathrm B\left(\mathrm d\boldsymbol{\Omega}\right)\vphantom{\dfrac{a}{b}}\:\:} \tag{16}\label{16} \end{equation} qed.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

(1) Note that an orthogonal matrix $\:\mathrm B\:$ with $\:\vert \mathrm B \vert\boldsymbol{=+}1\:$ represents a pure rotation, while an orthogonal matrix $\:\mathrm B\:$ with $\:\vert \mathrm B \vert\boldsymbol{=-}1\:$ represents a pure rotation plus a reflection.


(2) ADDENDUM

If $\mathrm M$ is a real $3\times 3$ orthogonal matrix then :
\begin{equation} \mathrm M^{\boldsymbol{-}1}\boldsymbol{=}\mathrm M^{\boldsymbol{\top}}\,,\quad \det(\mathrm M)\boldsymbol{=}\vert \mathrm M \vert \boldsymbol{=}\boldsymbol{\pm}1 \qquad (^{\boldsymbol{\top}}\boldsymbol{\equiv}\text{transpose}) \tag{A-01}\label{A-01} \end{equation} and \begin{equation} \boxed{\:\: \mathrm M\left(\mathbf a \boldsymbol{\times}\mathbf b\right)\boldsymbol{=}\vert \mathrm M \vert\cdot \left(\mathrm M\mathbf a \boldsymbol{\times}\mathrm M\mathbf b\right)\boldsymbol{=}\boldsymbol{\pm}\left(\mathrm M\mathbf a \boldsymbol{\times}\mathrm M\mathbf b\right)\quad \mathbf a,\mathbf b \in \mathbb R^{3} \vphantom{\dfrac{a}{b}}\:\:} \tag{A-02}\label{A-02} \end{equation}

Proof of identity \eqref{A-02}

If $\:\mathrm M\:$ is an invertible linear transformation in $\:\mathbb{C}^{3}\:$ represented by the $\:3\times 3\:$ complex matrix
\begin{equation} \mathrm M\boldsymbol{=} \begin{bmatrix} \mathrm m_{11} & \mathrm m_{12} & \mathrm m_{13} \\ \mathrm m_{21} & \mathrm m_{22} & \mathrm m_{23} \\ \mathrm m_{31} & \mathrm m_{32} & \mathrm m_{33} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{\rho}_{1} \\ \boldsymbol{\rho}_{2} \\ \boldsymbol{\rho}_{3} \end{bmatrix} \tag{A-03}\label{A-03} \end{equation} where $\:\boldsymbol{\rho}_{i}\; (i=1,2,3) \:$ denote its row complex 3-vectors \begin{equation} \boldsymbol{\rho}_{1}\boldsymbol{=}\left(\mathrm m_{11},\mathrm m_{12},\mathrm m_{13}\right)\quad \boldsymbol{\rho}_{2}\boldsymbol{=}\left(\mathrm m_{21},\mathrm m_{22},\mathrm m_{23}\right)\quad \boldsymbol{\rho}_{3}\boldsymbol{=}\left(\mathrm m_{31},\mathrm m_{32},\mathrm m_{33}\right) \tag{A-04}\label{A-04} \end{equation} then its inverse $\:\mathrm M^{\boldsymbol{-}1}\:$ is expressed in column complex 3-vector form as follows \begin{equation} \mathrm M^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \dfrac{\left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)}{\boldsymbol{\rho}_{1} \boldsymbol{\cdot} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)} & \dfrac{\left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)}{\boldsymbol{\rho}_{2} \boldsymbol{\cdot} \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)} & \dfrac{\left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)}{\boldsymbol{\rho}_{3} \boldsymbol{\cdot}\left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)} \end{bmatrix} \tag{A-05}\label{A-05} \end{equation} and since from linear algebra for the determinant of $\:\mathrm M\:$ we have \begin{equation} \det\left(\mathrm M\right) \boldsymbol{=}\vert \mathrm M \vert \boldsymbol{=}\boldsymbol{\rho}_{1} \boldsymbol{\cdot} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)\boldsymbol{=}\boldsymbol{\rho}_{2} \boldsymbol{\cdot} \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right) \boldsymbol{=}\boldsymbol{\rho}_{3} \boldsymbol{\cdot} \left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right) \boldsymbol{\neq} 0 \tag{A-06}\label{A-06} \end{equation}
equation \eqref{A-05} yields \begin{equation} \boxed{\:\: \mathrm M^{\boldsymbol{-}1}\boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert} \begin{bmatrix} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right) & \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right) & \left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right) \end{bmatrix}\:\:} \tag{A-07}\label{A-07} \end{equation}

Note that here the outer product $\:\mathbf{a}\boldsymbol{\times}\mathbf{b}\:$ of two complex 3-vectors $\:\mathbf{a}= \left(\mathrm{a}_{1},\mathrm{a}_{2},\mathrm{a}_{3}\right)\:$ and $\:\mathbf{b}= \left(\mathrm{b}_{1},\mathrm{b}_{2},\mathrm{b}_{3}\right)\:$ is defined as if they are real 3-vectors \begin{equation} \mathbf{a} \boldsymbol{\times}\mathbf{b}= \begin{bmatrix} \mathrm{a}_{2}\mathrm{b}_{3} - \mathrm{a}_{3}\mathrm{b}_{2} \\ \mathrm{a}_{3}\mathrm{b}_{1} - \mathrm{a}_{1}\mathrm{b}_{3} \\ \mathrm{a}_{1}\mathrm{b}_{2} - \mathrm{a}_{2}\mathrm{b}_{1} \end{bmatrix} \tag{A-08}\label{A-08} \end{equation} while the expression $\:\mathbf{a}\boldsymbol{\cdot}\mathbf{b}\:$ is defined by \begin{equation} \mathbf{a} \boldsymbol{\cdot}\mathbf{b}=\mathrm{a}_{1}\mathrm{b}_{1}+\mathrm{a}_{2}\mathrm{b}_{2}+\mathrm{a}_{3}\mathrm{b}_{3} \tag{A-09}\label{A-09} \end{equation} not to be confused with the usual inner product in $\:\mathbb{C}^{3}\:$ \begin{equation} \boldsymbol{\langle}\mathbf{a},\mathbf{b}\boldsymbol{\rangle}=\mathrm{a}_{1}\overline{\mathrm{b}}_{1}+\mathrm{a}_{2}\overline{\mathrm{b}}_{2}+\mathrm{a}_{3}\overline{\mathrm{b}}_{3} \tag{A-10}\label{A-10} \end{equation}

Now, in the special case of a real orthogonal matrix $\:\mathrm M\:$ we have $\:\mathrm M^{\boldsymbol{-}1}\boldsymbol{=}\mathrm M^{\boldsymbol{\top}}$, see eq.\eqref{A-01}. From eq.\eqref{A-03} \begin{equation} \mathrm M^{\boldsymbol{\top}}\boldsymbol{=} \begin{bmatrix} \mathrm m_{11} & \mathrm m_{21} & \mathrm m_{31} \\ \mathrm m_{12} & \mathrm m_{22} & \mathrm m_{32} \\ \mathrm m_{13} & \mathrm m_{23} & \mathrm m_{33} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \boldsymbol{\rho}^{\boldsymbol{\top}}_{1} & \boldsymbol{\rho}^{\boldsymbol{\top}}_{2} & \boldsymbol{\rho}^{\boldsymbol{\top}}_{3} \end{bmatrix} \tag{A-11}\label{A-11} \end{equation} Equating $\:\mathrm M^{\boldsymbol{-}1}\:$ and $\:\mathrm M^{\boldsymbol{\top}}\:$ as given by equations \eqref{A-07} and \eqref{A-11} respectively we have \begin{equation} \boldsymbol{\rho}_{1}\boldsymbol{=}\dfrac{\left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)}{\vert \mathrm M \vert}\quad \boldsymbol{\rho}_{2}\boldsymbol{=}\dfrac{\left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)}{\vert \mathrm M \vert}\quad \boldsymbol{\rho}_{3}\boldsymbol{=}\dfrac{\left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)}{\vert \mathrm M \vert} \tag{A-12}\label{A12} \end{equation} Now, \begin{equation} \mathrm M\left(\mathbf a \boldsymbol{\times}\mathbf b\right)\boldsymbol{=} \begin{bmatrix} \boldsymbol{\rho}_{1}\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \boldsymbol{\rho}_{2}\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \boldsymbol{\rho}_{3}\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \end{bmatrix} \boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert} \begin{bmatrix} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \end{bmatrix} \tag{A-13}\label{A-13} \end{equation} Using the identity \begin{equation} \left(\mathbf c\boldsymbol{\times}\mathbf d\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right)\boldsymbol{=}\left(\mathbf c\boldsymbol{\cdot}\mathbf a\right)\left(\mathbf d\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\mathbf c\boldsymbol{\cdot}\mathbf b\right)\left(\mathbf d\boldsymbol{\cdot}\mathbf a\right) \tag{A-14}\label{A-14} \end{equation} we have \begin{align} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) & \boldsymbol{=}\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf a\right) \tag{A-15.1}\label{A-15.1}\\ \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) & \boldsymbol{=}\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf a\right) \tag{A-15.2}\label{A-15.2}\\ \left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) & \boldsymbol{=}\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf a\right) \tag{A-15.3}\label{A-15.3} \end{align} and equation \eqref{A-13} yields \begin{align} \mathrm M\left(\mathbf a \boldsymbol{\times}\mathbf b\right) & \boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert} \begin{bmatrix} \left(\boldsymbol{\rho}_{2} \boldsymbol{\times}\boldsymbol{\rho}_{3}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \left(\boldsymbol{\rho}_{3} \boldsymbol{\times}\boldsymbol{\rho}_{1}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \\ \left(\boldsymbol{\rho}_{1} \boldsymbol{\times}\boldsymbol{\rho}_{2}\right)\boldsymbol{\cdot}\left(\mathbf a \boldsymbol{\times}\mathbf b\right) \end{bmatrix} \nonumber\\ & \boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert} \begin{bmatrix} \left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf a\right) \\ \left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf a\right) \\ \left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf a\right)\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf b\right)\boldsymbol{-}\left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf b\right)\left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf a\right) \end{bmatrix} \nonumber\\ & \boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert} %\begin{pmatrix} \underbrace{\begin{bmatrix} \left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf a\right)\\ \left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf a\right)\\ \left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf a\right) \end{bmatrix}}_{\mathrm M\mathbf a} \boldsymbol{\times} \underbrace{\begin{bmatrix} \left(\boldsymbol{\rho}_{1}\boldsymbol{\cdot}\mathbf b\right)\\ \left(\boldsymbol{\rho}_{2}\boldsymbol{\cdot}\mathbf b\right)\\ \left(\boldsymbol{\rho}_{3}\boldsymbol{\cdot}\mathbf b\right) \end{bmatrix}}_{\mathrm M\mathbf b} \vphantom{\dfrac{\dfrac{x}{y}}{\dfrac{x}{y}}} %\end{pmatrix} \boldsymbol{=} \dfrac{1}{\vert \mathrm M \vert}\left(\mathrm M\mathbf a \boldsymbol{\times}\mathrm M\mathbf b\right) \tag{A-16}\label{A-16} \end{align} Given that $\:\vert \mathrm M \vert^2\boldsymbol{=}1$, that is $\:1/\vert \mathrm M \vert\boldsymbol{=}\vert \mathrm M \vert$, we have finally the proof of identity \eqref{A-02}.


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How is the angular velocity vector $\vec{\omega}$ is transfered ?

Lets look at this equation:

$$\vec{v}=\vec{\omega}\times \vec{r}=\tilde{{\omega}}\,\vec{r}\tag 1$$

where

$$\tilde{{\omega}}= \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix}$$

a antisymmetric skew matrix.

we transformed the vector $\vec{v}$ and $\vec{r}$ with arbitrary orthogonal transformation matrix $S$ where $\det(S)\ne 0=\pm 1$

$\vec{v'}=S\,\vec{v}$ ,$\vec{r'}=S\,\vec{r}$

and get with equation (1)

$$\vec{v'}=S\,\vec{v}=S\,\tilde{\omega}\,\vec{r}=S\,\tilde{\omega}\,S^{T}\,\vec{r'} \overset{!}{=}\tilde{\omega'}\,\vec{r'} \quad $$

$\Rightarrow$

$$\tilde{\omega'}=S\,\tilde{\omega}\,S^{T }\tag 2$$

Example:

$$S=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

$\Rightarrow$ equation (2)

$$\begin{bmatrix} 0 & -\omega'_z & \omega'_y \\ \omega'_z & 0 & -\omega'_x \\ -\omega'_y & \omega'_x & 0 \\ \end{bmatrix}=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\,\begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix}\,\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}= \begin{bmatrix} 0 & \omega_z & -\omega_y \\ -\omega_z & 0 & -\omega_x \\ \omega_y & \omega_x & 0 \\ \end{bmatrix}$$ $\Rightarrow$ $$\vec{\omega'}=\begin{bmatrix} \omega_x \\ -\omega_y \\ -\omega_z\\ \end{bmatrix}=-S\,\vec{\omega}=\det(S)\,S\,\vec{\omega}$$

$$\boxed{\vec{\omega'}=\det(S)\,S\,\vec{\omega}}$$

obviously is the transformation of the angular velocity vector $\vec{\omega}$ , if the determinate of the transformation matrix $S$ not equal one ,different from the transformation of a "regular" vector, this is why the angular velocity vector is a pseudovector.

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  • $\begingroup$ $B\epsilon B^{-1}$ Goldstein book is wrong?. The result is in general not anti symmetric matrix, I correct it. $\endgroup$ – Eli Mar 13 at 7:25
  • $\begingroup$ Thank you for your answer. How can I prove that the minus sign of $\vec{\omega ’}=-S\vec{\omega}$ is in fact the determinant of the transformation $S$ in general case? $\endgroup$ – SH Lee Mar 13 at 7:36
  • $\begingroup$ Please see Goldstein book $\endgroup$ – Eli Mar 13 at 8:57
  • $\begingroup$ @Frobenius a little bit confusing because in this case $B^{-1}=B^{T}$ $\endgroup$ – Eli Mar 13 at 13:29
  • $\begingroup$ @Frobenius I edit the dokument so I think it is now correct?. Thank for your remarks $\endgroup$ – Eli Mar 13 at 13:44

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