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[NOTES/CM-08008] Proper Rotations and $SO(3)$

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The definition and properties of proper rotations are presented.

By \(R\) we denote the rotation matrix relating position vectors of a point w.r.t. a frame \(K\) and w.r.t to a rotated frame \(K\Prime\). The matrix \(R\) has an expression in terms of direction cosines of system of axes \(K\) w.r.t \(K\Prime\). The following properties of the matrix \(R\) may be verified directly using the properties of direction cosines.

  1. The matrix $R$ is an orthogonal matrix \begin{equation} R^{T}R=R R^{T}=I \Longrightarrow R^{-1}=R^T. \end{equation} Here $R^T$ denotes the transpose of the matrix $R$. Also $\det R=\vec{e}_1\cdot(\vec{e}_2\times\vec{e}_3)$ is the triple product $\vec{e}_1\cdot(\vec{e}_2\times\vec{e}_3)$ of unit vectors along the old axes, using their components in $K^{\,\prime}$. Since the axes $K$ and ${K}\Prime$ are right handed, \begin{equation} \det R = \vec{e}_1\cdot(\vec{e}_2\times\vec{e}_3)=1. \end{equation} Thus every rotation may be represented by an orthogonal matrix $R$ with $\det R=+1$. In general, a three by three orthogonal matrix can have determinant $\pm1$. The rotations correspond to the case $\det R=1$. The case of $\det R=-1$ corresponds to a rotation followed by an inversion $\vec{x}^{\,\prime} = - \vec{x}$.
  2. The orthogonality property $R^{T} R=I$, $R\cdot R^{T}=I$ may also be written as \begin{eqnarray} \sum_{k} R_{i k} R_{j k}=\delta_{i j}, \qquad \sum_{k} R_{k i} R_{k j}=\delta_{i j} . \label{EQ13} \end{eqnarray}
  3. The components of a vector $\vec{A}$ in two different frames are related by \begin{equation} \left[\begin{array}{c} {A_1}\Prime\\ {A_2}\\ {A_3}\Prime \end{array}\right]=R\left[\begin{array}{c} {A_1}\\{A_2}\\{A_3} \end{array}\right] \end{equation} which will also be written as $\acute{A}=R A$, or, \begin{equation} A_{l}\Prime=\sum R_{lm} A_m \label{EQ15} \end{equation}
  4. If we have two vectors $\vec{A}$ and $\vec{B}$, their components will depend on the choice of coordinate axes. But the scalar product $\vec{A} \cdot \vec{B}$ computed from $A_1 B_1+A_2 B_2+A_3 B_3$ is the same as computed from ${A_1}\Prime B_1\Prime+{A_2}\Prime B_2\Prime+{A_3}\Prime B_3\Prime$. To prove the above statement we consider \begin{equation}\label{EQ90} \therefore  {A_1}\Prime B_1\Prime+{A_2}\Prime B_2\Prime+{A_3}\Prime B_3\Prime =\sum_i A_{i}B_{i}= A_1 B_1+A_2 B_2+A_3 B_3, \end{equation}Here in  \eqref{EQ90} the orthogonality property, \eqref{EQ13}, has been used. The above result,\eqref{EQ90}, can be stated as a property of rotations that under a rotation scalar product of two vectors remains invariant. This implies that the length of a vector $\Vert A\Vert(\equiv\sqrt{(A\cdot A})$, and the angle between two vectors, \begin{equation} \cos\theta=\frac{\vec{A}\cdot \vec{B}}{\Vert\vec{A}\Vert\, \Vert\vec{B}\Vert}, \end{equation} are also invariant. Similarly, it can be proved that the volume of a parallelopiped with edges, given by $\vec{A},\vec{B},\vec{C}$, $$V=\vec{A} \cdot \vec{B} \times \vec{C},$$ does not change under rotation.
  5. Conversely, every orthogonal matrix $R$ with $\det R=+1$ represents a rotation. Knowing $R$ and a set of axis $K$, the new set of axes $\acute{K}$ is easily found; the three columns of $R$ give the components of unit vectors along the three new axis.
  6. Let $S$ be a $3 \times 3$ matrix such that $ \sum {A_k\Prime}^2=\sum {A_k}^2$ holds for every vector $\vec{A}$, where \begin{equation} \left[\begin{array}{c} {A_1}\Prime\\ {A_2}\Prime\\ {A_3}\Prime \end{array}\right]= S \begin{bmatrix}A_1\\A_2\\A_3 \end{bmatrix} . \end{equation} In other words, if the length of a vector $\vec{A}$ as computed from its components $(A_1,A_2,A_3)$ is equal to that computed from the transformed components $({A_1}\Prime,{A_2}\Prime,{A_3}\Prime)$, then $S$ is an orthogonal matrix: \begin{equation} S^{T} S=I.\label{EQ22} \end{equation} The equality $\det S=1$ does not follow from \eqref{EQ22}, it implies only $\det S=\pm 1$.
  7. Since a rotation matrix $R$ is $3\times3$ orthogonal matrix with determinant 1, the eigenvalues of $R$ are given by $1, e^{i\theta}$, $e^{-i\theta}$. The eigenvector corresponding to $1$ will give the axis of rotation.

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