Rotations

For our passive rotations, we will change from one orthonormal, right-handed basis to another orthonormal,

$$\hat{e}'_i \cdot \hat{e}'_j = \delta_{ij}$$

right-handed basis.

$$\hat{e_1}'_i \hat{e_2}'_j \hat{e_3}'_k \epsilon_{ijk}=1$$

Note that the unit vectors $\hat{e}'_i$ are vectors in the original (not primed) frame.

If we rotate the coordinate axes, we can compute a vector in the new (primed) coordinates from the original vector by multiplying by a Rotation Matrix.

$$\vec{x}'=R\vec{x}$$

$$x_i'=R_{ij} x_j$$

This is the normal way we write a matrix multiplication.

$$\begin{pmatrix}x' \cr y' \cr z' \cr\end{pmatr... ...31}&R_{32}&R_{33}\end{pmatrix}\begin{pmatrix}x \cr y \cr z \end{pmatrix}$$

With a little thought, we can compute the elements of the rotation matrix. Lets compute the vector $\vec{x}'$, the position vector in the rotated coordinates. To get the $i$ component of $x$, just take the vector and dot it into the new unit vector $\hat{e}_i'$, written in the original coordinates.

$$x_i'=\hat{e}_i'\cdot\vec{x}=\hat{e}_i'\cdot\hat{e}_j x_j$$

Note that in the above equation, $\vec{x}$ is a vector that doesn't depend on the coordinates. The components of the vector $x_i'$ in the primed frame do depend on the primed coordinate axes.

So the rotation matrix is just given by

$$R_{ij}=\hat{e}_i'\cdot\hat{e}_j$$

where $\hat{e}_i'$ is written in the original coordinates.

Thus, each element of the rotation matrix is simply the cosine of the angle between a new coordinate axis and an old coordinate axis.

It is easy to see physically that the product of two rotations is just some other rotation.

$$R^{(1)} R^{(2)}= R^{(3)}$$

This means that the rotations form a group. We can also see physically that rotations (like matrices) do not commute.

As an example, lets make a rotation through a small angle $\theta$ in the $xy$ plane, leaving the $z$ axis unchanged. The angle between the $x$ and the $x'$ axis is $\theta$. The angle between the $y$ and the $y'$ axis is also $\theta$. The angle between the $x$ and the $y'$ axis is ${\pi\over 2}+\theta$. The angle between the $y$ and the $x'$ axis is ${\pi\over 2}-\theta$. The angle between the $z$ and the $z'$ axis is 0. All the other angle with the $z$ axis are ${\pi\over 2}$.

So we can write the rotation matrix.

$$R(\theta)=\begin{pmatrix}\cos(\theta) & \cos(... ...ta) & 0 \cr -\sin(\theta) & \cos(\theta) & 0 \cr 0 & 0 & 1\end{pmatrix}$$

In three dimensions, a rotation in the $xy$ plane can be said to be a rotation about the $z$ axis. This is not true in four dimensions.

One can write the dot product between two vectors just in terms of the lengths of vectors.

$$\vec{u}\cdot\vec{v}={1\over 2}\left(\Vert\vec{u}+\vec{v}\Vert^2-\Vert\vec{u}\Vert^2-\Vert\vec{v}\Vert^2\right)$$

Therefore, preserving the length of vectors implies that dot products are invariant. The invariance of dot products implies that both the lengths of vectors and the angle between vectors are unchanged in a rotation.

Lets use the fact that dot products are invariant to derive a property of the rotation matrices.

$$x_i'v_i'=x_iv_i$$

$$R_{ij} x_j R_{ik} v_k=x_iv_i$$

$$R_{ij} R_{ik} x_j v_k=x_iv_i$$

$$R^T_{ji} R_{ik} x_j v_k=x_iv_i$$

$$R^T_{ji} R_{ik} = \delta_{jk}$$

$$R^TR=1$$

$$R^{-1}=R^T$$

The inverse of a rotation matrix is its transpose. We call these matrices Orthogonal Matrices. The rotations in three dimensions are a representation of the Special Orthogonal Group SO(3). These matrices have determinant 1. If we include parity inversions with rotations we have the larger Orthogonal Group O(3). These matrices have $\det R =\pm 1$.

Any rotation in three dimensions can be written as a rotation in some plane or as a rotation about an axis orthogonal to that plane.

We can describe a constantly rotating coordinate system as a rotation in a plane by and angle

$$\theta = \omega t$$

where $\omega$ is the angular velocity. We can also define $\vec{\omega}$ to be a vector orthogonal to the plane with magnitude equal to that angular velocity $\Vert\vec{\omega}\Vert=\omega$. The rotation matrix depends on time. Since $R^TR=1$, the time derivative of that product must be zero at any time so lets evaluate it at $t=0$.

$${d\over dt}(R^TR)\vert _{t=0}=\dot{R}^TR(0)+R^T(0)\dot{R}= \dot{R}^T+\dot{R}=0$$

This implies that the time derivative of $R$ is an antisymmetric matrix, thus representable by an axial vector.