Infinitesimal Rotation

An infinitesimal transformation of a Vector ${\bf r}$ is given by

$${\bf r}' = ({\hbox{\sf I}}+{\hbox{\sf e}}){\bf r},$$ (1)

where the Matrix ${\hbox{\sf e}}$ is infinitesimal and I is the Identity Matrix. (Note that the infinitesimal transformation may not correspond to an inversion, since inversion is a discontinuous process.) The Commutativity of infinitesimal transformations ${\hbox{\sf e}}_1$ and ${\hbox{\sf e}}_2$ is established by the equivalence of

$$({\hbox{\sf I}}+{\hbox{\sf e}}_1)({\hbox{\sf I}}+{\hbox{\sf ... ...e}}_2 \approx {\hbox{\sf I}}+{\hbox{\sf e}}_1+{\hbox{\sf e}}_2$$ (2)

$$({\hbox{\sf I}}+{\hbox{\sf e}}_2)({\hbox{\sf I}}+{\hbox{\sf ... ...}}_1 \approx {\hbox{\sf I}}+{\hbox{\sf e}}_2+{\hbox{\sf e}}_1.$$ (3)

Now let

$${\hbox{\sf A}}\equiv {\hbox{\sf I}}+{\hbox{\sf e}}.$$ (4)

The inverse ${\hbox{\sf A}}^{-1}$ is then ${\hbox{\sf I}}-{\hbox{\sf e}}$, since

$${\hbox{\sf A}}{\hbox{\sf A}}^{-1}= ({\hbox{\sf I}}+{\hbox{\s... ...}) = {\hbox{\sf I}}^2-{\hbox{\sf e}}^2 \approx {\hbox{\sf I}}.$$ (5)

Since we are defining our infinitesimal transformation to be a rotation, Orthogonality of Rotation Matrices requires that

$${\hbox{\sf A}}^{\rm T}={\hbox{\sf A}}^{-1},$$ (6)

but

$${\hbox{\sf A}}^{-1}={\hbox{\sf I}}-{\hbox{\sf e}}$$ (7)

$$({\hbox{\sf I}}+{\hbox{\sf e}})^{\rm T} = {\hbox{\sf I}}^{\r... ...+{\hbox{\sf e}}^{\rm T}={\hbox{\sf I}}+{\hbox{\sf e}}^{\rm T},$$ (8)

so ${\hbox{\sf e}}= -{\hbox{\sf e}}^{\rm T}$ and the infinitesimal rotation is Antisymmetric. It must therefore have a Matrix of the form

$${\hbox{\sf e}} = \left[{\matrix{ 0 & d\Omega_3 & -d\Omega_2... ..._3 & 0 & d\Omega_1\cr d\Omega_2 & -d\Omega_1 & 0\cr}}\right].$$ (9)

The differential change in a vector ${\bf r}$ upon application of the Rotation Matrix is then

$$d{\bf r} \equiv {\bf r}'-{\bf r} = ({\hbox{\sf I}}+{\hbox{\sf e}}){\bf r}-{\bf r}={\hbox{\sf e}}{\bf r}.$$ (10)

Writing in Matrix form,

$$d{\bf r} = \left[\begin{array}{c}x\\ y\\ z\end{array}\right]\left[\begin{a... ...2\\ -d\Omega_3 & 0 & d\Omega_1\\ d\Omega_2 & -d\Omega_1 & 0\end{array}\right]$$

$$= \left[\begin{array}{c} y\,d\Omega_3-z\,d\Omega_2\\ z\,d\Omega_1-x\,d\Omega_3\\ x\,d\Omega_2-y\,d\Omega_1\end{array}\right]$$ (11)

$$= (y\,d\Omega_3-z\,d\Omega_2)\hat {\bf x}+(z\,d\Omega_1-x\,d\Omega_3)\hat {\bf y}+(x\,d\Omega_2-y\,d\Omega_1)\hat {\bf z}$$

$$= {\bf r}\times d\Omega.$$ (12)

Therefore,

$$\left({d{\bf r}\over dt}\right)_{\rm rotation,\ body} = {\bf r}\times {d\Omega\over dt} = {\bf r}\times \omega,$$ (13)

where

$$\omega \equiv {d\Omega\over dt} = \hat {\bf n}{d\phi \over dt}.$$ (14)

The total rotation observed in the stationary frame will be a sum of the rotational velocity and the velocity in the rotating frame. However, note that an observer in the stationary frame will see a velocity opposite in direction to that of the observer in the frame of the rotating body, so

$$\left({d{\bf r}\over dt}\right)_{\rm space} = \left({d{\bf r}\over dt}\right)_{\rm body} +\omega \times {\bf r}.$$ (15)

This can be written as an operator equation, known as the Rotation Operator, defined as

$$\left({d\over dt}\right)_{\rm space}= \left({d\over dt}\right)_{\rm body}+\omega \times.$$ (16)

See also Acceleration, Euler Angles, Rotation, Rotation Matrix, Rotation Operator