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Infinitesimal Rotation

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

{\bf r}' = ({\hbox{\sf I}}+{\hbox{\sf e}}){\bf r},
\end{displaymath} (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
\end{displaymath} (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.
\end{displaymath} (3)

Now let
{\hbox{\sf A}}\equiv {\hbox{\sf I}}+{\hbox{\sf e}}.
\end{displaymath} (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}}.
\end{displaymath} (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},
\end{displaymath} (6)

{\hbox{\sf A}}^{-1}={\hbox{\sf I}}-{\hbox{\sf e}}
\end{displaymath} (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},
\end{displaymath} (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].
\end{displaymath} (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}.
\end{displaymath} (10)

Writing in Matrix form,

$\displaystyle d{\bf r}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}x\\  y\\  z\end{array}\right]\left[\begin{a...
...2\\  -d\Omega_3 & 0 & d\Omega_1\\  d\Omega_2 & -d\Omega_1 & 0\end{array}\right]$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{c} y\,d\Omega_3-z\,d\Omega_2\\  z\,d\Omega_1-x\,d\Omega_3\\  x\,d\Omega_2-y\,d\Omega_1\end{array}\right]$ (11)
  $\textstyle =$ $\displaystyle (y\,d\Omega_3-z\,d\Omega_2)\hat {\bf x}+(z\,d\Omega_1-x\,d\Omega_3)\hat {\bf y}+(x\,d\Omega_2-y\,d\Omega_1)\hat {\bf z}$  
  $\textstyle =$ $\displaystyle {\bf r}\times d\Omega.$ (12)

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

\omega \equiv {d\Omega\over dt} = \hat {\bf n}{d\phi \over dt}.
\end{displaymath} (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}.
\end{displaymath} (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.
\end{displaymath} (16)

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

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© 1996-9 Eric W. Weisstein