An infinitesimal transformation of a Vector
is given by
![\begin{displaymath}
{\bf r}' = ({\hbox{\sf I}}+{\hbox{\sf e}}){\bf r},
\end{displaymath}](i_489.gif) |
(1) |
where the Matrix
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
and
is established by the
equivalence of
![\begin{displaymath}
({\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}](i_492.gif) |
(2) |
![\begin{displaymath}
({\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}](i_493.gif) |
(3) |
Now let
![\begin{displaymath}
{\hbox{\sf A}}\equiv {\hbox{\sf I}}+{\hbox{\sf e}}.
\end{displaymath}](i_494.gif) |
(4) |
The inverse
is then
, since
![\begin{displaymath}
{\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}](i_497.gif) |
(5) |
Since we are defining our infinitesimal transformation to be a rotation, Orthogonality
of Rotation Matrices
requires that
![\begin{displaymath}
{\hbox{\sf A}}^{\rm T}={\hbox{\sf A}}^{-1},
\end{displaymath}](i_498.gif) |
(6) |
but
![\begin{displaymath}
{\hbox{\sf A}}^{-1}={\hbox{\sf I}}-{\hbox{\sf e}}
\end{displaymath}](i_499.gif) |
(7) |
![\begin{displaymath}
({\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}](i_500.gif) |
(8) |
so
and the infinitesimal rotation is Antisymmetric. It must therefore have
a Matrix of the form
![\begin{displaymath}
{\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}](i_502.gif) |
(9) |
The differential change in a vector
upon application of the Rotation Matrix is then
![\begin{displaymath}
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}](i_503.gif) |
(10) |
Writing in Matrix form,
Therefore,
![\begin{displaymath}
\left({d{\bf r}\over dt}\right)_{\rm rotation,\ body} = {\bf r}\times {d\Omega\over dt} = {\bf r}\times \omega,
\end{displaymath}](i_509.gif) |
(13) |
where
![\begin{displaymath}
\omega \equiv {d\Omega\over dt} = \hat {\bf n}{d\phi \over dt}.
\end{displaymath}](i_510.gif) |
(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
![\begin{displaymath}
\left({d{\bf r}\over dt}\right)_{\rm space} = \left({d{\bf r}\over dt}\right)_{\rm body} +\omega \times {\bf r}.
\end{displaymath}](i_511.gif) |
(15) |
This can be written as an operator equation, known as the Rotation Operator, defined as
![\begin{displaymath}
\left({d\over dt}\right)_{\rm space}= \left({d\over dt}\right)_{\rm body}+\omega \times.
\end{displaymath}](i_512.gif) |
(16) |
See also Acceleration, Euler Angles, Rotation, Rotation Matrix, Rotation Operator
© 1996-9 Eric W. Weisstein
1999-05-26