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===== ROTATIONS IN SPACE =====

==== INTRODUCTION ====

In Cartesian coordinates, the natural orthonormal basis is $\{\ii,\jj,\kk\}$,
where $\ii\equiv\xhat$, $\jj\equiv\yhat$, $\kk\equiv\zhat$ denote the unit
vectors in the $x$, $y$, $z$ directions, respectively.  The position vector
from the origin to the point ($x$,$y$,$z$) takes the form
$$\rr = x \,\ii + y \,\jj + z \,\kk$$
Note that $\ii$, $\jj$, $\kk$ are constant.

A moving object has a position vector given by
$$\rr(t) = x(t) \,\ii + y(t) \,\jj + z(t) \,\kk$$
Its velocity $\vv$ and acceleration $\aa$ are obtained by differentiation,
resulting in
\begin{eqnarray}
 \vv &=& \dot{\rr} = \dot{x} \,\ii + \dot{y} \,\jj + \dot{z} \,\kk \\
 \aa &=& \ddot{\rr} = \ddot{x} \,\ii + \ddot{y} \,\jj + \ddot{z} \,\kk
\end{eqnarray}
where dots denote differentiation with respect to $t$.

As in the plane, if we wish to describe things as seen from some point
$(p,q,s)$ other than the origin, all we have to do is replace $\rr$ by
$$\rr_\subM = \rr - \RR$$
where
$$\RR = p \,\ii + q \,\jj + s \,\kk$$

==== SPHERICAL COORDINATES ====

In spherical coordinates
\begin{eqnarray}
 x &=& r\sin\theta\cos\phi \\
 y &=& r\sin\theta\sin\phi \\
 z &=& r\cos\theta
\end{eqnarray}
the natural orthonormal basis is $\{\rhat,\that,\phat\}$, where
\begin{eqnarray}
\rhat
 &=& \sin\theta\cos\phi \>\ii + \sin\theta\sin\phi \>\jj + \cos\theta \>\kk \\
\that
 &=& \cos\theta\cos\phi \>\ii + \cos\theta\sin\phi \>\jj - \sin\theta \>\kk \\
\phat &=& -\sin\phi \>\ii + \cos\phi \>\jj
\end{eqnarray}
Again, this is a basis everywhere \textit{except} at the origin, since neither
$\theta$ nor $\phi$ are defined there.

Consider an observer located at the point ($R$,$\Theta$,$\Phi$), \textit{not}
at the origin, whose natural basis is just $\{\rhat,\that,\phat\}$.  With
respect to this observer, a moving object therefore has a \textit{relative}
position vector of the form
$$\rr_\subM(t) = Z(t) \,\rhat - Y(t) \,\that + X(t) \,\phat$$
for some functions $X$, $Y$, $Z$.

We will normally take our observer to be on the surface of the Earth, which we
will approximate as a sphere.  Roughly speaking, $\Theta$ and $\Phi$ give the
latitude and longitude of the observer, respectively; $R$ is the radius of the
Earth.

Note that the equator corresponds to $\theta={\pi\over2}$, and that $\theta$
\textit{decreases} as you approach the North Pole.  The functions
($X$,$Y$,$Z$) define \textit{Cartesian} coordinates for the observer: $Z$ is
altitude, $Y$ is distance to the \textit{north}, and $X$ is distance to the
\textit{east}; this explains the peculiar conventions in defining $X$, $Y$,
$Z$.

The ``true'' position is given by
$$\rr(t) = \RR + \rr_\subM(t)$$
where
$$\RR = R \,\rhat$$

==== ROTATING FRAME ====

An observer ``standing still'' on the surface of the Earth is really a
rotating observer whose position is given by
\begin{eqnarray}
 r &=& R = \hbox{constant} \\
 \theta &=& \Theta = \hbox{constant} \\
 \phi &=& \Phi = \Omega t
\end{eqnarray}
Observers in this frame perceive $\rhat$, $\that$, $\phat$ to be constant.
(The sun ``rises'' and ``sets''!)  They will therefore compute the relative
velocity and acceleration of a moving object with (relative) position vector
$\rr_\subM$ by taking derivatives of the \textit{coefficients}:
\begin{eqnarray}
 \vv_\subM(t) = \dot{Z} \,\rhat - \dot{Y} \,\that + \dot{X} \,\phat \\
 \aa_\subM(t) = \ddot{Z} \,\rhat - \ddot{Y} \,\that + \ddot{X} \,\phat
\end{eqnarray}

The basis vectors now take the form
\begin{eqnarray}
\rhat &=& \sin\theta\cos(\Omega t) \>\ii + \sin\theta\sin(\Omega t) \>\jj
 + \cos\theta \>\kk \\
\that &=& \cos\theta\cos(\Omega t) \>\ii + \cos\theta\sin(\Omega t) \>\jj
 - \sin\theta \>\kk \\
\phat &=& -\sin(\Omega t) \>\ii + \cos(\Omega t) \>\jj
\end{eqnarray}
so that
\begin{eqnarray}
\rhatdot &=& -\Omega\sin\theta\sin(\Omega t) \>\ii
 + \Omega\sin\theta\cos(\Omega t) \>\jj \\
\thatdot &=& -\Omega\cos\theta\sin(\Omega t) \>\ii
 + \Omega\cos\theta\cos(\Omega t) \>\jj \\
\phatdot &=& -\Omega\cos(\Omega t) \>\ii - \Omega\sin(\Omega t) \>\jj
\end{eqnarray}
Comparing these equations with the preceding ones, we see that
\begin{eqnarray}
\rhatdot &=& \Omega\sin\theta \,\phat = \om \times \rhat \\
\thatdot &=& \Omega\cos\theta \,\phat = \om \times \that \\
\phatdot &=& -\Omega\sin\theta \,\rhat - \Omega\cos\theta \,\that
 = \om \times \phat
\end{eqnarray}
where we have introduced the angular velocity
$$\om
  = \Omega \,\kk
  = \Omega \, ( \cos\theta \,\rhat - \sin\theta \,\that )
  $$

Thus, just as in the 2-dimensional case, for \textit{any} relative vector of
the form
$$\FF(t) = f(t) \,\rhat(t) + g(t) \,\that(t) + h(t) \,\phat(t)$$
we have
\begin{eqnarray}
\dot{\FF}
 &=& \left( \dot{f} \,\rhat + \dot{g} \,\that + \dot{h} \,\phat \right)
 + \left( f \,\rhatdot + g \,\thatdot + h \,\phatdot \right) \\
 &=& \left( \dot{f} \,\rhat + \dot{g} \,\that + \dot{h} \,\phat \right)
 + \left( f \,\om\times\rhat + g \,\om\times\that + h \,\om\times\phat
 \right) \\
 &=& \left( \dot{f} \,\rhat + \dot{g} \,\that + \dot{h} \,\phat \right)
 + \om \times \FF
\end{eqnarray}
As before, the first term is the ``naive'' derivative of $\FF$; this ``naive''
differentiation is precisely what was used to obtain $\vv_\subM$ and then
$\aa_\subM$ starting from $\rr_\subM$.

We are finally ready to compare the relative and ``true'' velocities and
accelerations.  
Differentiating
$$\rr = \RR + \rr_\subM$$
we obtain
\begin{eqnarray}
\vv = \dot{\rr}
 &=& \dot{\RR} + \dot{\rr}_\subM \\
 &=& \om\times\RR + \left( \vv_\subM + \om\times\rr_\subM \right)
\end{eqnarray}
Further differentiation yields
\begin{eqnarray}
\aa = \dot{\vv}
 &=& \om\times\dot{\RR}
 + \left( \dot{\vv}_\subM + \om\times\dot{\rr}_\subM \right) \\
 &=& 	\om\times(\om\times\RR)
 + \left(\aa_\subM + \om\times\vv_\subM \right)
 + \om\times\left( \vv_\subM + \om\times\rr_\subM \right) \\
 &=& \om\times(\om\times\RR)
 + \aa_\subM + 2\,\om\times\vv_\subM + \om\times(\om\times\rr_\subM) \\
\end{eqnarray}

Rewriting these expressions slightly, we obtain the following equations for
the relative velocity and acceleration:
\begin{eqnarray}
 \vv_\subM &=& \vv - \om\times\rr \\
 \aa_\subM &=& \aa - 2\,\om\times\vv_\subM - \om\times(\om\times\rr)
\end{eqnarray}
As before, the \textit{effective} acceleration $\aa_\subM$ therefore consists
of 3 parts: the ``true'' acceleration $\aa$, the \textit{centrifugal}
acceleration $-\om\times(\om\times\rr)$ and the \textit{Coriolis} acceleration
$-2\,\om\times\vv_\subM$.

On the surface of the Earth, we have
$$\rr \approx \RR$$
so that the centrifugal acceleration can be approximated as
$-\om\times(\om\times\RR)$.  In analogy with the planar case, the centrifugal
acceleration always points away from the axis of the Earth's rotation, and is
strongest at the equator.  This acceleration is perceived as a small (less
than~1\%) correction to the acceleration due to gravity: Straight down, as
defined by a plumb bob, does \textit{not} point towards the center of the
Earth!  ((Straight down is, however, perpendicular to the surface of the
Earth: The centrifugal force has deformed the Earth's surface, resulting in an
equatorial radius which is slightly (just over 20 km) greater than the polar
radius.))  For a more detailed discussion, see pages 390--391 of Marion and
Thornton.

Finally, \textit{for motion parallel to the surface of the Earth}, the
Coriolis acceleration always points to the \textit{right} of the direction of
motion $\vv_\subM$ in the Northern Hemisphere (and to the \textit{left} in the
Southern Hemisphere).