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===== THE FOUCAULT PENDULUM =====

Marion and Thornton gives the standard treatment of the Foucault pendulum in
Example 10.5 on pages 398--401.  However, there is an easier way to get the
same result.  The basic idea is to separate the problem into 2 parts: an
ordinary pendulum influenced by gravity, and a Coriolis-like effect acting on
the direction of motion of the pendulum.

Let $\qq$ denote the initial direction of motion of the pendulum.  We are not
concerned with the periodic nature of the motion, but only with the change in
$\qq$ from one period to the next.  With respect to an observer on the Earth
we have
$$\qq = X \,\phat - Y \,\that$$
where the signs are chosen such that $X$ measures distance to the East and Y
to the North.

Consider now the derivative of $\qq$:
\begin{eqnarray}
\dot{\qq}
 &=& \left( \dot{X} \,\phat - \dot{Y} \,\that \right)
 + \left( X \,\phatdot - Y \,\thatdot \right) \\
 &=& \left( \dot{X} \,\phat - \dot{Y} \,\that \right)
 - \Omega\cos\theta \left( X \,\that + Y \,\phat \right)
 - \Omega\sin\theta X \rhat
\end{eqnarray}
Now the last term involves vertical motion, which we will ignore --- gravity
ensures that the pendulum moves (nearly) orthogonal to $\rhat$.  But gravity
is the only force involved, so that the remaining terms should vanish.
Equivalently, the \textit{only} change in $\qq$ should be in the $\rr$ direction
\textit{with respect to a fixed observer}.  We therefore obtain the equations
\begin{eqnarray}
 \dot{X} - \Omega\cos\theta \,Y &=& 0 \\
 \dot{Y} + \Omega\cos\theta \,X &=& 0
\end{eqnarray}
These equations can be easily solved by introducing the variable
$$Z = X + iY$$
which results in
$$\dot{Z} + i\Omega\cos\theta \,Z = 0$$
This equation has the solution
$$Z(t) = Z(0) e^{-i(\Omega\cos\theta)t}$$
or equivalently
$$\pmatrix{X(t)\\ \noalign{\smallskip} Y(t)}
 = \pmatrix{~~~\cos(\psi t)&\sin(\psi t)\\
 \noalign{\smallskip}
 -\sin(\psi t)&\cos(\psi t)}
 \pmatrix{X(0)\\ \noalign{\smallskip} Y(0)}
$$
where $\psi=\Omega\cos\theta=\pm\Omega\sin\lambda$, and where the sign depends
on the hemisphere.  In the Northern Hemisphere, the initial direction of the
pendulum therefore rotates \textit{clockwise} by an angle $\psi t$.