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\begin{document}
\begin{center}
{\bf \large The Pendulum, Elliptic Functions and Imaginary Time\\}
\vspace{0.5cm}
{\small Math 241 Homework \\}
{\small John Baez}
\end{center}
The sine and cosine functions are analytic on the entire complex
plane, and also periodic in one direction. It's interesting to
look for nice functions that are periodic in {\it two} directions
on the complex plane. Such a function can't be analytic everywhere ---
it must have poles --- since otherwise it would be bounded. Apart
from that it can be very nice: it can be analytic except at poles
that form a lattice in the complex plane, like this:
\medskip
\centerline{\epsfysize=1.0in\epsfbox{lattice.eps}}
\medskip
\noindent
A function like this is called an {\bf elliptic function}.
Since the plane modulo a lattice is a torus,
you can also think of an elliptic function as a function from
the torus to the Riemann sphere. The complex
plane modulo a lattice is also called an {\bf elliptic curve};
these are important examples of Riemann surfaces.
Jacobi, Weierstrass and other mathematicians did a
lot of work on elliptic functions in the 1800s.
Elliptic functions and elliptic curves have many
applications to
number theory, ultimately leading to very deep results such as Wiles'
proof of Fermat's theorem. They also have lots of applications to
physics --- and here you will learn about one of the simplest!
If you're feeling less ambitious, do problems 1-11. If you're
feeling more ambitious, do problem 12.
Start with a pendulum where a particle of mass $m$ is constrained
by a rod to lie on a circle of radius $r$ in the $xz$ plane:
\medskip
\centerline{\epsfysize=2.0in\epsfbox{pendulum.eps}}
\medskip
\noindent
To keep things simple we'll neglect the mass of
the rod. The position of the pendulum is a function
\[ q \maps \R \to S^1 , \]
that is, a function of time taking values in the circle.
Concretely we will think of $q(t)$ as the angle counterclockwise
from the downwards $z$ axis.
However, we are allowed to treat this angle as a real number
only if we remember that two angles describe
the same position of the pendulum when they differ by an integral multiple
of $2\pi$. Thus $q(t)$ is really a member, not of $\R$, but of
the quotient group $\R/2\pi \Z$, which is a way of thinking of
the circle $S^1$.
Since the position of the pendulum at time
$t$ is really a point $q(t) \in S^1$, it follows that
$\dot q(t)$ is really an
element of the tangent space $T_{q(t)} S^1$.
Similarly, the corresponding momentum $p(t)$ lies in the
cotangent space $T_{q(t)}^* S^1$, and the state of the
pendulum is described by a point $(q(t),p(t))$ in $T^* S^1$.
However, we can treat the time derivative of an angle as a
real number using the isomorphism $S^1 \iso \R/{2\pi \Z}$,
so we get an isomorphism $T_{q(t)}S^1 \iso \R$, and thus an
isomorphism $T^*_{q(t)} S^1 \iso \R$. We thus have
\[ T^* S^1 \iso S^1 \times \R \]
and using this we can think of $(q(t),p(t))$ as a point in
$S^1 \times \R$. Of course, most physicists do all this without making
such a fuss about it!
\vskip 1em
\noindent
{\textit{\textbf{Now here's where you come in....}}}
\vskip 1em
\noindent
1. Using what you already know about the mechanics of
point particles, show that the kinetic energy of the pendulum
is
\[ K(\dot q) = {1 \over 2} mr^2 {\dot q}^2 \]
\vskip 1em
\noindent
2. Assuming the force of gravity is a vector pointing down with
magnitude $mG$, show that we can assume the potential energy of
the pendulum to be
\[ V(q) = -mG \cos q .\]
\vskip 1em
\noindent
3. Using what you know about classical mechanics
on a Riemannian manifold, show that the Hamiltonian of
the pendulum,
\[ H \maps T^* S^1 \to \R, \]
is given by
\[ H(q,p) = {p^2\over 2mr^2} - mG \cos q . \]
\vskip 1em
\noindent
4. Work out Hamilton's equations for the pendulum and
show that
\[ \dot q = {p \over mr^2} , \]
\[ \dot p = -mG \sin q ,\]
and thus
\[ \ddot q = -{G\over r^2} \sin q .\]
\vskip 1em
\noindent
{\it Digression 1: Note that $\dot q$ is really the
{\bf angular velocity} of the pendulum, while $p = mr^2 \dot q$
is really its {\bf angular momentum}. }
\vskip 1em
\noindent
{\it Digression 2: If the angle $q$ stays small,
we can use the approximation $\sin q \simeq q$ to approximate the
pendulum by a harmonic oscillator with
\[ \ddot q = -{G \over r^2} q . \]
However, when the angle becomes large the pendulum becomes
very different from the harmonic oscillator. For example,
if the pendulum starts out at $q = \pi, p = 0$ at time zero,
it will stay there for all times, balanced upside down! This is
an unstable equilibrium.}
\vskip 1em
\noindent
5. Plot the level curves of $H$ as a function of
$(q,p) \in (\R/2\pi\Z) \times \R$. Since energy is
conserved, the state $(q(t),p(t))$ must stay on one
of these level curves as it evolves in time. Use this to
qualitatively describe the behavior of the pendulum for
various different values of the energy. In particular,
find stable and unstable equilibria.
\vskip 1em
\noindent
6. Supposing that the pendulum's energy equals $E \in \R$,
show that
\be \dot q = \pm \sqrt{{2\over mr^2}(E + mG \cos q)} .
\label{thetadot} \ee
\vskip 1em
\noindent
7. To reduce the clutter and focus on essentials, switch
to units where $mr^2 = mG = 1$. Using equation (\ref{thetadot}),
and taking the positive square root, show that
\be
t = \int {dq \over \sqrt{2(E + \cos q)}} .
\label{int}
\ee
\vskip 1em
8. If we could do the integral in equation (\ref{int}),
we'd know $t$ as a function
of $q$. Then we could solve for $q$ as a function
of $t$ and we'd be done!
Unfortunately, this integral cannot
be done using elementary functions --- it's
a so-called {\bf elliptic integral}. To bring it into Jacobi's
favorite form, let's work not with $q$ but with
\[ x = \sqrt{2\over E+1} \; \sin(q / 2) \]
Show that
\[ \dot x = \sqrt{(1 - x^2)(1 - k^2x^2)} \]
where $k$, the so-called {\bf modulus}, is given by
\[ k = \sqrt{E+1\over 2}. \]
Conclude that
\be t = \int {dx \over \sqrt{(1 - x^2)(1 - k^2x^2)}}
\label{Jacobi}
\ee
This is {\bf Jacobi's elliptic integral of the first kind}.
When we solve for $x$ as a function of $t$,
we get an {\bf elliptic function}.
\vskip 1em
\noindent {\it
Digression: To do integrals involving the square root of a
quadratic function of $x$, you need inverse trig functions. However,
for integrals involving the square root of a cubic or quartic
function of $x$, you need inverse elliptic functions --- or
in other words, elliptic integrals. Why are they called `elliptic'?
Well, if you work out the circumference of the ellipse
\[ {x^2\over a^2} + {y^2 \over b^2} = 1 \]
you get $4a$ times this:
\[ \int_0^1 \sqrt{1 - k^2 x^2 \over 1 - x^2} \, dx \]
where $k^2 = 1 - b^2/a^2$. The stuff under the square root here is
not a quartic in $x$, but the integral is closely related to the
one we've been discussing: it's called {\bf Jacobi's elliptic integral
of the second kind}.}
\vskip 1em
\noindent
9. I've been a bit sloppy about the limits of integration
in equation (\ref{Jacobi}). Show that if we start our clock
so that $t = 0$ when our pendulum happens to be pointing straight down,
we have
\[ t = \int_0^x {dy \over \sqrt{(1 - y^2)(1 - k^2 y^2)}} . \]
If we now solve this for $x$ as a function of $t$ we get, by definition,
the elliptic function $\sn(t,k)$. In other words:
\[ x = \sn(t,k) \quad {\rm means} \quad
t = \int_0^x {dy \over \sqrt{(1 - y^2)(1 - k^2 y^2)}} \]
To make this more precise we'd need to worry about the branch
points in the integrand at $y = \pm 1$, $y = \pm 1/k$, but let's
not worry about those just yet.
\vskip 1em
\noindent
{\it Digression 3:
It's easy to see that when $k = 0$, the function $\sn(t,k)$ reduces
to the good old sine function. There is also an elliptic
function $\cn(t,k)$ that reduces to $\cos t$
when $k = 0$, and one called $\dn(t,k)$ that
reduces to $1$ when $k = 0$. They satisfy
identities like
\[ \sn^2(t,k) + \cn^2(t,k) = 1 \qquad
k^2 \sn^2(t,k) + \dn^2(t,k) = 1 \]
\[ {d\over dt} \sn(t,k) = \cn(t,k) \dn(t,k) \qquad
{d\over dt} \cn(t,k) = -\sn(t,k) \dn(t,k) \qquad
{d\over dt} \dn(t,k) = -k^2 \sn(t,k) \cn(t,k) , \]
so before you know it, you've got a whole world of generalized trig
formulas on your hands! Back in the 1800s, any decent mathematician
would know this stuff. You should too.}
\vskip 1em
\noindent
{\textit{\textbf{But now for the really cool part:}}}
\vskip 1em
\noindent
10. Using part 5, show that $q(t)$ and thus $\sn(t,k)$ is
periodic as a function of $t$.
\vskip 1em
\noindent
11. Show that making the replacement
\[ t \mapsto it \]
in Newton's law is equivalent to reversing the sign of all
forces.
\vskip 1em
\noindent
In the present problem, this amounts to
reversing the force of gravity, making it pull the pendulum up.
But an upside-down pendulum is just another pendulum. Therefore
the function $\sn(it,k)$ must {\rm also} be periodic as a function of
$t$. This suggests that $\sn(z,k)$, as a function of $z \in \C$,
is periodic in both the real and imaginary directions. And it's true!
\vskip 1em
\noindent
{\textit{\textbf{
So, the pendulum gives a physical explanation of the fact that
elliptic functions are periodic in two directions on the
complex plane!}}}
\vskip 1em
\noindent
12. Prove, as rigorously as you can, that $\sn(z,k)$ is
periodic in two directions. You can do this either by fleshing
out the above argument, or by studying the integral in equation
(\ref{Jacobi}) and worrying about those branch points. In fact we have
\[ \sn(z+4K,k) = \sn(z,k) , \qquad \sn(z+2iK',k) = \sn(z,k) \]
where for $0 < k < 1$
\[ K = \int_0^1 {dy \over \sqrt{(1 - y^2)(1 - k^2 y^2)}} \]
and
\[ K' = \int_1^{1/k} {dy \over \sqrt{(y^2 - 1)(1 - k^2 y^2)}} . \]
\end{document}