the cotangent space T∗
q(t)S1, and the state of the pendulum is described by a point (q(t), p(t)) in
T∗S1. However, we can treat the time derivative of an angle as a real number using the isomorphism
S1∼
=R/2πZ, so we get an isomorphism Tq(t)S1∼
=R, and thus an isomorphism T∗
q(t)S1∼
=R. We
thus have
T∗S1∼
=S1×R
and using this we can think of (q(t), p(t)) as a point in S1×R. Of course, most physicists do all this
without making such a fuss about it!
Now here’s where you come in....
1. Using what you already know about the mechanics of point particles, show that the kinetic energy
of the pendulum is
K( ˙q) = 1
2mr2˙q2
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.
3. Using what you know about classical mechanics on a Riemannian manifold, show that the
Hamiltonian of the pendulum,
H:T∗S1→R,
is given by
H(q, p) = p2
2mr2−mG cos q.
4. Work out Hamilton’s equations for the pendulum and show that
˙q=p
mr2,
˙p=−mG sin q,
and thus
¨q=−G
r2sin q.
Digression 1: Note that ˙qis really the angular velocity of the pendulum, while p=mr2˙qis really
its angular momentum.
Digression 2: If the angle qstays smal l, we can use the approximation sin q'qto approximate the
pendulum by a harmonic oscillator with
¨q=−G
r2q.
However, when the angle becomes large the pendulum becomes very different from the harmonic
oscillator. For example, if the pendulum starts out at q=π, p = 0 at time zero, it will stay there for
all times, balanced upside down! This is an unstable equilibrium.
5. Plot the level curves of Has a function of (q , p)∈(R/2πZ)×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
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