Student Seminar: Classical and Quantum Integrable Systems

Substituting these formula into the expression for the kinetic energy T = 1I 2 iΩ 2 i
obtain the kinetic energy in terms of the Euler angles.
we
Problem. By using Euler angles relate the angular momenta in the moving and the stationary
coordinate systems. The momentum M is directed along the Z axis of the stationary coordinate
system.
We have
M sin θ sin ψ = I 1 Ω 1 ,
M sin θ cos ψ = I 2 Ω 2 ,
M cos θ = I 3 Ω 3 .
From here
cos θ = I 3Ω 3
M , tan ψ = I 1Ω 1
I 2 Ω 2
.
Solution of the last problem allows one to find
√
I 3 (M
cos θ =
2 − 2EI 1 )
dn τ ,
M 2 (I 3 − I 1 )
√
I 1 (I 3 − I 2 ) cn τ
tan ψ =
I 2 (I 3 − I 1 ) sn τ .
Thus, both angles θ and ψ are periodic functions of time with the period T (the same
period as for Ω!). However, the angle φ does not appear in the formulas relating the
angular momenta in the moving and the stationary coordinate systems. We can find
it from
Ω 1 = ˙φ sin θ sin ψ + ˙θ cos ψ
Ω 2 = ˙φ sin θ cos ψ − ˙θ sin ψ .
Solving we get
˙φ = Ω 1 sin ψ + Ω 2 cos ψ
.
sin θ
This leads to the differential equation
dφ
dt = M I 1Ω 2 2 + I 2 Ω 2 2
.
I1Ω 2 2 1 + I2Ω 2 2 2
Thus, solution is given by quadrature but the integrand contains elliptic functions in
a complicated way. One can show that the period of φ, which is T ′ is not comparable
with T . This leads to the fact that the top never returns to its original state. The
periods T and T ′ are the periods of motion over the Liouville torus.
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