Student Seminar: Classical and Quantum Integrable Systems

while
i.e.
a =
p
1 − e = k
2 2|E| .
On the other hand, since the sectorial velocity C is constant we have
∫ T
0
C =
∫ T
0
dt dS
dt = S , =⇒ CT = J 2 T = S ,
T = 2S J =
2πk
( √ 2|E|) = √ 2π a 3 2 .
3 k
It is interesting to note that the total energy depends only on the major semi-axis
a and it is the same for the whole set of elliptical orbits from a circle of radius a
to a line segment of length 2a. The value of the second semi-axis do depend on the
angular momentum.
The Runge-Lenz vector and the Liouville torus. The phase space of the motion in the
central field is T ∗ R 3 , i.e. it is six-dimensional. There are four conserved integrals:
three components of the angular momentum J i and the energy E. This shows that
the motion happens on the two-dimensional manifold. In case of the bounded motion
it is the two-dimensional Liouville torus. Thus, there are two frequencies associated
and when they are not rationally commensurable the orbits are not closed but rather
dense on the torus. For the specific Kepler motion (with any sign of k) there is one
more non-trivial conserved quantity appears which is absent for a generic central
potential: The Runge-Lenz vector (for definiteness we assume that k > 0):
⃗R = ⃗v × ⃗ J − k ⃗r r .
Problem. Show that the Runge-Lenz vector is conserved.
Indeed, we have
˙⃗R = ˙⃗v ×
}{{}
J ⃗ −k ⃗v r + k⃗r(⃗v⃗r) r 3
m ⃗r×⃗v
= m ˙⃗v × (⃗r × ⃗v) − k ⃗v r + k⃗r(⃗v⃗r) r 3 .
On the other hand,
m ˙⃗v = − ∂U ⃗r
∂r r = −k ⃗r r 3
and, therefore,
˙⃗R = −k 1 r 3 ⃗r × (⃗r × ⃗v) − k⃗v r + k⃗r(⃗v⃗r) r 3
Further one has to use the formula
⃗r × (⃗r × ⃗v) = (⃗v⃗r)⃗r − r 2 ⃗v
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