Student Seminar: Classical and Quantum Integrable Systems

One has
∆φ circ =
∫ xmax
x min
dx
√
2E + 2k x
J β − x 2
β
E→0
→
∫ xmax
x min
dx
√
2k
x
J β − x 2
β
Rescale x = αy with α satisfying the relation 2k
J β α β = α 2 , then we get
∆φ circ =
∫ 1
We note that the result does not depend on J.
0
dy
√
yβ − y = π
2 2 − β .
Now we are ready to find the potentials for which all bounded orbits are closed. If
all bounded orbits are closed, then, in particular, ∆φ circ = 2π m = const. That means
n
that ∆φ circ should not depend on the radius, which is the case for the potentials
V (r) = ar α , α > −2 and V (r) = b log r .
In both cases ∆φ circ = √ π
2+α
. If α > 0 then lim E→∞ ∆φ circ (E, J) = π and therefore
2
α = 2. If α < 0 then lim E→0 ∆φ circ (E, J) = π . Then we have an equality
2+α
π
= √ π
2+α 2+α
which gives α = −1. In the case α = 0 we find ∆φ circ = √ π 2
which
is not commensurable with 2π. Therefore all bounded orbits are closed only for
V = ar 2 and U = − k .
r 2
2.2.2 The Kepler laws
For the original Kepler problem we have
and
Integrating we get
∫
φ =
V (r) = − k r + J 2
2r 2 .
Jdr
√
r 2 2(E + k − J 2
)
r 2r 2
φ = arccos
J
√ − k r J
.
2E + k2
J 2
An integration constant is chosen to be zero which corresponds to the choice of an
origin of reference for the angle φ at the pericenter. Introduce the notation
√
J 2
k = p , 1 + 2EJ 2
= e ,
k 2
This leads to
r =
p
1 + e cos φ
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