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Problem 3
A physical system has kinetic energy
1
T = m q̇12 + q̇22 q12 + q22 ,
2
and potential energy
U=
q12
α
.
+ q22
(a) Derive the Hamiltonian
and the Hamiltonian equations of motion
P
for this system. H(q, p) = pq̇ − L(q, q̇) where L = T − U .
pi = mq̇i q12 + q22 , for i = 1, 2
So the Hamiltonian is
2
1
p1 + p22
H= 2
+α .
2m
q1 + q22
Hamilton’s equations of motion are
1
p
i,
q12 + q22 m
2
2qi
p1 + p22
+
α
.
ṗi = 2
2m
q1 + q22
q̇i =
(b) Give an explicit expression for the phase space flow and Liouville
equation for this system.
~v = (q̇i , ṗi ) , phase space flow
∂ρ
~
0=
+ ~v · ∇ρ,
Liouville equation
∂t
2
∂ρ
1
pi ∂
p1 + p22
∂
0=
+
+ 2
+ 2α qi
ρ
∂t
m
∂pi
q1 + q22 m ∂qi
(c) Derive the Hamilton-Jacobi equation for this system. The HamiltonJacobi equation is
∂S
∂S
H qi ;
+
= 0,
∂qi
∂t
and since H does not depend explicitly on time we may take S = W − Et.
Subtituting H into the above equation yields
#
"
!
1
1
∂S 2
∂S 2
+
+α .
E= 2
∂q1
∂q2
q1 + q22 2m
7
Alternative Solution The q12 + q22 terms in the Hamiltonian are suggestive of a rotational symmetry. Make the following coordinate transformation:
√
θ
q1 = 2 r cos
,
2
√
θ
q2 = 2 r sin
.
2
With these new coordinates r, θ, the Hamiltonian becomes
H=
p2θ
p2r
β
+
+ ,
2
2m 2mr
r
where β = α/4. Thus, this problem is equivalent to the two-dimensional
Kepler problem.
8
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