Part I: Classical Dynamics–Solutions
Problem 1 (33 pts)
A particle moves on the x axis with Hamiltonian
H(x, px ) =
p2x x2 x4
+
− .
2
2
4
(a) (5 pts) The equations of motion (Hamilton’s equations) are:
ẋ =
∂H
= px ,
∂px
p˙x = −
∂H
= −x + x3 .
∂x
(b) (5 pts) The equilibrium points of the system are found by setting the
righthand sides of Hamilton’s equations in (a) equal to zero. We have a
stable equilibrium at x = 0, px = 0, H = 0, and unstable equilibria at x =
±1, px = 0, H = 41 . Stability/instability is determined by expanding H in
Taylor series about the respective equilibrium points. To degree 2, the local
H is a harmonic oscillator for a stable equilibrium, an inverted oscillator for
an unstable equilibrium.
(c) (8 pts)
2
1
0
-1
-2
-2
-1
0
1
2
√
(d) (5 pts) A particle is launched at x = 0 with initial velocity 1/ 2. The
initial point is on the separatrix curve H = 1/4, directly above the origin.
The initial momentum is positive and so the point moves to the right along
1
the separatrix, arriving at the unstable equilibrium point at x = 1, p = 0
after time
Z
0
1
Z 1
dx
dx
1 Z1
=
=∞
dx =
2
ẋ
0 2(1 − x )
0 2(1 + x)(1 − x)
(e) (5 pts) We make the following transformation of phase space coordinates:
(x, px ) 7→ (θ, J),
√
√
x = 2J sin θ,
px = 2J cos θ.
To prove that this is a canonical transformation, we show that the Poisson
bracket of x with px is equal to 1 if [θ, J] = 1.
[x, px ] =
∂x ∂px
∂x ∂px
−
= sin2 θ + cos2 θ = 1.
∂θ ∂J
∂J ∂θ
(f) (5 pts)
Ĥ(θ, J) = J − J 2 sin4 θ
To get the new Hamiltonian, we need only to average over θ:
¯ = J¯ − H̄(θ̄, J)
1 ¯2 Z 2π 4
3
J
sin θ = J¯ − J¯2 + O(2 )
2π
8
0
¯ with corrections of order 2 .
The new Hamiltonian H̄ is a function only of J,
The oscillation frequency, correct to first order in is given by
ω̄ =
∂ H̄
3
= 1 − J¯ + O(2 ).
¯
∂J
4
Problem 2 (34 pts)
A dynamical system with two angular degrees of freedom has the Hamiltonian
5
H(φ1 , φ2 , L1 , L2 ) = L1 + L2 + (3L1 + 2L2 )2 sin(2φ1 − 3φ2 ).
3
2
(a) (5 pts) Find a linear combination F of L1 and L2 which commutes, in the
sense of Poisson brackets, with H. Show that F is a constant of the motion.
It is sufficient to find a, b such that
[aL1 + bL2 , 2φ1 − 3φ2 ] = 0.
From the elementary Poisson brackets,
−2a + 3b = 0.
Hence we can take
2
F = c(L1 + L2 ),
3
where c is an arbitrary numerical constant. Since
[F, H] = 0.
∂F
∂t
= 0, we have Ḟ =
(b) (5 pts) Show that F/c and H are functionally independent (i.e their
phase-space gradient vectors are linearly independent) everywhere in the
phase space.
For dependence, all 2x2 minors of the rectangular matrix of partial derivatives,
!
∂H
∂L1
1
∂H
∂L2
2
3
∂H
∂φ1
∂H
∂φ2
0
0
must vanish. In particular, one of the minors is
∂H
2 5
2 ∂H
−
= − = −1,
3 ∂L1 ∂L2
3 3
and so the gradient vectors are linearly independent everywhere.
(c) (5 pts) Statements (a) and (b) imply that the topological structure of a
compact, connected phase-space manifold Mh,f on which H and F take the
respective values h and f must be that of a 2-torus.
(d) (6 points) We make a canonical transformation
(φ1 , φ2 , L1 , L2 ) 7→ (α1 , α2 , I1 , I2 )
such that
α1 = φ1 ,
2
α2 = φ2 − φ1 ,
3
3
This is accomplished by means of a generating function
2
S(φ1 , φ2 , I1 , I2 ) = I1 φ1 + I2 (φ2 − φ1 ),
3
which gives
2
∂S
= I1 − I2 ,
∂φ1
3
∂S
L2 =
= I2 ,
∂φ2
L1 =
so that
2
2
F = c(I1 − I2 + I2 ) = cI1 ,
3
3
as desired. Plugging the expressions for L1 , L2 , φ1 , φ2 into H gives
H̄(α1 , α2 , I1 , I2 ) = I1 + I2 − 9I12 sin 3α2 .
(d) (5 points) To verify that the system is separable with respect to the
coordinates α1 , α2 , I1 , I2 , we note that for fixed H̄ = h and I1 = F/c = f , we
have that I1 is a constant for 0 ≤ α1 < 2π, while I2 is a function of α2 for
0 ≤ α2 < 2π, namely
I2 = h − f + 9f 2 sin 3α2 .
The manifold is thus diffeomorphic to a product of circles, hence to a 2-torus,
as required by the Liouville-Arnold theorem.
(e) (8 points) We use the separability to introduce action-angle coordinates
on the manifold Mh,f . We have
J1 =
1 I
I1 dα1 = f,
2π
J2 =
1 I
I2 dα2 = h − f.
2π
The generating function for the CT to action-angle coordinates is
S(α, J) =
Z
I1 (α1 , J)dα1 +
Z
I2 (α2 , J)dα2 = J1 α1 + J2 α2 − 3J12 cos 3α2 .
and so we get
θ1 =
∂S
= α1 − 6f cos 3α2 ,
∂J1
∂S
θ2 =
= α2 .
∂J2
4
Part II: Continuum Mechanics -- Solutions
Problem 3 (33 pts)