HW9 - SteadyServerPages

Assignment 9
Goldstein 10.5 Show that the function
S=
1
mω (q 2 + α2 ) cot ωt − α mω q csc ωt
2
is a solution of the Hamilton-Jacobi equation for Hamilton’s principal function for the linear harmonic
oscillator with
1
H=
(p2 + m2 ω 2 q 2 )
2m
Show that this function generates a correct solution to the motion of the harmonic oscillator.
The first part of this is a straightforward problem in plug and chug. We need only verify that the
Hamilton-Jacobi equation is satisfied by the given S. In particular, we need
∂S
= mω q cot ωt − αmω csc ωt
∂q
cos ωt
∂S
1
1
+ αmω 2 q
= − mω 2 (q 2 + α2 ) 2
∂t
2
sin ωt
sin2 ωt
Substituting these into the left hand side of the Hamilton-Jacobi equation
∂S ∂S
H q,
+
=0
∂q
∂t
we find
∂S ∂S
i ∂S
1 h ∂S 2
+ m2 ω 2 q 2 +
H q,
+
=
∂q
∂t
2m
∂q
∂t
2 2 1
1
q 2 + α2
cos ωt
1 m ω
2
q cos ωt − α + mω 2 q 2 − mω 2
+ αmω 2 q
=
2
2m sin ωt
2
2
sin2 ωt
sin2 ωt
which, in fact, will vanish after a line or two more of algebra which I’m too lazy to write out. .
We now want to show that S gives a correct solution to the motion. From our earlier calculation we
know that
∂S
p=
= mω q cot ωt − αmω csc ωt
∂q
We also know that the “new” coordinates, Qi , (in this case, just Q) is a constant given by
Q=β=
∂S
∂α
and where P = α. Once we know (constant) Q (or β), we can invert to get q in terms of the constants α
and β. So we have
∂S
β=
= mω α cot ωt − mω q csc ωt
∂α
which we invert for q as
mω q = mω α cos ωt − β sin ωt
Using this form for q in the expression for p, we get
p = −β cos ωt − α mω sin ωt
and this is the (correct) solution for the motion. Initial conditions can be tied to α = q0 and β = −p0 .
1
Goldstein 10.8 Suppose the potential in a problem of one degree of freedom is linearly dependent upon
time, such that the Hamiltonian has the form
H=
p2
− mAtx
2m
where A is a constant. Solve the dynamical problem by means of Hamilton’s principal function, under the
initial conditions t = 0, x = 0 and p = mv0 .
The challenge with this problem is that the resulting Hamilton-Jacobi equation is not obviously separable, at least not in the simple additive way that we have been considering. The Hamilton-Jacobi equation
is
∂S
1 ∂S 2
− mAtx +
0=
2m ∂x
∂t
If we were to assume a form for Hamilton’s principal function of S(x, t) = S1 (x) + S2 (t), the middle term
coupling t to x would keep us from being able to separate the equation. So we have to be a bit more clever.
A seemingly stupid thing to try would be to try to cancel somehow the coupled piece thinking that though
it is not additively separable, the term does not seem that horribly difficult. Thus we make the assumption
that a part of S(x, t) = S1 (x, t) + S2 (x, t) is made to cancel this term. We’ll try S1 , namely
∂S1
= mAtx
∂t
which integrates to
S1 (x, t) =
1
mAt2 x + c(x)
2
Of course, the integration “constant,” c(x), can be absorbed into S2 (x, t). At this stage, we note that the
principal function takes a form that doesn’t seem that much simpler and we have a trade off in that the
squared term is going to look worse:
0=
2 ∂S
1
1 ∂S2
2
+ mAt2 +
2m ∂x
2
∂t
However, this is in fact, separable. We can put the x dependent stuff on one side and the t dependent stuff
on the other. In particular, we write
∂S2
1
= − mA t2 ±
∂x
2
r
−2m
∂S2
∂t
It doesn’t look pretty, but we can now assume that S2 is separable and takes the form S2 (x, t) = S3 (x)+S4 (t).
Substituting in, we can now separate the remainder of the Hamilton-Jacobi equation:
1
∂S3 (x)
= α = − mA t2 ±
∂x
2
r
−2m
∂S4 (t)
∂t
This integrates to
S3 (x) = αx
S4 (t) = −
1 2
1
1 2 2 5
α t + αmA t3 +
m A t
2m
3
20
which yields a Hamilton’s principal function of
S(x, t) =
1 2
1
1 2 2 5
1
α t + αmA t3 +
mA t2 x + αx −
m A t .
2
2m
3
20
2
It is worth pointing out, and a little thought will confirm it, that this was only possible because the coupling
term was linear in x. Had it not been, this method would not necessarily have worked.
Solving the dynamical problem entails finding x(t) and p(t) which we do from
1
∂S
= mA t2 + α
∂x
2
α
A
∂S
=x−
t − t3
Q=β=
∂α
m
6
p=
and by inverting the last relation:
x(t) = β +
α
A
t + t3
m
6
1
mA t2 + α
2
and relating the constants α and β to the initial conditions, namely α = mv0 and β = 0 so that we have
p(t) =
A 3
t
6
1
p(t) = mv0 + mA t2
2
x(t) = v0 t +
3
Goldstein 10.14 A particle of mass m moves in one dimension under a potential V = −k/|x|. For energies
that are negative, the motion is bounded and oscillatory. By the method of action-angle variables, find an
expression for the period of motion as a function of the particle’s energy.
The Hamiltonian is
H=
p2
k
−
2m |x|
and the corresponding Hamilton-Jacobi equation is
1 h ∂W 2 2km i
−
=α
2m
∂x
|x|
where α is a constant identified with the total energy of the system and we are using, W , Hamilton’s
characteristic function as the original Hamiltonian is time independent and therefore the time dependance
in Hamilton’s principal function, S, separates out: S(q, α, t) = W (q, α) − αt. We are interested in the
action-angle formulation of this problem and we note that
s
2km
∂W
= 2mα +
p=
∂x
|x|
which allows us to define the action variables
I
J=
p dx
The closed interval over which we want to do the corresponding integration can be understood in terms
of the bound, oscillatory trajectories we are assuming. In particular, the energy of the system is negative:
E = −|α| and the turning points of the motion will occur at E = V or x± = ±k/|α|. As the integrand will
be even, we can take the integral to be
Z x+ r
p
x+
J = 4 · 2m|α|
− 1 dx
x
0
Z 1r
p
1
= 4x+ 2m|α|
− 1 du
u
0
s
√
Z
2m ∞ v − 1
= 4k
dv
|α| 1
v2
where in the last line we have made the substitution v = 1/u. The integral can be evaluated in a variety of
ways. A nice one is to consider the contour integral
I √
z−1
dz
z2
C
around a “pac-man” like contour which avoids the branch point at z = 1 and encloses the double pole at
z = 0. This can be then be shown to be twice the integral we want and from the residue theorem, we find
√
I √
z−1
d h z − 1 2 i
dz = 2πi
z z2
dz
z2
C
z=0
=π
Thus we have
s
J = 2πk
4
2m
.
|α|
This is a constant as we would expect our action variable to be. Of course, this can be rewritten as
H(J) = E = −|α| = −
8π 2 mk 2
J2
so that the frequency of the oscillatory motion is given by
16π 2 mk 2
∂H
=ν=
∂J
J3
which, in turn, gives us a period for the motion
τ=
1
J3
=
ν
16π 2 mk 2
τ=
πk
2m
or, in terms of the energy,
|E|
2m
5
−3/2
Goldstein 10.17 Solve the problem of the motion of a point projectile in a vertical plane, using the
Hamilton-Jacobi method. Find both the equation of the trajectory and the dependence of the coordinates
on time, assuming the projectile is fired off at time t = 0 from the origin with the velocity v0 , making an
angle α with the horizontal.
The Hamiltonian is
p2y
p2x
+
+ mgy
2m 2m
Because it is time independent, we can write S = W − αt and use Hamilton’s characteristic function in the
Hamilton-Jacobi equation:
1 ∂W 2
1 ∂W 2
+
+ mgy = αE
2m ∂x
2m ∂y
H=
W and the equation are separable if we assume W = Wx (x) + Wy (y). We get
1 ∂Wx 2
= αx2
2m ∂x
1 ∂Wy 2
+ mgy = αy2
2m ∂y
where αx2 + αy2 = αE . These are easily integrated to yield:
W =
p
2mαx2 x −
i3/2
1 h
2
2m
α
−
mgy
y
3m2 g
The new coordinates are defined by
Qi = βi =
and calculation yields
∂S
∂W
∂αE
=
−
t
∂αi
∂αi
∂αi
√
2m x − 2αx t
q
2αy
2m αy2 − mgy − 2αy t
βy = −
mg
βx =
Inverting these gives
βx + 2αx t
√
2m
2
2
αy
g
y(t) =
−
βy + 2αy t
mg 8αy2
x(t) =
From above, we can read off the momenta
√
∂W
= αx 2m
∂x
q
∂W
py =
= 2m αy2 − mgy
∂y
mg
βy + 2αy t
=−
2αy
px =
The initial conditions will allow us to relate our constants from the Hamilton-Jacobi method to more
familiar quantities. In particular, because the projectile is fired from the origin at t = 0, we have
βx = 0
βy2
αy2
g
=
8αy2
mg
6
The initial conditions for the momenta are
p2x + p2y = 2m αx2 + αy2 = m2 v02
0
αy
py =
= tan α
px 0
αx
As a result of the last two relations, we can quickly show
r
1
mv02 cos α
2
r
1
αy =
mv02 sin α
2
αx =
and
√
βy =
2m 2
v0 sin2 α
g
The solution then can be rewritten as
x(t) = v0 cos α t
1
y(t) = v0 sin α t − gt2
2
px = mv0 cos α
py = mv0 sin α − mgt
7
Goldstein 10.20 Find the frequencies of a three-dimensional harmonic oscillator with unequal force constants using the method of action-angle variables. Obtain the solution for each Cartesian coordinate and
conjugate momentum as functions of the action-angle variables.
The Hamiltonian is
H=
p2y
p2x
p2
1
1
1
+
+ z + kx x2 + ky y 2 + kz z 2 ,
2m 2m 2m 2
2
2
is time independent, hence H = E, and we can define action variables
I
Ji =
pi dqi
Using Hamilton’s characteristic function in the Hamilton-Jacobi equation, we can verify that the latter is
separable with
αE =
1 h ∂Wx (x) 2 ∂Wy (y) 2 ∂Wz (z) 2 i 1 +
+
+
kx x2 + ky y 2 + kz z 2
2m
∂x
∂y
∂z
2
With the separation, we get
1 ∂Wx 2 1
+ kx x2 = αx2
2m ∂x
2
1 ∂Wy 2 1
+ ky y 2 = αy2
2m ∂y
2
1 ∂Wz 2 1
+ kz z 2 = αz2
2m ∂z
2
where αE = αx2 + αy2 + αz2 . Recalling that pi = ∂W/∂qi , our action variables are then
I r 1
Ji =
2m αi2 − ki qi2 dqi
2
s
Z π/2
q
2αi2
cos2 ψ dψ
= 4 · 2mαi2
ki 0
r
m
2
= 2π αi
ki
p
P
where we have used the substitution sin ψ = qi ki /2αi2 . The energy can now be written as E = αE = i αi2
or
r
r
r i
1 h
kx
ky
kz
H(Ji ) = E =
Jx
+ Jy
+ Jz
2π
m
m
m
From this, we can get
∂H
1
νi =
=
∂Ji
2π
r
ki
m
which are the frequencies of our oscillator.
Given the symmetry between the different directions, we need only calculate the solution in one direction.
We have the action variables and now need the angle variables. These are found as
wi =
8
∂W
∂Ji
so we have to be able to calculate this. To do this, we need the separated pieces of the characteristic function
in terms of the angle variables
Z r
1
2m αi2 − ki qi2 dqi
2
Z q p
1
=√
Ji mki − mki πqi2 dqi
π
Wi =
which, when differentiated with respect to the action variables, gives the angle variables
∂Wi
∂Ji
r
Z
dqi
1 mki
q √
=
2
π
Ji mki − mki πqi2
" s √
#
π mki
1
−1
sin
qi
=
2π
Ji
wi =
We can invert this to get qi in terms of the action-angle variables:
s
qi (Ji , wi ) =
J
√i
sin(2πwi )
π mki
We also need the momenta which we get from
∂W
∂qi
q p
1
√
=
Ji mki − mki πqi2
π
s
√
Ji mki
cos(2πwi )
=
π
pi =
While this certainly looks like the right solution and answers the question asked, namely to obtain the
solution in action-angle variables, it might not be in the most recognizable form. We might ask, for instance,
what happened to the time? Recall that our transformation to action-angle variables swallowed up the time
in the angle variables. They are the new coordinates:
Qi = wi = νi t + βi
Recall too that the action variables are related to the total energy. If we also make the substitution ωi2 =
ki /m = 2πνi (but don’t confuse this with wi ), we can rewrite our solution as
r
Ji
sin ωi t + 2πβi
πmωi
r
Ji mωi
cos ωi t + 2πβi
pi =
π
qi =
9

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