6 Class exercise sheet

6
Class exercise sheet
Hamilton - Jacobi:
For a general Hamiltonian H (q, p, t) we want to look for a canonical transformation which
will produce a new Hamiltonian K (Q, P, t) = 0. This means that all the coordinates and
momenta are constant since,
Ṗ = −
∂K
=0 ;
∂Q
Q̇ =
∂K
=0
∂P
(1)
(2)
We will search for a generating function S = F (q, α, t), where α = ∂S
= const (recall that
∂q
∂S
∂S
= p and that ∂α = Q), this will give us the following Hamiltonian
∂q
∂S
∂S
K = H q,
,t +
=0
(3)
∂q
∂t
This will give us a partial differential equation for S instead of two ODE for p and q. We
will also define that β = ∂S
∂α
Since both α and β are constant we have
α(q(t), p(t), t) = α (q(t = 0), p(t = 0), t = 0) = const
β(q(t), p(t), t) = β (q(t = 0), p(t = 0), t = 0) = const
(4)
(5)
this we can reverse and find q(t) and p(t).
For a Hamiltonian which is not time dependent we will guess the following generating function
S = W (q, α) − Et
(6)
to get rid of the time derivative.
Exercise 6.1: Particle in gravitational field - the H-J method
Solve the motion of a particle of mass m, moving in a uniform gravitational field g, using
the Hamilton-Jacobi method.
Solution:
For a particle in a gravitational field g the Hamiltonian is
1
H=
p2x + p2y + p2z + mgz
2m
1
(7)
The Hamilton-Jacobi equation is
1
2m
∂S
∂x
2
+
∂S
∂y
2
+
∂S
∂z
2 !
+ mgz +
∂S
=0
∂t
(8)
we will use separation of variable and guess
S = Wx (x) + Wy (y) + Wz (z) − Et
(9)
This will give us
1
2m
∂Wx
∂x
2
+
∂Wy
∂y
2
+
∂Wz
∂z
2 !
+ mgz = E
(10)
this will give us
1
2m
∂Wx
∂x
2
∂Wy
∂y
2
1
2m
1
2m
∂Wz
∂z
2
=
1 2
p = const
2m x
(11)
=
1 2
p = const
2m y
(12)
+ mgz = const
From this we can write our generating function
Z
Z
Z
∂Wy
∂Wz
∂Wx
dx +
dy +
dz − Et =
S =
∂x
∂y
∂z
2
Z p
p2y
px
= p x x + py y +
2m (Ez − mgz)dz −
+
+ Ez t
2m 2m
(13)
(14)
(15)
Since we also know that
∂S
px
=x− t
∂px
m
∂S
py
=
=y− t
∂py
m
r s
Z
p
∂S
∂
2 Ez
=
=
2m (Ez − mgz)dz − t = −
−z−t
∂Ez
∂Ez
g mg
β1 =
(16)
β2
(17)
β1
2
(18)
We can now reverse all three equations to get
px
t + β1
m
py
y =
t + β2
m
Ez
1
z =
− g (β3 + t)2
mg 2
x =
(19)
(20)
(21)
We can now plug in our initial conditions to receive
β1 = x(0)
β2 = y(0)
2Ez
2z(0)
β3 = ±
−
mg 2
g
(22)
(23)
(24)
so we just need to find Ez , this will be done from energy conservation
Ez =
p2z (0)
+ mgz(0)
2m
(25)
Using all the initial conditions we finally get the solution to the motion of the particle
px (0)
t
m
py (0)
y = y(0) +
t
m
pz (0)
1
z = z(0) −
t − gt2
m
2
x = x(0) +
3
(26)
(27)
(28)
Central force motion and the effective potential:
When encountering a central force problem (i.e Kepler’s problem) can formally reduce the
problem of the motion of to bodies to and equivalent one-body problem in which we must
determine only the motion of a ”particle” of mass µ in the central field described by the
m2
and r = r1 − r2 .
potential function U (r). Where µ ≡ mm11+m
2
In addition we can use the angular momentum conservation of the problem to further simplify the problem, using the angular momentum, L, we can rewrite the energy as a kinetic
term and an effective potential which consist of the centrifugal potential energy and the
potential energy of the problem. Note that the substitution of the L back is only done for
the energy, this cannot be done in the Lagrangian since it will break the principal on which
the E-L equation is based on.
Drawing the effective potential has many merits, it will usually give us really good understanding of possible orbits for different energies which the particle can take. The minimas
and maximas of the effective potential will tell us of the stable and unstable circular orbits.
Exercise 6.2: Motion in 1D
A particel of mass m moves in the one-dimensional potential
x2 − x
U (x) = U0 2 e a
a
(29)
1. Sketch U (x). Identify the location(s) of any local minima and/or maxima, and be sure
that your sketch shows the proper behavior as x → ±∞
2. Sketch the motions for a total energy E = 52 U0 . Do the same for E = U0
3. Derive an expression for the period T of the motions when |x| << a.
Solution:
1. In order to find the local minima and maxima we derive in accordance to x and set to
zero.
U0
x2 − x
0
U (x) = 2 2x −
e a =0
(30)
a
a
4
For finite x’s the solutions are:
x=0
;
x = 2a
Clearly x = 0 is the local minimum and x = 2a is a local maximum.
2. The black line is for E = 25 U0 and we have two types of motion:
(a) Bound orbit.
(b) Hyperbolic orbit.
The blue line is for E = U0 , this is just a hyperbolic orbit.
5
(31)
3. Expanding U (x) in a Taylor series about x = 0 we have
x3
x4
U0
2
+ 2 + ...
U (x) = 2 x −
a
a
2a
(32)
The leading order term is sufficient for |x| << a. The potential energy is then equiva0
. So the period is
lent to that of a spring, with spring constant k = 2U
a2
r
T = 2π
m
= 2π
k
s
ma2
2U0
(33)
Exercise 6.3: Hohmann transfer
A spacecraft is parked in a circular orbit around Earth, NASA want to send the spacecraft
to the Moon’s orbit. To do this they will use the Hohmann transfer orbit.
The Hohmann transfer orbit is an elliptical orbit used to transfer between two circular orbits
of different radii in the same plane.
We will want to find the total velocity change, ∆v, and the transfer time required for
such a maneuver.
1. Assuming that the potential energy is U (r) = − kr find the radius of the circular orbit,
rc , and the energy of the system, E(rc ) in that orbit (Hint: remember that L = ~r × p~).
2. Using the above item find the velocity of spacecraft in the circular orbit.
3. For and elliptical orbit things are a bit more complicated (but not too much). Find
the relation between rmin the shortest distance from one of the foci and rmax is the
longest distance from the same focus point.
6
4. Find the energy of the system for an elliptical orbit E(rmin , rmax ). Using
k
L2
1
k
L2
1
−
+
+
−
+
Ee =
2
2
2
rmin 2mrmin
2
rmax 2mrmax
(34)
should help you. Check to see that your answer is correct by taking a limit.
5. Find the velocity of the spacecraft in the elliptical orbit.
6. You can now calculate ∆v. Do it.
3
2)
7. Given Kepler’s third law T 2 = π 2 (r1 +r
find the time required for the maneuver.
8µ
Solution:
1. We first want to write down the energy as a function of r only and constants. Since this
is a two body problem we can reduce the problem to a one body problem on a plane.Let
us first write down the Lagrangian of the problem
k
k
1
1
L = m~v 2 + p
= m ẋ2 + ẏ 2 + ż 2 + p
2
2
x2 + y 2 + z 2
x2 + y 2 + z 2
we move to spherical coordinates and pick our plane θ = π2 ,
k
1 L = m ṙ2 + r2 φ̇2 +
2
r
We see that φ is a cyclic coordinate and so
L = mr2 φ̇
(35)
(36)
(37)
If we want to use this we must turn to the energy, which is constant, because if we
place it in our Lagrangian or Hamiltonian it will break the principal on which E-L is
based on. The energy is
1 2 k
m~v −
2
r
k
1
2
2 2
=
m ṙ + r φ̇ −
2
r
1 2
L2
k
1
=
mṙ +
− = mṙ2 + Vef f (r)
2
2
mr
r
2
E =
7
(38)
(39)
(40)
The stable circular orbit will be in the minimum of the effective potential
∂Vef f (r)
=0
∂r
⇒
rc =
L2
mk
(41)
Using this and the angular momentum we can find the energy of the system in the
circular orbit
Ec = −
k
2rc
(42)
2. We can now use both the found energy and angular momentum to get
k
1
Ec = m~v 2 −
2
rc
(43)
which will give us
r
~vc =
k
mrc
(44)
3. In this case we use the fact that the effective potential accounts for all the energy at
rmax and rmin when there is zero velocity in the r-direction. In this case:
Ee = −
k
rmin
+
k
L2
L2
=
−
+
2
2
2mrmin
rmax 2mrmax
(45)
1
We can treat this equation as a quadratic equation for rmin
and solve. One solution is
that rmax = rmin . This represents the circular orbit. The other solution is:
1
rmin
=
2km
1
−
2
L
rmax
(46)
4. Doing the algebra with the solution found above we get
Ee = −
k
rmax + rmin
(47)
we see that if we take rmax = rmin then we will get the energy of the circular orbit.
8
5. We again use both the found energy and angular momentum to get
s
2krmax
~vrmin =
mrmin (rmin + rmax )
s
2krmin
~vrmax =
mrmax (rmin + rmax )
(48)
(49)
6. WE can now calculate ∆v need to go from a lower orbit to a higher one. It is just the
sum of the two changes in velocity. Therefore
∆v = (ver1 − vcr1 ) + (vcr2 − ver2 ) =
s
s
r
r
2kr2
2kr1
k
k
−
+
=
−
mr1 (r1 + r2 )
mr2 (r1 + r2 )
mr2
mr1
(50)
(51)
7. Kepler’s third law give us the period of the elliptical orbit, but we only need half an
orbit, so
T
t= =π
2
r
m
k
9
r1 + r2
2
32
(52)