Download 4 Class exercise sheet

4
Class exercise sheet
Poisson brackets:
We define the Poisson brackets as follows:
[f, g] =
X ∂f ∂g
∂f ∂g
−
∂pk ∂qk ∂qk ∂pk
k
(1)
with this in mind we find that
[f, p] = −
∂f
∂q
;
[f, q] =
∂f
∂p
(2)
The equations of motion for any function f will be
df
∂f
= [f, H] +
dt
∂t
(3)
where H is the Hamiltonian.
Hamiltonian formalism:
We define the Hamiltonian as a function of qi and pi which are canonical variables.
The transformation from Lagrangian to Hamiltonian is as follows:
1. We first need to perform the Lagander transform
H=
X
q̇i
i
X
∂L
−L=
q̇i pi − L
∂ q̇i
i
(4)
∂H
∂pi
=−
∂qi
∂t
(5)
2. Find the canonical momenta.
3. Express q̇i in terms of pi .
4. Express H in terms of pi .
And the Hamilton EOM are
∂H
∂qi
=
∂pi
∂t
;
This gives us 2n degrees of freedom and 2n first order differential equations. (Instead of n
degrees of freedom and n second order differential equations in Lagrangian formalism).
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Exercise 4.1: Particle in a magnetic field
A charged particle is moving in a uniform magnetic field H. The vector potential is in the
form of Ay = xH and Ax = Az = 0.
~
Reminder the magnetic potential is −e~v · A
1. Write the Lagrangian.
2. Find the canonical momenta.
3. Find the Hamiltonian.
4. Solve the EOM for the Hamiltonian. Notice there is a way to write the Hamiltonian
which will simplify this.
Solution:
1. The Lagrangian is
1
L = m ẋ2 + ẏ 2 + ż 2 + eBxẏ
2
(6)
2. The canonical momenta are

 px = mẋ
∂L
pz = mż
⇒
pi =

∂ q̇i
py = mẏ + eBx
(7)
3. So we just need to follow the recipe given
X
m 2
1
H=
q̇i pi − L = mẋ2 + mż 2 + mẏ 2 + eBxẏ −
ẋ + ẏ 2 + ż 2 − eBxẏ = m ẋ2 + ẏ 2 + ż 2 (8)
2
2
using the momenta we found
px
= ẋ
(9)
m
pz
= ż
(10)
m
py eBx
−
= ẏ
(11)
m
m
2
so our Hamiltonian is
p2x + p2z
1
H=
+
(py − eBx)2
2m
2m
(12)
4. We can write this Hamiltonian just like a harmonic oscillator Hamiltonian
H=
1
p2
p2x
+ mω 2 (x − x0 )2 + z
2m 2
2m
2
(13)
2
py
where we defined x0 = eB
and ω 2 = emB2 .
The reason we can do this is that both z and y are cyclic coordinates which means that
their momenta is conserved.
5. So the solution for x is given for free
x = a cos (ωt + α) + x0
;
px = −mωa sin (ωt + α)
(14)
z is also obvious and simple
ż =
pz
∂H
=
∂pz
m
⇒
z=
pz t
+ z0
m
(15)
And for y we use the solution of x
1
∂H
eB py
=
(py − exB) =
− x = −aω cos (ωt + α)
∂py
m
m eB
y = −a sin (ωt + α) + y0
ẏ =
(16)
(17)
The particle moves along a spiral with it axis parallel to B. The generalized momentum
py determines the distance of that axis from the y − z plane.
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Exercise 4.2: Spherical Pendulum
φ
M
R
m
θ
A bead of mass m is free to slide along a friction less hoop of radius R and mass M .
1. Write down the Lagrangian for this system. Are there any cyclic coordinates?
2. Find the Hamiltonian of the system.
3. Check your answer to item 1 explicitly using Poisson brackets.
4. Find the Hamilton’s EOM for the angle θ.
5. What are the equilibrium positions?
6. What is the frequency of small oscillations about the stable equilibrium?
7. There is one value of ω that is rather special; what is it, and why is it special?
Now, assume the hoop rotates with constant angular velocity ω around a vertical diameter.
1. Find the Hamiltonian of the system.
2. Is H the energy? Is H conserved?
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Solution:
1. There are two DOF. To solve those it would be wise to move to spherical coordinates
and plug in the constraints of the problem r = R and φ̇ = ω into the Lagrangian to
receive
We have a rigid body whose center of mass is fixed and which is free to rotate only
about one axis, leaving one degree of freedom. The bead is free to move around the
ring, but its distance from the center is fixed and its angle in the horizontal plane is
determined by the angle of the ring. Working in spherical coordinates such that θ = 0
is the bottom for simplicity, we find that the Lagrangian is:
1 2 1 2
2 2
2
2
2
IΩ + m ṙ + r θ̇ + r sin θφ̇ + mgr cos θ
(18)
L =
2
2
1
1
=
M R2 φ̇2 + mR2 θ̇2 + sin2 θφ̇2 + mgR cos θ
(19)
4
2
φ is cyclic
2. We write the conjugate momentum,
∂L
= (I + mR2 sin2 θ)φ̇
∂ φ̇
= mR2 θ̇
pφ =
(20)
pθ
(21)
Using the Legendre transform the Hamiltonian is then:
H=
p2φ
p2θ
− mgR cos θ
+
2(I + mR2 sin2 θ) 2mR2
(22)
3. The equations of motion are given by
df
∂f
= [f, H] +
dt
∂t
If we want f to be a constant of motion it must obey
our case this means [pφ , H] = 0 giving us
[pφ , H] =
(23)
df
dt
= 0 giving us [f, H] = 0. In
∂H
∂H ∂pφ ∂H ∂pφ
−
=−
=0
∂pφ ∂φ
∂φ ∂pφ
∂φ
5
(24)
4. Hamilton’s EOM are
pθ
∂H
=
∂pθ
mR2
mR2 p2φ sin θ cos θ
∂H
=
= −
2 − mgR sin θ
∂θ
I + mR2 sin2 θ
θ̇ =
ṗθ
(25)
(26)
5. We are now looking for equilibrium, this means that θ̈ = 0. So we need to find an
equation θ̈, we can use Hamilton’s EOM to find that
g
θ̈ = sin θ cos θω 2 −
(27)
R
so for equilibrium
g
2
0 = sin θ cos θω −
R
(28)
This has three solutions: θ = 0, θ = π and cos θ = ω2gR .
Notice that there is a condition for cos θ = ω2gR because cos θ can only go between −1
and 1 meaning that ω 2 ≥ Rg , this means we have two cases:
(a) The case where ω 2 ≤ Rg : This means that the two extrema are θ = 0 and θ = π.
It is very intuitive that the stable extrema is θ = 0 but one can also check by taking
θ → π + δ if we expand for small δ we receive
g
δ
(29)
δ̈ = ω 2 +
R
This solution is exponential which means that this point is unstable, making θ = 0
the stable equilibrium point.
So to find the frequency of small oscillations we need to solve for small amplitude,
giving us
g
− ω2 θ
(30)
θ̈ = −
R
(b) The case where ω 2 ≥ Rg : The extremas are the same as item (a.) and cos θ0 = ω2gR .
This time around θ = 0 is also unstable (this is obvious because Rg − ω 2 is
negative now giving us an exploding solution). So the stable point is now cos θ0 =
g
.
ω2 R
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To find the frequency of small oscillations we expand around θ = θ0 + δ which will
give us
r
g2
ω sin θ0 = ω 2 − 2 2
(31)
ω R
p
6. The special frequency is ω = Rg , this is the critical frequency above which the mass
wants to move away from the bottom of the hoop.
We now assume that ω is constant
1. The Lagrangian is
1
1
L = M R2 φ̇2 + mR2 θ̇2 + sin2 θφ̇2 + mgR cos θ
4
2
(32)
Our constraint is now φ̇ = ω which we will plug at the Lagrangian level to receive
1 2
1
1
2 2
2
(33)
L = mR θ̇ + R m sin θ + M ω 2 + mgR cos θ
2
2
2
This will give us the following Hamiltonian
H=
(34)
2. The Hamiltonian is not the energy, because the Cartesian coordinates (x, y) in the
horizontal plane are related to θ by (x, y) = R sin θ(cos ωt, sin ωt), up to a phase. Since
this relation involves t, the Hamiltonian is not the energy. But H is in fact conserved,
because there is no t dependence in L.
The Hamiltonian differs from the energy due to the minus sign in the second term.We
will simplify the problem by taking M → 0. So this means that the energy equals
H + mR2 ω 2 sin2 θ. But as noted above, H is conserved. So the energy takes the form of
a constant plus mR2 ω 2 sin2 θ. That is, the energy is larger when mass is farther away
from the axis of rotation. Where does this additional energy come from? It comes from
the work that the hoop does on the bead, due to the normal force perpendicular to the
plane of the hoop. We can be quantitative about this as follows.
To find the normal force, note that the angular momentum around the vertical axis is
given by m (R sin θ)2 ω. The time derivative of this (which equals the torque) is
τ = 2mR2 ω sin θ cos θθ̇ = (R sin θ) 2mRω cos θθ̇
(35)
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Since the ”lever arm” is R sin θ, the normal force must be 2mRω cos θθ̇. (You can also
derive this by using the fact that the normal force must balance the Coriolis force in
the rotating frame.)
In a small time dt, the mass moves a distance (R sin θ)(ωdt) horizontally, so the work
done by the normal force during this time is (2mRω cos θθ̇)(R sin θωdt). The total work
done by the time the mass reaches an angle θ is therefore
Z θ
Z
dθ
2mRω cos θR sin θωdθ = mR2 ω 2 sin2 θ
(36)
(2mRω cos θ )(R sin θωdt) =
dt
0
which agrees with the additional energy we found above.
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