Download Lecture 19: Calculus of Variations II

Lecture 19: Calculus of Variations II - Lagrangian
1. Key points
Lagrangian
Euler-Lagrange equation
Canonical momentum
Variable transformation
Maple
VariationalCalculus package
EulerLagrange
2. Newton's method vs Lagrange's method
In the Newton's theory of motion, the position of a particle is determined by an ODE,
(2.1)
In general we find the trajectory by solving this ODE with an initial conditions
.
and
In the Lagrange's method, we try to find the trajectory by minimizing a functional. In this case, we
use a boundary condition: the initial position
and the final position
. The
functional is given in a form:
(2.2)
where is called Lagrangian.
The trajectory
that minimizes is determined by the Euler-Lagrange equation
(2.3)
The Newton's method and the Lagrange's method should give the same trajectory. Therefore, Eq.
(2.3) should be equivalent to (2.1). If the force is given by a potential energy function
, then
satisfies the requirement, where is kinetic energy.
Proof
Consider a particle of mass
in a one-dimensional potential field
. The Newton equation is
(2.1.1)
Using the Lagrangian
The Euler-Lagrange equation is given by
(2.1.3)
3. Coordinate transformations
Consider a coordinate transformation from
to
. where are in general a function
of
. So, we write it as
What is the Newton equations in the new coordinates? The result is usually very complicated. (See
the example below). However, Euler-Lagrange equation has exactly the same mathematical form.
The new Lagrangian and old one are related by the coordinate transformation as
(3.1)
The Euler-Lagrange equation under the new coordinates is
(3.2)
This is because
.The minimum of should not depend on
the choice of the coordinates.
Example: Cartesian vs Spherical
Cartesian coordinates
Spherical coordinates
(3.1.1)
(3.1.4)
(3.1.1)
(3.1.4)
(3.1.2)
(3.1.5)
(3.1.3)
(3.1.6)
(3.1.7)
(3.1.11)
(3.1.8)
(3.1.12)
(3.1.9)
(3.1.13)
(3.1.14)
(3.1.10)
(3.1.14)
(3.1.10)
4. Canonical momentum and its conservation
We define a canonical momentum as
(4.1)
Then, the Euler-Lagrange equation is written as
(4.2)
For Cartesian coordinates, the canonical momentum corresponding to the coordinate is given by
0
which is linear momentum in the direction.
For the spherical coordinates, the canonical momentum corresponding to the coordinate is given by
0
which is angular momentum
If the Lagrangian does not depend on a coordinate , the right hand side of (4.2) vanishes. Therefore,
. This indicates that the corresponding canonical momentum conserves in time.
5. Energy conservation
When the Lagrangian does not depend on time explicitly, the following quantity conserves in time.
(5.1)
where
is a canonical momentum corresponding to the coordinate .
Proof
where we used
5. Examples
5.1
A horizontal disk of radius is spinning about its
center in the counterclockwise with a constant
angular velocity . An end of a massless string of
length is fixed to the edge of the disk. A mass
is attached to the other end of the
string. The mass horizontally oscillates with
respect to the suspension point. The angle shown
in the figure can be used as a generalized
coordinate. Show that the equation of motion for
is mathematically equivalent to that of a
pendulum except for the gravity is replaced with
something else.
First, we express the position of the mass
and
in terms of the generalized coordinate
(6.1.1)
(6.1.2)
and the corresponding velocity
(6.1.3)
(6.1.4)
Since there is no potential energy, Lagrangian has only kinetic energy terms
(6.1.5)
simplify
=
0
Taking into account
simplified to
, the Lagrangian is further
(6.1.6)
Now, we derive the equation of motion using Euler-Lagrange equation
.
(6.1.7)
simplify
=
This equation is mathematically equivalent to the equation of motion for a usual pendulum with gravity
Answer:
.
5.2
Four massless rigid rods of length are connected by
hinges so that they form a rhombus (see Figure.) The
top corner A is fixed. Two objects of mass and
another of mass are attached to the hinges as
shown in Figure. The mass can move only
vertically along the vertical axis. The whole system
rotates around the vertical axis with angular velocity
of . Then, the system has only one degree of
freedom. For example, the angle shown in Figure
1.
.
and the corresponding velocity uniquely determine
the state of the system.
1. Express the Lagrangian of the system using
$\theta$ as a coordinate.
2. Find the equation of motion for $\theta$.
3. Find an equilibrium angle $\theta^*$ as a
function of $\Omega$.
Define positions
0
Define velocities
0
Construct kinetic energy for each mass
Construct potential energy for each mass
(1) Lagrangian
(6.2.1)
Simplifying the expression,
(6.2.2)
(2) Equation of motions
(6.2.3)
(3) Equilibrium position
To find an equilibrium position, we set =0.
(6.2.4)
(6.2.5)
Hence, there are three equilibrate (note that arccos has two values) if
only one if
and
.
Answer:
(a)
(b)
(c) When
equilibrium angles
,
is a stable equilibrium. If
,
becomes unstable. New
are stable.
5.3
Two identical masses connected by a string
hang over two massless pulleys as shown in
Figure. While the left mass moves only in a
vertical line, the right one is free to swing in
the plane defined by the pulleys and the left
mass. In order to describe the motion of
masses, we introduce generalized coordinates
constant, which is a constraint in this problem.
1. Find the Lagrangian for the system using
the generalized coordinates.
2. Find the canonical momentum for each
generalized coordinates.
3. Find the equations of motion.
Let the position of the left mass be
Since the length of string is fixed,
and . Immediately, we know
.
(6.3.1)
where
is a constant. The position of the right mass is given by
Their velocities are
0
The Lagrangian for the system is
(6.3.2)
simplify
=
(6.3.3)
(2) Using the definition of canonical momentum, we obtain:
(6.3.4)
(6.3.5)
(3) Using Euler-Lagrange equation,
(6.3.6)
simplify
=
(6.3.7)
(6.3.8)
simplify
=
(6.3.9)
Therefore, the equations of motion are
(6.3.10)
and
(6.3.11)