(2.3)(2.3)

(2.1)(2.1)

(2.2)(2.2)

Lecture 19: Calculus of Variations II - Lagrangian

1. Key pointsLagrangianEuler-Lagrange equationCanonical momentumVariable transformation

MapleVariationalCalculus packageEulerLagrange

2. Newton's method vs Lagrange's methodIn the Newton's theory of motion, the position of a particle is determined by an ODE,

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:

where is called Lagrangian.The trajectory that minimizes is determined by the Euler-Lagrange equation

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.

ProofConsider a particle of mass in a one-dimensional potential field . The Newton equation is

(3.1.13)(3.1.13)

(3.1)(3.1)

(3.1.7)(3.1.7)

(3.2)(3.2)

(3.1.8)(3.1.8)

(2.1.1)(2.1.1)

(2.1.3)(2.1.3)

(3.1.9)(3.1.9)

(3.1.1)(3.1.1)

(3.1.14)(3.1.14)

(3.1.6)(3.1.6)

(3.1.4)(3.1.4)

(3.1.5)(3.1.5)(3.1.2)(3.1.2)

(3.1.12)(3.1.12)

(3.1.11)(3.1.11)

(3.1.3)(3.1.3)

(3.1.10)(3.1.10)

Using the Lagrangian

The Euler-Lagrange equation is given by

3. Coordinate transformationsConsider 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

The Euler-Lagrange equation under the new coordinates is

This is because .The minimum of should not depend on

the choice of the coordinates.

Example: Cartesian vs Spherical

Cartesian coordinates Spherical coordinates

(3.1.13)(3.1.13)

(4.1)(4.1)

(3.1.7)(3.1.7)

(3.1.8)(3.1.8)

(2.1.1)(2.1.1)

(3.1.9)(3.1.9)

(3.1.1)(3.1.1)

(3.1.14)(3.1.14)

(3.1.6)(3.1.6)

(3.1.4)(3.1.4)

(3.1.5)(3.1.5)(3.1.2)(3.1.2)

(3.1.12)(3.1.12)

(3.1.11)(3.1.11)

(3.1.3)(3.1.3)

(3.1.10)(3.1.10)

4. Canonical momentum and its conservation

We define a canonical momentum as

Then, the Euler-Lagrange equation is written as

(6.1.1)(6.1.1)

(4.2)(4.2)

(2.1.1)(2.1.1)

(5.1)(5.1)

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 conservationWhen the Lagrangian does not depend on time explicitly, the following quantity conserves in time.

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 oflength is fixed to the edge of the disk. A mass

is attached to the other end of thestring. The mass horizontally oscillates with respect to the suspension point. The angle shownin 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

(6.1.6)(6.1.6)

(6.1.2)(6.1.2)

(6.1.3)(6.1.3)

(2.1.1)(2.1.1)

(6.1.5)(6.1.5)

(6.1.7)(6.1.7)

(6.1.4)(6.1.4)

and the corresponding velocity

Since there is no potential energy, Lagrangian has only kinetic energy terms

= simplify

12

Taking into account , the Lagrangian is further simplified to

Now, we derive the equation of motion using Euler-Lagrange equation .

= simplify

This equation is mathematically equivalent to the equation of motion for a usual pendulum with gravity .

Answer:

(2.1.1)(2.1.1)

.

5.2

Four massless rigid rods of length are connected byhinges so that they form a rhombus (see Figure.) Thetop 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 velocityof . Then, the system has only one degree of freedom. For example, the angle shown in Figure and the corresponding velocity uniquely determine the state of the system.

Express the Lagrangian of the system using $\theta$ as a coordinate.

1.

Find the equation of motion for $\theta$.2. Find an equilibrium angle $\theta^*$ as a function of $\Omega$.

3.

Define positions

Define velocities

(6.2.1)(6.2.1)

(6.2.2)(6.2.2)

(6.2.3)(6.2.3)

(2.1.1)(2.1.1)

Construct kinetic energy for each mass

Construct potential energy for each mass

(1) Lagrangian

Simplifying the expression,

(2) Equation of motions

(3) Equilibrium position

(6.2.5)(6.2.5)

(2.1.1)(2.1.1)

(6.2.4)(6.2.4)

To find an equilibrium position, we set =0.

Hence, there are three equilibrate (note that arccos has two values) if and

only one if .

Answer:

(a)

(b)

(c) When , is a stable equilibrium. If , becomes unstable. New

equilibrium angles 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.

Find the Lagrangian for the system using the generalized coordinates.

1.

Find the canonical momentum for each generalized coordinates.

2.

Find the equations of motion.3.

(6.3.5)(6.3.5)

(6.3.4)(6.3.4)

(6.3.3)(6.3.3)

(6.3.6)(6.3.6)

(6.3.1)(6.3.1)

(2.1.1)(2.1.1)

(6.3.2)(6.3.2)

Let the position of the left mass be and . Immediately, we know . Since the length of string is fixed,

where is a constant. The position of the right mass is given by

Their velocities are

The Lagrangian for the system is

= simplify

(2) Using the definition of canonical momentum, we obtain:

(3) Using Euler-Lagrange equation,

(6.3.11)(6.3.11)

(6.3.6)(6.3.6)

(6.3.10)(6.3.10)

(2.1.1)(2.1.1)

(6.3.8)(6.3.8)

(6.3.7)(6.3.7)

(6.3.9)(6.3.9)

= simplify

= simplify

Therefore, the equations of motion are

and