Simple examples

In order to introduce the concept of generalized coordinates, let us first consider a few simple examples.

Consider, first, the problem of a simple pendulum moving in the x-y plane. The pendulum has a length l and moves under the action of gravity, so that its potential energy is mgh. The system is illustrated in the figure below.

We could use Cartesian coordinaets x and y to describe the location of the pendulum bob, but x and y are not independent. In fact, since the length of the pendulum is constant, they are related by

This condition would need to be imposed as a constraint on the system, which can be inconvenient. It is more natural to use the angle that the pendulum makes with respect to the vertical to describe the motion. But what would be the equation of motion for ? In order to find out what this is, we only need to express the Lagrangian in terms of . Now, the Lagrangian in terms of x and y is givn by

where we have introduced a general potential function, however, for this example, we know that the potential is given by U(x,y) = -mgy.

The Cartesian coordinates x and y are related to by a set of transformation equations:

In order to transform the kinetic energy, we need the time derivatives of the transformation equations:

Substituting the transformations and their derivatives into the Lagrangian gives

Now, given the Lagrangian, we just turn the crank on the Euler-Lagrange equation and derive the equation of motion for :

so that the equation of motion is

As another example, consider again a particle moving in the x-y plane subject to a potential U that is a function only of the distance of the particle from the origin of the coordinate system. This distance is , so that U = U(r). An example would be a radial harmonic potential , which is a ``bowl'' potential shown below:

Of course, we can choose to work in Cartesian coordinates, x, and y. In this case, we would write down the Lagrangian

where the potential U(r) is expressed as a function of x and y through the dependence of r on x and y. Then, the equations of motion for x and y can be computed with forces obtained via the chain rule

which is perfectly correct. However, it obviscates some of the important physics of the problem, which can be revealed by working in polar coordinates r and , which are more natural for this problem. These are related to x and y via the transformations:

with time derivatives

Substituting into the kinetic energy gives, after some algebra

so that the Lagrangian can be expressed as

There will be two equations of motion for r and given by

Thus, all we need to do to determine the equations of motion is turn the crank. For r, we have

We see immediately that if , then , and the particle will remain in circular motion with a centripetal acceleration . For , we have

which can also be written as

The last line is in the form of a conservation law, which states that the quantity, . This is known as the conservation of areal velocity. The constant is usually defined such that

or

Substitution of this conservation law into the radial equation gives an uncoupled equation for r alone:

The conservation law and uncoupled radial equation would have been difficult to decude directly in terms of Cartesian coordinates, however, the Lagrangian formulation allows these results to be derived rather easily using the machinery of the Euler-Lagrange equation.

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