On this page I'll try to solve the two-dimensional pendulum problem in various ways, including using rational trigonometry.

I'll start with the simple pendulum, and then maybe the double one and arbitrary long ones.

# The Simple Pendulum

## Lagrangian

The Lagrangian for the simple pendulum is $$L = \frac{1}{2}mv^2 - mgh$$

With an appropriate choice of length, mass and time units, this can be written: $$L = \frac{v^2}{2} - h$$

## Using trigonometric functions

With trigonometric functions, the speed $$v$$ and elevation $$h$$ become respectively $$\dot\theta$$ and $$1 - \cos(\theta)$$.

The Lagrangian then becomes: $$L = \frac{\dot{\theta}^2}{2} + \cos(\theta) - 1$$

The Euler-Lagrange equation then gives:

$$\ddot{\theta} = -\sin{\theta}$$

which we can turn into a first order derivative by considering $$\dot{\theta}$$ as an additional variable.

## Using only rational numbers

We're going to use the following rational parametrization of the circle:

$$\mathrm{M}(\mu) = (\frac{1-\mu^2}{1+\mu^2}, \frac{2\mu}{1+\mu^2})$$

The squared speed becomes: $$v^2 = \frac{4\dot{\mu}^2}{(1+\mu^2)^2}$$

Thus the Lagragian is now: $$L = \frac{2\dot{\mu}^2}{(1+\mu^2)^2} + \frac{1-\mu^2}{1+\mu^2}$$

And the Euler-Lagrange equation:

$$\frac{d}{dt}(\frac{4\dot{\mu}}{(1+\mu^2)^2}) = -\frac{8\dot{\mu}^2\mu}{(1+\mu^2)^3} - \frac{4\mu}{(1+\mu^2)^2}$$

...turns into:

$$\frac{4\ddot{\mu}}{(1+\mu^2)^2} - \frac{16\dot{\mu}^2\mu}{(1+\mu^2)^3} = -\frac{8\dot{\mu}^2\mu}{(1+\mu^2)^3} - \frac{4\mu}{(1+\mu^2)^2}$$ $$\ddot{\mu} = \frac{2\dot{\mu}^2\mu}{1+\mu^2} - \mu$$

### Dealing with the exceptional case

The naive approach described above has a major issue : it can not deal with full swings. Indeed the point $$\lim_{\mu\to\infty}\mathrm{M}(\mu) = (-1, 0)$$ can never be reached with that parametrization.

A simple solution is to switch to a different parametrization when it is expedient. $$\mathrm{N}(\nu) = (-\frac{1-\nu^2}{1+\nu^2}, -\frac{2\nu}{1+\nu^2})$$

The dynamics should not be much different. Indeed there is only one sign change: $$\ddot{\nu} = \frac{2\dot{\nu}^2\nu}{1+\nu^2} + \nu$$

We then need a relation between $$\mu$$ and $$\nu$$. That is, we must find the solution to $$\mathrm{M}(\mu) = \mathrm{N}(\nu)$$

It is not too hard to see that $$\nu = -\frac{1}{\mu}$$ is a solution, as long as $$\mu\ne 0$$.

From this we can derive the relation between the generalized velocities $$\dot{\nu} = \frac{\dot\mu}{\mu^2},\qquad \dot{\mu} = \frac{\dot\nu}{\nu^2}$$ Your browser can't display canvas.