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 $$ \dot{\nu} = \frac{\dot\mu}{\mu^2},\qquad \dot{\mu} = \frac{\dot\nu}{\nu^2} $$