The simple plane pendulum $$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin{\theta} = 0$$

has the very perdy phase portrait

perdy phase portrait Meanwhile, a domain coloring of $\sin(z)$ in the complex plane is

even perdier

Why are these so similar?


The trajectories of the differential equation satisfy the implicit equation (with $v = dy/dt$) $$f(\theta,v) = \frac{v^2}{2} - \frac{g}{l} \cos \theta = A, \ A \ge -\frac{g}{l}$$

I think the white curves are the level curves of $|\sin(x+iy)|$. These are given by $$g(x,y) = \cosh (2y) - \cos(2x) = B,\ B \ge 0$$

The two are related by the change of variables $B = Al/g + 1$, $\theta = 2 x$, $v = 2 \sqrt{g/l} \sinh y$.

  • $\begingroup$ The implicit equation is just conservation of energy, yes? $\endgroup$ – AndrewG Oct 25 '12 at 0:32
  • $\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Oct 28 '12 at 23:06

The equations of the phase curves in the phase portrait of the simple plane pendulum actually correspond to different energy conservation relations: $$ \dot{\theta}^2 - \frac{g}{l}\cos(\theta) = C_0 $$

And in the colored graph of $\sin(z)$ in the complex plane the lines are the lines of constant magnitude: $$ \|\sin(x+yi)\|^2 = C $$ which can be transformed into another form by the steps below $$ \begin{align} \|\sin(x)\cosh(y) + i\cos(x)\sinh(y)\|^2 &= C \\ \sin(x)^2\cosh(y)^2 + \cos(x)^2\sinh(y)^2 &= C \\ (\sin(x)^2 + \cos(x)^2)\frac{e^{2y}+e^{-2y}}{2} + \sin(x)^2-\cos(x)^2 &= C \\ \frac{e^{2y}+e^{-2y}}{2} -\cos(2x) &= C \end{align} $$ when $y$ is not far from $0$, $\frac{e^{2y}+e^{-2y}}{2} \approx 4y^2 = (2y)^2$,so if we replace $(x,y)$ by $(u,v)$ with $u=2x, \, v=2y$, then the equation becomes $$ v^2 -\cos(u) = C. $$ I think this is why the two plots look so similar. When $y$ goes far from $0$, their forms may no longer be such similar.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.