Notes
2026-10-30
So, I have the dynamical equation \(\dot{\textbf{x}}=\mathbf{f}(\mathbf{x})\). This results in a trajectory \(\textbf{x}(\textbf{x}_0,t)\). This trajectory passes through the poincaré section \(\sufsec\) at \(\textbf{x}(\textbf{x}_0,\tau_i)\equiv\mathbf{x}_i\), for all \(i\in\mathbb{N}\), let \(T: \sufsec \rightarrow \sufsec\).
Resulting in the equation \[T(\textbf{x}_i)=\mathbf{x}_{i+1},\] naturally, \(T^n(\mathbf{x}_i)=\mathbf{x}_{i+n}\). If there exists \(n\in\mathbb{N}\) such that \(T^n(\textbf{x})=\mathbf{x}\) we call the trajectory periodic.
So lets say we want to find all periodic orbits, what does the algorithm look like to find these?
I am trying to find the symmetries of the Henon Heiles potential. How do I determine the symmetry group of the Henon Heiles potential.
\[\Epot(\textbf{x})=\VHH\]
On can easily see that the parity \(P_x: x \mapsto -x\) results in \[P_x\Epot(x,y)=\Epot(-x,y)=\Epot(x,y)\] due to \(x\) appearing only as quadratic contributions which leads to this symmetry transformation.
The other symmetry transformations need to be varyfied, i.e. being an equilateral (also need to be varyfied).
So lets look at our points of equilibrium that satisfy \(\dot{\mathbf{x}}=0\) resulting with the constraint \(p_x=p_y=0\) as well as \(\dot{p}_x=\dot{p}_y=0\)
\[\frac{\partial\Epot}{\partial x} \vec{e_x} + \frac{\partial\Epot}{\partial y} \vec{e_y}=0\]
I should look at the individual terms so that I can find out if the if the potential carries the same symmetry group of an equilateral triangle (\(m=1, \alpha=1, \omega=1\)):
\[ \begin{align} \frac{\partial\Epot}{\partial x} &= x+2xy \overset{!}{=}0 \\ \Rightarrow x&=0 \wedge y \text{ or } x \wedge y=-\frac{1}{2} \end{align} \]
Let us consider \((0,y)\) in \(\tfrac{\partial \Epot}{\partial y}\)
\[ \begin{align} \frac{\partial\Epot}{\partial y} &= y+x^2-y^2 &\overset{!}{=}0 \\ (0,y)\,\Rightarrow y-y^2 &=0 \\ \Leftrightarrow y &=y^2 & \Rightarrow& y_{1,2}=0,1 \\ (x, -\frac{1}{2})\Rightarrow -\frac{1}{2} + x^2 -\frac{1}{4} &=0 \\ \Leftrightarrow x^2 &= \frac{3}{4} \\ \Leftrightarrow x &= \pm \frac{\sqrt{3}}{2} \end{align} \]
As expected our stable points for of the Hénon-Heiles potential are \[\mathcal{E}=\Bigg\{\vec{0}=\begin{align}\pmatrix{0\\0}, \,\vec{A}=\pmatrix{0\\1},\, \vec{B}=\pmatrix{-\frac{\sqrt{3}}{2}\\-\frac{1}{2}}, \,\vec{C}=\pmatrix{\frac{\sqrt{3}}{2}\\-\frac{1}{2}} \end{align}\Bigg\}.\]
Where \(\vec{0}\) is the center of our potential, by linearization of \(\Epot(\mathbf{x})\) around \(\vec{0}\) we obtain a 2D harmonic potential:
\[T_{\mathbf{x}_i\in\mathcal{E}}[f](\mathbf{x})=\mathbf{f}(\mathbf{x}_i)+\sum_{k=1}^4 \frac{\partial \mathbf{f}(\mathbf{x})}{\partial \mathbf{x}(k)}\Bigg|_{\mathbf{x}=\mathbf{x}_i}\,(\mathbf{x}_{i}(k)-\mathbf{x}(k)) + \mathcal{O}(x_k x_p),\] with \(\mathbf{x}_i\in\mathcal{E}\) being our stable points; The notation \(\mathbf{x}(k)\) comes from the definition of a vector as the map from an alphabet \(\mathcal{A}\) to a field \(\mathbb{F}\) (commonly irrational / complex numbers \(\mathbb{R}\) / \(\mathbb{C}\)), i.e. \(\mathbf{x}:\mathcal{A}\rightarrow \mathbb{F}\). The alphabet \(\mathcal{A}\) can be any kind of ordered index \(\mathcal{A}=(1,2,\dots,n)_{n\in\mathbb{N}}\) or a set of unordered symbols \(\mathcal{A}=(a,b,c,d)\). So if we take \(\vec{A}\in\mathcal{E}\) defined above then \(\mathcal{A}=\{1,2\}\) and \(\mathbb{K}=\mathbb{R}\) results in \(\mathbf{x}(1)=0\) as well as \(\mathbf{x}(2)=1\) being the components of the vector labeled by the alphabet \(\mathcal{A}\).
To obtain the dynamics of our system around this point we should take our dynamical system and linearize \(\mathbf{f}(\mathbf{x})\) at \(\mathbf{x}=\in \mathcal{E}\). This can be done to a later point as well.
Here we want to find the discrete rotations by an angle \(\varphi\in(0,2\pi)\), \(R(\varphi)\Epot(\mathbf{x})=\Epot(\mathbf{x})\). It seems sensible to transform into spherical coordinates \(x=r\, \cos(\theta)\) and \(y=r\, \sin(\theta)\), \(\theta\in(0,2\pi)\) and \(r\in \mathbb{R}: r\geq0\) positive real values.
This results in the following potential that can be simplified as well,
\[\begin{align}\Epot(r,\theta)&=\tfrac{r^2}{2}+r^3\Big(\cos^2(\theta)\sin(\theta)+\tfrac{1}{3}\sin^3(\theta)\Big), \\ &= \tfrac{r^2}{2}+\tfrac{r^3}{4}\Big(\sin(3\theta)+\sin(\theta)+\tfrac{1}{3}(\sin(3\theta)+\sin(\theta))\Big). \end{align}\]
Now we require \(R(\varphi)\Epot(r,\theta)=\Epot(r,\theta')\overset{!}{=}\Epot(r,\theta)\).