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I have a hamiltonian of the form $$H(\phi,z) = (1-z^2)\cos(2\phi) + \chi z^2$$ with position $\phi$ and conjugate momentum $z$. It has this form provided that $z \in [-1,1]$ and we have natural representation of the dynamics on sphere with $z$ being distance on the $z$-axis and $\phi$ - azimuthal angle.

Now, I am looking for equilibrium points (fixed points) and one of them includes $z = 1$ or $z=-1$ and some $\phi$. I am not sure if I can call such points fixed points cause phase $\phi$ is not defined on poles. How to find the approximate dynamics around poles (it depends on $\chi$) ?

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They are fixed points because the location is well defined.

You think that it is a problem only because you are represented it in $(z,\phi)$. If you write it in Cartesian coordinate, the fix points will be clear to you.

For the dynamics around a fixed point, first you should calculate the dynamic equation $(z,\dot{\phi})$ and linearize it around the point. Then you can get an approximate solution by solving two coupled linear differential equations.

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  • $\begingroup$ I have a problem with that cause, if You look from a plane perspective you may find around pole that azimuthal angle linearly runs away, but on a sphere point rotates around pole (this is not actually a fixed point when $\chi > 1$, but dynamics is "different") $\endgroup$ – WoofDoggy Jan 9 '15 at 11:47
  • $\begingroup$ @Nex_Friedrich It is what you should expected. If $\phi(t) = k t$ a linear equation, which mean the motion is around the pole in a circle. It really just a problem of coordinate transformation. $\endgroup$ – unsym Jan 10 '15 at 0:45
  • $\begingroup$ and stability - if something is not bounded on the plane it becomes so on the sphere $\endgroup$ – WoofDoggy Jan 10 '15 at 18:01
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Hint :

\begin{equation} H(\phi,z) = (1-z^2)\cos2\phi + \chi z^2 \tag{01} \end{equation}

\begin{equation} \left. \begin{cases} \overset{\boldsymbol{.}}{z}=\boldsymbol{-}\dfrac{\partial H}{\partial \phi}=\boldsymbol{+}2\left(1-z^{2}\right)\sin2\phi \vphantom{\dfrac{\dfrac{a}{b}}{b}}\\ \overset{\boldsymbol{.}}{\phi}=\boldsymbol{+}\dfrac{\partial H}{\partial z}=\boldsymbol{-}2\left(\cos2\phi-\chi\right)z\vphantom{\dfrac{\dfrac{a}{b}}{b}} \end{cases} \right\} \tag{02} \end{equation} To find fixed points the two equations in (03) must be fulfilled simultaneously \begin{equation} \left. \begin{cases} \overset{\boldsymbol{.}}{z}=0 \vphantom{\dfrac{a}{b}}\\ \overset{\boldsymbol{.}}{\phi}=0 \vphantom{\dfrac{a}{b}} \end{cases}\right\} \Longrightarrow \left. \begin{cases} \left(1-z^{2}\right)\sin2\phi=0 \vphantom{\dfrac{a}{b}}\\ \left(\cos2\phi-\chi\right)z=0 \vphantom{\dfrac{a}{b}} \end{cases}\right\} \tag{03} \end{equation}

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