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I am performing a numerical simulation of the example in this page, but am having problems because the Hamiltonian (specifically the $\phi$ momentum part of the kinetic energy) is undefined (infinite?) at $\theta = 0$.
$$H = \frac{P_\theta^2} {2 m l^2} + \frac{P_\phi^2} {2 m l^2 \sin^2 \theta} - m g l \cos \theta$$

Have I made a mistake or missed something, or is this a known and accepted feature of certain simple physical systems? Of course, on the same page, it can be seen that the Lagrangian is well behaved there, but I am using a symplectic integrator so I need the Hamiltonian form.

To put it another way; with the pendulum in its motionless free-hanging state, I understand that the Hamiltonian is degenerate. I am surprised there is no mention or discussion of this on the Wikipedia page.

m4r35n357
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  • If $\theta$ is zero, there can be no momentum $P_{\phi}$. So, I guess you have to interpret that term of the Hamiltonian to be zero. It is therefore not a real singularity. – Raskolnikov Oct 03 '22 at 12:01
  • PS: Maybe add the formula of the Hamiltonian in the body of your text as well. – Raskolnikov Oct 03 '22 at 12:01
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    @Raskolnikov done, but I had to use an external search engine to find the mathjax guide (I thought it was Tex) . . . I can't find it in SE help – m4r35n357 Oct 03 '22 at 12:10
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    @Raskolnikov if $\theta$ is near zero, I think $P_\phi$ can be very high. Computer numerical algorithms are not good at "interpreting" ;) I understand that this is some kind of "coordinate issue", but not for the Lagrangian! Could the Hamiltonian be written in Cartesian coordinates? – m4r35n357 Oct 03 '22 at 12:14
  • Good point, but still, it's not an essential singularity, therefore there should be a better coordinate system. But why do you insist on solving with the Hamiltonian if, as you point out, the Lagrangian does not have this issue? – Raskolnikov Oct 03 '22 at 12:16
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    @Raskolnikov because I want to use my symplectic integrator. I have a Taylor series integrator that can work with Lagrangians (e.g. double pendulum), but unavoidable energy drift makes it not worth the effort. https://en.wikipedia.org/wiki/Energy_drift – m4r35n357 Oct 03 '22 at 12:19
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    @Raskolnikov thinking about this a bit more, that Hamiltonian is not even separable (specifically the "problem" term!), so my explicit symplectic integrators are not valid here anyway ;) – m4r35n357 Oct 05 '22 at 09:04

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