Suppose I have an integrable Hamiltonian system $H(q_{1}, p_{1},..., q_{n}, p_{n})$, with first integrals $F_{1} = H, F_{2},..., F_{n}$. Excluding certain singular level sets (i.e. separatrices), one may convert to action-angle variables $(I, \varphi)$, (at least locally in the complement of singular level sets), and write $H(I)$.

My question is, how exactly are actions $I$ related to first integrals $F$? We know that generally, $I_{i} =I_{i}(F_{1},..., F_{n})$, for $i=1,..,n$, i.e. $I$'s are functions of $F$'s. Can we say anything further about the specific form of these functions?

More specifically, for what type of Hamiltonians can we take $I_{i} = F_{i}$?

Edit. My motivation comes from the following Hamiltonian: $$H = H(q_{1}, p_{1},..., q_{j}, p_{j}, F_{k}(q_{j+1}, p_{j+1}), q_{j+2}, p_{j+2},..., q_{n}, p_{n})$$

where $F_{k}$ is one of the first integrals, that depends only on $(q_{j+1}, p_{j+1})$. Can one, in this case, take $I_{k} = F_{k}$?

As a further motivation, consider a one-dimensional oscillator $H = p^{2} + q^{2}$; here one may simply take $I = H$ (up to an adjustment of a constant factor). How does this generalise to the type of Hamiltonian given above?

  • $\begingroup$ Comment to the post (v3): The last appearance of the word first integrals seems to play the role of a separation function. $\endgroup$ – Qmechanic Mar 1 '18 at 19:34
  • $\begingroup$ 1. If the separation functions $F_1, \ldots, F_n$ are functionally independent, then they are automatically Poisson commuting and constants of motion, cf. my Phys.SE answer here. 2. The interplay between constants of motion and action variables is considered in my Phys.SE answer here. $\endgroup$ – Qmechanic Mar 2 '18 at 18:41

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