Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a simple question concerning how to make a linear change of variables without destroying the symplectic structure of the Hamiltonian?

For example suppose I have a Hamiltonian in action-angle variables given by

\begin{eqnarray} \dot{\theta_1} &=& r_1 +r_2 \\ \dot{\theta_2} &=& r_1 \\ \dot{r_1} &=& \theta_2 \\ \dot{r_2} &=& \theta_1 \\ \end{eqnarray}

Then I want to define a new variable say $\rho= r_1 +r_2$. What will be the associated angle? I am pretty sure I cannot just pick $\theta_1$ or $\theta_2$

share|cite|improve this question
this is not my specialty, but there is such a thing as symplectic diffeomorphisms (or change of variables) : – Glougloubarbaki Jun 21 '12 at 14:27

2 Answers 2

up vote 1 down vote accepted

With new action variables $R_1 = r_1 + r_2$ and $R_2 = r_2$, you can take the new angles as $\phi_1 = \theta_1$ and $\phi_2 = \theta_2 - \theta_1$. In this way you preserve the canonical one-form, $$R_1 \,d\phi_1 + R_2 \, d\phi_2 = (r_1 + r_2) \, d\theta_1 + r_2 \, (d\theta_2 - d\theta_1) = r_1 \, d\theta_1 + r_2 \, d\theta_2,$$ and therefore also the symplectic form. To learn more, read about canonical transformations and their generating functions.

share|cite|improve this answer

Any admissible hamiltonian change of coordinates must preserve the Poisson brackets

On web the page of Giovanni Gallavotti contains a lot of materials (also free books) on the subject. At the moment it is not accessible.

share|cite|improve this answer
That wasn't my question. Suppose I have a transform for my action variables can I find a corresponding transformation for my angle variables such that I preserve the canonical form? – Novo Jun 21 '12 at 17:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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