The background of my question is a well known fact: a Hamiltonian system with $n$ degrees of freedom with $n$ constants of motion is integrable.

My question is about the case in which there are only two constants of motion, one being the Hamiltonian $H$ itself and the other a given function $J$:


Books only make examples and do not provide theorems (e.g. Goldstein "Classical mechanics", third edition, sections 8.2 and 8.3, from page 343). One of the examples is when $J$ is a conserved momentum. In such examples, by means of a suitable canonical transformation, the Hamiltonian is taken to the form:

$$H(p_1, q_1, \dots, p_{n-1}, q_{n-1}, P)\tag{1}$$

in which $J=P$. Here $H$ still has $n$ degrees of freedom, i.e. the space is $2n$ dimensional, although $H$ does not depend on one of the $2n$ variables (let us call it $Q$). The result is that the variable $Q$ is decoupled from the system, and another variable, $P$, is a constant.

The same can be expressed by saying that the system is rewritten in $n-1$ degrees of freedom, or $2(n-1)$ variables, with a reduced Hamiltonian

$$H^r_J(p_1, q_1, \dots, p_{n-1}, q_{n-1})$$

Here the reduced Hamiltonian has $J$ as a parameter.

The books often lead the readers to think that this is a general fact. Does anyone knows if this is a theorem? And where do I find the precise statement and proof? (edited: the answer should tackle the problem from the global point of view).

I think that the form of Eq. 1 is not general, i.e. I think that the presence of an additional constant of motion does not always lead to an expression like Eq. 1, unless some additional hypothesis is given. A counter-example is:

$$H(p_1, q_1, \dots, p_n, q_n) = H_a(p_1, q_1, \dots, p_m, q_m) + H_b(p_{m+1}, q_{m+1}, \dots, p_n, q_n)$$

where $H_a$ and $H_b$ are two Hamiltonians with only one constant of motion each (i.e. $H_a$ and $H_b$). Clearly, $H_a$ is a constant of motion for $H$, but $H$ cannot be taken to a form as in Eq.1.

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    $\begingroup$ I think that in general you would have to find another function on phase space, say $f$, that satisfies $\{f, J\} = 1$. This is not always possible, although it is possible when $J$ is angular momentum ($f$ is then the angle $\theta$). For example, I don't think there's any $f$ such that $\{f, H\} = 1$ when $H = p^2 + q^2$. $\endgroup$ Apr 6, 2019 at 20:58
  • $\begingroup$ Good point, thank you! Do you know any generalization of the form of Eq.1 for generic $J$? $\endgroup$ Apr 7, 2019 at 11:16
  • $\begingroup$ What approach you use depends upon what $J$ denotes. I'm guessing $q_{n}$ is a cyclic coordinate which implies $p_{n}=J$ - but it appears you're ignoring $q_{n}$. And if $J$ denotes angular momentum then there could be several constants of motion. $\endgroup$ Apr 26, 2019 at 22:21
  • $\begingroup$ And if you're going to make a reference to Arnold, you should provide the textbook name and page number. We can look at the page on Amazon. $\endgroup$ Apr 26, 2019 at 22:30
  • $\begingroup$ I added the requested citation (Goldstein). However, it is a citation of a book where only examples are made and the theorem is not given. I've not found it anywhere. But asking to colleagues, they are sure it exists. You can also read the answer below, where it is said to be a corollary of MWM theorem, although this is not written in any book. $\endgroup$ Apr 27, 2019 at 16:28

2 Answers 2


Let there be given a $2n$-dimensional symplectic manifold $(M,\omega)$ with 2 globally defined functions $$H,J: M\to \mathbb{R},$$ such that $$ \{H, J\}~=~0. $$ Then it is an easy exercise to prove the following two propositions.

  • Proposition 1. Let $p_n\!\equiv\! J$. If there is given an open neighborhood $U\subseteq M$ and $2n-1$ functions $$q^1,\ldots, q^n, p_1,\ldots, p_{n-1}:U \to \mathbb{R}$$ such that $$(q^1,\ldots, q^n, p_1,\ldots, p_{n-1}, p_n|_U),$$ is a local canonical/Darboux coordinate system, then $$\frac{\partial H}{\partial q^n}~=~0. $$

  • Proposition 2. Let $p_1\!\equiv\! H$ and $p_2\!\equiv\! J$. Given a point ${\rm pt}\in M$. If $$\mathrm{d}H({\rm pt})\wedge\mathrm{d}J({\rm pt})~\neq~ 0 ,$$ then (according to the Caratheodory–Jacobi–Lie theorem) there exists an open neighborhood $U\subseteq M$ of the point ${\rm pt}\in M$ and $2n-2$ functions $$q^1,\ldots, q^n, p_3,\ldots, p_n:U \to \mathbb{R}$$ such that $$(q^1,\ldots, q^n, p_1|_U,p_2|_U,p_3, \ldots, p_n),$$ is a local canonical/Darboux coordinate system.

  • $\begingroup$ If I correctly understand, this answer only applies locally. The question aimed at a global point of view. I edited the question to make this more explicit. @Qmechanic: is it possible to learn something globally relevant from your answer? $\endgroup$ Apr 11, 2019 at 9:30
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    $\begingroup$ Proposition 2 is a local result. FWIW, proposition 1 is global if $U\!=\!M$. $\endgroup$
    – Qmechanic
    Apr 11, 2019 at 10:35
  • $\begingroup$ I see... However, the hypothesis of proposition 1 are quite implicit. My main goal is to find a generalization, a form that is always valid. A proposition with a more explicit hypothesis, with the corresponding proof, could also help. $\endgroup$ Apr 11, 2019 at 12:03

The specific theorem you are referring to in the background part of the question is the Liouville-Arnold theorem where an $n$-dimensional Hamiltonian system with $n$ involutory constants-of-motion is integrable. I've personally found it pretty tough to apply in practice because it usually just states a fact, but doesn't give you the recipe on how to find the explicit system. The Frobenius theorem is essentially a different version at the vector field level with generalizations to other topological spaces. There's one for Hamiltonian systems, or rather symplectic and Poisson manifolds.

The theorem you are asking about in your question is instead a corollary of the theorem on the Marsden-Weinstein-Meyer reduction. Essentially, for a $2n$-dimensional Hamiltonian system, you can remove double the number of involutory constants-of-motion. If you have $m$ constants of motion, you reduce to $2n - 2m$ dimensions. So say you have $n$ constants-of-motion, then $2n - (2)(n) = 0$. A $0$-dimensional system has been integrated completely so that agrees with Liouville-Arnold.

Moving on to your specific example where $\{H,J\} = 0$, and then

$$ H_{\text{new}}(q_1, p_1, \cdots, q_{n-1}, p_{n-1}, J) $$

You're absolutely correct that this isn't general. This can easily be seen by counting the number of variables. There's an odd number which means this system isn't Hamiltonian at all. However, it doesn't mean it's wrong. Since $\{J, H\} = 0$, then $\dot{J} = 0$ and therefore it shouldn't be included in the dynamical dimension of the system. It's an arbitrary parameter that affects how dynamics evolve, but is not a state itself.

Finally for the counter-example where $H = H_a(q_1, p_1, \cdots, q_m, p_m) + H_b(q_{m+1}, p_{m+1}, \cdots, q_n, p_n)$, you have

$$ 0 = \{H_a, H_b\} $$

This is really easy to show by just looking at what coordinates are involved in either Hamiltonian. The issue you're having is that the canonical transformation has to take place over the specific coordinates contained in the $H_i$ terms. This specific example, with $H_a = J_a$ and $H_b = J_b$, might look something like

$$ H_{\text{new}}(q_1, n_1, \cdots, q_{m-1}, p_{m-1}, J_a, q_m, p_m, \cdots, q_{n-1}, p_{n-1}, J_b) $$

I personally prefer to write

$$ \begin{aligned} H_{\text{new}} : \mathbb{R}^{2(n-c)} \times \mathbb{R}^{c} &\rightarrow \mathbb{R} \\ (Q_1, P_1, \cdots, Q_{n-c}, P_{n-c}, J_1, \cdots, J_c) &\mapsto (\text{the actual function}) \end{aligned} $$

Where $c$ is the number of $J$'s you have. The first line in that equation tells you that, yes indeed we have an even dimensional system and therefore this is truly Hamiltonian since $2(n-c)$ is always even. It also says that we have $c$ arbitrary parameters. Your case has $c=2$, but this should hold for any number of $c$. The second line tells us that $q_i = Q_i$ and $p_i = P_i$ might not necessarily be true because our transformation might have been very complicated, which they almost always are.

Example with $c=1$

Apologies in advance for this not being a "pure" physics problem nor keeping everything in a Poisson language. I have very few real examples of symplectic reduction. Applying this procedure to the Brachistochrone problem, we start with the system

$$ \begin{aligned} \min_{\theta} J &= \int_{0}^{t_f} 1 dt \\ \text{Subject to:} \; \dot{x} &= v\cos(\theta) \\ \dot{y} &= v\sin(\theta) \\ \dot{v} &= g\sin(\theta) \end{aligned} $$

plus some boundary conditions, but they are unimportant to symplectic reduction. Dualizing, or turning this into a Hamiltonian problem, we have

$$ \begin{aligned} H &= \lambda_x v \cos(\theta) + \lambda_y v \sin(\theta) + \lambda_v g \sin(\theta) \\ \dot{\textbf{x}} &= \frac{\partial H}{\partial \textbf{p}}, \dot{\textbf{p}} = -\frac{\partial H}{\partial \textbf{x}}, \frac{\partial H}{\partial \theta} = 0 \end{aligned} $$

where $\textbf{x} = [x,y,v]$ and $\textbf{p} = [\lambda_x, \lambda_y, \lambda_v]$. The total dimension of $\textbf{x}$ and $\textbf{p}$ here is 6. $\theta$ is what's called a control variable, and you can somewhat safely ignore this variable for the purpose of reduction. By crunching the dynamic equations, you can readily see that

$$ \begin{aligned} \dot{\lambda}_x &= 0 \\ \dot{\lambda}_y &= 0 \end{aligned} $$

So $c=2$, but let's just focus on $\int\ \dot{\lambda}_x = J_1$ for now. The prohibitive inversion in symplectic reduction is solving $J_1$ for some state, any state really. Here it is easy, we can solve $J_1$ for $\lambda_x$.

$$ \lambda_x = J_1 $$

That's step 1. Step 2 is to eliminate $x$ from the entire system. Because $x$ is also a symmetry, by definition it doesn't appear in the equations-of-motion and therefore we don't need to do any heavy lifting to remove it. Our final system becomes

$$ \begin{aligned} H &= J_1 v \cos(\theta) + \lambda_y v \sin(\theta) + \lambda_v g \sin(\theta) \\ \dot{\textbf{x}} &= \frac{\partial H}{\partial \textbf{p}}, \dot{\textbf{p}} = -\frac{\partial H}{\partial \textbf{x}}, \frac{\partial H}{\partial \theta} = 0 \\ \dot{x} &= v \cos(\theta) \end{aligned} $$

where $\textbf{x} = [y, v]$ and $\textbf{p} = [\lambda_y, \lambda_v]$. The total dimension of $\textbf{x}$ and $\textbf{p}$ here is 4.

Wait a second, but $\dot{x}$ is still in the equations? Isn't this 5 dimensional? Yes and no. There is no closed form solution for $x$ because it is a function of other quantities that have no closed form solution. However, it is not a part of the dynamics, and can therefore be integrated using numerical quadrature rules like Simpson's rule, instead of propagators like RK45 or Hermite-Simpson collocation. This has a significant impact on computational resources since quadrature can be computationally parallelized whereas it is impossible to achieve complete parallelization on RK45 or collocation-based schemes. $J_1$ is a constant value that affects how the equations of motion evolve, but is also not an equation-of-motion, which is why I made the distinction above in $H_{\text{new}}$.

Slightly More Complicated $J$

In a different problem, say for instance, $J_1 = x^2 + y^2$. This represents the radius of a circle which is constant. Inverting $J_1$ to solve for any state, we get

$$ x = \pm \sqrt{J_1 - y^2} $$

Clearly, you would be better off choosing radius and angle coordinates, however that requires prior knowledge about the structure of the solution.

Counterargument to your Counterargument

Consider two Hamiltonians, each representing a circle

$$ \begin{aligned} H_1 &= x^2 + y^2 \\ H_2 &= u^2 + v^2 \end{aligned} $$

Combining the system gives you

$$ H = x^2 + y^2 + u^2 + v^2 = H_1 + H_2 $$

where $\textbf{x} = [x, u]$ and $\textbf{p} = [y, v]$. In this case, you can choose action angle coordinates for both Hamiltonians, or even worse, you can do

$$ \begin{aligned} x &= \pm\sqrt{J_1 - y^2} \\ u &= \pm\sqrt{J_2 - v^2} \end{aligned} $$

There is nothing preventing me from using the above transformation to bring the dimension down to a 0-dimensional, albeit horrible system. This process can be refined to use a purely Poissonian language, though I'm not that skilled.

  • $\begingroup$ Interesting to know about the Marsden-Weinstein-Meyer reduction... @Michael, could you please suggest where I can find a version that can be easily read by a layman like me? The versions I see are extremely technical. About the expression of $H_{new}$, actually it does not represent the case I was proposing: in my question, $H_a$ and $H_b$ do not have any other function $J$ as constant of motion, so I think that the discussion in the answer does not apply. $\endgroup$ Apr 11, 2019 at 9:38
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    $\begingroup$ @DorianoBrogioli I hear you. I'm an engineer, not a physicist, so learning MWM reduction was self taught and pretty difficult for me. Start with Stephanie Singer's Symmetry in Mechanics, then once you grow out of that move to either Vladimir Arnold's Mathematical Methods of Mechanics or Marsden's Introduction to Mechanics. $\endgroup$ Apr 11, 2019 at 15:34
  • $\begingroup$ And for your second point, I think my case is still valid. I modified my comment to show $H_a = J_a$, etc. Sorry I got a bit sloppy with my naming conventions. $\endgroup$ Apr 11, 2019 at 15:35
  • $\begingroup$ The hard mathematics of the cited paper is difficult also for a physicist like me! So I will follow the suggested learning. About the second point, now it is more clear, but I still think it is not always possible. The situation I was asking about is with c=2 (no other constant besides $H_a$ and $H_b$). I have clues saying that $H_{new}$ cannot be obtained. Do you have the proof of your claim? It would nicely fit in the answer. Else I can provide my clues and we can discuss them. $\endgroup$ Apr 11, 2019 at 20:07
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    $\begingroup$ Quick question before I can form a decent response. After Eqs. 21-22, you say evolution is the "same" for two Hamiltonians. I see $\{H_s, P_i\}$ in the first and $\{H_s, Q_i\}$ in the second equation. To me, the $F_2/F_1$ coefficient is from choice of coordinate transformation, and the Poisson bracket with $H_s$ is the reduced dynamics. While I do not know yet if these equations are "correct", they certainly do not look wrong to me. Can you clarify what you think is wrong here? $\endgroup$ Apr 15, 2019 at 15:02

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