# How many constants-of-motion for a given Hamiltonian?

## Main Question or Discussion Point

I am using Jose & Saletan's "Classical Dynamics", where they introduce a rather contrived Hamiltonian in the problem set: $$H(q_1,p_1,q_2,p_2) = q_1p_1-q_2p_2 - aq_1^2 + bq_2^2$$ where a and b are constants. This Hamiltonian has several constants-of-motion, including f = q1q2, as can be easily checked. In fact, at this point I am aware of four "functionally independent" constants of motion.

Since this Hamiltonian is a function of four variables, is there some theorem that says there are at most four functionally independent constants of motion? If not, then how would I know when I have found enough to form a basis?

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Note: The authors define functions f and g to be functionally independent if both functions can be written as functions of a third function. It would seem that this is a relatively obscure topic.

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Khashishi
That definition of functionally independent isn't clear enough.
If $E$ and $L$ are constants of motion, then $E^L$ and $2E-L$ and any other combinations are also constants of motion.
If $K$ is a constant of motion, then so is $E+K$. But is $E+K$ functionally independent with $E^L$?

That definition of functionally independent isn't clear enough.
If $E$ and $L$ are constants of motion, then $E^L$ and $2E-L$ and any other combinations are also constants of motion.
If $K$ is a constant of motion, then so is $E+K$. But is $E+K$ functionally independent with $E^L$?
Thanks Khashishi. I have assumed that functional independence extends to more than two functions in the analogous way that linear independence does. That is, if $E$ and $L$ are constants of the motion, $\{E,L,E^L\}$ is not a functionally independent set.

Said differently, what is the least number of constants-of-motion that can be combined to form all other constants of motion? In my case, is this four? Is there a general theorem?

Generally to integrate a system $\dot x=v(x),\quad x\in\mathbb{R}^m$ you need m-1 independent first integrals $f_k(x),\quad k=1,\ldots,m-1$. And this is the maximal system: any other first integral depends on $f_k(x),\quad k=1,\ldots,m-1$.
There is a useful fact about the Hamiltonian systems. If the Hamiltonian has the form $H=H(f(p_1,\ldots,p_s,q_1,\ldots,q_s),p_{s+1},\ldots,p_m,q_{s+1},\ldots,q_m)$ then $f$ is a first integral. So in your case the functions $q_1p_1-aq_1^2,\quad -q_2p_2+bq_2^2$ are the first integrals and the Hamiltonian depends on these functions. Due to the specific of Hamiltonian systems, It is sufficient to integrate this system explicitly. For details see https://loshijosdelagrange.files.wordpress.com/2013/04/v-arnold-mathematical-methods-of-classical-mechanics-1989.pdf

The set of functions $f_j(x)$ is called independent in a domain $D$ if the vectors $\nabla f_j$ are linearly independent at each point $x\in D$

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• vanhees71 and Undoubtedly0
Thank you, wrobel. This was exactly what I was looking for. I have found it simple to show that there are at most $2n$ functionally independent constants-of-motion for a Hamiltonian of $2n$ freedoms.