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

  • #1

Main Question or Discussion Point

I am using Jose & Saletan's "Classical Dynamics", where they introduce a rather contrived Hamiltonian in the problem set: [tex] H(q_1,p_1,q_2,p_2) = q_1p_1-q_2p_2 - aq_1^2 + bq_2^2 [/tex] 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.
 

Answers and Replies

  • #2
Khashishi
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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##?
 
  • #3
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?
 
  • #4
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##
 
Last edited:
  • #5
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.
 

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