1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conserved quantities

  1. Aug 20, 2014 #1
    Hi all,

    I am preparing for my "second chance exam" in analytical mechanics.
    It is a graduate course i.e. based on geometry. (Our course notes are roughly based on Arnold's book).

    I was able to find some old exam questions and one of those has me stumped, completely.
    The question gives 3 general Hamiltonians, no details whatsoever and asks to find as many conserved quantities as possible.
    *Warning* The second one really hurts my brain.

    ##\mathcal{H}_1 = \mathcal{H}_1\left( f_1(q^1,p_1), \ldots, f_N(q^N,p_N)\right)##
    ##\mathcal{H}_2 = g_N( g_{N-1}(\ldots g_2(g_1(q^1,p_1),q^2,p_2)\ldots ,q^{N-1},p_{N-1}),q^N,p_N)##
    ##\mathcal{H}_3 = \sum_{i=1}^N \left(\dot{q}^i(t)\right)^2+V\left( \sum_{i=1}^N \left(q^i(t)\right)^2\right)##

    Hamiltonian (1) isn't terribly complicated at first sight, however without further information about the functions ##f_i## I'm not sure what I can deduce. As far as I can tell, I cannot claim anything to be conserved except the Hamiltonian itself. (Which is easy to see because there is no explicit time-dependence so ##dH_1/dt = \{H,H\} = 0##)

    The second Hamiltonian, I don't even understand why the instructor would do such a thing to us. It is horrible and unless someone knows a neat trick or has some cool information about this, I suggest we all pretend that doesn't exist. My guess is that only the conservation of the Hamiltonian follows from this, once again.

    Definition (3) seems like a nice expression. But there is a strange thing going on here, we have ##\dot{q}^i## in there but no generalized moments.
    I could state that the moments are cyclic but that would be a trap I think.
    This Hamiltonian hasn't been "fully transformed" from the Lagrangian.
    I believe the "real" Hamiltonian might be invariant under rotations in the coordinates ##\vec{q}##.
    My (very short and vague) motivation is that the positions appear squared only, no mixing either.

    So does anybody have some general remarks/ideas/resources to help me with this kind of stuff?

  2. jcsd
  3. Aug 22, 2014 #2
    So I made some 'progress', the first 2 are definitely based on the use of (time independent) Hamilton-Jacobi.
    Even though this is not much to go by, I suppose it might help some people with the same problems.

    I might come back to this later and go into some detail if I can find the time.
  4. Aug 22, 2014 #3


    User Avatar
    Science Advisor

    [tex]\dot{ f }_{ i } = \{ H , f_{ i } \} = \sum_{ n = 1 }^{ N } \left( \frac{ \partial H }{ \partial p^{ n } } \frac{ \partial f_{ i } }{ \partial q_{ n } } - \frac{ \partial H }{ \partial q_{ n } } \frac{ \partial f_{ i } }{ \partial p^{ n } } \right) .[/tex]
    Then, use
    [tex]\frac{ \partial f_{ i } }{ \partial q^{ n } } = \frac{ \partial f_{ i } }{ \partial p^{ n } } = 0 , \ \ \mbox{ for all } \ n \neq i .[/tex]

    Try to calculate
    [tex]\dot{ g }_{ 1 } = \{ H , g_{ 1 } \} .[/tex]

    Nothing prevent you from setting [itex]\dot{ q }^{ i } = p^{ i }[/itex]. Then, you can show that the orbital angular momentum is conserved
    [tex]\frac{ d L_{ i } }{ d t } = \{ H , \epsilon_{ i j k } q^{ j } p^{ k } \} = 0 .[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook