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Hamiltonian mechanics

  1. Dec 9, 2006 #1
    1. The problem statement, all variables and given/known data

    Using spherical coordinates [tex](r, \theta, \phi)[/tex], obtain the Hamiltonian and the Hamilton equations of motion for a particle in a central potential V(r).
    Study how the Hamilton equations of motion simplify when one imposes the initial conditions [tex]p_{\phi}(0) = 0[/tex] and [tex]\phi (0) = 0[/tex]


    3. The attempt at a solution

    I have obtained a Hamiltonian

    [tex]H = \frac{1}{2m} \left( p_r^2 + \frac{p_{\theta}^2}{r^2} + \frac{p_{\phi}^2}{r^2 \sin^2{(\theta)}} \right) + V(r)[/tex]

    and from this also the equations of motion

    [tex]\dot{r} = \frac{p_r}{m}[/tex]
    [tex]\dot{\theta} = \frac{p_{\theta}}{mr^2}[/tex]
    [tex]\dot{\phi} = \frac{p_{\phi}^2}{r^2 \sin^2{(\theta)}}[/tex]

    [tex]m \ddot{r} = \frac{1}{m} \left( \frac{p_{\theta}^2}{r^3} + \frac{p_{\phi}^2}{r^3 \sin^2{(\theta)}} \right) - \frac{\partial V}{\partial r}[/tex]
    [tex]m^2 \left( 2r \dot{r} \dot{\theta} + r^2 \ddot{\theta} \right) = \frac{p_{\phi}^2 \cos{(\theta)}}{r^3 \sin^3{(\theta)}}[/tex]
    [tex]2 \dot{r} \sin{(\theta)} \dot{\phi} + r \left( 2 \cos{(\theta)}\dot{\theta} \dot{\phi} + \sin{(\theta)} \ddot{\phi} \right) = 0[/tex]

    But how should I proceed with the last part of the problem?
     
  2. jcsd
  3. Dec 9, 2006 #2

    cristo

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    I'm not too sure how you've got the last 3 eqns. It would be easier to just use the fact that [tex] \dot{p_i}=-\frac{\partial H}{\partial q_i} [/tex] Try doing it this way, then you will have the momentum terms on the LHS. It might be clear from the phi eqn obtained in this way how to use the initial conditions.
     
  4. Dec 10, 2006 #3
    That's EXACTLY what my equations are. ;)
     
  5. Dec 10, 2006 #4

    cristo

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    Ok, well consider the last equation. It can be written [tex] \dot{p_\phi} = -\frac{\partial H}{\partial \phi} = 0 [/tex] Integrating this, and using the initial conditions will allow you to simplify the other equations
     
  6. Dec 10, 2006 #5
    Integrate with respect to what? I don't see this at all..
     
  7. Dec 10, 2006 #6

    cristo

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    This equation can also be written as [tex] \frac{d p_\phi}{dt}=0 [/tex]. Can you integrate this?
     
  8. Dec 10, 2006 #7
    That would take me back to

    [tex]p_\phi = mr^2 \sin^2{\theta} \dot{\phi} = 0[/tex]?
     
  9. Dec 10, 2006 #8

    cristo

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    Well... we get the equation [tex] p_\phi= C [/tex] where C is a constant of integration. Then, the intial conditions will imply that [tex] p_\phi [/tex] is zero for all t. Do a similar thing for [tex] \phi [/tex]. Remember that you are trying to show how the equations *simplify* on imposing the initial conditions
     
  10. Dec 10, 2006 #9
    I don't get it. How does this simplify the equations?
     
  11. Dec 10, 2006 #10

    cristo

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    Try putting [tex] p_\phi = 0 [/tex] into the equations you've posted in your question. You'll see it simplifies the equations! Then do a similar thing for the third equation, using the inital condition for phi
     
  12. Dec 10, 2006 #11
    But what about the [tex]\phi[/tex]-part?
     
  13. Dec 10, 2006 #12

    cristo

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    I'm not sure what else I can say without doing it for you! What equation do you get when you sub [tex] p_{phi}=0 [/tex] into the third equation? How can you solve this, and what is the answer given the initial conditions? [hint.. posts #6 and #8]
     
  14. Dec 10, 2006 #13
    Then [tex]\phi = const. = 0[/tex]? Aha! ;)
     
  15. Dec 10, 2006 #14

    cristo

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    Correct. Now the equations will look a lot simpler!
     
  16. Dec 10, 2006 #15
    Yeah thanks for your help. I'm blaming on the fact that it's sunday today. ;)
     
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