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Can a Hamiltonian be formed from this Lagrangian?

  1. Jun 17, 2010 #1
    [tex]L=\frac{1}{2}m(\dot{q}_1-\dot{q}_2)^2-V(q_1,q_2)[/tex]

    Because if we put

    [tex]p_1=\frac{\partial L}{\partial \dot{q}_1}[/tex]
    [tex]p_2=\frac{\partial L}{\partial \dot{q}_2}[/tex]

    we get

    [tex]p_1=-p_2=m(\dot{q}_1-\dot{q}_2)[/tex]

    We can't invert to get [tex]\dot{q_1}[/tex] in terms of the two momenta. We can still write down a Hamiltonian of sorts since

    [tex]H=p_1\dot{q}_1+p_2\dot{q}_2-L[/tex]

    [tex]=p_1(\dot{q}_1-\dot{q}_2)-L[/tex]

    [tex]=\frac{p_1^2}{m}-L[/tex]

    [tex]=\frac{p_1^2}{2m}+V(q_1,q_2)[/tex]

    or equivalently

    [tex]=\frac{p_2^2}{2m}+V(q_1,q_2)[/tex]

    or

    [tex]=\frac{p_1^2}{4m}+\frac{p_2^2}{4m}+V(q_1,q_2)[/tex]


    The main thing then is that we can't get an equation of motion which looks like

    [tex]\dot{q_j}=\frac{\partial H}{\partial p_j}[/tex]

    What do we do with Lagrangians like this? Does the Hamiltonian method just fail?
     
  2. jcsd
  3. Jun 17, 2010 #2

    K^2

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    Science Advisor

    I think you'll find that the only potential that doesn't cause a problem is the one where [itex]V(q_1,q_2)=V(q_1-q_2)[/itex] Otherwise, your problem is not well-defined.

    The q1 and q2 can increase/decrease simultaneously with zero generalized momentum. If that makes change in potential, you effectively have a mode with zero mass and finite force. That's bad.

    That's really what you are seeing near the end. You really only have one degree of freedom: q1-q2. You tried to write the equation as if there are two. You ended up with a degeneracy. So naturally, the Hamiltonian cannot be used to distinguish two degenerate degrees of freedom. Define the new coordinate q=(q1-q2), and the problem resolves itself.
     
  4. Jun 17, 2010 #3

    Mute

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    Homework Helper

    Try changing your lagrangian's basis. Let [itex]Q_1 = q_1 - q_2[/itex], [itex]Q_2 = q_1 + q_2[/itex]. Take derivatives with respect to these new variables.

    [tex]\mathcal L = \frac{\dot{Q_1}^2}{2m} - V\left(\frac{Q_1+Q_2}{2},\frac{Q_2-Q_1}{2}\right)[/tex]
     
  5. Jun 17, 2010 #4
    Quite right. Thanks, both.
     
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