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!

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

    User Avatar
    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

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook