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Lagrangians giving the same equations of motion

  1. Oct 7, 2012 #1
    Hi,

    I'm trying to clear up a confusing point in the book by José and Saletan, concerning equivalent Lagrangians (in the sense that they give you the same dynamics). It is clear that if
    [itex] L_1 - L_2 = \frac{d\phi ( q,t )}{dt},[/itex]
    then [itex]L_1[/itex] and [itex]L_2[/itex] will have the same equations of motion. However, what about the inverse problem?

    It seems that it is sometimes assumed that two Lagrangians giving the same equations of motion must differ by such a total time derivative (Jose and Saletan do this in problem 2.4 and in the beginning of section 2.2.2). However, as they write later in that section, the two Lagrangians
    [itex]L_1 = \dot{q}_1 \dot{q}_2 - \omega^2 q_1 q_2[/itex]
    [itex]L_2 = \frac{\dot{q}_1^2}{2}+\frac{\dot{q}_2^2}{2} - \frac{\omega^2}{2} q_1^2 - \frac{\omega^2}{2} q_2^2[/itex]
    quite clearly give the same equations of motion, but aren't related by a total time derivative of a function of position and time.

    Is this simply because [itex]L_1[/itex] is non-local? And does the inverse property then always hold for local Lagrangians on the standard form T-V?
     
  2. jcsd
  3. Oct 7, 2012 #2

    mfb

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    Interesting system. I cannot answer your question, a coordinate transformation in L1 gives something close to L2: ##q_1 = \frac{1}{\sqrt{2}}(p_1+p_2)## and ##q_1 = \frac{1}{\sqrt{2}}(p_1-p_2)## in L1 gives the structure of L2, just with flipped signs for one coordinate. As both coordinates in L2 are independent, the equations of motion will stay the same.
     
  4. Oct 8, 2012 #3

    vanhees71

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    I just looked up the passage in the textbook by Jose and Saletan. He's deriving very nicely, when two Lagrangians are equivalent in the sense of giving the same dynamics. One possibility is that they differ only by the total derivative of a function of the generalized coordinates and explicitly on time. This is a sufficient but not a necessary condition, which is nicely demonstrated by your example.

    As far as I understand the reason is that Hamilton's principle of least action is not only invariant under point transformations, i.e., under changes from one set of generalized coordinates to another one but under canonical transformations in the Hamiltonian formulation.

    Let's look at your example from the point of view of the Hamiltonian formulation. Let's start with
    [tex]L_1=\dot{q}_1 \dot{q}_2-\omega^2 q_1 q_2.[/tex]
    To go over to the Hamilton formalism we first need the canonical momenta:
    [tex]p_1=\frac{\partial L_1}{\partial \dot{q}_1}=\dot{q}_2, \quad p_2=\frac{\partial L_1}{\partial \dot{q}_2}=\dot{q}_1.[/tex]
    Then the Hamiltonian reads
    [tex]H_1=\dot{q}_j p_j-L=2p_1 p_2-(p_1 p_2-\omega^2 q_1 q_2)=p_1 p_2 + \omega^2 q_1 q_2.[/tex]
    For the second Lagrangian we have (I change the names of the variables to [itex](Q_1,Q_2)[/itex])
    [tex]L_2=\frac{1}{2}(\dot{Q}_1^2+\dot{Q}_2^2)-\frac{\omega^2}{2} (Q_1^2+Q_2^2).[/tex]
    The canonical momenta are
    [tex]P_1=\dot{Q}_1, \quad P_2=\dot{Q}_2[/tex]
    and the Hamiltonian
    [tex]H_2=\frac{1}{2}(P_1^2+P_2^2)+\frac{\omega^2}{2} (Q_1^2+Q_2^2).[/tex]
    Now we see that the transformation
    [tex]q_1=(Q_1+\mathrm{i} Q_2)/\sqrt{2}, \quad q_2=(Q_1-\mathrm{i} Q_2)/\sqrt{2}, \quad p_1=(P_1-\mathrm{i} P_2)/\sqrt{2}, \quad p_2=(P_1-\mathrm{i} P_2)/\sqrt{2}[/tex]
    is canonical since
    [tex]\{q_j,q_k \}=\{p_j,p_k \}=0, \quad \{q_j,p_k \}=\delta_{jk}[/tex]
    with the Poisson bracket defined as
    [tex]\{f,g \}=\frac{\partial f}{\partial Q_k} \frac{\partial g}{\partial P_k} - \frac{\partial f}{\partial P_k} \frac{\partial g}{\partial Q_k}.[/tex]
    To find the generating function, we best choose the form
    [tex]q_k=\frac{\partial g}{\partial p_k}, \quad P_k=\frac{\partial g}{\partial Q_k}, \quad H_2(Q,P,t)=H_1(q,p,t)+\partial_t g.[/tex]
    It's easy to see that we can set
    [tex]g=\frac{1}{\sqrt{2}} \left [(P_1-\mathrm{i} P_2) q_1 + (P_1+ \mathrm{i} P_2) q_2 \right ].[/tex]
    Then we find indeed that
    [tex]H_2(Q,P)=H_1[q(Q,P),p(Q,P)],[/tex]
    which is consistent with the generating function, because [itex]\partial_t g=0[/itex].

    Thus, in fact the two Lagrangians are related through a canonical transformation in the equivalent Hamiltonian formulation of the problem, and that's why the two Lagrangians describe the same dynamics.
     
  5. Oct 8, 2012 #4
    Thanks to both of you for your replies. This punchline

    is particularly illuminating, and makes a lot of sense. It pays to think of the physical motion (on the configuration manifold) and the Lagrangians (as defined on the tangent bundle) separately then, something I might have neglected at times. (Another nice example of this is that the angular momentum is conserved for both Lagrangians, but it's associated with different symmetries.) Interestingly, the two Hamiltonians supposedly give very different theories when quantized, but that's another story.

    It seems that it all boils down to Jose and Saletan being a bit sloppy when they wrote problem 2.4, then. Thanks again.
     
  6. Oct 9, 2012 #5

    vanhees71

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    The interesting point is that the n-dimensional isotropic harmonic oscillator has not only the quite obvious SO(N) symmetry but a larger symmetry group that is equivalent to the larger group SU(N).
     
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