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Theoretical mechanics question

  1. Dec 16, 2007 #1
    Here's a simple question. Anyone who knows the answer can save me from some effort by telling the answer.

    For arbitrary canonical transform [itex](q,p)\mapsto (Q,P)[/itex] in the Hamilton's formulation, does there always exist a corresponding transformation [itex]q\mapsto Q[/itex] in the Lagrange's formulation? By corresponding transformation, I mean such transformation, that when we move from the Lagrange's system [itex]L(Q,\dot{Q})[/itex] to the Hamilton's formulation, we get the same [itex]K(Q,P)[/itex] as from the original canonical transform.

    Another question (or perhaps the same): I know that we can always move from Lagrange's formulation to the Hamilton's formulation, but is the converse true. Do such Hamiltonian systems exist, that you cannot achieve by starting from some Lagrangian system?
  2. jcsd
  3. Dec 30, 2007 #2
    The answer generally is No!

    Transformations [itex]q\mapsto Q[/itex] (point transformations) are a subclass of the canonical transformations [itex](q,p)\mapsto (Q,P)[/itex].
    After all, canonical transformations are between [itex] 2\,n[/itex] variables [itex](q,p)[/itex], while point tranformations are between [itex] n[/itex] variables [itex] q[/itex].

    Take or example the 1D problem, of a particle moving with the aid of a central force

    [tex]f(x)=-\frac{k}{x^2}\Rightarrow V(x)=\frac{k}{x}[/tex]

    Then the Lagrangian and Hamiltonian functions are

    [tex]\mathcal{L}=K-V\Rightarrow\mathcal{L}=\frac{1}{2}\,m\,\dot{x}^2-\frac{k}{x}, \quad \mathcal{H}=K+V\Rightarrow\mathcal{H}=\frac{1}{2\,m}\,p^2+\frac{k}{x}[/tex]

    Perform now the non-point canonical transformation

    [tex](q,p)\mapsto (-P,Q), \quad \mathcal{H} \mapsto \mathcal{H}', \quad \mathcal{L} \mapsto \mathcal{L}' [/tex]

    and you will see that there is no function [tex] q=f(Q)[/tex] which achieves the transformation [tex]\mathcal{L} \mapsto \mathcal{L}' [/tex].

    Of course, if the canonical transformation happens to belong to the subclass of point transformations then the answer is obviously yes.

    The Lagrangian and Hamiltonian descriptions are equivalent, you can pass from one to the other by Legendre's transformation.
  4. Jan 9, 2008 #3
    There is something that doesn't make sense in this. If the set of coordinate transformations in the Hamiltonian formalism is larger than the set of coordinate transformations in the Lagrangian formalism, how could all Hamiltonian systems have a corresponding Lagrangian system?
  5. Jan 10, 2008 #4
    Or then I got it. If we move from Lagrangian formulation to Hamiltonian formulation, perform a non-pointwise coordinate transformation, and move back to Lagrangian formulation, we just get completely new Lagrangian?
  6. Jan 10, 2008 #5
    That's the point I was trying to establish at my 1st answer! :smile:
  7. Jan 10, 2008 #6


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    I don't think one can always move from the Lagrangian to the Hamiltonian system. The condition for which it's possible to make a legendre transformation to the hamiltonian framework is when the hessian matrix represented by

    [tex]\left(\frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j}\right)[/tex]

    is non-singular.

    For example, consider the Lagrangian

    [tex]\mathcal{L} =\frac{1}{2} (\dot{q}_1-\dot{q}_2)^2} + V(q_1,q_2)[/tex]
    Last edited: Jan 10, 2008
  8. Jan 10, 2008 #7
    By strange coincidence, me and Rainbow Child were just talking about system like this in the other thread Fermion oscillator
  9. Jan 11, 2008 #8
    And yes it was a mistake from me to say that we can always move from L to H (in the standard manner).
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