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Unproved statement from my textbook

  1. Jul 31, 2012 #1
    I've included a pic of a section of Arnold's Mathematical Methods of Classical Mechanics. The relevant passage is: "Since [itex]h_*^s[/itex] preserves [itex]L[/itex], the translation of the solution, [itex]h^s\circ \mathbf{\varphi}:\mathbb{R}\rightarrow M[/itex] also satisfies Lagrange's equation for any s."

    Maybe Arnold thought it was too obvious to prove, but for whatever reason I'm having trouble seeing why it's true. I tried proving it with chain rule (I was sure the proof will be a simple cal exercise), but to no avail. It's probably really trivial to prove, but my brain doesn't seem to be working. Could someone help?
     
  2. jcsd
  3. Jul 31, 2012 #2
    Forgot my pic.
     

    Attached Files:

  4. Aug 2, 2012 #3
    Anyone?
     
  5. Aug 2, 2012 #4

    Ben Niehoff

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    Attached image is unreadable. I have the book, though, what page is this?
     
  6. Aug 2, 2012 #5
    Page 88. What's wrong with the image? Too small?
     
  7. Aug 4, 2012 #6
    Maybe this'll make the image more readable: http://imgur.com/Emv3s

    Also, in case it wasn't know, Lagrange's equation is [itex]\frac{d}{dt}\frac{\partial L}{\partial \mathbf{\dot{q}}} = \frac{\partial L}{\partial \mathbf{q}}[/itex], where L is a function of [itex]\mathbf{q}[/itex] and [itex]\mathbf{\dot{q}}[/itex], both of which are part of R^n. In the above equation equation, it's understood that [itex]\mathbf{q}[/itex] and [itex]\mathbf{\dot{q}}[/itex] are replaced by [itex]\mathbf{q}(t)[/itex] and [itex]\mathbf{\dot{q}}(t)[/itex] such that [itex]\mathbf{\dot{q}}(t) = \frac{d}{dt} \mathbf{q}(t)[/itex].
     
    Last edited: Aug 4, 2012
  8. Aug 6, 2012 #7
    Okay, so I'll show you my attempt at a solution. Something must have went terribly wrong somewhere; maybe one of you can find my mistake.

    We want to show that if [itex]\mathbf{Q}(\mathbf{q},t)[/itex] is the result of applying a transformation to the system's generalized coordinates [itex]\mathbf{q}[/itex]at time [itex]t[/itex], and [itex]L(\mathbf{Q},\dot{\mathbf{Q}},t) = L(\mathbf{q},\dot{\mathbf{q}},t)[/itex], and [itex]\mathbf{q}(t)[/itex] satisfies Lagrange's equations, then so does [itex]\mathbf{Q}(\mathbf{q}(t),t)[/itex].

    [tex]\frac{\partial}{\partial \mathbf{q}}L(\mathbf{q},\dot{\mathbf{q}},t) =\frac{\partial}{\partial \mathbf{q}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q},\dot{\mathbf{q}},t),t)= \frac{\partial}{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \mathbf{Q}}{\partial \mathbf{q}}[/tex]

    [tex]\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{q},\dot{\mathbf{q}},t) \right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q},\dot{\mathbf{q}},t),t)\right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \dot{\mathbf{Q}}}{\partial \dot{\mathbf{q}}}\right)[/tex]

    Now, the two expressions above equal each other because [itex]\mathbf{q}(t)[/itex] satisfies Lagrange's equation. From this, I'm supposed to conclude that
    [tex]\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\right)=\frac{\partial }{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)[/tex]
    but I don't see how that's possible.
     
  9. Aug 9, 2012 #8
    If there's something unclear about how I formulated the question -- or if you think I should make a new question from scratch because of how scrambled this question has become -- let me know please.
     
  10. Aug 9, 2012 #9
    This is most trivially seen from the principle of least action. h does not change the Lagrangian, so it does not change the functional, so hv is a stationary point if v is stationary.
     
  11. Aug 9, 2012 #10
    I think Q = Q(q), not Q = Q(q,t). That map isn't time dependent.
     
  12. Aug 9, 2012 #11
    Yes, thank you! Now it makes sense.

    But that said, I'm now left wondering why my (attempted) proof using the Lagrangian directly didn't work. Any ideas?

    I asked whether the map needs to be time independent or not in another PF thread, and the answer I got was that it could be time dependent without affecting Noether's theorem.

    Can someone else confirm who's right?
     
  13. Aug 9, 2012 #12
    Well if it's time dependent, I think we could map a solution q(t) onto pretty much any path Q(t) that we wanted. There would be no reason to expect a Q(t) so constructed to satisfy the equations of motion.
     
  14. Aug 9, 2012 #13
    One problem is that you seem to confuse the formal argument q with a particular trajectory, also denoted by q.

    The derivatives Lagrange's equation are taken by the formal argument, but are evaluated at a particular trajectory. The equation should really be written this way: [tex]\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)} - \frac {d}{dt} \left[\frac {\partial L}{d \dot{q}}\right]_{q = Q(t)} = 0[/tex]

    It is true.
     
  15. Aug 9, 2012 #14
    I suspected my mistake had something to do with that. But where do I go from there?

    I can't say that
    [tex]\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)}=\left[\frac {\partial L} {\partial q}\right]_{q = q(t)}[/tex]
    because it isn't true. There's probably a manipulation I can do with chain rule, but I don't see it.
     
  16. Aug 9, 2012 #15
    The point here is not mapping an arbitrary trajectory to another trajectory. The point is being able to construct a non-trivial map that preserves the Lagrangian.
     
  17. Aug 9, 2012 #16
    The map h is a map from M to M, and so it can be used to map trajectories into other trajectories - hence the "translation of the solution" mentioned.

    Note that the configuration space M has no time coordinate defined on it - so h is time-independent by definition. At least by Mr Arnold's definition, if I read him correctly.

    Example - the trajectory x = 2t is a solution for a free-particle Lagrangian. If we translate it, x = 2t + 5, it's still a solution (because the free particle Lagrangian has the appropriate symmetry).

    But if we use a time-dependent translation, to get say x = 2t + sin(t), it's not a solution any more.
     
  18. Aug 9, 2012 #17
    I think you could look at variations of q, then the derivatives of L can be represented as limits of variations of L, where you could apply L(hv) = L(v).
     
  19. Aug 9, 2012 #18
    He does not consider non-holonomic constraints, either.

    And it does not preserve the Lagrangian, too. But if you find one that does, even in a time dependent fashion, then it will necessarily map a solution to a solution.
     
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