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Lagrangian a function of?

  1. Jul 13, 2008 #1
    Hmm how does one prove that a lagrangian is a function of just the generalized coordinates and the generalized speed and not the generalized "Accelaration"?
  2. jcsd
  3. Jul 14, 2008 #2


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    In general, one assumes that it does not depend on [itex]\ddot q_i[/itex] (with [itex]q_i[/itex] the generalized coordinates) because in most systems, it doesn't. But one cannot prove this independence, and indeed one can derive the Euler-Lagrange equations for a Lagrangian with does depend on them in the same way as usual.

    One should check, however, that the "acceleration" is really independent, otherwise one cannot consider [itex]q[/itex] and [itex]\ddot q[/itex] as independent coordinates and one would have to impose a constraint (e.g. the original Euler-Lagrange equation for the Lagrangian which does not depend on the acceleration).

    But in principle, I think this argument works:
    Suppose we have a Lagrangian [itex]\mathcal L(q, \dot q, \ddot q)[/itex]. Then a variation of the action gives
    [itex]\delta\left( \int \mathcal L \, \mathrm dt \right) = \int \left( \frac{\partial \mathcal L}{\partial q} \delta q + \frac{\partial \mathcal L}{\partial \dot q} \delta \dot q + \frac{\partial \mathcal L}{\partial \ddot q} \delta \ddot q \right) \, \mathrm dt. [/itex]
    Using that [itex]\delta\dot q = \frac{d(\delta q)}{dt}[/itex], etc. we get by partial integration (once on the second term, twice on the third term)
    [itex]\delta\left( \int \mathcal L \, \mathrm dt \right) = \int \left( \frac{\partial \mathcal L}{\partial q} - \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot q} + \frac{d^2}{dt^2} \frac{\partial \mathcal L}{\partial \ddot q} \right) \delta q \, \mathrm dt [/itex]
    assuming that all the boundary terms from the partial integrations vanish. Now for this to vanish for arbitrary variations (under these conditions), the bracketed term must be zero and we find a new "Euler-Lagrange equation",
    [tex]\frac{\partial \mathcal L}{\partial q} + \frac{d^2}{dt^2} \frac{\partial \mathcal L}{\partial \ddot q} = \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot q} [/tex]
  4. Jul 14, 2008 #3
    hmm ya i cud smell that but just wanted to confirm if i was missing out something
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