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Time dependent lagrangian

  1. Aug 3, 2011 #1
    Suppose I have a mechanical system with l + m degrees of freedom and an associated lagrangian

    [itex]L(\alpha,\beta,\dot{\alpha},\dot{\beta},t)[/itex]

    where [itex]\alpha\in\mathbb{R}^l[/itex] and [itex]\beta\in\mathbb{R}^m[/itex].
    Now suppose I have a known [itex]\mathbb{R}^l[/itex]-valued function f(t) and define a new lagrangian

    [itex]M(\beta,\dot{\beta},t)=L(f(t),\beta,\dot{f}(t), \dot{\beta},t)[/itex]

    Do the equations that derive from M correctly describe the motion of the initial mechanical system, where the first l degrees of freedom are constrained to the motion f(t) (by means of an external force)?
     
  2. jcsd
  3. Aug 3, 2011 #2

    BruceW

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    [tex]\frac{\partial L}{\partial q} = \frac{d}{dt} \frac{\partial L}{\partial \dot{q} } [/tex]
    where q is any one of the coordinates of [itex]\alpha[/itex] or [itex]\beta[/itex] (i.e. there are l+m separate equations).
    From this, I would assume that [itex]M(\beta , \dot{\beta} , t) [/itex] would be correct, because this simply gives the m equations, which correspond to [itex]\beta[/itex].
    (I know I've not done a rigorous proof or anything, but this seems to make sense to me).
     
  4. Aug 4, 2011 #3
    Uhm, please don't let me write the formula, but when you take the derivative with respect to t of the momentum dM/dv, don't you get extra terms due to f(t) and df(t)/dt?
     
  5. Aug 4, 2011 #4

    Bill_K

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    Yes, this is correct. In fact it's done all the time: write down T and V as if the particles were entirely free, and then impose the constraints on them. The Lagrangian M you get from this will describe the motion of the constrained system.

    What you cannot do is the other way around: trying to substitute f(t) into the equations of motion you originally derived from L.
     
  6. Aug 4, 2011 #5
    Yes, it sounds so obvious. I feel stupid now... :)
     
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