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Lagrange equations with constraints

  1. Dec 30, 2006 #1
    When we seek the extreaml value of the functional [tex]\Phi(\gamma) = \int_{t_0}^{t_1} L(x(t),\dot{x}(t),t)dt[/tex] where x can be taken from the entire E^n then we come to the well-known Lagrange equations.

    Now when we are given a constraint, that [tex]x \in M[/tex], where M is a differentiable manifold and when the coordinates on this manifold are [tex]q_i[/tex], then the Lagrange equations look "almost" the same, only the coordinate x is "replaced" by q:
    [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0[/tex]

    So now to the question:
    I understand how we obtain the Lagrange equations without constraints, but I can't find any proof of the equations with constraints. How is this done? Is the proof difficult (assuming only some basic knowledge of differential geometry).
  2. jcsd
  3. Dec 30, 2006 #2
    try lagrange multiplier? What kind of constraint are you given?
  4. Dec 30, 2006 #3
    Your question is meaningless. The action principle is well posed if and only if [itex]x\in\mathcal{M}[/itex]. What you call a "constraint" is not actually a constraint at all but simply a statement of a necessary property of the action principle.

    In case you're interested, constraints arise in the Lagrangian formulation of the action principle if and only if the Lagrangian is singular. This is an extremely well understood subject and should be familiar to anyone who has, for example, taken an undergraduate course in analytical mechanics.
  5. Dec 31, 2006 #4
    Now I'm a bit confused.
    What I meant is:
    When you for example take a particle in some force field and the particle is free i.e. its configuration space is the entire E^3, so it can be theoretically anywhere in the space, then using the principle of least action you're looking for the extremal of the functional [tex]\Phi(\gamma) = \int_{t_0}^{t_1} L(x(t),\dot{x}(t),t)dt[/tex], where L is a function defined on E^3 and in this case L = T - U.
    Now it can be easily shown, that if a curve is an extremal of the functional it must satisfy the equation [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0[/tex]

    But now:
    if we take a particle which is not free, for example it's on a rod and it must satisfy the equation [tex]x_1^2 + x_2^2 + x_3^2 = R^2[/tex]
    and now we're looking again on the extremal of the functional [tex]\Phi(\gamma) = \int_{t_0}^{t_1} L(x(t),\dot{x}(t),t)dt[/tex], but in this case x isn't from the entire E^3 - it must lie on the sphere.
    Now I've been taught that the procedure here is "almost" the same as in the former example - we take new coordinates - for example spherical, we write the Lagrangian in the new coordinates and then solve the Lagrange equations. The solution will be the extremal of the functional [tex]\Phi(\gamma)[/tex].
    But my question is why can we proceed like this? I miss some kind of proof of this procedure.
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