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Symplectic manifold question

  1. Jul 13, 2007 #1

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    I've been reading about the abstract formulation of dynamics in terms of symplectic manifolds, and it's amazing how naturally everything falls out of it. But one thing I can't see is why the generalized momenta should be cotangent vectors. I can see why generalized velocities are tangent vectors, making the Lagrangian a function on the tangent bundle of configuration space. But then the book I'm reading claims that since:

    [tex] p_i = \frac{\partial L}{\partial \dot q_i} [/tex]

    it follows that pi is clearly a cotangent vector.

    To me it seems like what pi is is a function assigning numbers to vectors in the "tangent space of the tangent space of a point in the manifold". Note that by what's in the quotes, I don't mean "the tangent space of the tangent bundle" (which would put the momentum in [itex]T^*(TM)[/itex] ), because it doesn't depend on the change in configuration.

    But it also shouldn't be in [itex]T^*M[/itex] unless the Lagrangian is a linear function of velocity, so that its partial derivative with respect to velocity doesn't depend on your location in the tangent space, and so can be taken as a uniform linear functional on the tangent space. What am I missing here?
     
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  3. Jul 13, 2007 #2

    George Jones

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    The [itex]\dot{q}^i[/itex] are, at the point [itex]q[/itex] of [itex]M[/itex] labeled by the generalized coordinates [itex]q^i[/itex], components of the tangent vector

    [tex]v = \dot{q}^i \partial_{q^i}[/tex]

    of a curve in configuration space [itex]M[/itex].

    The [itex]p_i[/itex] are, at [itex]q[/itex], components of a particular covector [itex]\omega[/itex] with respect to a particular basis of covectors. This covector [itex]\omega[/itex] is the covector naturally associated with the tangent vector [itex]v[/itex] by the metric [itex]g[/itex] that comes from the kinetic energy

    [tex]T = \frac{1}{2} g_{ij} \dot{q}^i \dot{q}^j.[/tex]

    In other words,

    [tex]\omega \left( u \right) = g \left(v,u\right) = g_{ij} \dot{q}^i u^j[/itex]

    for all tangent vectors [itex]u[/itex] at [itex]q[/itex].

    Setting [itex]\omega = p_i dq^i[/itex] gives that [itex]\omega_j = g_{ij} \dot{q}^i.[/itex] But, if

    [tex]L = \frac{1}{2} g_{jk} \dot{q}^j \dot{q}^k - U \left( q \right),[/tex]

    then

    [tex]\frac{\partial L}{\partial \dot{q}^j} = g_{jk} \dot{q}^k.[/tex]
     
    Last edited: Jul 13, 2007
  4. Jul 13, 2007 #3

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    Ok, my problem was that, unless the dependence of L on velocity is linear, the covector depends on [itex]\dot q[/itex]. An ordinary covector field is normally defined at each point on the manifold, not on the tangent bundle. But in fact with the momentum you do want a covector field on the entire tangent bundle, or more specifically, a map from the tangent bundle to the cotangent bundle, allowing you to replace the tangent manifold from Lagrangian mechanics with the cotangent manifold from Hamiltonian mechanics. I think I have it now, but please correct me if that sounds wrong. Thanks.
     
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