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Hamiltonian Noether's theorem in classical mechanics

  1. Jul 18, 2014 #1
    How does one think about, and apply, in the classical mechanical Hamiltonian formalism?

    From the Lagrangian perspective, Noether's theorem (in 1-D) states that the quantity

    [tex]\sum_{i=1}^n \frac{\partial \mathcal{L}}{\partial ( \frac{d y_i}{dx})} \frac{\partial y_i^*}{\partial \varepsilon} - \left[\sum_{j=1}^n \frac{\partial \mathcal{L}}{\partial ( \frac{d y_j}{dx})} \frac{d y_j }{\partial x} - \mathcal{L}\right]\frac{\partial x^*}{\partial \varepsilon}[/tex]

    is conserved if the Lagrangian [itex]\mathcal{L}(x,y_i,y_i')[/itex] is invariant under a continuous one-parameter group of infinitesimal transformations of the form

    [tex]T(x,y_i,\varepsilon) = (x^*,y_i^*) = (x^*(x,y_i,\varepsilon),y_i^*(x,y_i,\varepsilon)).[/tex]

    From the action perspective, Noether's theorem states the equality of the 1-forms:

    [tex]\mathcal{L}(x,y_i,y_i')dx = \sum_{j=1}^n p_i d y_j - \mathcal{H}dx = \mathcal{L}(x^*,y_i^*,y_i'^*)dx^* = \sum_{i=1}^n p_i d y_i^* - \mathcal{H}dx^*[/tex]

    which can be used to determine (additive) symmetries nicely.

    How do I use this formalism to understand the Hamiltonian Noether theorem in a general context? I'll usually see a claim that [itex]dA/dt = [H,A][/itex] is the Hamiltonian Noether's theorem, and I can't make sense of this in the context of my description of Noether above. This appears to derive the Poisson brackets as part of Noether from what I've developed above, but I can't make much sense of it to be honest I'm sure the answer is supposed to link the local Lie algebra tangent vector structure to the global Lie group transformation in the Lagrangian, but saying that in words is one thing, in math it's another, thanks.
  2. jcsd
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