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Conserved Noether charge and gravity

  1. Dec 5, 2011 #1

    haushofer

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    If one considers the Lagrangian of a non-relativistic particle in a gravitational field,

    [tex]
    L = \frac{m}{2}(\delta_{ij}\dot{x}^i \dot{x}^j + 2 \phi(x^k) )
    [/tex]

    it transforms under

    [tex]
    \delta x^i = \xi^i (t), \ \ \ \ \delta \phi = \ddot{\xi}^i x_i
    [/tex]

    as a total derivative:

    [tex]
    \delta L = \frac{d}{dt}(m \dot{\xi}^i x_i)
    [/tex]

    My question is: why is the corresponding Noether charge

    [tex]
    Q = p_i \xi^i - m \dot{\xi}^i x_i
    [/tex]

    not conserved if one uses the equations of motion? I'm staring at the problem now for quite some time, missing something obvious, but I can't see it :)
     
  2. jcsd
  3. Dec 5, 2011 #2

    haushofer

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    It's confusing, because the EOM for phi is the Poisson equation, which is not dynamical; it could be obtained by adding

    [tex]
    \lambda[\partial_i \partial^i \phi - 4 \pi G \rho]
    [/tex]

    to the Lagrangian, where lambda is a Lagrange multiplier. But this doesn't change the situation.
     
  4. Dec 5, 2011 #3

    haushofer

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    So I more or less have the answer to my question; the variation [itex]\delta L[/itex] also contains a

    [tex]
    \frac{\partial L}{\partial \phi}\delta \phi = m \ddot{\xi}^i x_i
    [/tex]

    term, which I don't want. So how to get the corresponding Noether charges?
     
  5. Dec 6, 2011 #4

    haushofer

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    Does somebody know where this problem is treated?
     
  6. Dec 8, 2011 #5

    haushofer

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    I think I have the answer, so nevermind.
     
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