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Finding Noether Charges from Action

  1. Nov 20, 2008 #1

    malawi_glenn

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    Giving an action, a general one:

    [tex]S = \int dt L(q^i,\dot{q}^i,t)[/tex]

    now assume this action is invariant under a coordinate transformation:

    [tex]q^i \rightarrow q^i + \epsilon ^a (T_a)^i_jq^j[/tex]

    Where T_a is a generator of a matrix Lie group.

    Now one should be able to find the consvered quantities, the "Noether Charges", and how those relate to the matrix lie group.

    BUT HOW?

    I have never done so much in school about actions, just lagrangians and hamolitonians.

    For instance, if one only considered translation: [tex]q^i \rightarrow q^i + \epsilon q^i[/tex], and if the Lagrangian/hamiltonian is invariant under translations -> we know that the linear momentum is conserved. But how do we show it with the action and noether currents/charges?

    Now this is a quite general question, I have never quite understood this, and is related to what I asked a couple of days ago in the math section about Lie Subgroups. I am trying to appreciate Group Theory, in perticular Lie Groups. This is a "general example" on its application to classical mechanics which I found somewhere, but then I found out that I am totally lost when it comes to performing the "searches" for noether charges.
     
  2. jcsd
  3. May 29, 2009 #2

    Hao

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    Have you found the answer to this question?
     
  4. May 31, 2009 #3
    I'm going to use [itex]\mathcal{L}\left(\phi,\partial_\mu\phi,t\right)[/itex] as my lagrangian as this is the notation I'm used to typing :) But this doesn't change the process.

    Ok since you know that the action is invariant under this transformation we have that [itex]\delta\mathcal{L}=0[/itex], so that:
    [tex]
    \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi_i}\delta_{\phi_i}+\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta\left(\partial_\mu\phi_i\right)
    [/tex]
    And we can then change [itex]\delta\left(\partial_\mu\phi_i\right)[/itex] to [itex]\partial_\mu\left(\delta\phi_i\right)[/itex].
    Using the Euler-Lagrange equation:
    [tex]
    \frac{\partial\mathcal{L}}{\partial\phi_i}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\right)
    [/tex]
    We can see that our first equation's first term can be replaced with the RHS of the E-L giving:
    [tex]
    \delta\mathcal{L}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\right)\delta_{\phi_i}+\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\partial_\mu\left(\delta\phi_i\right)
    [/tex]
    Which we can bring into one term by the product rule:
    [tex]
    \delta\mathcal{L}=\partial_\mu\left(\frac{\partial_\mu\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)
    [/tex]
    Since as we stated [itex]\delta\mathcal{L}=0[/itex] we can make the identification:
    [tex]
    \partial_\mu\underbrace{\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)}_{J^\mu}
    [/tex]
    As this fulfils [itex]\partial_\mu J^\mu=0[/itex]

    We can then look at the [itex]\delta\phi_i[/itex] terms under an infinitesimal change:
    [tex]
    \delta\phi_i&=\phi_i-\phi'_i=\phi_i-\left(\phi_i+\epsilon^a\left(T_a\right)^i_j\phi^j\right)\\
    &=-\epsilon^a\left(T_a\right)^i_j\phi^j
    [/tex]
     
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