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Variation of Lagrangian under Lorentz transformations

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that under an infinitesimal Lorentz transformation: [tex] x^\mu \to x^\mu+\omega^\mu_\nu x^\nu[/tex] so: [tex]\phi\to\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi[/tex] the Lagrangian varies as:

    [tex]\delta \mathcal{L}=-\partial_\mu(\omega^\mu_\nu x^\nu \mathcal{L})[/tex]



    3. The attempt at a solution

    The new Lagrangian will be:

    [tex]
    \mathcal{L}(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi,\partial_\sigma(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi))[/tex]

    The variation is:

    [tex]
    \delta\mathcal{L}=\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi+\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\partial_\sigma \delta \phi=-\omega^\mu_\nu \left[x^\nu\partial_\mu \phi \frac{\partial \mathcal{L}}{\partial \phi}+\partial_\sigma(x^\nu\partial_\mu\phi)\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\right][/tex]


    If the last term in the sum vanished somehow, I would arrive to the solution, but I can't see how it is zero. Or have I made a mistake before?

    I think I've done some progress:

    [tex]-\omega^\mu_\nu \left[\partial_\sigma(x^\nu\partial_\mu\phi)\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\right]=-\omega^\mu_\nu \left[ \partial_\mu\phi\frac{\partial \mathcal{L}}{\partial (\partial_\nu\phi)}+x^\nu \partial_\mu\phi\mathcal{L}\right]=[/tex]

    Could I use Euler-Lagrange equations to make this zero?
     
    Last edited: Jun 6, 2014
  2. jcsd
  3. Jun 6, 2014 #2
    Seems to me that there is a simple answer: what kind of field is the lagrangian density? How do these fields transform? And then with a property of ##\omega## it seems ok.

    In your calculations I think there is something wrong with ##\partial_{\sigma}##: it's not a scalar.
     
  4. Jun 6, 2014 #3
    Could you develop

    [tex]\delta \mathcal{L}=-\partial_\mu(\omega^\mu_\nu x^\nu \mathcal{L})[/tex]
     
  5. Jun 6, 2014 #4
    I'm going to bed...

    $$ \partial_{\sigma}\phi \to \partial_{\sigma}\phi-\omega^\mu_\nu x^\nu\partial_\mu\partial_\sigma\phi+\omega^\mu_\sigma\partial_\mu\phi $$
     
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