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Conformal group, infinitesimal transformation

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    In order to determine the infinitesimal generators of the conformal group we consider an infinitesimal coordinate transformation:
    [itex]x^{\mu} \to x^\mu+\epsilon^\mu[/itex]
    We obtain [itex]\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial\cdot\epsilon)\eta_{\mu\nu}[/itex] where d is the dimension of spacetime.
    Derive [itex](\eta_{\mu\nu}\Box+(d-2)\partial_\mu \partial_\nu)\partial\cdot\epsilon=0[/itex]

    2. Relevant equations
    [itex]\eta_{\mu\nu}\eta^{\mu\nu}=d[/itex]

    3. The attempt at a solution
    I think I am really close to the solution, but somehow I don't arrive there.
    [itex]\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial\cdot\epsilon)\eta_{\mu\nu}[/itex]
    [itex]\Rightarrow \partial^\nu\partial_\mu\epsilon_\nu+\partial^\nu{\partial}_\nu\epsilon_\mu=\frac{2}{d}\partial^\nu({\partial}\cdot\epsilon)\eta_{\mu\nu}[/itex]
    [itex]\Rightarrow \partial_\mu(\partial\cdot\epsilon)+\Box{\epsilon}_\mu=\frac{2}{d}\partial_\mu(\partial\cdot\epsilon)[/itex]
    [itex]\Rightarrow (d-2)\partial_\mu(\partial\cdot\epsilon)+d\cdot{\Box}{\epsilon}_\mu=0[/itex]
    [itex]\Rightarrow (d-2)\partial_\mu\partial_\nu(\partial\cdot\epsilon)+d\partial_\nu\Box{\epsilon}_\mu=0[/itex]
    I think I need to use [itex]d=\eta_{\mu\nu}\eta^{\mu\nu}[/itex] now, but I don't get the right result.
    Can somebody help me?

    Best regards, physicus
     
  2. jcsd
  3. Jun 1, 2012 #2

    fzero

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    The first term in (**) is symmetric in ##\mu\nu##, so we can write (**) as

    $$(d-2)\partial_\mu\partial_\nu(\partial\cdot\epsilon)+\frac{d}{2} \Box ( \partial_\nu{\epsilon}_\mu + \partial_\mu{\epsilon}_\nu)=0.$$

    Alternatively, you can take the divergence of (*) with ##\partial^\mu## and add it to (**) to get this equation. It should be clear what to do next.
     
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