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Quantum current density in general relativity

  1. Aug 10, 2011 #1
    Hi All,

    As shown here


    one can modify the derivative operator in the Minkowski Klein-Gordon equation to generate the GR case for an arbitrary metric.

    My question is, what would the quantum current density j be in this case? Given here
    is the classical current density in the EM case but that's not really what i'm after.

    Furthermore, if i generate a radial wave equation from the GR KG equation in some metric and transform it in a Schrodinger like form, can I just use the standard tools of non-rel QM and use [itex] \psi^* \frac{d}{dx} \psi [/itex] for j?

  2. jcsd
  3. Aug 10, 2011 #2
    I think a lagrangian like
    \mathcal{L}=\bar{\psi}\gamma^\mu \partial^\nu \psi g_{\mu\nu}
    would get you essentially
    \bar{\psi}\gamma^\mu \partial^\nu \psi=\theta^{\mu\nu}
    when varied by the metric, where the current corresponds to the [itex]0\nu[/itex] part.

    But I kinda made that up, so we should keep that tentative lol.
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