Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum current density in general relativity

  1. Aug 10, 2011 #1
    Hi All,

    As shown here

    http://en.wikipedia.org/wiki/Klein–Gordon_equation#Gravitational_interaction

    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
    http://en.wikipedia.org/wiki/Four-current
    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?

    Cheers!
     
  2. jcsd
  3. Aug 10, 2011 #2
    I think a lagrangian like
    [tex]
    \mathcal{L}=\bar{\psi}\gamma^\mu \partial^\nu \psi g_{\mu\nu}
    [/tex]
    would get you essentially
    [tex]
    \bar{\psi}\gamma^\mu \partial^\nu \psi=\theta^{\mu\nu}
    [/tex]
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quantum current density in general relativity
Loading...