Quantum current density in general relativity

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SUMMARY

The discussion focuses on modifying the Minkowski Klein-Gordon equation to derive the quantum current density in the context of general relativity (GR). The user references the classical current density from the four-current concept and questions its applicability in the quantum regime. They propose using a Lagrangian formulation, specifically \(\mathcal{L}=\bar{\psi}\gamma^\mu \partial^\nu \psi g_{\mu\nu}\), to relate the quantum current density to the energy-momentum tensor, suggesting that the 0ν component corresponds to the desired current density.

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  • Understanding of the Klein-Gordon equation in quantum mechanics
  • Familiarity with general relativity and metric tensors
  • Knowledge of Lagrangian mechanics and field theory
  • Basic concepts of quantum current density and energy-momentum tensors
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FunkyDwarf
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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 Schrödinger 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!
 
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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.
 

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