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A general linear transformation is given by

\begin{align} \psi'(x) \to g \psi(x) g^{-1}, \end{align}

The gauge-covariant derivative associated with this transformation is

\begin{align} D_\mu \psi=\partial_\mu \psi -[iqA_\mu, \psi]. \end{align}

Finally, the field is given as

\begin{align} R_{\mu\nu}= [D_\mu,D_\nu], \end{align}

where, $R_{\mu\nu}$ is the Riemann tensor.

From here, what steps lead me to the Einstein field equation?

Qmechanic
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Anon21
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1 Answers1

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None. Nothing of this has anything to do with gravity. Your equation for "R" defines the gauge field curvature $F_{\mu\nu}$, not the Riemann curvature $R_{\alpha \beta\mu\nu}$.

mike stone
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  • I am using the general linear group for the gauge. – Anon21 Jul 05 '22 at 17:32
  • You are still not going to get gravity as GR is not a gauge theory (fibre bundle) in the usual sense. There is a formulation of (non super) suopergravity that gauges the Poincare group, but each fibre conatins the whole of space, so again it's not a conventional gauge theory. – mike stone Jul 05 '22 at 17:52