3
$\begingroup$

In the book "Quantum Gravity in 2+1 dimensions" by Steven Carlip he writes down a possible modification to the Einstein-Hilbert Action in 3d (eq. 1.16 to eq. 1.18)

\begin{equation} I_{GCS}=-\frac{1}{32\pi G\mu}\int d^3x \epsilon^{\lambda \mu \nu}\Gamma ^{\rho}_{\lambda \sigma }\Big[\partial_{\mu}\Gamma^{\sigma}_{\rho \nu}+\frac{2}{3} \Gamma_{\mu\tau}^{\sigma}\Gamma_{\nu\rho}^{\tau}\Big]. \end{equation}

Without much explanation he says that the equations of motion for this action (plus the usual Einstein-Hilbert term) are

\begin{equation} G^{\mu \nu}+\frac{1}{\mu}C^{\mu \nu}=0 \end{equation}

where

\begin{equation} C_{\mu \nu}=\frac{1}{\sqrt{-g}}\epsilon^{\mu\rho\sigma}\nabla_{\rho}\Big[R^{\nu}_{\sigma}-\frac{1}{4}\delta^{\nu}_{\sigma}R\Big] \end{equation}

Where $R^{\nu}_{\sigma}$ is the Ricci tensor and $R$ is the curvature scalar.

My question is: is there any trick to solve this variation of the action? the math seems inhumanly hard and I'm not seeing any easy way this could be done.

$\endgroup$
  • $\begingroup$ See my post (link below), it is basically the same action as yours, except in the vielbein formalism. You can convert it to the metric formalism using the torsion free condition. The way to compute the variation is described in a footnote on page 438 (page 30 from title) of [1]. $$~$$ physics.stackexchange.com/q/411224 $$~$$ [1]: S. Deser, R. Jackiw, S. Templeton; Topologically Massive Gauge Theories (1982). $\endgroup$ – NormalsNotFar Dec 13 '18 at 11:31
  • $\begingroup$ It might help to rewrite your action in the form $$I_{GCS}\propto \int d^3x \epsilon \bigg\{ R\Gamma +\frac{1}{3}\Gamma\Gamma\Gamma\bigg\}.$$ The coefficient $\frac{1}{3}$ might be wrong, I don't remember. $\endgroup$ – NormalsNotFar Dec 13 '18 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.