The action and the source normalization

Displayed formulas in this passage

• VI@1#eq:action · line 379

  $$S[\gamma ,\varphi ]=\frac{2}{{\kappa }^2}\int \sqrt{-\gamma }R[\gamma ]{\mathrm{d}}^{4}x-\frac{1}{2}\int \sqrt{-\gamma }\left({\gamma }^{\mu \nu }{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi +m^2{\varphi }^2\right){\mathrm{d}}^{4}x.$$

• VI@1#eq:G1 · line 390

  $${G^{(1)}}_{\mu \nu }(a)=-\frac{1}{2}{\square }_{\eta }{\bar{a}}_{\mu \nu }+{\partial }_{(\mu }D(a{)}_{\nu )}-\frac{1}{2}{\eta }_{\mu \nu }{\partial }^{\alpha }D(a{)}_{\alpha }.$$

• VI@1#eq:full-equations · line 415

  $$G_{\mu \nu }(\gamma )=\frac{{\kappa }^2}{4}{T}_{\mu \nu }(\gamma ,\varphi ),({\square }_{\gamma }-m^2)\varphi =0.$$

• unlabeled · line 399

  $$G^{(1)}K=0,{\partial }^{\mu }{G^{(1)}}_{\mu \nu }(a)=0.$$

• unlabeled · line 405

  $$T_{\mu \nu }(\gamma ,\varphi )={\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -\frac{1}{2}{\gamma }_{\mu \nu }\left({\gamma }^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi +m^2{\varphi }^2\right).$$