# Stress-energy tensor of Jordan-Fierz-Brans-Dicke action

I have a question about the JFBD action, in particular the part that involves a scalar field. It is given by $$S_\phi = \int d^4x \sqrt{-g} \left( \frac{1}{2} g^{\mu\nu}(\nabla_{ \mu}\phi)(\nabla_{\nu}\phi)-V(\phi)-\frac{\xi}{2}R\phi^2 \right).$$ The energy-stress tensor is defined as $$T_{\mu\nu}^{\phi}:=\frac{-2}{\sqrt{-g}}\frac{\delta S_{\phi}}{\delta g^{\mu\nu}}$$. I should get the following: $$T_{\mu\nu}^{\phi}=(\partial_{\mu}\phi)(\partial_{\nu}\phi)-g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta}(\partial_{\alpha}\phi)(\partial_{\beta}\phi)-V(\phi) \right] -\xi G_{\mu\nu}\phi^2-\xi[\nabla_{\mu}\nabla_{\nu}-g_{\mu\nu}\square]\phi^2,$$ where $$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$$, $$\square=g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}$$ and $$R$$ the Ricci scalar. Now the terms up and including to the one containing the Einstein tensor I can figure out, but then I get into trouble. I used that $$\delta R=R_{\mu\nu} \delta g^{\mu\nu} + \nabla_{\rho}(g^{\mu\nu}\delta \Gamma^{\rho}_{\mu\nu}-g^{\mu\rho} \delta \Gamma_{\nu\mu}^{\nu})$$ and here the second term does not vanish by stokes because it is coupled to the field $$\phi$$ which depends on spacetime. But I am not sure how to treat it and prove that $$\nabla_{\rho}(g^{\mu\nu}\delta \Gamma^{\rho}_{\mu\nu}-g^{\mu\rho} \delta \Gamma_{\nu\mu}^{\nu})=[\nabla_{\mu}\nabla_{\nu}-g_{\mu\nu}\square]\delta g^{\mu\nu}$$ or something else in this direction. Any hints or solutions would be very welcome.

• How did you obtain last term in stress-energy tensor? Feb 24, 2020 at 21:28
• You can mention that last term in the action looks like gravity action times $\phi^2$, so it gives us only Einstein tensor. Did I missing something? Feb 24, 2020 at 21:32
• The last term is the action is the nonminimal coupling, and the fact that it is multiplied by $\phi^2$ makes it different because $\phi=\phi(x)$ a field that depends on spacetime. Feb 24, 2020 at 21:48

$$\delta\Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\lambda}(\nabla_\nu\delta g_{\mu\lambda}+\nabla_\mu\delta g_{\nu\lambda}-\nabla_\lambda\delta g_{\mu\nu}).$$
One can check, that: $$\nabla_{\rho}(g^{\mu\nu}\delta \Gamma^{\rho}_{\mu\nu}-g^{\mu\rho} \delta \Gamma_{\nu\mu}^{\nu})=[\nabla_{\mu}\nabla_{\nu}-g_{\mu\nu}\square]\delta g^{\mu\nu}$$