# Field equation of $R$-squared gravity in Einstein frame

The action in $$R^2$$ gravity is given by $$\begin{equation} S[g^*_\mathrm{\mu\nu}] = \frac{1}{16\pi}\int d^4x \sqrt{-g^*}(R^*+aR{^*}^2) \end{equation}$$ Where $$g^* = \det(g^*_\mathrm{\mu\nu})$$ and $$a$$ is a constant. For my problem I neglect the matter term. I've allready transformed this action in the Einstein frame by using a conformal transformation of the metric $$g^*_\mathrm{\mu\nu}=A^2g_\mathrm{\mu\nu}$$ $$\begin{equation} S[g_\mathrm{\mu\nu},\phi] = \frac{1}{16\pi}\int d^4x \sqrt{-g}(R-2\partial_\mu\phi\partial^\mu\phi-V(\phi)) \end{equation}$$ Here is $$\phi$$ a scalar field and $$V(\phi)$$ a potential. Now I want to derive the filed equation from this action in the Einstein frame. For this I define $$f(R) = R-2\partial_\mu\phi\partial^\mu\phi-V(\phi)$$. So the variation respect to the metric is \begin{align} \delta S &= \int d^4x \left(f(R)\delta\sqrt{-g}+\sqrt{-g}\delta f(R)\right)\\ &= \int d^4x \left(f(R)\delta\sqrt{-g}+\sqrt{-g}\frac{df}{dR}\delta R\right) = 0 \end{align} After some calculations there are those two terms in the action $$\begin{equation} \delta S \propto \int d^4x\sqrt{-g}\frac{df}{dR}\left(g_\mathrm{\mu\nu}g^\mathrm{\gamma\alpha}\nabla_\alpha\nabla\gamma\delta g^\mathrm{\mu\nu}-\nabla_\mu\nabla_\nu\delta g^\mathrm{\mu\nu}\right) \end{equation}$$ The problem is that the derivatives relate to the variation in the metric. To work around this problem I want to integrate by parts twice. So that the derivatives are on $$\frac{df}{dR}$$. But what is this term $$\frac{df}{dR}$$? I think it is equal to one. But the derivatives on this constant vanishes, right? Nevertheless, In the final field equations there are these terms $$\begin{equation} R_\mathrm{\mu\nu}-\frac{1}{2}g_\mathrm{\mu\nu}R = 2\partial_\mu\phi\partial_\nu\phi-g_\mathrm{\mu\nu}\partial^\sigma\phi\partial_\sigma\phi-\frac{1}{2}V(\phi)g_\mathrm{\mu\nu} \end{equation}$$ $$\\$$

Edit: The variation of the action respect to the metric is \begin{align} \delta S &= \int d^4x \sqrt{-g}\left[(R_\mathrm{\mu\nu}-\frac{1}{2}g_\mathrm{\mu\nu}R)\delta g^\mathrm{\mu\nu}+\frac{1}{2}g_\mathrm{\mu\nu}V(\phi)\delta g^\mathrm{\mu\nu}+g_\mathrm{\mu\nu}\partial_\gamma\phi\partial^\gamma\phi\delta g^\mathrm{\mu\nu}\right]\\ &+\int d^4x \sqrt{-g}\left(g_\mathrm{\mu\nu}\Box\delta g^\mathrm{\mu\nu}-\nabla_\mu\nabla_\nu\delta g^\mathrm{\mu\nu}\right) \end{align} where I set $$\frac{df}{dR}=1$$. I don't know how to calculate the second integral.

• You missed the term $-\partial_{\mu} \phi \partial_{\nu} \phi/2$ in your action and the second integral is the contribution at infinity (see for example en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action or any standard GR textbook, and many question on this topic here). It's $g^{\mu\nu}\delta R_{\mu\nu}$ after you use the definition of the variation of the Ricci tensor. Mar 16, 2021 at 16:10