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

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

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. Commented Mar 16, 2021 at 16:10

## 1 Answer

You can vary your new Einstein frame action the sandard way. By varying with respect to the metric the Ricci scalar will give the Einstein tensor and the scalar field terms the energy momentum tensor of a self interacting scalar field i.e the desired equation.