= Solution
For the canonical scalar action and the stress-tensor convention in the paper, varying the inverse metric and volume element gives the <stress-energy tensor of a canonical scalar field>
$$
T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-\widetilde g_{\mu\nu}
\left[\frac12\widetilde g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi-V(\phi)\right].
$$
The vacuum <f(R) gravity> equation, divided by $F$, is
$$
G_{\mu\nu}-\frac1F\nabla_\mu\nabla_\nu F+\frac1F g_{\mu\nu}\Box F
=-\frac{RF-f}{2F}g_{\mu\nu}.
$$
Insert this in the transformed <Einstein tensor> from part (a). Since $g_{\mu\nu}=-\widetilde g_{\mu\nu}/F$, its algebraic term becomes $\widetilde g_{\mu\nu}(RF-f)/(2F^2)$. Also $\phi=\sqrt{3/2}\log(-F)$ gives $\partial_\mu\phi=\sqrt{3/2}\,\partial_\mu F/F$. The remaining derivative tensor is exactly the kinetic part of $T_{\mu\nu}$: the combination $g_{\mu\nu}(\nabla F)^2$ equals $\widetilde g_{\mu\nu}(\widetilde\nabla F)^2$. Thus
$$
\boxed{\widetilde G_{\mu\nu}=T_{\mu\nu},\qquad V(\phi)=\frac{Rf'(R)-f(R)}{2f'(R)^2}.}
$$
This establishes the <Einstein-frame representation of metric f(R) gravity> in the specified normalization. To express the potential as a function of $\phi$, choose a branch where $R\mapsto f'(R)$ is locally invertible; the conformal metric already requires $f'<0$.
Back to article page
