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In the mostly-minus convention with vacuum gravitational action proportional to $\int \sqrt{-g}f(R)$ and $f^{\prime }<0$, the displayed conformal transformation turns its vacuum equation into ${\tilde{G}}_{\mu \nu }=T_{\mu \nu }(\varphi )$, with potential $V=(R{f}^{\prime }-f)/(2f^{\prime 2})$ and the stress-energy tensor of a canonical scalar field. The conformal Ricci formula cancels the second derivatives of $f^{\prime }$ against the original field equation. Its remaining derivative tensor is ${\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -{\tilde{g}}_{\mu \nu }(\tilde{\nabla }\varphi {)}^2/2$. Expressing $V$ as a function of $\varphi$ requires a locally invertible $R\mapsto f^{\prime }(R)$. When $f^{\prime \prime }=0$, this scalar-tensor parametrization degenerates and introduces no new propagating scalar degree of freedom.

The classical stress-energy tensor of a canonical scalar field has a gradient-product term and a metric-proportional term. Contracting with a real null vector removes the latter, leaving the displayed square for a real field. This proves the null energy condition for a minimally coupled canonical scalar, including the massless case. It is not a statement about arbitrary curvature couplings, wrong-sign kinetic terms or a renormalized quantum stress-energy tensor.

Write $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. Vary the covariant metric while holding the scalar fixed. The supplied determinant and inverse-metric variations giveBy the definition $\delta S=\frac{1}{2}\int \sqrt{-g}{T}^{\mu \nu }\delta g_{\mu \nu }$, lowering its indices yields the stress-energy tensor of a canonical scalar field:The sign in the determinant variation is positive for varying $g_{\mu \nu }$ and would be negative for varying the inverse metric; matching that convention is essential.

Let $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. In the metric variation specified here, the covariant metric tensor $g_{\mu \nu }$ is varied while the scalar field is held fixed. Differentiating the inverse and the determinant givesThus $\delta \mathcal{L}=\frac{1}{2}{\partial }^{\mu }\varphi {\partial }^{\nu }\varphi \delta g_{\mu \nu }$, and the action varies asThe functional derivative therefore gives $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$. Lowering both indices yields the stress-energy tensor of a canonical scalar field:The determinant variation has a positive sign because the variable is $g_{\mu \nu }$; varying the inverse metric tensor instead would reverse that intermediate sign.

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 fieldThe vacuum f(R) gravity equation, divided by $F$, isInsert this in the transformed Einstein tensor from part (a). Since $g_{\mu \nu }=-{\tilde{g}}_{\mu \nu }/F$, its algebraic term becomes ${\tilde{g}}_{\mu \nu }(RF-f)/(2F^2)$. Also $\varphi =\sqrt{3/2}\log (-F)$ gives ${\partial }_{\mu }\varphi =\sqrt{3/2}{\partial }_{\mu }F/F$. The remaining derivative tensor is exactly the kinetic part of $T_{\mu \nu }$: the combination $g_{\mu \nu }(\nabla F{)}^2$ equals ${\tilde{g}}_{\mu \nu }(\tilde{\nabla }F{)}^2$. ThusThis establishes the Einstein-frame representation of metric f(R) gravity in the specified normalization. To express the potential as a function of $\varphi$, choose a branch where $R\mapsto f^{\prime }(R)$ is locally invertible; the conformal metric already requires $f^{\prime }<0$.