= Averaged null energy condition for a nonminimally coupled scalar
Fix the curvature-coupling convention by the classical action
$$
S=\frac12\int\sqrt{-g}\left[(1-\xi\Phi^2)R-(\nabla\Phi)^2\right]d^4x
$$
in units $8\pi G=1$ and signature $(-,+,+,+)$. Here $R$ is the <scalar curvature>. Write the metric equation in Einstein form $G_{ab}=T^{\rm eff}_{ab}$, so the <stress-energy tensor> includes the curvature-coupling contribution rather than only the minimally coupled kinetic term.
Let $k^a$ be the tangent to an affinely parametrized <null geodesic>, use prime for $d/d\lambda$ and put $F=1-\xi\Phi^2$. The null-contracted metric equation is
$$
T^{\rm eff}_{ab}k^ak^b=\frac{(\Phi')^2-\xi(\Phi^2)''}{F}
=\frac{(1-2\xi)(\Phi')^2-2\xi\Phi\Phi''}{F}.
$$
An <integration by parts>, with $\Phi\Phi'/F$ vanishing at the endpoints, yields
$$
\int T^{\rm eff}_{ab}k^ak^b\,d\lambda
=\int\frac{1+\xi(4\xi-1)\Phi^2}{(1-\xi\Phi^2)^2}(\Phi')^2\,d\lambda.
$$
For $\xi<0$, $F>0$ and the numerator is positive for every real $\Phi$. The <averaged null energy condition> therefore holds along a complete affine ray when the stated boundary and integrability conditions hold. This is a result for the displayed coupled classical model and its normalization; the scalar equation alone does not determine a test field's averaged stress on an arbitrary fixed metric.
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