Along the affine null geodesic write $\Phi'=U^a\nabla_a\Phi$. Contracting the scalar stress tensor with $U^aU^b$ removes every term proportional to $g_{ab}$. Since $G_{ab}=T_{ab}$ and
$$
U^aU^b\nabla_a(\Phi\nabla_b\Phi)
=\frac d{d\lambda}(\Phi\Phi')
=(\Phi')^2+\Phi\Phi'',
$$
one obtains
$$
(1-\xi\Phi^2)T_{ab}U^aU^b
=(1-2\xi)(\Phi')^2-2\xi\Phi\Phi''.
$$
Put $D=1-\xi\Phi^2$. An integration by parts gives
$$
\int_{-\infty}^{\infty}T_{ab}U^aU^b\,d\lambda
=\left[-\frac{2\xi\Phi\Phi'}D\right]_{-\infty}^{\infty}
+\int_{-\infty}^{\infty}
\frac{1+\xi(4\xi-1)\Phi^2}{D^2}(\Phi')^2\,d\lambda.
$$
The stated endpoint condition kills the boundary term. If $\xi<0$, then $D>0$ and $\xi(4\xi-1)>0$, so the remaining integrand is nonnegative. Consequently this nonminimally coupled scalar satisfies the <averaged null energy condition>:
$$
\boxed{\int_{-\infty}^{\infty}T_{ab}U^aU^b\,d\lambda\geq0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 3 / c / i / Solution

{title2=ANEC}

The averaged null energy condition requires
$$
\int_{-\infty}^{\infty}T_{ab}k^ak^b\,d\lambda\geq0
$$
along every complete affinely parametrized null geodesic. It is weaker than the pointwise null energy condition.


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