Because $R=O(\epsilon)$ and
$$
f(0)=0,
\qquad f'(0)=1,
\qquad f''(0)=0,
$$
the <Taylor series> gives $f(R)=R+O(\epsilon^3)$ and $f'(R)=1+O(\epsilon^2)$. The extra derivative terms in the f(R) equation are therefore at least second order. At first order the theory reduces to the <Linearized Einstein equations>
$$
G^{(1)}_{\mu\nu}=8\pi\mathcal T_{\mu\nu}.
$$
In <harmonic coordinates>, equivalently the <Lorenz gauge in linearized gravity> $\partial_\mu\bar h^\mu{}_\nu=0$, the <trace-reversed metric perturbation>
$$
\bar h_{\mu\nu}=h_{\mu\nu}-\frac12\eta_{\mu\nu}h
$$
satisfies
$$
\boxed{\partial^\rho\partial_\rho\bar h_{\mu\nu}
=-16\pi\mathcal T_{\mu\nu}},
\qquad
\boxed{\partial_\mu\bar h^\mu{}_\nu=0}.
$$
Thus <Linearized f(R) gravity about Minkowski spacetime> has the same weak-field predictions as general relativity under these assumptions. Differences can first appear through nonlinear curvature terms or in backgrounds about which $f''(R)$ is nonzero.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 3 / b / iii / Solution

{c}

If $f(0)=0$, $f'(0)=1$, and $f''(0)=0$, then $f(R)=R+O(R^3)$ near zero. The field equations linearized about <Minkowski spacetime> are therefore the <Linearized Einstein equations>; deviations first appear beyond linear order.

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 Incoming links: Linearized f(R) gravity about Minkowski spacetime