Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 3 / b

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

The pure gravitational action is invariant under diffeomorphisms. An infinitesimal diffeomorphism generated by a compactly supported vector field $\xi$ changes the metric by its tensor Lie derivative,Using symmetry of $E^{ab}$ and the variational expression already derived,As ${\xi }_b$ is arbitrary, the off-shell metric divergence identity isThis Noether identity follows for every metric tensor without imposing either the gravitational field equation or the matter equations. It is a consequence of the metric action's diffeomorphism invariance, not a conclusion requiring a long component calculation.

In four dimensions, Lovelock's theorem restricts a local natural symmetric divergence-free metric tensor with at most second derivatives of the metric to a constant linear combination of the Einstein tensor and the metric. A generic nonlinear f(R) gravity equation contains ${\nabla }_a{\nabla }_b{f}^{\prime }(R)$: the Ricci scalar already has second metric derivatives, so this term generally introduces fourth metric derivatives. It therefore violates the second-order hypothesis, although it remains covariant, symmetric and divergence-free.

The exception must be stated. For $f(R)=aR+b$, the same tensor reduces towhich is precisely of Lovelock form. In particular, $f(R)=R$ is a counterexample to a blanket claim that the theorem never applies. The intended exclusion concerns generic nonlinear $f$, with $f^{\prime \prime }$ not identically zero. Restricting attention to a special constant-curvature solution of a nonlinear theory does not turn its off-shell equations into a universally second-order metric tensor.

A four-dimensional vacuum Einstein solution with the fixed cosmological constant $\Lambda$ has $R_{ab}=\Lambda g_{ab}$ and $R=4\Lambda$, constant. Thus every derivative of $f^{\prime }(R)$ vanishes andThe metric tensor is nondegenerate, so this is zero precisely whenThis is the necessary and sufficient Einstein metric condition in f(R) gravity for the specified $\Lambda$, and works for every such Einstein metric, even when its Weyl tensor is nonzero. There is no need to divide by $f^{\prime }(4\Lambda )$; the degenerate case where both $f$ and $f^{\prime }$ vanish at that curvature is included. At $\Lambda =0$, the condition is simply $f(0)=0$.

If the intention is to demand the inclusion for every real $\Lambda$ simultaneously, the stronger functional condition is $R{f}^{\prime }(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=C{R}^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives $f(R)=C{R}^2$, including $C=0$. This stronger reading is distinct from fixing one cosmological constant.