$f^{\prime}(R)=0$ in $f(R)$ gravity

Suppose in a certain $f(R)$ gravity theory, $f^{\prime}(R)=0$ for some finite value of $R$. (e.g. let $f(R)=R+\alpha R^2$ with $\alpha<0$. $f^{\prime}(R)=0$ at $R=-\frac{1}{2\alpha}$.)

Also suppose I am considering the flat FLRW metric where thr Ricci scalar $R=6(\dot{H}+2H^2)$ with $H$ the Hubble parameter. The $f(R)$ field equations are given by

\begin{eqnarray} 3H^2&=&\frac{\kappa}{f^{\prime}}(\rho+\rho_{curv}) \\ \dot{H}&=&-\frac{\kappa}{2f^{\prime}}(\rho +p+\rho_{curv}+p_{curv}) \end{eqnarray}

where

\begin{eqnarray} \rho_{curv}&=&\frac{Rf^{\prime}-f}{2\kappa}-\frac{3Hf^{\prime\prime}\dot{R}}{\kappa} \\ p_{curv}&=&\frac{\dot{R}^2f^{\prime\prime\prime}+2H\dot{R}f^{\prime\prime}+\ddot{R}f^{\prime\prime}}{\kappa}-\frac{Rf^{\prime}-f}{2\kappa} \end{eqnarray}

Clearly, when $f^{\prime}(R)=0$, $H^2,\dot{H}\longrightarrow\infty$. So we should have $R=6(\dot{H}+2H^2)\longrightarrow\infty$. This is a contradiction because we started with the assumption that $f^{\prime}(R)=0$ for some finite $R$.

Can someone point out where am I going wrong?

• is there a physical reason to be interested in such a system? :)
– Ilja
Apr 18 '16 at 14:37
• The $R+\alpha R^2$ model is the simplest $f(R)$ model that assumes a bouncing solution without violating the strong energy condition in the matter sector(contrary to GR, where strong energy condition must be violated for a bouncing solution). In such a scenario, the above question can be physically relevant @Ilja
– Neel
Apr 18 '16 at 14:55

$f(a)$ is not an independent degree of freedom from $R$ or from $a$. You have a system of equations here, and if you solve for the exact form of $a(t)$, you are not likely to admit solutions that have $f(R(a(t))) = 0$ for finite $t$. If you do, however, it's likely an indication of geodesic incompleteness of the solution.