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}


\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?

  • $\begingroup$ is there a physical reason to be interested in such a system? :) $\endgroup$
    – Ilja
    Apr 18 '16 at 14:37
  • $\begingroup$ 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 $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.