 Solution

The supplied superhorizon evolution equation isFor an adiabatic cosmological perturbation, $\delta P=({\bar{P}}^{\prime }/{\bar{\rho }}^{\prime })\delta \rho$, so $\delta P_{\mathrm{nad}}=0$ and the comoving curvature perturbation $\mathcal{R}$ is conserved on superhorizon scales.

For photons and cold dark matter, put $A={\bar{\rho }}_{\gamma }$, $C={\bar{\rho }}_c$, and use $\bar{P}=A/3$, $A^{\prime }=-4\mathcal{H}A$, and $C^{\prime }=-3\mathcal{H}C$. ThenThe cosmological entropy perturbation $S={\delta }_c-3{\delta }_{\gamma }/4$ implies ${\delta }_c=S+3{\delta }_{\gamma }/4$, so the adiabatic part cancels andSince $\bar{\rho }+\bar{P}=C+4A/3$, the curvature evolves asIn the sign convention requested in the question, $d\mathcal{R}/d\ln a=-fS$, this means