= Solution
The supplied superhorizon evolution equation is
$$
\frac{d\mathcal R}{d\ln a}
=-\frac{\delta P_{\rm nad}}{\bar\rho+\bar P},
\qquad
\delta P_{\rm nad}=\delta P-
\frac{\bar P'}{\bar\rho'}\delta\rho.
$$
For an <adiabatic cosmological perturbation>, $\delta P=(\bar P'/\bar\rho')\delta\rho$, so $\delta P_{\rm nad}=0$ 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 $\bar P=A/3$, $A'=-4\mathcal H A$, and $C'=-3\mathcal H C$. Then
$$
\frac{\bar P'}{\bar\rho'}=\frac{4A}{3(4A+3C)}.
$$
The <cosmological entropy perturbation> $S=\delta_c-3\delta_\gamma/4$ implies $\delta_c=S+3\delta_\gamma/4$, so the adiabatic part cancels and
$$
\delta P_{\rm nad}
=-\frac{4AC}{3(4A+3C)}S.
$$
Since $\bar\rho+\bar P=C+4A/3$, the curvature evolves as
$$
\boxed{\frac{d\mathcal R}{d\ln a}
=\frac{4\bar\rho_\gamma\bar\rho_c}
{(4\bar\rho_\gamma+3\bar\rho_c)^2}S}.
$$
In the sign convention requested in the question, $d\mathcal R/d\ln a=-fS$, this means
$$
\boxed{f(\bar\rho_\gamma,\bar\rho_c)
=-\frac{4\bar\rho_\gamma\bar\rho_c}
{(4\bar\rho_\gamma+3\bar\rho_c)^2}}.
$$
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