difference of ${\cal R}$ of two frames, which is: \[difference\] -[R]{}&=&--+q \
&=&+-q , where we have used the background equations. Note also that $\delta q$ is defined from the equation $\delta T^0_i=\partial_i\delta q$, so one has: &=&-(\^+F\^-F\_0\^-F) ,\
q&=&-(+e\^[2b]{}) , respectively.

Although the right hand side of Eq. (\[difference\]) looks some complicated, it only contains terms that involves $\delta\phi$ and $\delta\chi$. Actually in such a two-field system (or $f(R)$+single field in Jordan frame), there are only two degrees of freedom and $\delta\phi$ and $\delta\chi$ can become a complete set which can present everything. Therefore, after some straightforward calculation and making use of the inverse transformation of Eq. (\[QsigmaQs\]), we finally express $\tilde{\cal R}-{\cal R}$ in terms of $\delta\phi$ and $\delta\chi$ (or $\cal R$ and $\cal S$, because of their one-to-one correspondance) as: \[differenceR\] -[R]{} , where higher order terms are omitted and as it is perturbation on large scales, Eq. (\[eqlargescale\]) is used.

From the above formula one can see that, the curvature perturbation ${\cal R}$ in the two frames differs by a quantity proportional to the isocurvature perturbation, ${\cal S}$. That means some parts of isocurvature perturbations has now been transferred into the adiabatic ones during frame transformation. 