Going to the <PERSON> frame in f(R) theories
First of all thank you for your time!
I have a question that I can't solve. In every review that I read, I find that when you want to go to the <PERSON> frame in a $f(R)$ theory what you have to do is basically a <PERSON> transformation going from $f(R)$ to $F \cdot R - U$ where $U=F \cdot R-f(R)$ (notice that I am missing some constants with dymensions over there, but the idea is to do this <PERSON> transformation that changes our variable $R$ by the new variable $F$ (like a canonical momentum comes to sustitute the velocities in the Hamiltonian) and put the final result in a hybrid form that keeps both variables together. Until there everything is clear. The problem is that when we want to go to the <PERSON> frame we need to do a conformal transformation, and when you do that, you choose the conformal factor to cancel the $F$ in front of $R$ and, according to the literature, thats all, you don't have to worry abount the term $U$, it is not affected at all by the conformal transformation. And also you don't have to transform $F$.
I can't understand that, in the same sense as $p$ (the momentum) depends on $v$ (the velocity), $F$ here will depend on $R$ and it should be transformed by the conformal transformation (I could write the dependence but there's no point in doing that). And of course, is obvious that $f(R)$ in $U$ depends on $R$. 651
As much as I read the different reviews that exist on the literature about $f(R)$ theories and different papers in that area, I always find that $U$ and $F$ remains the same after the conformal transformation, it only affects $R$ in $F \cdot R - U$ (and of course the determinant of the metric that goes in the action). Could some of you explain me why is that possible?
EDIT: It is relevant to notice that in $f(R)$ you only have one degree of freedom, $R$, that must be fixed by choosing a metric. But in the literature, when you do this trick, at the end of the day they do some change of variables in order to define an scalar field $\phi$ as a function of $R$ that has a canonical kinetic term. The potential $U$ will depend on $\phi$ too. And at this point they claim that they have 2 degrees of freedom, $R$ and the new scalar field $\phi$. But I can't understand this, because if you track $\phi$ you can see that it depends on $R$ always, and you can't go from 1 degree of freedom to 2.
P.D.: if someone want me to be more specific I can edit this with some steps and some references, but if you work with $f(R)$ theories, this should be bread and butter for you, and even the notation is very standard in the literature that I have read. 651