Higgs non-minimal coupling to gravity: Jordan and Einstein frame In this paper they consider the Higgs non-minimally coupled to the Ricci scalar. I am trying to recalculate the steps from equation (5) until equation (10).
Let's start with (5):
$$\int d^4x\sqrt{-g} \left[f(h) R-\frac{1}{2}g^{\mu\nu}\partial_\mu h\partial_\nu h -U(h)\right]$$
where $f(h)=(M_p^2+\xi h^2)/2$ and $U(h)=\frac{\lambda}{4}(h^2-v^2)^2$. They then proceed with the transformation of the metric:
$$g_{\mu\nu}\to \tilde{g}_{\mu\nu}=g_{\mu\nu}\Omega^2,~~\tilde{g}^{\mu\nu}=g^{\mu\nu}\Omega^{-2},.$$
If I plug this in I get:
$$\int d^4x\sqrt{-\tilde{g}\frac{1}{\Omega^8}} \left[f(h) R-\frac{1}{2}g^{\mu\nu}\partial_\mu h\partial_\nu h -U(h)\right]$$
$$=\int d^4x\sqrt{-\tilde{g}} \left[\frac{f(h)}{\Omega^2} \frac{R}{\Omega^2}-\frac{1}{2}\frac{\tilde{g}^{\mu\nu}}{\Omega^2}\partial_\mu h\partial_\nu h -\frac{1}{\Omega^4}U(h)\right]$$
From this they identify
$$\frac{f(h)}{\Omega^2}=\frac{1}{2}M_p^2\text{, i.e. }\Omega^2=1+\frac{\xi h^2}{M_p^2}.$$
This now looks almost as the end result in equation (10) of the paper that I cite at the beginning.
From the book of Sean Carroll (Appendix G) I get the relation for $n=4$ dimensions:
$$ \frac{R}{\Omega^2}=\tilde{R}+6\frac{g^{\alpha\beta}}{\Omega^3}\partial_\alpha\partial_\beta\Omega .$$
Evaluating the last piec gives me
$$\partial_\alpha\partial_\beta\Omega=\left[-\frac{1}{\Omega^3}\left(\frac{\xi h}{M_p^2}\right)+\frac{1}{\Omega}\frac{\xi}{M_p^2}\right]\partial_\alpha h\partial_\beta h+\frac{1}{\Omega}\frac{\xi h}{M_p^2}\partial_\alpha\partial_\beta h$$
This is almost what I want, but only almost. The term $\partial_\alpha\partial_\beta h$ is too much and also the second term in the square brackets is too much. Any ideas on that?
 A: To answer the second part of your question about the kinetic term, if we're starting with the kinetic term, call it $T$, with
$$ T = \frac{1}{2}g^{\mu \nu} (\partial_{\mu}h) (\partial_{\nu}h),  $$
then first notice that under the conformal transformation
$$g_{\mu \nu} \mapsto \tilde{g}_{\mu \nu} = \Omega^2 g_{\mu \nu},$$
the inverse metric transforms like
$$g^{\mu \nu} \mapsto \tilde{g}^{\mu \nu} = \Omega^{-2} g^{\mu \nu}$$
giving the desired
$$T = \frac{1}{2}\Omega^{2} \tilde{g}^{\mu \nu} (\partial_{\mu}h) (\partial_{\nu}h)$$
which when combined with the $\Omega^{-4}$ factor coming from the volume element gives the term listed in the paper. If you apply the transformation to both the coordinates and the metric like you have done in your example, the contributions cancel. As I understand it, in a conformal transformation like this you transform the metric but not the coordinates, which is where I believe your factors of $\frac{\partial \tilde{x}}{\partial x}$ are not in line with the paper's derivation.
The transformation of the Ricci tensor under conformal transformations is carried out in an appendix of Carroll's introduction to general relativity. I hope that reference is sufficient to answer your first question.
