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Given the conformal transformation of a metric \begin{equation} g^*_\mathrm{\mu\nu} = A^2 g_\mathrm{\mu\nu} \end{equation} This results in the transformation of the ricci scalar \begin{equation} R^* = A^{-2}R+(D-4)(1-D)A^{-4}\partial_\mu\partial^{\mu}A+2(1-D)A^{-3}g^{\mu\nu}\nabla_\mu\partial_\nu A \end{equation} Here $D=4$ denotes the dimensions and $\nabla_\mu$ is the covariant derivative. With $D=4$ the second summand cancel out and this results in \begin{equation} R^* = A^{-2}R-6A^{-3}g^{\mu\nu}\nabla_\mu\left(-\frac{1}{\sqrt{3}}A\partial_\nu\phi\right) \end{equation} Here $A$ is defined as \begin{equation} A^2 = e^{-\frac{2}{\sqrt{3}}\phi} \end{equation} where $\phi$ denotes a scalar field. The final transformation for the ricci scalar I want is \begin{equation} R^* = A^{-2}R-2A^{-2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi \end{equation} Does anyone have any idea how to get from my transformation of the ricci scalar to the one I am looking for? I don't know how to handle with the covariant derivative, since then the Christoffel symbols appear there.

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  • $\begingroup$ I have done the calculation and I have an excess term proportional to $g^{\mu\nu}\nabla_\mu (\partial_\nu \phi)$. Any clue if this should be zero? $\endgroup$
    – AFG
    Mar 10 at 12:48
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    $\begingroup$ @AFG This term is a total derivative so vanishes in an action with appropriate boundary conditions. This should align with my answer below, although I think I have an accidental factor of 3. $\endgroup$
    – Eletie
    Mar 10 at 13:16
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The only way I see of getting the expression you're looking for is by using the Divergence theorem. So I assume this expression is in an integral and we're after some equations of motion in the end. Doing this schematically (as I don't know the context), it'd be some like something along the lines of $$\begin{align} \int-6 A^{-3} g^{\mu \nu} \nabla_{\mu}(\partial_{\nu}A) &= \int-6 A^{-3} \nabla_{\mu}(\partial^{\mu}A) \\ &= \big[-6 A^{-3} \,\partial^{\mu}A \big]_{\partial{\mathcal{M}}} -\int \partial^{\mu}A \,\nabla_{\mu}\big(-6 A^{-3}\big) \ . \end{align} $$ Ignoring the boundary term (which I'm assuming won't contribute to some E.o.M), we have $$\begin{align} -g^{\mu \nu}\partial_{\mu}A \,\nabla_{\mu}\big(-6 A^{-3}\big) &= \frac{+1}{\sqrt{3}}A \partial^{\mu}\phi\ \partial_{\mu}(-6 A^{-3}) \\ &= \frac{-6}{\sqrt{3}}A \partial^{\mu}\phi {\sqrt{3}} A^{-3} \partial_{\mu} \phi \\ &= -6 A^{-2}\partial^{\mu} \phi\partial_{\mu} \phi \ . \end{align} $$

Edit: Actually I get a factor of $3$ different, so you can try spot the mistake if you know the final transformation you need is correct (but the method using the divergence theorem is the way to go).

Edit 2: To get the factor you're looking for you probably want to include the metric determinant in the integral (which also includes conformal factors), but the context would help with this.

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