Conformal transformation of a metric - Ricci scalar

Given the conformal transformation of a metric $$$$g^*_\mathrm{\mu\nu} = A^2 g_\mathrm{\mu\nu}$$$$ This results in the transformation of the ricci scalar $$$$R^* = A^{-2}R+(D-4)(1-D)A^{-4}\partial_\mu\partial^{\mu}A+2(1-D)A^{-3}g^{\mu\nu}\nabla_\mu\partial_\nu A$$$$ Here $$D=4$$ denotes the dimensions and $$\nabla_\mu$$ is the covariant derivative. With $$D=4$$ the second summand cancel out and this results in $$$$R^* = A^{-2}R-6A^{-3}g^{\mu\nu}\nabla_\mu\left(-\frac{1}{\sqrt{3}}A\partial_\nu\phi\right)$$$$ Here $$A$$ is defined as $$$$A^2 = e^{-\frac{2}{\sqrt{3}}\phi}$$$$ where $$\phi$$ denotes a scalar field. The final transformation for the ricci scalar I want is $$$$R^* = A^{-2}R-2A^{-2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi$$$$ 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.

• 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?
– AFG
Mar 10 at 12:48
• @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. Mar 10 at 13:16

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).