Definition of conformal flatness I was trying to prove that the Ricci scalar $R$ is not invariant under conformal transformations and when we talk about conformal transformations we have the relation:
$$ \hat g_{\mu\nu} = \Omega(x) \eta_{\mu\nu}$$
Is this always the case or can we have some other metric $g_{\mu\nu}$ instead of $ \eta_{\mu\nu}$? I was thinking about this because its a lot easier if the metric is $ \eta_{\mu\nu}$, when computing the Christoffel symbols we can just take them out of the derivatives since all the elements of $ \eta_{\mu\nu}$ are constants.
 A: The presence of Minkowski metric tensor $ \eta_{\mu \nu}$ means, that you are considering the particular class of the metrics, related to the $\eta_{\mu \nu}$ by conformal factor $\Omega(x)$. In general case, one cannot obtain the given metric $g_{\mu \nu}$ from Minkowski metric by a conformal transformation.
However, if you are speaking in the context of string theory, one also has a differomorphism invariance, such that the initial metric can be brought to the form, $\Omega(x) \eta_{\mu \nu}$ by a suitable reparametrisation - it is in fact fixing of gauge.
Non-invariance of the Ricci curvature $R$ under the conformal transformations can be established readily due to the fact, that $R$ involves two derivatives and has a mass dimension of two. Hence, the rescaling with the constant factor $\lambda$ is bound to produce a factor $\lambda^2$.
However, if you are interested in the exact equation how does the $R$ transform - there would be a certain derivation. Assumption of the conformally equivalent to Minkowski metric simplifies it significantly.
A: The metric $\eta_{\mu\nu}$ is just one flat metric, if you use spherical coordinates $(u^0,u^1,u^2,u^3) = (t,r,\theta,\phi)$ in Minkowski space you have:
$$\text{d}s^2 = \hat{g}_{\mu\nu} \text{d}u^\mu \text{d}u^\nu = -\text{d}t^2 + \text{d}r^2 + r^2\text{d}\theta^2 + r^2\sin^2\theta\text{d}\phi^2$$
This metric has a zero Riemann tensor: $\text{Rm}(\hat{g}) = (R_{\alpha\beta\gamma\delta}) = 0$, so every metric of the form:
$$\tilde{g}_{\mu\nu} = \Omega(t,r,\theta,\phi)\ \hat{g}_{\mu\nu}$$
will be "conformally flat" (i.e. its Weyl tensor will be zero).
