Does every set of orthogonal coordinates in 2D Euclidean space yield a conformally-separable metric? Consider the 2D Euclidean plane in Cartesian coordinates which has the metric
$$ds^2 = dx^2 + dy^2$$
If we transform into polar coordinates, we obtain the metric in the form
$$ds^2 = r^2 (\frac{1}{r^2} dr^2 + d\vartheta^2)$$
If we transform into parabolic coordinates, we obtain the metric in the form
$$ds^2 = (\sigma^2 + \tau^2)(d\sigma^2 + d\tau^2)$$
Etc. etc. In any case, it seems that every orthogonal coordinate transformation into coordinates $w,z$ transforms the metric into a "conformally-separable" form
$$ds^2 = \Omega^2(w,z) (W(w) dw^2 + Z(z) dz^2)$$
That is, each of the diagonal metric components are, up to a common conformal factor $\Omega^2$, functions only of the respective coordinates. (This means that e.g. the Laplace equation will be separable in these coordinates.)
Is this the case for any analytic 2D orthogonal coordinates? (Can you give a nice canonical reference for further reading?)

My attempts:
I tried to approach this question by first investigating the question: does every orthogonal coordinate transformation generated by a holomorphic function yield such a metric?
Consider a holomorphic function $f(z)$ with $x=\Re (f(z)),\, y=\Im (f(z))$ and the transformed coordinates are $w=\Re (z),\,v=\Im (z)$. This gives us the coordinate transform as $x=x(w,v),\,y=y(w,v)$ The holomorphic nature of $f(z)$ means that it fulfills the Cauchy-Riemann equations which, in this case, can be written as
$$\partial_w x= \partial_v y$$
$$\partial_v x= -\partial_w y$$
By transforming the Euclidean metric using this relation we obtain
$$d s^2 = [(\partial_w x)^2 + (\partial_w y)^2] d w^2 + [(\partial_v x)^2 + (\partial_v y)^2] d v^2$$
which can be also rewritten as
$$d s^2 = |\partial_w f|^2 d w^2 + |\partial_v f|^2 d v^2$$
However, it is not clear how does this amount to the conformal-separable form...
 A: Your attempt is nearly there. Note that using the Cauchy-Riemann equations
$$d s^2 = [(\partial_w x)^2 + (\partial_w y)^2] d w^2 + [(\partial_v x)^2 + (\partial_v y)^2] d v^2 $$
$$= [(\partial_w x)^2 + (-\partial_v x)^2] dw^2 + [(\partial_v x)^2 + (\partial_w x)^2] d v^2$$
$$= [(\partial_w x)^2 + (\partial_v x)^2] ( dw^2+dv^2)$$
I think what you are looking for is the notion of conformal maps. Or more specifically in this case the fact that the conformal maps in $\mathbb{R}^2 \equiv \mathbb{C}$ are exactly the holomorphic (and anti-holomorphic, if you do not care about orientations) functions.
A: I do not think that the question posed in the title has been answered for general orthogonal coordinates in the Euclidean plane, i.e. when F and G do NOT obey the Cauchy-Riemann equations.  Take for example these orthogonal curves in the 2D Euclidean half plane using cylindrical coordinates $(\rho,z)$ in the $\phi=constant$ plane  :$$\nabla F(\rho,z)\cdot\nabla G(\rho,z)=0 \\  F_{\rho}G_{\rho}+F_{z}G_{z} = 0$$but with\begin{eqnarray}F_{\rho}=+\rho  e^{H(\rho,z)} G_z\\F_z=-\rho e^{H(\rho,z)} G_{\rho}\end{eqnarray}The above equations would have to be transformed so that they are in the form of the C-R equations to yield a conformally separable metric, which looks very unlikely in general.  Or can that be  done?  So it seems to me that only some orthogonal coordinate systems in 2D Euclidean space can yield a conformally separable metric.
