# 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...

• What is a orthogonal transformation? – coconut Jan 18 '17 at 16:17
• @coconut In Cartesian coordinates on the Euclidean plane it is a transformation $w(x,y), z(x,y)$ such that $\partial_x w \partial_y w= -\partial_x v \partial_y v$. On a general Riemannian manifold a set of orthogonal coordinates is such that the tangent vectors to the coordinate lines are orthogonal to each other. – Void Jan 18 '17 at 16:23
• @coconut But yes, you are right that it makes more sense to talk about coordinates rather than transformations, I made appropriate edits. – Void Jan 18 '17 at 16:30

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.
• Haha, that was just around the corner :) Just one quick question for a full answer - is every orthogonal coordinate system generated by a conformal map (plus an additional coordinate rescaling)? E.g. in the case of polar coordinates we take $f(z)=\exp(z)$, but then we need to rescale one of the coordinates $\vartheta=v,\, r= \exp(w)$. – Void Jan 20 '17 at 10:38
• Taking your definition, it should be. As W only depends on w and Z only depends on z, you can transform it into the definition of a conformal map by transforming with something along the line of $w \mapsto \int_0^w \sqrt{W(s)} ds$ (or the other way round, I haven't checked) and similiar for z. – mlk Jan 20 '17 at 11:31