# Coordinate transformation bringing $ds^2 = y^2 dx^2 + x^2 dy^2$

In Wolfgang Rindler's book "Essential Relativity" there is an unsolved exercise about coordinate change in two dimensions (pp 140-141). The author proposes as example four infinitesimal distances, three of them accounting for the flat plane (in some coordinate system) and one more for a two dimensional space with curvature. Two for the flat plane get its coordinate transformation in the book but not the third. Further, Rindler write: "The reader will not guess in a hurry (and should not try) how the third arises, though it too results from transforming the Cartesian". The problem I've tried to solve without success is the following:

Find the coordinate transformation from Cartesian (where $ds^2 = d\bar{x}^2 + d\bar{y}^2$) that brings the following infinitesimal distance: $$ds^2 = y^2 dx^2 + x^2 dy^2$$

Let's denote $$x=x(x',y'),\\ y=y(x',y'),$$ then $$dx=\frac{\partial x}{\partial x'}dx'+\frac{\partial x}{\partial y'}dy',\\ dy=\frac{\partial y}{\partial x'}dx'+\frac{\partial y}{\partial y'}dy'.$$ Now $dx^2+dy^2=y'^2dx'^2+x'^2dy'^2$ implies $$\left(\frac{\partial x}{\partial x'}\right)^2+\left(\frac{\partial y}{\partial x'}\right)^2=y'^2,\\ \left(\frac{\partial x}{\partial y'}\right)^2+\left(\frac{\partial y}{\partial y'}\right)^2=x'^2,\\ \left(\frac{\partial x}{\partial x'}\right)\left(\frac{\partial x}{\partial y'}\right)=-\left(\frac{\partial y}{\partial x'}\right)\left(\frac{\partial y}{\partial y'}\right).$$ Now we can try $\left(\frac{\partial x}{\partial x'}\right)=\left(\frac{\partial y}{\partial x'}\right)$ (another one can be fixed correspondingly) which turns out to doesn't work. Then we can try $\left(\frac{\partial x}{\partial x'}\right)=\left(\frac{\partial y}{\partial y'}\right)$, we finally obtain the following transformation: $$x=\frac{\sqrt{2}}{2}x'y';\\ y=\frac{\sqrt{2}}{4}y'^2-\frac{\sqrt{2}}{4}x'^2.$$ Now we can check that $$dx^2+dy^2=y'^2dx'^2+x'^2dy'^2.$$
• They are parabolic coordinates and the scale factors are both the same $C\sqrt (x'^2 + y'^2)$ – Rafa Budría Aug 20 '16 at 19:23
$$\begin{cases} \bar x=\frac{1}{2}xy\cos(\ln(y/x))-\frac{1}{2}xy\sin(\ln(y/x)) \\ \bar y=\frac{1}{2}xy\sin(\ln(y/x))+\frac{1}{2}xy\cos(\ln(y/x)) \\ \end{cases}$$