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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: $$ds^2 = y^2 dx^2 + x^2 dy^2$$

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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 $$dx^2+dy^2=y'^2dx'^2+x'^2dy'^2.$$

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  • $\begingroup$ Thank you very much. Is there some, say, heuristics to approach to the solution? I tried comparing several ones to get cues and got nothing. $\endgroup$ – Rafa Budría Aug 20 '16 at 19:10
  • $\begingroup$ @RafaBudría, I will update my answer in a moment. $\endgroup$ – Wein Eld Aug 20 '16 at 19:12
  • $\begingroup$ They are parabolic coordinates and the scale factors are both the same $C\sqrt (x'^2 + y'^2)$ $\endgroup$ – Rafa Budría Aug 20 '16 at 19:23
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The question has been answered in the stackexchange maths forum, here, thanks to the collaboration of three kind users. The details about how the solution was given are worth to see.

This is the transformation from the coordinates with that line element to cartesian:

$$ \begin{cases} \bar x=\frac{1}{2}xy\cos(\ln(y/x))-\frac{1}{2}xy\sin(\ln(y/x)) \\ \bar y=\frac{1}{2}xy\sin(\ln(y/x))+\frac{1}{2}xy\cos(\ln(y/x)) \\ \end{cases} $$

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