Harald Kirsch

genug Unfug.


Eskimo Mites (I.5.2, p. 79)*



In this exercise we are asked to prove that the coordinate transformation $$\begin{align} x &= \theta \sin(\phi) \label{eq1}\\ y &= \theta \cos(\phi) \label{eq2}\\ \end{align}$$ leads to the line segment \begin{equation} \label{solution} ds^2 = (1-y^2/3)dx^2 + (1-x^2/3)dy^2 + 2/3\, xy\, dx\, dy+\dots \end{equation} In addition we know that \begin{equation}\label{ds2} ds^2 = d\theta^2 + \sin^2(\theta)d\phi . \end{equation}


To get the $ds$ in terms of $dx$ and $dy$, we need to solve for the latter. Start by squaring \ref{eq1} and \ref{eq2} and adding them to see that \begin{equation*} x^2+y^2 = \theta^2(\sin^2(\phi)+\cos^2(\phi)) =\theta^2 \end{equation*} or \begin{equation} |\theta| = \sqrt{x^2+y^2} \end{equation} Differentiation results in \begin{align} d\theta &= \nabla(\sqrt{x^2+y^2}) \pmatrix{dx\cr dy} \\ &=\frac{1}{2\sqrt{x^2+y^2}}(2x,2y)\pmatrix{dx\cr dy}\\ &=\frac{x\,dx + y\,dy}{\sqrt{x^2+y^2}}\\ &=\frac{x\,dx + y\,dy}{|\theta|}. \end{align} To solve for $d\phi$ divide \ref{eq1} by \ref{eq2} and differentiate both sides, using the quotient rule: \begin{align} d\frac{x}{y} &= d\frac{\sin(\phi)}{\cos(\phi)}\\ \frac{dx\,y - x\,dy}{y^2} &=\frac{(\cos^2(\phi)+\sin^2(\phi)) d\phi}{\cos^2(\phi)} \end{align} From equation \ref{eq2} we see that $\cos(\phi)=y/\theta$, plug it in into the denominator and solve for $d\phi$: \begin{equation} d\phi = \frac{y\,dx - x\,dy}{\theta^2} \end{equation} Now that we have $d\theta$ and $d\phi$ we can use it in equation \ref{ds2}, which will become slighly cumbersome: \begin{equation}\label{awkward} ds^2 = \frac{(x\,dx + y\,dy)^2}{\theta^2} +\sin^2(\theta) \frac{(y\,dx - x\,dy)^2}{\theta^4} \end{equation} From the hint in the excercise text that we should work out up to 2nd orders, we can speculate that a power series expansion of $\sin^2(\theta)$ needs to be used, namely $\theta^2 -\theta^4/3 + O(\theta^6)$. We see the suspicious $1/3$ appear already. We insert it into equation \ref{awkward}: \begin{align} ds^2 &= \frac{(x\,dx + y\,dy)^2}{\theta^2} + (\theta^2 -\theta^4/3 + O(\theta^6)) \frac{(y\,dx - x\,dy)^2}{\theta^4}\\ &= \frac{(x\,dx + y\,dy)^2}{\theta^2} + (1 -\theta^2/3 + O(\theta^6)) \frac{(y\,dx - x\,dy)^2}{\theta^2}\\ \end{align} Consider first the parts with a $\theta^2$ in the denominator: \begin{align} \qquad &(x\,dx + y\,dy)^2 + (y\,dx - x\,dy)^2\\ &= (x\,dx)^2 + x y\,dx\,dy + (y\,dy)^2 +(y\,dx)^2 - x y\,dx\,dy + (x\,dy)^2\\ &= (x^2+y^2)(dx^2+dy^2)\\ &= \theta^2(dx^2+dy^2) \end{align} We can reduce the $\theta^2$ with the denominator and arrive at: \begin{align} ds^2 & = dx^2+dy^2 + (-\theta^2/3 + O(\theta^6))\frac{(y\,dx - x\,dy)^2}{\theta^2}\\ &= dx^2+dy^2 +\frac{-y^2 dx^2 +2xy\,dx\,dy - x^2 dy^2}{3}\\ &\qquad+ (y\,dx - x\,dy)^2 O((x^2+y^2)^2)\\ &= (1-\frac{y^2}{3})dx^2 + (1-\frac{x^2}{3})dy^2 +2/3 xy\,dx\,dy \\ &\qquad+ (y\,dx - x\,dy)^2 O((x^2+y^2)^2) \end{align} We see that this is the solution as stated in equation \ref{solution}. The last term in the sum has only powers of $x$ and $y$ greater $4$.