We assign coordinates ($\rho, \phi$) to each point in the surface of a sphere, where $\rho$ is the distance from the south pole of the sphere to the point where a straight line passing through both the south pole and the point in question intersects the tangent plane to the north pole, and $\phi$ is the usual azimuthal angle. Show that the line element for the surface of the sphere in these coordinates is $ds^2=\frac{d\rho^2}{(1+\frac{\rho^2}{a^2})^2}+\frac{\rho^2d\phi^2}{1+\frac{\rho^2}{a^2}}$.
My attempt was as follows:
In cartesian coordinates, $ds^2=dx^2+dy^2+dz^2$. I placed the origin of coordinates in the center of the sphere and differenciated the equation defining the sphere (we assume radius a): $x^2+y^2+z^2=a$ to get $2xdx+2ydy+2zdz=0$. From this last one I got a constraint over $dz$ and I plugged it into the equation for $ds$, obtaining the following expression: $ds^2=dx^2+dy^2+\frac{(xdx+ydy)^2}{a^2-(x^2+y^2)}$.
Once there, I tried looming for a relation between x, y and $\rho$, which I found to be by using a bit of trigonometry and finding the intersection point of the sphere surface and the line coming from the pole: $x=(\frac{2a}{\rho})^2\sqrt{\rho^2-(2a)^2}cos(\phi)$ and $y=(\frac{2a}{\rho})^2\sqrt{\rho^2-(2a)^2}sin(\phi)$, but differenciating these and plugging them into the expression I had found for $ds^2$ leads me to some terrible algebra which I think is not taking me to the correct solution…
Best Answer
OK, this problem is a mess. Now that we have one of the standard stereographic projection mappings, we know that the mapping is conformal, and so the formula as given must be incorrect.
Let $a$ be the diameter of the sphere, and let $(\psi,\phi)$ be the usual spherical coordinates (with $\psi$ the angle from the vertical and $\phi$ the polar angle, as given in the problem). Then we all know that $$ds^2 = \big(\frac a2\big)^2(d\psi^2 + \sin^2\psi\,d\phi^2).$$ We observe that we have a right triangle with legs $a$ and $\rho$, and angle $\psi/2$ at the south pole. Thus, $$\rho = a\tan\big(\frac\psi 2\big) \quad\text{and}\quad \sin\psi = 2\sin\big(\frac\psi 2\big)\cos\big(\frac\psi 2\big)=\frac{2a\rho}{a^2+\rho^2}.$$ Now easy computation gives $$\frac{d\rho}{1+\big(\frac\rho a\big)^2}=\frac a2\,d\psi.$$ Thus, \begin{align*} ds^2 &= \big(\frac a2\big)^2(d\psi^2 + \sin^2\psi\,d\phi^2) = \frac{d\rho^2}{\big(1+(\frac\rho a)^2\big)^2} + \frac{\rho^2\,d\phi^2} {\big(1+(\frac\rho a)^2\big)^2} \\ &= \frac1{\big(1+(\frac\rho a)^2\big)^2}(d\rho^2 + \rho^2\,d\phi^2), \end{align*} which shows that the metric is conformally equivalent to the usual metric on the tangent plane at the north pole, as expected.