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...

  • $\begingroup$ I haven't worked it out, but my immediate suggestion would be to start with the line element in usual spherical coordinates and then relate the usual $d\theta$ (assuming you're using usual European or physicists' spherical coordinates) to $d\rho$. $\endgroup$ Mar 3, 2021 at 23:47
  • $\begingroup$ That sounds like a good idea! I thought of it, but relating $d\theta$ to $d\rho$ turned out a bit difficult and didn't want to lose much time on it without being sure... Now I'll definitely give it a second chance! $\endgroup$ Mar 4, 2021 at 11:04
  • $\begingroup$ On further reflection, I think you have the definition of $\rho$ wrong (and I suspect that the diameter of the sphere should be $a$, not the radius). Try using $\rho$ to be the distance from the north pole to the point in the tangent plane. $\endgroup$ Mar 4, 2021 at 18:17
  • $\begingroup$ Having done a bit of computation, I just do not believe the formula with either definition of $\rho$. I can get the first part easily with my suggested $\rho$, but the second does not come out right. It's quite possible I'm making a silly error. Where did this come from? $\endgroup$ Mar 4, 2021 at 20:20
  • $\begingroup$ Actually you're right! I emailed my professor regarding the definition of $\rho$ and it's actually the distance from the north pole to the intersection point IN THE TANGENT PLANE, so you're right. I still weren't able to finish the exercise though, so if you were kind enough to let me know how did you went through your computation, the help would be much appreciated. Thank you very much! $\endgroup$ Mar 5, 2021 at 9:43

1 Answer 1


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.

  • $\begingroup$ Thank you very much! This was of much help!! $\endgroup$ Mar 5, 2021 at 19:21

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