Let $w=\frac1{\sqrt3}(1,-1,1)$. Find the area of the spherical circle C($w, \frac{\pi}6$) under the stereographic projection $\phi(x,y,z)=\left(\frac{x}{1-z}, \frac{y}{1-z}\right)$.

Unfortunately, I have no idea how to do it. Would appreciate some advice.


The image of the circle is an ellipse and to find its area we just need to find its axes. To make things easier, it is convenient to rotate circle $C$ by 45° around $z$-axis, in order to have its axis on the $x-z$ plane: $w={1\over\sqrt3}(\sqrt2,0,1)$. The center of the circle is then $o=w\cos{\pi\over6}={1\over2}(\sqrt2,0,1)$ and its radius is $\sin{\pi\over6}={1\over2}$.

Any point $p$ of the circle can be written as $p=o+r$, where $r$ is any vector perpendicular to $w$ and of length ${1\over2}$. In general, we can write $r$ as $r=(t,s,-\sqrt2t)$, where $s$ and $t$ must satisfy $3t^2+s^2={1\over4}$ in order to have $||r||={1\over2}$. We have then: $$ p=\left({\sqrt2\over2}+t,\,s,\,{1\over2}-\sqrt2t \right), \quad\hbox{with:}\quad 3t^2+s^2={1\over4}. $$ The stereographic projection of $p$ is then $$ p'=\left({\sqrt2/2+t\over1/2+\sqrt2t},\,{s\over1/2+\sqrt2t}\right). $$ To find the endpoints of the major axis of the projected ellipse is easy, because they lie on $x$-axis and correspond to points on the circle with maximum and minimum value of $z$, which is to say with maximum and minimum value of $t$. Plugging $t=\pm{1\over\sqrt{12}}=\pm{1\over2\sqrt3}$ into the formula for $p'$ yields then $$ x_{\max-\min}=2\sqrt2\pm\sqrt3, $$ which corresponds to a semi-major axis of $\sqrt3$ and a center of the ellipse located at $(2\sqrt2,0)$.

To find the major axis, observe that its endpoints have the same $x$ as the center of the ellipse, and $x=2\sqrt2$ corresponds to $t=-{1\over3\sqrt2}$ and consequently to $s=\pm{1\over2\sqrt3}$. Plugging these into the formula for $p'$ yields $$ y_{\max-\min}=\pm\sqrt3. $$ It turns out that semi-minor axis is $\sqrt3$ too, and our ellipse is just another circle! Its area, of course, is then $3\pi$.


The above answer is more involved than necessary, as it turns out that the image of a circle under stereographic projection is always a circle. It is then enough to find the endpoints of that diameter of the circle on the sphere which is the intersection of the circle with the plane passing through $z$-axis and the center of the circle. The images of those points are the endpoints of a diameter of the image circle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.