How to parametrize circles on a sphere by the distortion of the equator?

Question

I guess am having a very silly problem right now. Considering a unit sphere $S^2$ and, for example, a curve, in spherical coordinates, $c(t)=(1, \frac{\pi}{2},t)$ that goes around the equator how can I parametrize a family of curves $c(t,s)$ which correspond to rotations of the equator and satisfy $c(0,s)=c(0)$ and $c(\pi,s)=c(\pi)$?

Answer 1

You know that for a suitable choice of coordinate system, spherical coordinates $(r,\theta,\varphi)$ correspond to cartesian coordinates $(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)$. Specifically, for your equator you have cartesian coordinates $(\cos t,\sin t,0)$. Now you want to rotate that, while keeping $(1,0,0)$ and $(-1,0,0)$ fixed. So you rotate around the $x$ axis, and get something like $(\cos t,\sin t\cos s,\sin t\sin s)$. Now turn that back into spherical coordinates. You have $r=1$ and

\begin{align*} \cos\theta&=\sin t\sin s \\ \tan\varphi&=\frac{\sin t\cos s}{\cos t}=\tan t\cos s \end{align*}

So one possible parametrization would be

$$c(t,s)=\bigl(1,\arccos(\sin t\sin s),\arctan(\tan t\cos s)\bigr)$$

except that the $\arctan$ in there only determins an angle modulo $180°$, so it would often end up in the wrong quadrant. You'd have to take the same quadrant as $t$, and one hackish way to express this would be

$$c(t,s)=\bigl(1,\arccos(\sin t\sin s),\arctan(\tan t\cos s)+t-\arctan(\tan t)\bigr)$$

since $t-\arctan(\tan t)$ is exactly the difference between the actual angle $t$ and the angle computed by $\arctan$. You still have to deal with the infinities if $t=(2k+1)\pi/2$.

There might be some clever formula to combine the $\arctan$ and the $\tan$ inside, even though there is an extra factor inside. There might be a way to simplify things choosing a parametrization of the curve which is not constant speed. One might also try to simplify things by expressing the rotation not as an angle, but as something else, e.g. the tangent of half the angle or whatever. I haven't explored these options in depth.