In the book Elementary Differential Geometry, by Andrew Pressley, the author proves that a geodesic will cross every parallel of a surface of revolution. I'm not understanding why.
Let's parametrise the surface of revolution by $\sigma(u,v)=(f(u)\cos v,f(u) \sin(v), g(u))$.
Let the surface be at a distance $> \Omega$ from the axis of rotation.
Let $u_0$ be the least upper bound of u on a geodesic, and let $\Omega+2\epsilon$, with $\epsilon>0$, be the radius of the parallel at $u=u_0$. $\Omega$ is the 'angular moment' on the geodesic, which is shown to be constant everywhere in the geodesic.
We're able to deduce that for $u$ sufficiently close to $u_0$, we have $|\dot u| \geq \sqrt{ 1-\left(\frac{\Omega}{\Omega+\epsilon}\right)^2}>0 $.
How can we can state the geodesic (other than a meridian or a parallel) will cross every parallel with this information?
Best Answer
If the geodesic fails to cross every parallel, then either (a) it intersects a "maximum/minimum" parallel $u=u_0$ tangentially or (b) it approaches a parallel $u=u_0$ asymptotically. In either event, we must have $\dot u$ either equal to or approaching $0$ on the geodesic. This contradicts the positive lower bound on $|\dot u|$ given in the text.