Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the $xz$-plane of $\mathbb R^3$, consider the closed non-singular curve $\gamma$ which is the image of the function $$t\mapsto (1+2\sin^2(t))(\cos(t),0,\sin(t)).$$ (Note that $\gamma$ is invariant under $\pi$ rotation about the $z$-axis.)

Form a surface of revolution $S\subset \mathbb R^3$ by revolving $\gamma$ about the $z$-axis. I am curious about the behavior of all unit speed geodesics on $S$ (with respect to the Riemannian metric on $S$ induced from the usual Euclidean metric on $\mathbb R^3$) which pass through the point $(\sqrt2,0,\sqrt2)$.

Looking forward to see your opinions. Thanks in advance.

share|cite|improve this question
up vote 1 down vote accepted

The behavior is pretty much the same as that for the torus of revolution. Your curve in the $xz$ plane is $$ (x^2 + z^2)^3 = (x^2 + 3 z^2)^2. $$ As you have doubtless noticed, there are horizontal geodesics on the surface of revoltuion at $z = 0, \sqrt 2, \; - \sqrt 2.$

The surface is sort of a dumbbell or peanut shape. If a geodesic leaves $z = \sqrt 2$ with $z$ increasing, the curve goes up and then down, intersecting $z = \sqrt 2$ in a downwards direction with the same angle with which it left. This is one outcome of Clairaut's relation.

Next, if a geodesic leaves $z = \sqrt 2$ at a downwards angle $\alpha,$ and intersects $z = 0,$ it does so with a nonzero angle $\beta,$ continues on to hit $z = -\sqrt 2$ with angle $\alpha,$ bounces back up to cross $z = -\sqrt 2$ upwards with angle $\alpha,$ hits $z = 0,$ it does so with angle $\beta$ again, reaches and crosses $z = \sqrt 2$ at an upwards angle $\alpha,$ and so on forever. For a countable set of angles $\alpha$ the curve runs over the same path over and over again.

If a geodesic leaves $z = \sqrt 2$ at a small angle $\alpha,$ the curve does not intersect $z = 0.$ Instead, the curve reaches a minimum $z$ value, then starts back up. This is shown by Clairaut's relation in terms of the radius $r$ away from the $z$ axis, see below. The relation that gives the minimum value of $r$ is $$ \sqrt 2 \; \cos \alpha = r_{\mbox{min}} \; . $$ As long as $\sqrt 2 \; \cos \alpha > 1$ this is what happens.

In between, there is a critical angle, call it $\alpha_0,$ such that a geodesic leaving $z = \sqrt 2$ downwards at angle $\alpha_0$ neither reaches a minimum $z$ value nor does it reach $z=0.$ Instead, the geodesic wraps around and around $z=0,$ getting closer and closer to it. A picture for this behavior is Exercise 18 and figure 4-22, pages 262-263 of Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo. Clairaut's relation is on page 257. A picture of the case when $z$ reaches a minimum is on page 259. I see; the critical angle $\alpha_0 = \frac{\pi}{4},$ also by Clairaut's relation. How easy! The relation for a geodesic leaving $z = r = \sqrt 2$ at a downwards angle $\alpha_0 = \frac{\pi}{4}$ is $$ r \cos \theta = 1. $$ If this geodesic ever reached the geodesic at $$ z=0, \; \; r = 1, $$ it would be with angle $\theta = 0,$ but it is not possible for distinct geodesics to be tangent. So $r > 1$ and $z > 0.$ This says that $\theta$ is never $0,$ and by continuity keeps the same $\pm$ sign, the geodesic continues travelling downwards, never achieving a minimum $z.$ Somebody put a good deal of effort into constructing this, making the numbers tractable.

The differing behavior is based on the difference between positive and negative Gauss curvature. In positive curvature, geodesics parting an intersection at an angle then get closer together compared with a pair of rays in the Euclidean plane and meet again later. In negative curvature, they diverge farther apart compared with rays in the plane. The magic word for this behavior is something called Toponogov's Theorem. For positive curvature, it is definitive. For negative curvature in a surface with no closed loops at all, it is also definitive. The very important exception is the subject of Kleinian groups and orientable surfaces of higher genus with constant curvature $-1.$

In this surface, Gauss curvature switches between positive and negative, with positive $z,$ at $$ z = 0.343145750..., \; \; \sqrt{x^2 +y^2} = 1.120193731... $$ which is not a geodesic.

I met do Carmo once at a conference. He later sent me a copy of Celso Costa's dissertation on minimal surfaces.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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