Geodesics on the twisted pseudosphere (Dini's surface)

I wonder how difficult it is to compute geodesics on Dini's Surface, a twisted pseudosphere?

Here is one parametrization, from Alfred Gray's Modern Differential Geometry of Curves and Surfaces, p.495: \begin{eqnarray} x(u,v) &=& a \cos (u) \sin (v)\\ y(u,v) &=& a \sin (u) \sin (v)\\ z(u,v) &=& a \left[\cos (v)+\log \left(\tan \frac{v}{2}\right)\right]+b u \end{eqnarray} Dini's surface has constant curvature of $\frac{-1}{a^2+b^2}$.

And here is an image, for $a=1,\; b=\frac{1}{12}$, with $u \in [0,8\pi]$ (The curve defined by $v=\pi/2$ is shown green): What I especially wonder is if there is a geodesic that spirals down through every turn, which would be kinda cool. :-)

The green curve below might be one of Robert Bryant's geodesics—the computation is complicated enough that I am quite uncertain. Have to leave it there for the nonce... • The first and second fundamental forms are known, Geodesic curvatures of twisted and untwisted surface geodesics remain the same.This could help to chart their course.But for an untwisted axisymmetric case a single geodesic Clairaut's constant $r \sin \psi = r_o$ should be given. Depending on this constant and start angle at cuspidal equator there are two ways how geodesics propagate on negative surfaces 1) oscillating/returning geodesics and 2) geodesics asymptotically approaching $r_o$ This behavior is invariant in twisted surface as well. – Narasimham Jul 6 '16 at 14:27

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t>0$ and $b=r\sin t>0$ and make the change of variables $$v=\arcsin\bigl((\tan t)(\tan \theta)\bigr) \qquad\text{and}\qquad u= \frac{\log\rho - f(\theta)}{\sin t}$$ where $0<\theta< \tfrac12\pi{-}t$ and $0 < \rho < \infty$ and where $f$ (an elementary function, but not a nice one, apparently) is defined on $0<\theta<\tfrac12\pi{-}t$ so that $$f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta},$$ then the induced metric on the lower nappe of the surface (which is what Joseph drew) becomes $$ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right),$$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

• I am happy to learn the word nappe!: "1. a sheet of rock that has moved sideways over neighboring strata as a result of an overthrust or folding. 2. A sheet of water flowing over a dam or similar structure." – Joseph O'Rourke Nov 26 '13 at 2:13
• @Joseph: Well, that wasn't what I had in mind when I used the word, but, OK. If you like words, might I suggest that Dini's surface, rather than being described as a 'twisted pseudosphere', might be more aptly described as 'unfurled'. – Robert Bryant Nov 26 '13 at 4:35
• Thank you, Robert! At some point in the future, I'd like to compute one of the geodesics curling down to $z=-\infty$, just for the aesthetic pleasure. – Joseph O'Rourke Nov 26 '13 at 11:52
• This must be a problem with choosing the right branch of some analytic continuation. Obviously, $f'$ is real in the range in question, so its integral will be, too. On the other hand, I see now, that I was wrong when I said that $f$ is not elementary. I made an incorrect change of variables when I attempted to integrate the given $f'$ and wound up with an elliptic integral. I'll fix this. – Robert Bryant Nov 27 '13 at 4:29
• I think you get the right branch (even in Mathematica) if you integrate $f'$ from $\theta$ up to $\pi/2-t$. When I do this (here $h=\theta$), I get the expression (in Mathematica notation) f[h_]:=1/2 (Log-2 Log[Sqrt Cos[h] + Sqrt[Cos[2 t] + Cos[2 h]]] + 2 Log[Sin[t]] + Abs[Cos[t]] (2 ArcTanh[(Sqrt Abs[Cos[t]] Cos[h])/Sqrt[Cos[2 t] + Cos[2 h]]] - Log[-Sec[t] Sqrt[-Sin[2 t]]] + Log[Sec[t] Sqrt[-Sin[2 t]]])) and this is a real function for $0<h<\pi/2-t$. This $f$ diverges at 0 (c.f. the "unreachable" geodesic on the curve $\theta=0$) and vanishes at $\pi/2-t$ (the cusp). – j.c. Nov 27 '13 at 14:06