13
$\begingroup$

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):
       DiniCurve
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...
       RobertGeodesic

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Narasimham Jul 6 '16 at 14:27
13
$\begingroup$

To answer Joseph's questions:

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 $z=+\infty$.

$\endgroup$
  • $\begingroup$ 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." $\endgroup$ – Joseph O'Rourke Nov 26 '13 at 2:13
  • $\begingroup$ @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'. $\endgroup$ – Robert Bryant Nov 26 '13 at 4:35
  • $\begingroup$ 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. $\endgroup$ – Joseph O'Rourke Nov 26 '13 at 11:52
  • 1
    $\begingroup$ 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. $\endgroup$ – Robert Bryant Nov 27 '13 at 4:29
  • 1
    $\begingroup$ 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]-2 Log[Sqrt[2] Cos[h] + Sqrt[Cos[2 t] + Cos[2 h]]] + 2 Log[Sin[t]] + Abs[Cos[t]] (2 ArcTanh[(Sqrt[2] 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). $\endgroup$ – j.c. Nov 27 '13 at 14:06

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.