0
$\begingroup$

Sine Gordon Equn for K>0

Continuing the above in this post parametric equations of a geodesic on psedosphere are derived to dispel/discuss a view that geodesics encountering boundary at the cuspidal equator of a pseudosphere surface in $(\mathbb R^3 )$ are confined and cannot exist beyond.

An intersting take from this exrcise is that the geodesics cut the circular cusp at polar angle $ \theta_{max}= \cot \alpha \,$ where $ \psi=\alpha.$

To find polar angle $\theta$

Considering differential length elements for any surface of revolution and applied for the tractrix in particular

$$ \sin \phi = \frac{ dr \tan \psi } { r d\theta} = r/r_o ;$$

on the pseudosphere of DE ($ \phi $ is slope angle between arc and symmetry axis, $\psi$ is angle geodesic makes to meridian, $r_o$ minimum radius where geodesics are tangential to a circle smaller than equator $ r_o<a $

$$ \sin \phi = \frac{ dr \tan \psi } { r d\theta} = r/r_o ;$$

Clairaut's Law, $r_o$ is minimum radius where lines are tangential and turn back on either side.

$$ r_o = r \sin \psi$$

plug into above and simplify

$$ \frac{d \theta}{a r_o}= \frac{dr}{r^2 \sqrt{r^2-r_o^2}} $$

Integrating with BC $ \theta=0 , r= r_o$ we have

$$\theta = a \sqrt{ 1/r_o^2 - 1/r^2 } \tag1$$

To find the $z$ coordinate

$$ \frac{dz}{dr}= \cot \phi = \frac{\sqrt{a^2-r^2} dr}{r} \tag2 $$

substitution

$$ r \cosh u= a , dr= sech u \, \tanh u \, dz= a \tanh^{2} du = a ( 1- sech^{2} u) du\tag3 $$

integrating with BC $ u=0, z=0$

$$ z = a \cosh^{-1}\frac{a}{r} - \sqrt{a^2-r^2} \tag4$$

The parametrization is

$$ (r,\theta,z) =( r, a \sqrt{ 1/r_o^2 - 1/r^2 }, \, a \cosh^{-1}\frac{a}{r} - \sqrt{a^2-r^2}) \tag5 $$

Visualization done in Mathematica also confirms that the geodesics enter the second mirrored nappe across the cuspidal equator, as it should, by virtue of $\pm$ in front of square root sign for $z,r$ co-ordinates.

Not a Confined Space before Cusp of pseudosphere

a = 1; ro = 0.18 a;
th[r_] = Sqrt[(a/ro)^2 - (a/r)^2]; 
z[r_] = a ArcCosh[a/r] - Sqrt[a^2 - r^2]
g1 = ParametricPlot3D[{r Cos[th[r]], r Sin[th[r]], z[r]}, {r, .1, 1}, 
   PlotStyle -> {White, Thick}];
g2 = ParametricPlot3D[{r Cos[th[r]], r Sin[th[r]], -z[r]}, {r, .1, 1},
    PlotStyle -> {White, Thick}];
g3 = ParametricPlot3D[{r Cos[th[r] + v], 
   r Sin[th[r] + v], +z[r]}, {r, .1, 1}, {v, -Pi, Pi}, 
  PlotStyle -> {Yellow, Opacity[0.95]}, Mesh -> {20, 20}]
g4 = ParametricPlot3D[{r Cos[th[r] + v], 
   r Sin[th[r] + v], -z[r]}, {r, .1, 1}, {v, -Pi, Pi}, 
  PlotStyle -> {Yellow, Opacity[0.95]}, Mesh -> {20, 20}]
Show[{g1, g2, g3, g4}, PlotRange -> All, Axes -> None, Boxed -> False]

Also iirc, work by Jacques Hadamard about geodesics on constant negative curvature surfaces was mentioned (in his biological sketch); can someone please point to the references?

$\endgroup$
  • $\begingroup$ My suggestion is to edit your questions in a separate tex editor and then post when ready. This way, there is no need to apologize. You can also delete your question, then edit and then hit "undelete" button. $\endgroup$ – Moishe Kohan Mar 5 '18 at 17:51
  • $\begingroup$ Thanks, done the above accordingly. $\endgroup$ – Narasimham Mar 6 '18 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.