3
$\begingroup$

For any smooth continuous surface it is known that only lines of curvature $ (\psi=0, \pm \pi/2, \pi...) $ have zero geodesic torsion, ( time being keeping aside great circles on a sphere) :

$$\tau_g= (\kappa_1- \kappa_2) \sin \psi \cos\psi $$

For a minimal surface we have $$\kappa_1<0,\,2 H =(\kappa_1+ \kappa_2)=0, \, \tau_g = H\, \sin 2 \psi =0 $$

So on a minimal surface all geodesics have zero Torsion, a result extremely easy to establish.

For torsion

$$\tau_g= (\kappa_1- \kappa_2) \sin \psi \cos\psi \rightarrow (-\kappa_1- \kappa_2) \sin \psi \cos\psi, $$

and likewise when the latter vanishes we have two real possibilities for Torsion-free geodesics on minimal surfaces:

$$ H=0 ,\tau_g=0, \kappa_g=0 $$

Since $H$ is a factor of $\tau_g$ both vanish together.

Torsionless geodesics on a catenoid of revolution:

CatnGeodTorsnZero

Is this statement true in general? If so, what are Differential Geometry textbook references? If not, what could be conceptually incorrect in this context?

Thanks for your comments or answers.

They are indeed not torsionless. However, not deleting background of this, what appears to be a pitfall.

For normal curvature

We have all possible curvatures $(\kappa_n, \tau_g) $ positive and negative combinations nicely depicted on a Mohr's circle scalar diagram applicable for either positive or negative signs of Gaussian curvature $ K=\kappa_1 \kappa_2$.

$$ \kappa_n= (\kappa_1\cos^2\psi + \kappa_2\sin ^2\psi ) \rightarrow (-\kappa_1 \cos^2\psi + \kappa_2\sin ^2\psi) ,$$

When the latter vanishes we have two real directions of asymptotes defined by $ \psi_{1,2}= \tan^{-1}\sqrt{\kappa_1/\kappa_2}$.The parameter lines are either parallel or perpendicular to arc for principal directions of curvature, defined by this angle.

So changing sign of $\kappa_1$ in this manner is a valid.

In addition, these are valid for $\tau$ depiction in bi-polar coordinate representations relevant to curvature. iso-$\tau$ bipolar coord patches

Representation of $(\kappa_n, \tau_g ) $ for negative Gauss curvature is sketched below using Mohr's circles of curvature.

H=0 & geodesics seems to discuss the topic but not clear where it is of direct relevance.

EDIT1:

K>0 & K<0 Mohr Circles

In this discussion about sign of curvature,combination of geodesics and new a scalar curvature $ \kappa_{n Meta} $ I am defining as:

$$ \kappa_{n Meta} = -\kappa_1 \cos^2 \psi + \kappa_2 \sin^2 \psi $$ result in revolved cusped surface Barrels ( different initial conditions) imaged below.They have an important common invariant $ a^2= K\cdot r^4$ whose significance is not yet clear to me.

knMeta_Geods_SOR

$\endgroup$
  • 4
    $\begingroup$ You made a sign mistake (who didn't do it at least once?). In the formula for the geodesic torsion we have the difference of principal curvatures, and in the mean curvature their sum. $\endgroup$ – Ivan Izmestiev Jun 2 '17 at 9:42
  • $\begingroup$ Suggestion: it is really hard to figure out what exactly is your question here. You ask "Is this statement true in general?" But I am entirely unsure to which statement you refer. And how general is general? Are you looking for statements about curves in (minimal) submanifolds of a manifold? $\endgroup$ – Willie Wong Jun 9 '17 at 18:26
  • $\begingroup$ Statement put in bold face. For minimal surfaces at least in R^3 whether geodesics have no torsion? $\endgroup$ – Narasimham Jun 9 '17 at 21:15
  • 2
    $\begingroup$ Seems to me that even on the catenoid, most geodesics have torsion. Think about taking the circular geodesic about the neck and peturbing the starting direction a small amount - the resulting geodesic should look locally like a helix. $\endgroup$ – Anthony Carapetis Jun 13 '17 at 5:28
  • 1
    $\begingroup$ Yes, principal curvatures change the sign if you take the opposite unit normal. But they do this simultaneously. $\endgroup$ – Ivan Izmestiev Jun 14 '17 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.