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$ \kappa = \cos s, \tau = \sin s $ and passing through (1,0,0), TNB triad identity matrix?

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When numerically computed it looks like a catenoid surface of revolution for all appearances. ( unit curvature at extremes and like an symptote at symmetrical plane). Its parametrization is not mentioned in Differential Geometry books that I could come across.

EDIT 1:

  1. Can it at least be proved that the line lies on a surface of revolution?

  2. In general for a surface of revolution, what functional relation/s can exist between $ \kappa, \tau $ ?

EDIT 2:

Initial values for (T,N,B) taken as an identity matrix to fix rotation.

EDIT 3:

The lines given by achille hui are described on an equiangular rotational hyperboloid inclined at $\pi/4$ to one axis, result of initial point $(0,0,0)$ and starting $T,N,B$ an identity matrix.

May I request someone to extend it further or integrate the locus incorporating constants for generality ? like

$$ (a \kappa, a \tau) = ( \cos s/b , \sin s/b) $$

An image of a hyperboloid of revolution for $ (a= 1/6, b =2 ) $:

enter image description here

Motivation of this post was in fact to readily plot these lines on a surface of revolution using scalar curvatures properly.

EDIT 4:

In hindsight it slightly puzzles (me)... Although we started off with two intrinsic curvature invariants in 3-space, when combined it forces the arc to belong to 1- sheet hyperboloid $ of \, revolution $ in $ \mathbb R^2 $ that was neither specified in the input nor expected (by me). Is there some explanation or insight for it to so happen?

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    $\begingroup$ What do you mean? There are uncountably many such space curves with different functions $\kappa$, $\tau$ satisfying the equation $\kappa^2 + \tau^2 =1$ $\endgroup$ – Robert Israel Jan 7 '16 at 21:58
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Instead of a curve passing through $(1,0,0)$, choose the coordinate system such that at time $t = 0$, the curve start at $(0,0,0)$ with $\vec{T}, \vec{N}, \vec{B}$ in the direction of $x, y, z$ respectively.

It turns out for $\kappa(s) = \cos(s), \tau(s) = \sin(s)$, one can convert the ODE for Fernet-Serret frame to an ODE over quaternion and solve it explicitly. The derivation is pretty messy and I won't reproduce it here. Feel free to throw the expression of space curve below to a CAS and verify it works.

In any event, the space curve $\vec{X}(s) = (x(s),y(s),z(s))$ is given by the formula:

$$\begin{cases} x(s) &= \frac{3}{\sqrt{2}}\sin(\sqrt{2}s)\cos(s) - 2\cos(\sqrt{2}s)\sin(s)\\ y(s) &= -\frac12(3\cos(\sqrt{2}s) + 1)\cos(s) - \sqrt{2}\sin(\sqrt{2}s)\sin(s) + 2\\ z(s) &= +\frac12(3\cos(\sqrt{2}s)-1)\cos(s) + \sqrt{2}\sin(\sqrt{2}s)\sin(s) - 1 \end{cases} $$

With a little bit of algebra (or an CAS), one can verify this curve lies on a one-sheet hyperboloid:

$$x^2 - 2(y-2)(z+1) = 4$$

Update

For the general case $\kappa(s) = \rho\cos(\alpha s), \tau(s) = \rho\sin(\alpha s)$ where $\rho, \alpha$ are constants, if I didn't make any mistake, the space curve is given by

$$\begin{cases} x(s) &= \frac{\rho^2}{2\beta} S(s)\\ y(s) &= y_0 -\frac{\rho^2}{2\beta^2}\left( \frac{2}{\rho}\cos(\alpha s) + \rho C(s)\right)\\ z(s) &= z_0 -\frac{\rho^2}{2\beta^2}\left( \frac{2}{\alpha}\cos(\alpha s) - \alpha C(s)\right) \end{cases} $$ where $\beta = \sqrt{\alpha^2+\rho^2}$, $y_0 = \frac{2}{\rho}$, $z_0 = \frac{\rho^2-2\alpha^2}{a\rho^2}$ and $\displaystyle\; \begin{cases} C(s) &= \frac{\cos((\beta-\alpha)s)}{(\beta-\alpha)^2} + \frac{\cos((\beta+\alpha)s)}{(\beta+\alpha)^2}\\ S(s) &= \frac{\sin((\beta-\alpha)s)}{(\beta-\alpha)^2} + \frac{\sin((\beta+\alpha)s)}{(\beta+\alpha)^2}\\ \end{cases} $.

Furthermore, the space curve lies on the one-sheet hyperboloid:

$$\rho^2\beta^2 x^2 + \rho^2(\alpha(z-z_0) - \rho(y-y_0))^2 - \alpha^2(\rho(z-z_0) + \alpha(y-y_0))^2 = \frac{4\alpha^2\beta^2}{\rho^2}$$

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  • $\begingroup$ In the equation for $y(s)$, should it be $3\cos(\sqrt2s)$ ? $\endgroup$ – bubba Jan 9 '16 at 2:58
  • $\begingroup$ @bubba, fixed. it should be $s$. $\endgroup$ – achille hui Jan 9 '16 at 3:26
  • $\begingroup$ Nice and important result this : $ \kappa(s) = \rho\cos(\alpha s), \tau(s) = \rho\sin(\alpha s) $..a one sheet hyperboloid of revolution. May be also connected with hyperbolic geometry. $\endgroup$ – Narasimham Jan 10 '16 at 10:05
  • $\begingroup$ Can $ \kappa(s) = \rho\sin(\alpha s), \tau(s) = \rho\cos(\alpha s) $ lead to a similar result? $\endgroup$ – Narasimham Feb 12 '16 at 18:15
  • $\begingroup$ @achillehui math.stackexchange.com/questions/1721396/… Do you think I am right about relevance of Vishesh's question here? $\endgroup$ – Narasimham Mar 31 '16 at 11:20
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Since curvature and torsion are invariant under Euclidean transformations, you can take any curve with given $\kappa(s), \tau(s)$ and revolve around any line, obtaining a surface of revolution made up of curves with the same $\kappa(s)$ and $\tau(s)$. But given an initial point on the curve where, to keep things simple, $\kappa$ and $\tau$ are nonzero, the initial $(T,N,B)$ could point in any three orthogonal directions, so the curves don't all lie on a surface, they fill up a region of space.

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  • $\begingroup$ Yes, forgot mentioning the assumed initial $(T,N,B) $ identity matrix also, as the fundamental theorem freezes the rigid/intrinsic equation defined object with the two Euclidean motions arrested. I like to look at the simplest/elegant parametrization in closed form, am really indifferent to any convenient initial conditions that may be assumed. $\endgroup$ – Narasimham Jan 8 '16 at 21:15

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