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When curvature and torsion are given a curve is fully defined (upto Euclidean motions) in 3-space.

$ k=const , \tau = 0 $ represents a circle in a plane ;

But what does the space curve

$$ k =0 , \tau= const,$$

represent?

The center line $ (u=0) $ of a right handed twisted helicoid with parametrization $( u \cos v, u \sin v, c \;v ) $ is a good example.

Curvature/Torsion of $u=0$ line of helicoid.

Clearly u=0 is a straight line at the helicoid mid with zero curvatures ( both normal (asymptotic $ k_n=0$) and geodesic $k_g=0$ ) as valid for a full straight line.

Using Enneper-Beltrami theorem torsion of the central parametric line at $\; u=0$ is found constant:

Evaluating Gauss curvature K

$$ K= \dfrac{-c^2}{(c^2+u^2)^2}, \tau = \sqrt{-K}= \pm \dfrac {1}{c}$$

The sign for the torsion of right helicoid is positive and, for the left handed helicoid it is negative.

A physical example is of a long human hair that can be twisted right or left with constant torsion even if the twist is not clearly visible. Other examples include long straight portions of DNA and other polymer molecules which inhabit such a surface.

EDIT1:

In another example the straight line parameterized by $$(x,y,z)= (a, b t, c t) $$ has zero curvature and non-zero torsion in this example when it becomes asymptotic on certain (arbitrary?) surfaces surfaces of negative Gauss curvature.

Moda

EDIT2:

What I meant by torsion without curvature is shown in the first figure Twist of Helicoid's straight/geodesic Spine. The special asymptotic line intrinsically characterizes how twist occurs during parallel transport in tangent spaces.

Narasimham
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    Curvature is zero only for any straight line curves. – Somos Dec 05 '19 at 23:52
  • Given are : curvature and torsion – Narasimham Dec 06 '19 at 00:00
  • You have not contradicted my comment. Give one example of a curve with zero curvature and is not a straight line. – Somos Dec 06 '19 at 00:17
  • I am looking for a curve whose both scalar properties are specified. – Narasimham Dec 06 '19 at 00:23
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    If curvature is zero, then torsion is undefined and is useless since the curve must be a straight line. – Somos Dec 06 '19 at 00:24
  • The question needed to be posed better involving surface embedment. For time being I am closing it . And thanks anyway for your comments. – Narasimham Dec 07 '19 at 11:12
  • @Somas: How should then the fundamental theorem be stated for generality? In special case when $\kappa=0, $ torsion $\tau $ should not be specified at all? – Narasimham Dec 07 '19 at 12:56
  • Exactly. You can not specify what which does not exist. – Somos Dec 07 '19 at 15:05
  • The theorem implies that for real $\tau, \kappa$ (both exist, as real independent functions) which together define a 3D space curve. In fact I was misled by that statement that does not exclude or warn about a zero $\kappa$ case. If it seems trivial/pedantic , sorry. And, thanks once again. – Narasimham Dec 07 '19 at 16:10
  • Yes, you were misled. It is not pedantic. The whole theory of curvature and torsion of "curved lines" is based on the implicit assumption that curvature is not zero except for isolated points. This should have been stated explicitly. – Somos Dec 07 '19 at 17:30
  • Thanks to you.. – Narasimham Dec 07 '19 at 17:33
  • Let us continue this discussion in chat. – Somos Dec 07 '19 at 18:12
  • If curvature is zero, then torsion is undefined and is useless since the curve must be a straight line. No !. When normal and geodesic curvatures of a straight line on a surface vanish.. it has its torsion well defined and is determined by the rotation of its bi-normal. Do you agree? – Narasimham Feb 03 '21 at 18:10
  • Actually that was the crux of my question. – Narasimham Mar 22 '21 at 08:24
  • Now you are asking about "geodesic curvature". The question was about a "space curve". Not a curve on a surface. I suggest you ask a different question if you wish to pursue this matter. – Somos Mar 22 '21 at 11:09
  • Sorry if that was still not clear. The picture also shows a (tube) straight line. on a surface where $k_g, k_n $ etc. are both zero, in $\mathbb R^3 $ like a laser through vacuum. – Narasimham Mar 23 '21 at 13:54
  • In you EDIT 2, you link to the torsion tensor, which is not the same thing as the torsion of a curve. – Didier Apr 24 '21 at 10:08
  • I was referring particularly to the geodesic torsion sketched in the figure. – Narasimham Apr 24 '21 at 10:18

3 Answers3

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If a 'curve' $\gamma:\>s\mapsto{\bf r}(s)$ is parametrized with respect to arc length then by definition $\kappa(s):=|\ddot{\bf r}(s)|$. The assumption $\kappa(s)\equiv0$ then implies $\ddot{\bf r}(s)\equiv{\bf 0}$, and this leads to ${\bf r}(s)={\bf a}+s {\bf u}$, where ${\bf u}$ is a unit vector.

Now, Torsion measures the failure of a curve to be planar. If $\gamma$ has zero torsion, it lies in a plane. Hence for $\kappa=0 \implies \tau = 0$ corresponding to a line. Lines look very much like lines, and they are certainly planar.

2

The whole theory of curvature and torsion of "curved lines" is based on the implicit assumption that the curvature is not zero except for isolated points. It is zero on a segment if and only if the curve segment is a straight line.

As the Wikipedia article Torsion of a curve states:

Let $\bf{C}$ be a space curve parametrized by arc length $s$ and with the unit tangent vector $\bf{t}$. If the curvature $\kappa$ of $\bf{C}$ at a certain point is not zero then the principal normal vector and the binormal vector at that point are the unit vectors $$ \bf{n}=\frac{\bf{t}'}{\kappa}, \quad \bf{b}=\bf{t}\times\bf{n}, $$ where the prime denotes the derivative of the vector with respect to the parameter $s$. The torsion $\tau$ measures the speed of rotation of the binormal vector at the given point. If is found from the equation $$ \bf{b}' = \tau\,\bf{n}. $$

Note carefully that the definition of $\bf{n}$ involves dividing by the curvature. Hence, if the curvature is $0$, $\bf{n}$ is not defined. Because this vector is not defined, this implies that the torsion can not be defined either since its definition uses $\bf{n}$.

Of course, you could just pick any fixed unit vector $\bf{n}$ which is perpendicular to the tangent $\bf{t}$ and then define $\bf{b}$ the usual way which leads to a constant binormal and hence by the definition of torsion we find $\,\tau=0.\,$ Obviously, the normal vector $\,\bf{n}\,$ is not unique, but in any case torsion is forced to be $\,0.\,$

Somos
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  • To me $(\tau,\kappa)$ appear to be two sides of the same (curvature) coin. In no way is the latter heavily tilted over the former. Had that been the case, the Fundamental Theorem should state an asymmetric situation somewhat like " A 3d curve is defined by $ \kappa(s), \tau(s) $ (barring Euclidean motions) as long as $\kappa$ is non-zero, in which case $\tau$ is indeterminate... This to me does n't sound good. But I can of course always be corrected. Strictly speaking Frenet-Serret have it that $\dfrac{t'}{\kappa} =\dfrac{b'}{\tau}=n $. – Narasimham Dec 08 '19 at 18:38
  • @narasimham Believe what you like. Facts matter. – Somos Dec 08 '19 at 19:29
  • http://webmath2.unito.it/ and other sources also exclude $ \kappa=0$ for 3d curves,needing to concede to the classic situation of course. Maybe I got overly weighed in by asymptotic curves on a surface $\kappa_n= 0$ existing for any prescribed $\tau_g$...( But then I can't understand why $\tau_n$ is meaningless in 3d. Anyway it needs more read/understand :)... – Narasimham Dec 08 '19 at 20:30
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This is a late response. But I wanted to point out that this question is not entirely devoid of merit. These are subtle pathological cases that arise in modeling problems in kinematics, mechanics and curve design and therefore justifiably dismissed by mainstream mathematicians. The notion of a twisted straight line with zero curvature but non zero torsion is mentioned in Nutbourne and Martin's book "Differential geometry applied to curve and surface design". See this link here. I have a copy of the book. Maybe it would be unwise to use the Frenet frame to tackle this issue.

But as an aside, if one takes a circular helix of radius $r$ and pitch $p$, then as $r \to 0$, it is easy to observe that the curvature goes to $0$, but the torsion goes to $1/p$. Indeed the straight axis of the helicoid can probably be treated this way.

Vishesh
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  • Sorry, I could n't reply before. When asking the question I had in mind a special generator of a right helicoid $ (u \cos(t), u \sin(t), v t) $ that you in other words referred to in second para of your comment. Center line $ u=0 $ of helicoid has constant torsion and zero curvature. ( e.g., locus of a propeller center of an advancing aircraft ). – Narasimham Apr 29 '23 at 15:43
  • contd.. The line comes about as a special 1-parameter case out of this 2 parameter surface. Yes, it cannot be single parameter space curve definable through T,N,B of Frenet/Serret alone. In addition I wanted to know if there would be any necessary association with asymptotic lines and hyperbolic geometry. – Narasimham Apr 29 '23 at 15:43
  • @Narasimham I guess any straight line on a surface would be asymptotic and geodesic. But the book link does seem to say something about what you are interested in. – Vishesh Apr 30 '23 at 02:49
  • But not on any surface. Since a straight line cannot be contained on a surface of constant non-zero $K=\kappa_1 \kappa_2$ , the surface should necessarily be a plane. My question includes torsion that is implicitly involved in this exceptional case and on surfaces similar to it – Narasimham Apr 30 '23 at 07:38