Can someone please explain this proof to me. I know that a circle is planar and has nonzero constant curvature, so this must be an exception, but I am a little lost on the proof. Thanks!

  • 3
    $\begingroup$ You are getting torsion and curvature confused I think. en.m.wikipedia.org/wiki/Torsion_of_a_curve $\endgroup$ – ClassicStyle Sep 20 '14 at 18:48
  • $\begingroup$ See Prove that curve with zero torsion is planar. Or is there a specific prove you don't understand? $\endgroup$ – Florian Sep 21 '14 at 6:20
  • $\begingroup$ There are no exceptions. Torsion is a movement out of the plane of curve. A helix is like drawing a circle, except instead of staying in the plane, it has torsion that brings it out of the plane spiraling outwards. $\endgroup$ – Kainui Sep 21 '14 at 8:13

A circle is indeed planar, and has constant nonzero curvature, but the torsion of a circle is zero; it's not an exception.

Having said this, let $\gamma(s)$ be a regular planar curve in $\Bbb R^3$ parametrized by arc length, say $\gamma:I \to \Bbb R^3$, where $I \subset \Bbb R$ is an open interval. $\gamma(s)$ regular means

$\dot{\gamma}(s) = \dfrac{d\gamma(s)}{ds} \ne 0 \tag{1}$

for any $s \in I$. If $P \subset \Bbb R^3$ is any plane passing through the point $\vec p_0 \in \Bbb R^3$, then the points $\vec r = (x, y, z) \in P$ satisfy an equation of the form

$(\vec r - \vec p_0) \cdot \vec n = 0, \tag{2}$

where $\vec n$ is the unit normal vector to $P$; that $P$ may be so described is well-known, and will be taken so here without further deomonstration. If we insert $\gamma(s) = (\gamma_x(s), \gamma_y(s), \gamma_z(s))$ into (2), we see that

$(\gamma(s) - \vec p_0) \cdot \vec n = 0, \tag{3}$

and differentiating (3) we obtain

$\dot \gamma(s) \cdot \vec n = 0. \tag{4}$

We next recall that

$\dot \gamma(s) = \vec T(s), \tag{5}$

where $\vec T(s)$ is the unit tangent vector to $\gamma(s)$, $\vec T(s) = \dot \gamma(s)$ (since $s$ is arc-length); then by (4)

$\vec T(s) \cdot \vec n = 0; \tag{6}$

furthermore, by the Frenet-Serret equation

$\dot {\vec T}(s) = \kappa \vec N(s), \tag{7}$

applied to (6) we have

$\kappa(s) \vec N(s) \cdot \vec n = \dot {\vec T}(s) \cdot \vec n = 0; \tag{8}$

from (8) we see that, as long as $\kappa(s) \ne 0$, that is, as long as $\vec N(s)$ may be defined, we have

$\vec N(s) \cdot \vec n = 0 \tag{9}$

holding as well as (6); thus both $\vec T(s)$ and $\vec N(s)$ are normal to $\vec n$ as long as they are defined. Now $T(s)$ and $N(s)$ form an orthonormal system; that is $\Vert \vec T(s) \Vert = \Vert \vec N(s) \Vert = 1$ and $\vec T(s) \cdot \vec N(s) = 0$, and since the unit binormal vector along $\gamma(s)$, $\vec B(s) = \vec T(s) \times \vec N(s)$ also satisifies

$\vec B(s) \cdot \vec T(s) = (\vec T(s) \times \vec N(s)) \cdot \vec T(s) = 0, \tag{10}$

$\vec B(s) \cdot \vec N(s) = (\vec T(s) \times \vec N(s)) \cdot \vec N(s) = 0, \tag{11}$

$\vec B(s) \cdot \vec B(s) = 1, \tag{12}$

we may conclude from the continuity of

$\vec B(s)$ that $\vec B(s) = \pm \vec n, \tag{13}$

a constant; thus from

$\dot {\vec B}(s) = -\tau(s) \vec N(s), \tag{14}$

the Frenet-Serret equation for $\dot {\vec B}(s)$, we may infer that

$\tau(s) \vec N(s) = 0 \Rightarrow \tau(s) = 0, \tag{15}$

since $\vec N(s) \ne 0$ wherever it is defined; we have shown that the torsion $\tau(s)$ of any plane curve $\gamma(s)$ vanishes.

Of course, there are a couple of caveats in the above argument, most notably the assumptions of regularity (so that $\vec T(s)$ exists), and non-vanishing curvature (so that $\vec N(s)$ exists); but I think these can be covered pretty easily; I'll defer the discussion until after we have handled the "if" direction of the assertion's logic.

So suppose $\tau(s) = 0$; that is, that $\gamma(s)$ is a regular curve in $\Bbb R^3$ with vanishing torsion. Then $\vec B(s)$ must be constant along $\gamma(s)$, by (14); choosing $s_0 \in I$ we have $\vec B(s_0) = \vec B(s)$ for all $s \in I$; thus by (10)-(11) $\vec T(s)$ and $\vec N(s)$ both belong to the subspace $V \subset \Bbb R^3$ with $\vec B(s_0) \bot V$; indeed, we may take this subspace to be spanned by $\vec T(s_0)$, $\vec N(s_0)$, since they form an orthonormal pair in $V$; $V = \text{span} \{ \vec T(s_0), \vec N(s_0) \}$. This being the case, we may write

$\dot {\gamma}(s) = \vec T(s) = \langle \vec T(s), \vec T(s_0) \rangle \vec T(s_0) + \langle \vec T(s), \vec N(s_0) \rangle \vec N(s_0); \tag{16}$

upon integrating (16) we find

$\gamma(s) - \gamma(s_0) = \int_{s_0}^s \dot {\gamma}(u) du = \int_{s_0}^s \vec T(u) du$ $= (\int_{s_0}^s \langle \vec T(u), \vec T(s_0) \rangle du) \vec T(s_0) + (\int_{s_0}^s \langle \vec T(u), \vec N(s_0) \rangle du) \vec N(s_0), \tag{17}$

which implies that

$(\gamma(s) - \gamma(s_0)) \cdot \vec B(s_0)$ $ =(\int_{s_0}^s \langle \vec T(u), \vec T(s_0) \rangle du) \langle \vec T(s_0), \vec B(s_0) \rangle$ $+ (\int_{s_0}^s \langle \vec T(u), \vec N(s_0) \rangle du) \langle \vec N(s_0), \vec B(s_0) \rangle = 0 \tag{18}$

for all $s \in I$; but the equation of the plane normal to $\vec B(s_0)$ passing through the point $\gamma(s_0)$ is in fact

$(\vec r - \gamma(s_0)) \cdot \vec B(s_0) = 0, \tag{19}$

where $\vec r = (x, y, z)$. Thus $\gamma(s)$ lies in this plane. QED.

We can actually take things a step further and present concise formulas for $\vec T(s)$ and $\vec N(s)$ in terms of $\int_{s_0}^s \kappa(u)du$ as follows: When $\tau(s) = 0$, the Frenet-Serret equations become

$\dot{\vec T}(s) = \kappa(s) \vec N(s), \tag{20}$

$\dot{\vec N}(s) = -\kappa(s) \vec T(s), \tag{21}$


$\dot {\vec B}(s) = 0. \tag{22}$

(22) implies $B(s)$ is constant; inspecting (20)-(21) reveals they may be written in combined form by introducing the six-dimensional column vector $\vec \Theta(s)$:

$\vec \Theta(s) = (\vec T(s), \vec N(s))^T, \tag{23}$

so that

$\dot {\vec \Theta}(s) = (\dot {\vec T}(s), \dot {\vec N}(s))^T; \tag{24}$

with this convention, (20)-(21) may be written

$\dot {\Theta}(s) = \begin{bmatrix} 0 & \kappa(s)I_3 \\ -\kappa(s)I_3 & 0 \end{bmatrix} \vec {\Theta}(s) = \kappa(s) J \vec{\Theta}(s), \tag{25}$

where $I_3$ is the $3 \times 3$ identity matrix and

$J = \begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix}; \tag{26}$

here it is understood that $J$ is presented in the from of $3 \times 3$ blocks. It is easy to see that

$J^2 = \begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix}\begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix} = \begin{bmatrix} -I_3 & 0 \\ 0 & -I_3 \end{bmatrix} = -I_6, \tag{27}$

$I_6$ being the $6 \times 6$ identity matrix. Careful scrutiny of (25) suggests that

$\Theta(s) = e^{(\int_{s_0}^s \kappa(u) du)J} \Theta(s_0) \tag{28}$

might be its unique solution taking the value $\Theta(s_0)$ at $s = s_0$; indeed, we may differentiate (28) with respect to $s$ to obtain

$\dot {\Theta}(s) = \dfrac{d}{ds}(\int_{s_0}^s \kappa(u)du)Je^{(\int_{s_0}^s \kappa(u) du)J} = \kappa(s)J \Theta(s_0) = \kappa(s) J \Theta(s), \tag{29}$

showing that (28) satisfies (26); furthermore (28) is consistent with the initial condition at $s = s_0$;

$\Theta (s_0) = e^{(\int_{s_0}^{s_0} \kappa(u) du)J} \Theta(s_0) = e^{0J} \Theta(s_0) = \Theta(s_0). \tag{30}$

It is worth pointing out that the reason (28) works as a solution is basically that the $s$-derivative of the matrix $e^{(\int_{s_0}^s \kappa(u) du)J}$ follows the scalar pattern

$\dfrac{d}{ds}e^{u(s)} = \dfrac{du(s)}{ds}e^{u(s)}, \tag{31}$


$\dfrac{d}{ds}e^{(\int_{s_0}^s \kappa(u) du)J} = \dfrac{d(\int_{s_0}^s \kappa(u) du)J}{ds}e^{(\int_{s_0}^s \kappa(u) du)J} = \kappa(s)Je^{(\int_{s_0}^s \kappa(u) du)J}. \tag{32}$

(32) applies by virtue of the fact that $\int_{s_0}^s \kappa(u) du)J$ and its derivative $\kappa(s) J$ commute with one another, being scalar function multiples of the same matrix $J$; for general matrix functions $A(s)$, it is not true that $A'(s)A(s) = A(s)A'(s)$, and the evaluation of $(d/ds)A(s)$ becomes much more complicated; we do not in general have

$\dfrac{d}{ds}e^{A(s)} = \dfrac{A(s)}{ds}e^{A(s)} \tag{33}$

in parallel with the scalar formula (31); the interested reader may consult my answer to this question (especially the material surrounding equations (15)-(20)) for a more detailed discussion. However, under the special circumstances that $A(s) = f(s)B$ for a constant matrix $B$, then $A'(s) = f'(s)B$ and $A(s)A'(s) = f(s)f'(s)B^2 = A'(s)A(s)$; $A(s)$ and its derivative always commute in this special case, which is what we have here. (32) applies and thus we have that (28) solves (25).

We examine the matrix $e^{(\int_{s_0}^s \kappa(u) du)J}$ occurring in (28) with an eye to determining its structure, and the structure of the solutions to (25). That $J^2 = - I_6$ has been noted. Thus we have

$J^2 = -I_6; \; \; J^3 = J^2J = -J; \;\; J^4 = J^3J= -J^2 = I_6,$ $J^5 = (J^4)J = I_6J = J, \tag{34}$

and in general,

$J^{4n + p} = J^{4n}J^p = (J^4)^nJ^p = (I_6)^n J^p = J^p, \tag{35}$

which shows that all cases of $J^m$, $m \in \Bbb Z$, are in fact covered by (34), i.e. for $0 \le p \le 3$. If we expand the matrix $e^{(\int_{s_0}^s \kappa(u) du)J}$ as a power series

$e^{(\int_{s_0}^s \kappa(u) du)J} = \sum_0^\infty \dfrac{((\int_{s_0}^s \kappa(u) du)J)^n}{n!} = \sum_0^\infty \dfrac{(\int_{s_0}^s \kappa(u) du)^nJ^n}{n!}, \tag{36}$

To be continued/completed; stay tuned!!?

  • $\begingroup$ The restriction to a regular plane curve rules out cases where, for example, the curve spirals in to a smaller and smaller radius, correct? $\endgroup$ – nwsteg Jun 14 '19 at 15:16
  • 1
    $\begingroup$ No, consider the clothoid in which $\kappa \sim s$, where $s$ is arc-length from a point on the curve. See math.stackexchange.com/questions/3097067/…. Also known as Euler spirals. Check it out! Cheers! $\endgroup$ – Robert Lewis Jun 14 '19 at 16:04

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.