# Curve where torsion and curvature equal arc length

I study differential geometry independently in my free time as an undergraduate. I am using the book by Do Carmo.

I recently read the section and local theory of curves and learned about torsion and curvature.

My question is, does there exist a curve that has both torsion and curvature equal to arc length? I have tried deriving such a curve, but I’ve failed.

I speculate that it must be somewhat helical in nature.

Standard equation of helix is given by $$\alpha(s)=(a\cos(s/c),a\sin(s/c),b)$$. The curvature of such a curve is $$\kappa(s)=\frac{a}{a^2+b^2}$$ and torsion is $$\tau(s)=\frac{b}{a^2+b^2}$$.

Clearly $$a=\frac{1}{2}s^{-1}=b$$.

I’m not sure if this is the right approach to take. I feel as though the curve’s normal ought to trace out a curve on a sphere, but it doesn’t.

Any help is appreciated.

• For any nondegenerate curve $|{\bf N}(s)| = 1$, so ${\bf N}(s)$ traces a curve on the unit sphere. It's true for curves satisfying $\kappa(s) = \tau(s) = s$, however, that ${\bf N}(s)$ has image contained in a great circle on that sphere. See my answer for an explicit solution $\gamma(s)$. – Travis Willse Mar 26 '19 at 4:58
• The planar curve that has curvature proportional to the arc length is known as the clothoid. It resurfaces in the solution by Travis. – Yves Daoust Mar 26 '19 at 9:39
• For future reference, any curve with $\tau/\kappa$ constant is a generalized helix. – Ted Shifrin Mar 27 '19 at 0:18

In fact, given any functions $$\kappa, \tau : (a, b) \to \Bbb R$$ satisfying $$\kappa(s) > 0$$ for all $$s \in (a, b)$$,

1. there is a curve $$\gamma(s)$$ parameterized by arc length whose curvature is $$\kappa(s)$$ and whose torsion is $$\tau(s)$$, and
2. any two such curves $$\gamma_1, \gamma_2$$ are unique up to rigit motions of $$\Bbb R^3$$, that is, there is a rigid motion $$A$$ of $$\Bbb R^3$$ such that $$\gamma_2 = A \circ \gamma_1$$.

This appears in $$\S$$ 1.5 of do Carmo's text, where it's called the Fundamental Theorem of the Local Theory of Curves; see also the appendix to $$\S$$ 4. In Clelland's excellent From Frenet to Cartan: The Method of Moving Frames, this is Corollary 4.15, where it's presented as a motivating special case of more general result that applies far beyond Euclidean geometry.

Setting the curvature and torsion of a curve $$\gamma$$ to prescribed functions $$\kappa, \tau$$ results in a nonlinear, third-order system in three functions (the components of $$\gamma$$), so for general $$\kappa, \tau$$ one shouldn't expect to find explicit, closed-form solutions $$\gamma$$.

On the other hand, the conditions $$\kappa(s) = \tau(s) = s$$ are tractable enough to find an explicit solution. Substituting in the usual Frenet equations in matrix form gives \begin{align*} \pmatrix{{\bf T}'(s)&{\bf N}'(s)&{\bf B}'(s)} &= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \pmatrix{\cdot&-\kappa(s)&\cdot\\\kappa(s)&\cdot&-\tau(s)\\\cdot&\tau(s)&\cdot} \\ &= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \cdot s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot} . \end{align*} Rearranging gives $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)}^{-1} \frac{d}{ds}\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot},$$ and solving formally yields $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = \pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)} \exp \left[\frac{1}{2} s^2\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot}\right] .$$ Since all solutions are the same up to rigid motions, we may as well take $$\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)}$$ to be any (special orthogonal) matrix we like, and it turns out to be convenient to take (cf. J.M.'s comment) $$\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)} = \pmatrix{ \frac{1}{\sqrt{2}}&\cdot&-\frac{1}{\sqrt{2}}\\ \cdot&1&\cdot\\ \frac{1}{\sqrt{2}}&\cdot&\frac{1}{\sqrt{2}}. }$$

We can also compute the matrix exponential explicitly, and putting this all together gives $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = \pmatrix{ \frac{1}{\sqrt{2}} \cos \frac{1}{\sqrt{2}} s^2&\ast&\ast\\ \frac{1}{\sqrt{2}} \sin \frac{1}{\sqrt{2}} s^2&\ast&\ast\\ \frac{1}{\sqrt{2}} &\ast&\ast } .$$ For a curve parameterized by arc length, $${\bf T}(s) = \gamma'(s)$$, so we can recover an explicit formula for a solution $$\gamma(s)$$ by integrating $${\bf T}(s)$$. Taking the initial condition $$\gamma(0) = (0, 0, 0)$$ yields the solution $$\color{#df0000}{\boxed{\gamma(s) = \pmatrix{ \frac{1}{\sqrt{2}} \int_0^s \cos \frac{1}{\sqrt{2}} \tau^2 d\tau\\ \frac{1}{\sqrt{2}} \int_0^s \sin \frac{1}{\sqrt{2}} \tau^2 d\tau\\ \frac{1}{\sqrt{2}} s \\ }}}.$$ Optionally, we can rewrite $$\gamma$$ in terms of the Fresnel integrals, $$C(x) := \int_0^x \cos t^2 \,dt$$ and $$S(x) := \int_0^x \sin t^2 \,dt$$, as $$\gamma(t) = \pmatrix{\frac{1}{\sqrt{2}} C\left(\frac{1}{\sqrt{2}} s\right)\\\frac{1}{\sqrt{2}} S\left(\frac{1}{\sqrt{2}} s\right)\\\frac{1}{\sqrt{2}} s} .$$

A plot of our solution $$\gamma(s)$$, $$-12 \leq s \leq 12$$: • For reference, here's the expression in terms of Fresnel integrals: $$\left(\frac{s}{2}+\frac{\sqrt{\pi}}{2 \sqrt{2}}C\left(\frac{\sqrt{2} s}{\sqrt{\pi }}\right)\quad\frac{\sqrt{\pi }}{2^{3/4}}S\left(\frac{\sqrt{2} s}{\sqrt{\pi }}\right) \quad\frac{s}{2}-\frac{\sqrt{\pi }}{2 \sqrt{2}}C\left(\frac{\sqrt{2} s}{\sqrt{\pi }}\right)\right)^\top$$ tho this uses the convention where the Fresnel integrals have an extra $\pi/2$ factor. As for the Do Carmo reference... – J. M. is a poor mathematician Mar 26 '19 at 6:48
• Since I can't edit the last comment anymore: a clockwise rotation about the $y$-axis gives a simpler set of parametric equations: $$\left(\frac{\sqrt{\pi }}{2^{3/4}}C\left(\frac{\sqrt{2} s}{\sqrt{\pi }}\right)\quad \frac{\sqrt{\pi }}{2^{3/4}}S\left(\frac{\sqrt{2} s}{\sqrt{\pi }}\right)\quad \frac{s}{\sqrt{2}}\right)^\top$$ – J. M. is a poor mathematician Mar 26 '19 at 6:56
• Thanks for the helpful comments. I've added the reference to do Carmo and (I hope you don't mind) tweaked the choice of initial frame in my answer to yield your nicer parametric equation for $\gamma$. – Travis Willse Mar 26 '19 at 7:52
• Not at all! Choosing a different initial Frenet frame is entirely equivalent to rotating the curve, of course. :) – J. M. is a poor mathematician Mar 26 '19 at 7:55
• It's true that as $s \to \infty$ the curve is asymptotic to a straight line. For the convenient frame used in my answer, this line is $x = y = 2^{-7 / 4} \sqrt\pi$. Whether this means the curve becomes "more straight" is a matter of definition. Indeed, the angle between $\bf T$ and this asymptote is a constant $\frac{\pi}{4}$---this is possible because as $s$ increases the curve becomes more tightly wound, owing to the quadratic polynomial inside the arguments of $\cos$ and $\sin$. – Travis Willse Mar 26 '19 at 8:55