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.
In fact, given any functions $\kappa, \tau : (a, b) \to \Bbb R$ satisfying $\kappa(s) > 0$ for all $s \in (a, b)$,
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 $$ \gamma(t) = \pmatrix{ \zeta C\left(\zeta s\right)\\ \zeta S\left(\zeta s\right)\\ \zeta^2 s } , $$ where $\zeta := \frac{1}{\sqrt[4]{2}}$.
Remark The trace $s \mapsto (\zeta C\left(\zeta s\right), \zeta S\left(\zeta s\right))$ of our curve in the $xy$-plane is an Euler spiral, that is, a curve whose curvature at a point is proportional to its arc length from a reference point.
A plot of our solution $\gamma(s)$, $-12 \leq s \leq 12$: