# Curves of constant curvature

In planar case, curves of constant curvature are lines and circles.

In spatial case, if torsion is also constant, then it must be circular helix. However, if torsion is arbitrarily given, such as $\tau(s)=e^s$, can we solve it explicitly?

If not, I wonder what characteristic properties it satisfies. (For example, $\tau/\kappa$ is constant iff general helix.)

Any help will be appreciated.

Usually, when studying spacial curves, the Frenet-Serret frame representation is very useful. My rule of thumb is that when you have to deal with a problem about curves in space, look at the Frenet-Serret equations first. They fully encode all the geometric information of the curve.

Let $r : (a,b) \to \mathbb{R}^3$ be a smooth curve $r=r(s)$ in three-space parametrized by archlength $s$ (important detail) and of non-zero curvature (otherwise, the frame and the equations are not defined, but usually these are exceptional points or the curve has a very special geometry). I write derivatives with respect to $s$ as dots $$\dot{r}(s) = \frac{dr}{ds}(s).$$ Then let \begin{align} T(s) &= \frac{dr}{ds}(s) = \dot{r}(s)\\ N(s) &= \frac{1}{|\dot{T}(s)|} \frac{dT}{ds}(s) = \frac{1}{|\dot{T}(s)|}\dot{T}(s)\\ B(s) & = T(s) \times N(s) \end{align} bet the unit tangent $T$ (i.e. $|T(s)| = 1$ equivalent to the fact that $s$ is archlength parameter), the unit normal $N$ and the unit bi-normal $B$ of the curve $r$. They form an orthonormal frame moving along the curve. The frame satisfies the following system of linear ordinary differential equations \begin{align} \dot{T} &= \kappa(s) \, N \\ \dot{N} &= -\, \kappa(s) \, T + \tau(s) \, B\\ \dot{B} & = -\tau(s)\, N \end{align} where $\kappa=\kappa(s)$ is the normal curvature of the curve $r$ and $\tau=\tau(s)$ is its torsion. These equations determine the curve uniquely up to Euclidean isometry (up to rotation and translation in three-space). This means that

Theorem. Given the curvature $\kappa=\kappa(s)$ and the torsion $\tau=\tau(s)$, the curve $r(s)$ is uniquely determined up to some rotation and translation in three-space. In particular, it's geometry is fully determined.

This theorem is the Fundamental Theorem of Differential Geometry of curves in three-space. Solution of these Frenet-Serret equations always exists (unique up to rotation and translation), as long as some natural regularity requirements of $\tau$ and $\kappa$ are satisfied (continuity, smoothness and stuff like that). However, if the curvature and the torsion are arbitrary functions, solving the system explicitly, in terms of known formulas and functions, might be impossible. But if the torsion and curvature are somewhat more special, then you get explicit results.

For example, assume $\tau(s) \equiv 0$. Then the equations reduce to \begin{align} \dot{T} &= \kappa(s) \, N \\ \dot{N} &= - \,\kappa(s) \, T\\ \dot{B} & = 0 \end{align} which means the binormal $B(s) \equiv B_0$ is a constant vector. Then the tangent $T$ and the normal $N$ are always orthogonal to this constant vector $B_0$, so the curve $r$ lies in a plane orthogonal to the vector $B_0$, i.e. $r$ is a planar curve.

If in addition to that the curvature is constant $\kappa(s) \equiv \kappa_0$, then the equations reduce to

\begin{align} \dot{T} &= \kappa_0 \, N \\ \dot{N} &= - \,\kappa_0 \, T\\ \dot{B} & = 0 \end{align} which means that if I differentiate the first equation once $$\ddot{T} = \kappa_0 \, \dot{N}$$ and then I plug the second equation about $\dot{N}$ I get $$\ddot{T} = -\kappa^2_0 T$$ Let's not forget that $T = \dot{r}$ so $\ddot{T} = \dddot{r}$ leading to the equation $$\dddot{r} = -\kappa^2_0 \dot{r}$$ which I can integrate once with respect to $s$ and obtain the equation $$\ddot{r} = \kappa^{2}_0 \, x_0 -\kappa^2_0 \, {r}$$ Furthermore, $\dot{r}= T$ and $\ddot{r} = \dot{T} = \kappa_0 \, N$ so $$\kappa_0 \, N = \kappa^{2}_0 \, x_0 -\kappa^2_0 \, {r}$$
$$\frac{1}{\kappa_0} \, N = x_0 - {r}$$ which after taking the norm and observing that $|N|=1$ implies that $$\frac{1}{\kappa_0} \,| N| = |x_0 - {r}|$$ i.e. $$|r(s) - x_0| = \frac{1}{\kappa_0}$$ for all values of the parameter $s$, meaning $r(s)$ traverses a circle centered at $x_0$ and of radius $\frac{1}{\kappa_0}$.

Can you figure out why when the torsion is constant $\tau(s) \equiv \tau_0$ and the curvature is constant $\kappa(s) \equiv \kappa_0$, the resulting curve is a helix, i.e. it is a curve whose tangent vector forms a constant angle with a special fixed direction, which is the direction of its helical axis?

In particular, the case you propose, $\tau(s) = e^s$ is most likely solvable explicitly, so you can actually find the curve with this torsion and constant curvature. Simply plug it in the Frenet-Serret equations and see what you can do.

• I was able to get Mathematica to solve the equations with $\kappa=1$ and $\tau = e^s$. I personally have no desire to work that hard. However, I would comment that as $s\to-\infty$, the curve of course approaches a circle, and as $s\to\infty$, it is a loosely wound spiral that looks more and more like a straight line. – Ted Shifrin Oct 1 '16 at 4:31
• @TedShifrin Ok, so what is the point of your question? What kind of answer do you expect? – Futurologist Oct 1 '16 at 19:32
• I didn't ask a question. I was suggesting that your comment "is most likely solvable explicitly" might be a bit glib. I also intended to provide a bit of qualitative insight. – Ted Shifrin Oct 1 '16 at 20:24