Curves with constant curvature and constant torsion

Question

Describe all curves in $\mathbb{R}^3$ which have constant curvature $κ > 0$ and constant torsion $τ$.

Any ideas what we can do to describe all such curves?

Do we have to use the formulas of the curvature and of the torsion?

Answer 1

As had been explained, you may choose to either solve the differential equation given by Frenet's identities in the space of matrices, or do the smart trick suggested by Ted Shiffrin; I shall choose the latter approach.

For my typing convenience, I shall also omit all the "mathhb" formatting, therefore the normal will just be $n$. Also, since $t$ is the tangent, I shall use $s$ for the time parameter on the curve.

Starting with $\ddot n = - (k^2 + \tau ^2)n$, we shall approach it in the usual way: first we consider its associated algebraic equation $\lambda ^2 = -(k^2 + \tau ^2)$, which has the roots $\pm \Bbb i \sqrt{k^2 + \tau ^2}$. Next, we know from the general theory of linear equations that this gives two fundamental solutions to the above equation, namely $u \sin rs$ and $v \cos rs$ where $r^2 = k^2 + \tau ^2$ and $u,v \in \Bbb R ^3$ (note that $r \ne 0$ because $k>0$ by assumption). Therefore, $n(s) = u \sin rs + v \cos rs$ for some constant vectors $u,v \in \Bbb R ^3$.

May $u,v \in \Bbb R ^3$ be really arbitrary? Not quite: note that $n(0) = v$ and since $\| n (s) \| = 1 \ \forall s$ (the curve is parametrized by arc-length and the Frenet frame is taken to be orthonormal), then $\| v \| = 1$. Next, if you derive this once, you get $\dot n (0) = ru$; on the other hand, $\dot n (0) = -k t(0) + \tau b (0)$, so $r u = -k t(0) + \tau b (0)$, so $r^2 \| u \| ^2 = \| -k t(0) + \tau b (0) \| ^2 = k^2 \| t (0) \| ^2 - 2 k \tau \langle t(0), b(0) \rangle + \tau ^2 \| v \| ^2$. Since $\| t \| = \| b \| = 1$ and $\langle t, b \rangle = 0$, the last expression is exactly $k^2 + \tau ^2 = r^2$, which implies $\| u \| = 1$. Finally, $\| n \| ^2 = 1$ implies $\| u \| ^2 \sin ^2 rs + 2 \sin rs \cos rs \langle u, v \rangle + \| v \| ^2 \cos ^2 rs = 1$ which, taking into consideration all of the above, implies $\langle u, v \rangle = 0$. To conclude, $u$ and $v$ must be orthogonal, and of length $1$.

Next, if your curve is $s \mapsto x(s) \in \Bbb R ^3$ ($x$ is a vector, not the first coordinate of a vector!), then using the equation $\dot t = k n$ and the fact that $t = \dot x$, you will get $\ddot x = kn = k (u \sin rs + v \cos rs)$, that you will have to integrate twice. Integrating once gives $\dot x = -\dfrac k r u \cos rs + \dfrac k r v \sin rs + w$, with $w \in \Bbb R ^3$, and integrating once more gives $x = - \dfrac k {r^2} u \sin rs - \dfrac k {r^2} v \cos rs + ws + x_0$, with $x_0 \in \Bbb R^3$.

Similarly to the discussion about $u$ and $v$, may $w$ be arbitrary? No, it may not: first, $\langle t, n \rangle = 0$ implies $\langle w, u \sin rs + v \cos rs \rangle = 0 \ \forall s$, which implies that $w \perp \text{span} \{u, v \}$. Second, using this and the fact that $\| \dot x \| ^2 = \| t \| ^2 = 1$ implies that $\| w \| = \dfrac {| \tau |} r$. There is no constraint, on the other hand, on $x_0$.

If $\tau = 0$ (so your curve is a plane curve), after a possible orthogonal transformation you may take $u = (-1, 0, 0)$ and $v = (0, -1, 0)$, so your curve will look like $x(s) = (\dfrac k {r^2} \sin rs, \dfrac k {r^2} \cos rs, 0) + x_0$ which, after a translation by $x_0$ will have the final nice form $x(s) = (\dfrac k {r^2} \sin rs, \dfrac k {r^2} \cos rs, 0)$, so it will be a circle.

If $\tau \ne 0$, since $\{ -u, -v, w \frac r {\ |\tau| }\}$ have been shown to form an orthonormal frame (in fact, we have worked with $u$ and $v$, but absorbing a minus sign into them doesn't change orthonormality), after a possible orthogonal transformation you may take them to form the familiar frame $\{ (1, 0, 0), (0, 1, 0), (0, 0, 1) \}$, in which case you curve looks like $x(s) = (\dfrac k {r^2} \sin rs, \dfrac k {r^2} \cos rs, \frac {| \tau |} r s) + x_0$. If you also translate your curve by $x_0$ (again, this doesn't change your curve), it will have the final nice form $x(s) = (\dfrac k {r^2} \sin rs, \dfrac k {r^2} \cos rs, s\dfrac { |\tau | } r)$, which is a helix. Note that this equation reduces to the one of the circle for $\tau \to 0$.

Therefore, the only curves that satisfy your problem are circles and helices. (Had you allowed $k=0$ too, you would have also obtained (segments of) straight lines).

Answer 2

HINT: If you don't want to solve the matrix differential equation, I recommend that you get a second-order differential equation for the principal normal $n$. You should know solutions of this by sight (at least, component by component).

Answer 3

From the Fresnet-Serret formulas, we can derive

$$\ddddot\gamma(s)=\kappa\ddot n(s)=-\kappa(\kappa^2+\tau^2)n(s)=-a^2\ddot\gamma(s),$$ with $a=\sqrt{\kappa^2+\tau^2}$.

The general solution for this fourh order linear equation is $$\color{green}{\gamma(s)=C\cos(as)+S\sin(as)+C_1s+C_0},$$ which describes an elliptic helix.

Anyway, computing the curvature for any $s$ (such as $s=0,s=\dfrac\pi{2a}$),

$$\kappa=\|\dot t(s)\|=\|\ddot\gamma(s)\|=a^2\|C\cos(as)+S\sin(as)\|,$$

we must have $$\|C\|=\|S\|=\frac\kappa{a^2}\text{, and }C\cdot S=0$$ for the curvature to be constant (the normal vector describes a circle).

Similarly,

$$\kappa\tau=\|\kappa\dot n(s)+\kappa^2t(s)\|=\|\ddot t(s)+\kappa^2t(s)\|=\|-a\tau^2(C\cos(as)+S\sin(as))+\kappa^2C_1\|$$ requires $$C\cdot C_1=S\cdot C_1=0\text{, and }\|C_1\|=\frac{\tau\kappa}a$$ (the tangent vectors describes a circular cone).

Hence the curve is a circular helix.