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I am in the process of proving that if a space curve (in R^3) has infinite length and the curvature tends towards 0 as the natural parameter s tends to infinity, the curve must be unbounded - i.e. not contained in any sphere of finite radius. This seems correct intuitively, but I have no guarantee it is correct, unless I am missing something obvious. One way to prove my hunch, I have deduced, is to use a lemma that any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly 2π, but it could be greater for all I know).

How might one go about proving such an upper bound exists, or if it exists? It might also be nice to know what the bound specifically is, too. I've thought it might be possible to pose this as a variational problem - maximizing length - and then reducing it into a simpler problem, but that appears to be hellishly complicated. Thoughts?

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If your curvature goes to zero with increasing arclength, in the limit you have something that looks like a straight line, which is unbounded. I'll leave it up to other people to make this argument rigorous. – J. M. Nov 30 '10 at 0:47
That's basically my thoughts. But it only locally looks like a line. Technically you could have a curve spiral around and never asymptotically approach being a line but still have curvature -> 0. And that doesn't even bother to touch on what torsion could do to the situation. – chroma Nov 30 '10 at 0:58
My gut feeling is that torsion can't make things any worse. Given a space curve $\gamma_1$ with curvature $\kappa(t)$ and torsion $\tau(t)$, suppose you construct another curve $\gamma_2$ whose torsion is identically zero, and whose curvature is $\lvert\tau(t)\rvert$. In other words, you're stuffing $\gamma_1$ into a plane while preserving the magnitude of its curvature. I conjecture that $\gamma_2$ is contained in a ball of no larger radius than any ball containing $\gamma_1$. If true, this would reduce the scope of the problem to purely plane curves. – Rahul Nov 30 '10 at 1:20
That's false. Take helices, for instance, which have both constant curvature and constant torsion. Curves with constant curvature and zero torsion are simply circles. The hypothetical helix can be any size while the circle's size is invariant to extending the helix. (Although being able to reduce this problem to plane curves would be vastly helpful.) – chroma Nov 30 '10 at 1:47
To be clear, the bounty I set is for showing that "any curve contained in the open unit ball with curvature always less than one must have a finite upper bound on its length (possibly 2π, but it could be greater for all I know)." It is not for the motivating problem of proving that the curve is unbounded, which already follows from my weaker answer. – Jonas Meyer Jan 31 '11 at 1:03

Let ${\mathbf{x}}(s)$ be a curve in ${\mathbb{R}}^3$ with natural parameter $s$. We will need the following lemma, the proof of which is given at the end of this answer.

Lemma: Choose a fixed point ${\mathbf{y}}$. Then the curvature $\kappa$ satisfies $$ \kappa \ge \left|\dot{\theta} + \frac{1}{r}\sin{\theta}\right|, $$ where $r = \lVert\mathbf{x} - \mathbf{y}\rVert$ and $\theta$ is the angle between the velocity $\dot{\mathbf{x}}$ and $\mathbf{x} - \mathbf{y}$.

Take $\mathbf{y} = \mathbf{x}(0)$; then $r(0)=0$ and $\theta(0)=0$. The function $r(s)$ is monotonically increasing as long as $0 \le \theta < \pi/2$ (since $\dot{r}(s) = \cos{\theta}$), so $\theta$ and $\kappa$ can be considered as single-valued functions of $r$ until that point. The above lemma then gives $$ \kappa(r) \ge \left|{\theta}'(r)\dot{r} + \frac{1}{r}\sin{\theta(r)}\right| = \left|{\theta}'(r)\cos{\theta(r)} + \frac{1}{r}\sin{\theta(r)}\right| = \left|\frac{1}{r}(r \sin\theta(r) )'\right|. $$

Until the first turning point of the motion (where the velocity becomes perpendicular to the radius), we have $$ R\sin\theta(R) \le \int_{0}^{R} r \kappa(r) dr. $$ If the curvature is strictly below a fixed value (say, $\kappa < K$), then the integral is less than $\frac{1}{2}KR^2$ for $R>0$, and we have the result that $$ \sin{\theta(r)} < \frac{1}{2}Kr $$ for $r>0$. A turning point is reached when $\theta=\pi/2$; this equation shows that the first such turning point must be at a radius greater than $2/K$, and hence the curve cannot be confined within a ball of diameter $2/K$. Finally, the arclength before reaching a given radius $R$ is bounded by $$ \begin{eqnarray} s(R) &=& \int_{0}^{R} \frac{ds}{dr}dr \\ &=& \int_{0}^{R} \frac{dr}{\cos\theta(r)} \\ &<& \int_{0}^{R} \frac{dr}{\sqrt{1 - \frac{1}{4}K^2 r^2}} \\ &=& \frac{2}{K}\sin^{-1}\left(\frac{1}{2}KR\right) \end{eqnarray} $$ for $R \le 2/K$. We conclude that any curve contained in the open unit ball with curvature $\kappa < 1$ must have length $s(2) < 2\sin^{-1}(1) = \pi$. Moreover, this bound is tight, since a circular arc joining the points at $\pm (1-\epsilon^2)\hat{\mathbf{z}}$ and the point at $(1-\epsilon)\hat{\mathbf{x}}$ has length approaching $\pi$ as $\epsilon \rightarrow 0$.

Proof of Lemma:

We will work in spherical coordinates centered at $\mathbf{y}$; then $$ \begin{eqnarray} {\mathbf{x}} &=& r\hat{\mathbf{r}}, \\ {\dot{\mathbf{x}}} &=& \dot{r}{\hat{\mathbf{r}}} + r\dot{\hat{\mathbf{r}}} \\ &=& \dot{r}{\hat{\mathbf{r}}} + r v_{\perp} \hat{\mathbf{v}}_{\perp}. \end{eqnarray} $$ Here $\hat{\mathbf{r}}$ is the unit vector from the origin to ${\mathbf{x}}$, and $\dot{\hat{\mathbf{r}}} = v_{\perp} \hat{\mathbf{v}}_{\perp}$ is its rate of change. Because $\hat{\mathbf{r}}$ has constant length, we have $\hat{\mathbf{v}}_{\perp}\cdot \hat{\mathbf{r}} = 0$. Taking the time derivative of this gives $$ 0 = \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}} + \hat{\mathbf{v}}_{\perp}\cdot \dot{\hat{\mathbf{r}}} = v_{\perp} + \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}}, $$ which we will use later. Now, because $s$ is a natural parameter, $$\lVert\dot{\mathbf{x}}\rVert^2 = \left(\dot{r}\right)^2 + \left(rv_{\perp}\right)^2 = 1;$$ so we can define $\theta \in [0,\pi]$ such that $\dot{r} = \cos{\theta}$ and $rv_{\perp} = \sin{\theta}$. The velocity and acceleration become $$ \begin{eqnarray} {\dot{\mathbf{x}}} &=& \left(\cos{\theta}\right){\hat{\mathbf{r}}} + \left(\sin{\theta}\right)\hat{\mathbf{v}}_{\perp}, \\ {\ddot{\mathbf{x}}} &=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\dot{\hat{\mathbf{r}}} + \left(\cos{\theta}\dot{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp} \\ &=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp}, \end{eqnarray} $$ and the acceleration has (two of its three) components $$ \begin{eqnarray} {\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}} &=& -\left(\sin{\theta}\dot{\theta}\right) + \left(\sin{\theta}\right)\left(\dot{\hat{\mathbf{v}}}_{\perp} \cdot \hat{\mathbf{r}}\right) \\ &=& -\left(\sin{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right), \\ {\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp} &=& +\left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right). \end{eqnarray} $$ This brings us to the result that the squared curvature $$ \kappa^2 = \lVert\ddot{\mathbf{x}}\rVert^2 \ge \left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}}\right)^{2} + \left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp}\right)^{2} = \left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)^{2}, $$ where $\theta$ is the angle between the velocity and the outward radial vector. The lemma follows by taking the square root of both sides.

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This looks very nice, and thanks for the answer. I may not get a chance to study it thoroughly for a day or two. – Jonas Meyer Feb 3 '11 at 3:13
Perhaps I am being dense, but doesn't your lemma require $y$ to be disjoint from the image of the curve? – Glen Wheeler Feb 3 '11 at 11:58
@Glen: Strictly speaking, yes, it only applies for ${\mathbf{x}}\neq{\mathbf{y}}$. The initial conditions $r(0)=\theta(0)=0$ can be replaced by $r(\epsilon)\rightarrow 0$ and $\theta(\epsilon)\rightarrow 0$ as $\epsilon\rightarrow 0^{+}$ without affecting the argument. – mjqxxxx Feb 3 '11 at 13:51
@Jonas: Thanks, good eye. I edited to fix those. – mjqxxxx Feb 5 '11 at 20:18

This is true, and follows from a theorem of Fary. Fary's theorem says that for closed curves inside a ball, the average curvature of the curve is at least as large as the curvature of the boundary of the ball.

Your curve isn't closed (or finite in length), so you have to think about what the theorem says in this context. But it will work: the average curvature is zero, and so it cannot fit in any (finite) ball.

I learned about Fary's theorem from Serge Tabachnikov; there are several proofs (of the 2-dimensional case, though most generalize to higher dimensions, as he describes at the end) in his beautiful short paper:

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Nice. I took the question to be more about what is in the title, and not the motivating problem that you seem to be addressing, but I think I now see how this implies the desired "lemma" too. If you had a curve of unbounded length in a unit ball with curvature less than 1, it seems you could connect 2 sufficiently "distant" (in terms of arclength along the curve) points to get a closed curve that still has average curvature less than 1 (although I haven't thought through the details). But I don't see what the supremum referred to in the title of the question would be. Do you? – Jonas Meyer Jan 31 '11 at 1:13
I don't know how to compute the actual supremum, no. But the strategy I had in mind is what you suggest. – aaron Jan 31 '11 at 13:12
I don't know how to compute the actual supremum, no. But the strategy I had in mind is what you suggest: by Fary's theorem, if we close up the curve, then the total length won't be more than the total curvature. My guess is that the supremum is $\pi$, coming from circular arcs that approximate half a circumference. (I imagine trying to stuff a rigid wire into a spherical paper lamp.) – aaron Jan 31 '11 at 13:22

Here is something along these lines, coming from estimates using elementary calculus, but it is much weaker than the lemma you want to prove.

Let $\gamma$ be a regular $C^2$ curve in $\mathbb{R}^3$ parametrized with respect to arclength $s$ (I will assume for simplicity that $s$ starts at $0$). If the curvature $\|\gamma''(s)\|$ is always less than $K$ and $\gamma$ is contained in a ball of radius $R\leq\frac{1}{4K}$, then the length of $\gamma$ is less than $\frac{1}{K}(1-\sqrt{1-4KR})$.

To see this, write $\gamma(s)=\gamma(0)+s\gamma'(0)+\int_0^s(s-t)\gamma''(t)dt$ (derived as in this Wikipedia article). Moving $s\gamma'(0)$ to the left and $\gamma(s)$ to the right and applying the triangle inequality, $$ \begin{align} s &\leq \|\gamma(s)-\gamma(0)\|+\|\int_0^s(s-t)\gamma''(t)dt\| \\ &\leq 2R +\int_0^s\|(s-t)\gamma''(t)\|dt \\ &\lt 2R +K\int_0^s(s-t)dt \\ &=2R+\frac{K}{2}s^2. \end{align} $$

For this inequality to always hold, $s$ must remain smaller than the smallest root of $\frac{K}{2}x^2-x+2R$. Thus, $s\lt\frac{1}{K}(1-\sqrt{1-4KR})$ as claimed.

You could still apply this to prove the original result as stated, but hopefully someone comes along with a stronger result.

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That does prove my original thought of k->0 implying unboundedness. So it turns out my inspired lemma is a more powerful conjecture! I hope we can find a way to prove that one too. – chroma Nov 30 '10 at 5:31
Yes, me too. It seems like such a natural problem that the answer should be "well-known" to the right reader. But I don't have any ideas. – Jonas Meyer Nov 30 '10 at 5:47

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