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Is there an intrinsic way of defining the arc length of a curve in $\mathbb{R}^{3}$, that is without resorting to a parametrization of the curve?

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    $\begingroup$ Yes, approximate it by a piecewise linear curve and take the limit (if it exists). $\endgroup$ – Qiaochu Yuan May 18 '15 at 5:48
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    $\begingroup$ @Cusp probably not. Such an approximation only requires a bunch of points on the curve, and one can define the error of the approximation via the sum of the maximum and the minimum of the length of the line segments in the approximation. Presumably for rectifiable curves as this 'error' approaches zero the usual distance (L^1 metric, say) approaches zero. $\endgroup$ – David Roberts May 18 '15 at 7:03
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    $\begingroup$ If you don't want to use a parametrization, do you only want to work with the trace of the curve? In that case you cannot distinguish between a line traversed once and a line traversed several times if you assume no injectivity. Oh, and how about the one dimensional Hausdorff measure of the trace? That requires no parametrization but does measure length. $\endgroup$ – Joonas Ilmavirta May 18 '15 at 7:17
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    $\begingroup$ Am I the only one doubting the research-level relevance of the question (at least in its present form)? $\endgroup$ – Loïc Teyssier May 18 '15 at 7:48
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    $\begingroup$ There's the Cauchy-Crofton formula for the arc length of a curve. It does not use a parametrization. It also generalizes to formulas for surface areas, etc. en.wikipedia.org/wiki/Crofton_formula The Crofton formulas also have versions that work for parametrized curves (arc length) vs. just the image/trace of the curve. $\endgroup$ – Ryan Budney May 18 '15 at 16:45
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When a definition applies to a much larger class of spaces $\ X\subseteq \mathbb R^3,\ $ then such definition rather is not directly related to any parametrization for the smaller class of the parametrized curves (which can be rectified). Several of variants of Hausdorff dimension for dimension 1 would be a possible answer. Another could be

$$\lim_{r\rightarrow +0}\ \frac {V(X+B(r))}{\pi\cdot r^2}$$

where $\ X+B(r)\ =\ \{y\in\mathbb R^3:\ \exists_{x\in X}\ d(x\ y)\le r\},\ \ d\ $ is the euclidean distance, and $\ V\ $ stands for volume (see also, optionally, the detailed terminology below). This will certainly work properly for the piecewise $\ C^1$-curves, but most likely for all rectifiable curve (see the justification below); and of course--by definition--for many other spaces in the Euclidean space. Again, variations of this definition are possible, e.g. using cubes with edges $\ \left[\frac k{2^n};\frac{k+1}{2^n}\right],\ $ etc.

TERMINOLOGY

  • $\ B(r)\ :=\ \{x\in\mathbb R^3: |x|\le r\}\ $ is the closed ball of radius $\ r,\ $ centered at the origin of $\ \mathbb R^3$;
  • $\ X+Y\ :=\, \ \{x+y:\ x\in X\ \ \&\ \ y\in Y\}\quad $ for $\ X\ Y\subseteq\mathbb R^3$.

JUSTIFICATION of the volume formula for the length

In the case of a finitely piece-wise linear curve, the above volume is a sum of the respective cylinders around the intervals plus/minus a negligible error when the radius approaches $\ 0.\ $ The general case of rectifiable curves is obtained by $\ \epsilon/\delta\ $ (:-) which I am ready to provide if asked to.

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  • $\begingroup$ Does your formula give the same value as the usual definition of arc length? How is the + to be interpreted? As a Minkowski sum? Thanks. $\endgroup$ – Felix Goldberg May 18 '15 at 13:41
  • $\begingroup$ @FelixGoldberg -- I've described in my answer the whole expression $\ X+B(r);\ $ That's sufficient to follow my answer. However, I will add detailed terminology too. And, yes, the $\lim$ formula in my answer gives the standard length of a curve (when it is defined, i.e. for rectifiable curves). I'll add the detailed proof for the piecewise linear curves. The rest is perhaps not trivial but still routine (Archimedes would take care of it in a blink of an eye). $\endgroup$ – Włodzimierz Holsztyński May 18 '15 at 22:50
  • $\begingroup$ Joonas, I am sorry to overlook your mentioning of 1-dim Hausdorff in a comment above, posted less than 1h. before my answer (I am a slow OM-writer thus most likely I didn't have a chance to see the relevant comment at the time). $\endgroup$ – Włodzimierz Holsztyński May 19 '15 at 0:23
  • $\begingroup$ Does this tube formula work for cusps such as $y^3=x^3$? It is piecewise $C^1$ and rectifiable. $\endgroup$ – Liviu Nicolaescu May 19 '15 at 10:24
  • $\begingroup$ @LiviuNicolaescu -- a good idea about cusps (for contradicting my argument) but my volume formula does work for cusps anyway since with the radius approaching $0$ the overlap near the cusp gets small (proportion wise). BTW, I disregarded your formula above since it seems mistyped, I think--you described simply a plane (not a curve) $\ x=y\ $ or I am confused. $\endgroup$ – Włodzimierz Holsztyński May 19 '15 at 11:28
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Assuming the curve $S$ is ``reasonable'' say, it is semialgebraic or, more generally, definable in an $o$-minimal category, then you can define the length using Crofton's formula. $\DeclareMathOperator{\Graff}{Graff}$ $\newcommand{\bR}{\mathbb{R}}$

Define $\Graff_2(\bR^3)$ to be the set of affine planes in $\bR^3$. Up to a multiplicative constant, there is a unique measure on $\Graff_2(\bR^3)$ invariant under the group of isometries of $\bR^3$. Pick one such measure $\mu$. Then

$$ C(\mu) {\rm length}\;(C)= \int_{\Graff_2(\bR^3)} \#(L\cap C) \mu(dL), $$

where $C(\mu)$ is a universal, explicit, positive constant that depends linealrly on the invariant measure $\mu$.

Remark. Federer's book on geometric measure theory contains varies concepts of measure and dimension one can associate to a subset of $\bR^n$: Hausdorff measure, integralgeometric measure (using Crofton's formula). For ``reasonable'' sets these concepts coincide. In particular, for reasonable $1$-dimensional sets they yield various definitions of length that do not use parametrizations of the curve.

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You can use the Darboux sums (named after the mathematician Gaston Darboux). If you have a totally ordered metric space $M$ with a minimum and a maximum ($\alpha$ and $\omega$) (this is, in particular, the case of a curve with an injective parametrization, but you do not have to refer to it). Call subdivision a sequence of points $s=(x_i)_{0\leq i\leq n}$ linearly ordered $(\forall i<n)(x_i<x_{i+1})$ and joining the endpoints $x_0=\alpha,\ x_n=\omega$. For every subdivision $s$, form the Darboux sum \begin{equation} l(s)=\sum_{i=1}^n d(x_i,x_{i-1})\ . \end{equation} Take the usual order of refinement between subdivisions, if these quantities converge for the net of refinement order, the limit the length of the linearly ordered metric space.

$$ length(M)=lim_{s\nearrow} l(s)\ . $$

Call rectifiable a (linearly ordered) metric space such that $length(M)<+\infty$. This notion is elementary, encompasses all others I know and has very nice properties :

  • if a (linearly ordered) metric space is rectifiable then all its intervals $[u,v]$ are so
  • length is additive : if $u<v<w$, then $$ length([u,w])=length([u,v])+length([v,w]) $$
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The most general definition of arc length will involve summing the distances between nearby points (taken as a limit, obviously) along the section of the curve. The arc length will be independent of parameterization chosen, so any parameterization can be chosen without loss of generality. Explicit parameterization can only be avoided if the curve is actually a function on one axis, over the section in question. (Or piecewise.) For example you could use $\intop\sqrt{1+\left(\frac{dy}{dx}\right)^2+\left(\frac{dz}{dx}\right)^2}dx$ but only if y = f(x), z = g(x) are (single-valued) functions of x. My advice is to not be afraid of parameterization; just use it. It's more general. So I think the short answer to your question is "no".

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    $\begingroup$ I don't think your contribution answers the question asked (the OP never said he was scared of parameterizations…) and he probably knows all this. He explicitly stated « without resorting to a parametrization», which is the sole content of your answer. The question was probably intended at a foundational level. $\endgroup$ – Loïc Teyssier May 18 '15 at 7:53
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    $\begingroup$ @LoïcTeyssier Indeed. $\endgroup$ – Felix Goldberg May 18 '15 at 8:44

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