Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Following the usual definition of length of a curve, the curve have to be rectifiable to have finite length. However for example the curve

$ \gamma \colon [0,1] \to \mathbb{R}, t \to t \cos^2(\pi/t) $

is not rectifiable. However I would say it makes perfectly sense to say it has length $1$.

So my question is, if there is a more general definition of length of a curve such that my example above has length $1$ (however such that for example the graph of the function $\gamma$ above has infinite length).

share|cite|improve this question
Why would you say that? – Qiaochu Yuan Apr 29 '11 at 10:01
Since it is just a parametrization of the interval [0,1] – student Apr 29 '11 at 10:07
Check out the following rectifiable parametrizations: $\gamma \colon [0,1] \to \mathbb{R}, t \to a\cdot t$. Do they all have length $1$ ? – Raskolnikov Apr 29 '11 at 10:21
@Raskolnikov, They have length $|a|$, however (suppose $a > 0$) it is $\gamma([0,1]) = [0,a]$. So length $a$ makes sense. – student Apr 29 '11 at 10:38
@user3445: but in this example Raskolinikov is measuring the length of the range, whereas in yours you are measuring the length of the domain. The domain has length 1. That was the point of his question. – Ross Millikan Apr 29 '11 at 13:02

The length of a curve is one of the simplest examples of a geometric quantity, that is, something which is invariant under reparametrisations. You can (and should) check that the usual notion of length is invariant under reparametrisations.

Your idea of length is certainly not invariant under reparametrisations, and so intuitively it is rejected as unnatural.

share|cite|improve this answer
"something which is invariant under reparametrisations." - can't be emphasized enough. +1. – J. M. Apr 29 '11 at 11:07

You have to distinguish between the ${\it curve}$, i.e., an equivalence class of maps $f:\ [0,1]\to {\mathbb R}^n$, and its "trace", i.e., the ${\it set}$ $f\bigl([0,1]\bigr)\subset{\mathbb R}^n$. If you declare your curve $\gamma$ to have length $1$ you would have to assign a twice surrounded circle the length $2\pi$. In this example the length of the "trace" is indeed still $2\pi$, but the length of the actual curve is $4\pi$.

share|cite|improve this answer
Ok. How would one define the length of the trace $\Gamma$? May I say: if there is an injective, regular map $f\colon [0,1] \to \mathbb{R}^n$ such that $f([0,1]) = \Gamma$ the length of $\gamma$ is defined as the length of the curve $f$? – student Apr 29 '11 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.