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I have a quick question that has been bothering me since I took my calc 2 final earlier. The question dealt with arc length and was pretty much plug and chug using the the arc length formula. My concern is my final answer left me with a negative length with didn't seem right to me, but I couldn't figure out any place where I went wrong.

Is a negative arc length a possible answer?

Travis Willse
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user231507
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2 Answers2

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The arc length of a curve cannot be negative, just as the distance between two points cannot be negative. But one can use the usual arc length formula to define an arc length parameterization of a curve, and this parameter can take on negative values (just as one can talk about negative displacements in some coordinate system.)

(The above prohibitions do not hold, by the way, for "indefinite signature metrics", like the Lorentzian spacetimes of general relativity.)

Travis Willse
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  • Thanks that's what I figured. I want to say the function I was given was f(x)=sqrt(x) evaluated from 0 to 1. The one sure thing I remember is after I computed everything I got .25-.5 – user231507 May 07 '15 at 05:36
  • Im pretty sure I got lost somewhere with the derivative being raised to power of 2. – user231507 May 07 '15 at 05:41
  • @Travis: "The above prohibitions do not hold for "indefinite signature metrics", like the Lorentzian spacetimes of general relativity." -- This seems intuitive by looking at the signs of spacetime intervals. However, I've struggled to find or establish a formal expression of this. Standard arc length would have to be replaced by $$\int_{\gamma}~\text{d}t~\text{sgn}[~g[~\gamma', \gamma'~]~]~\sqrt{|[~g[~\gamma', \gamma'~]~]|};$$ and the word "distance" (or "metric") might be substituted by a suitable generalization (my suggestion: "hypometric"). – user12262 Mar 16 '16 at 07:17
  • My comment about the indefinite case was a hedge to exclude exactly the issue you mention. For that same reason, in the indefinite case it's often better to work with the quantity $\langle \gamma', \gamma' \rangle$ that a signed square root (informally the "length-squared", though it need not be positive), just as it's often better to work with $X \cdot Y$ instead of the angle between $X$ and $Y$, as the usual definition (of angle) runs into trouble when $X$ or $Y$ is isotropic. – Travis Willse Mar 16 '16 at 14:02
  • Travis: "in the indefinite case it's often better to work with the quantity $\langle \gamma', \gamma' \rangle$ tha[n] a signed square root (informally the "length-squared", though it need not be positive)" -- Could you please tell me the formal, "technical" name of this quantity $\langle \gamma', \gamma' \rangle$, and perhaps sketch how "to work with it" when considering arc length? (I know about Lorentzian distance; but that's of definite sign, of course.) – user12262 Mar 18 '16 at 22:31
  • Travis: "often better to work with $X \cdot Y$ [...] as the usual definition (of angle) runs into trouble when $X$ or $Y$ is isotropic" -- Which sort of objects $X, Y$ do you mean there? p.s. Since you've been mentioning "the usual definition (of angle)": Considering three events $A, P, Q$ in flat spacetime with $$s^2[~A,P~] > 0,~~~s^2[~A,Q~] < 0,~~~s^2[~P,Q~] = 0,$$ then $$\frac{(2~s^2[~A,P~]~s^2[~A,Q~] + 2~s^2[~A,P~]~s^2[~P,Q~] + 2~s^2[~A,Q~]~s^2[~P,Q~] - (s^2[~A,P~])^2 - (s^2[~A,Q~])^2 - (s^2[~P,Q~])^2)}{(4~s^2[~A,P~]~s^2[~A,Q~])} \ge 1;$$ thus, it seems, indefinite arcs are not smooth. – user12262 Mar 18 '16 at 22:32
  • I don't know know a shorter name for that quantity that "the inner product of $\gamma'$ with itself" (but N.B. the some people reserve 'inner product' for definite bilinear forms). My comment about "how to work with it" was informal---what I really mean is that in practice one generally doesn't work with the square root of this quantity. As for smoothness, I don't quite understand your argument, but note that smoothness of a curve doesn't depend on a metric structure (and hence doesn't depend on whether a curve is isotropic), it just depends on a smooth structure. – Travis Willse Mar 18 '16 at 23:30
  • Travis: "people reserve 'inner product' for definite bilinear forms" -- Aha. (Unfortunately I'm not well-versed in such terminology conventions. Apparently, any metric tensor $g : T \times T \rightarrow \mathbb R$ is a dot product (a.k.a. inner product); but perhaps this doesn't apply to pseudo-metric tensors.) – user12262 Mar 19 '16 at 06:55
  • Travis: "your argument" --$$\text{Sin}[~\angle PAQ~] :=$$ $$\left( \frac{(2~s^2[~A,P~]~s^2[~A,Q~]+2~s^2[~A,P~]~s^2[~P,Q~]+2~s^2[~A,Q~]~s^2[~P,Q~]-(s^2[~A,P~])^2-(s^2[~A,Q~])^2-(s^2[~P,Q~])^2)}{(4~s^2[~A,P~]~s^2[~A,Q~])} \right).$$(Recall Heron and "trignometry".) "smoothness of a curve doesn't depend on a metric structure" -- But surely a notion of "smoothness" can be defined from metric structure; or its generalizations. (And I care about what's compatible with that.) – user12262 Mar 19 '16 at 07:03
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The arc length formula is: $s $ $\displaystyle = \int_a^b \sqrt{1+\left(\frac{\mathrm d y}{\mathrm d x}\right)^2} \operatorname d x \\[1ex] \displaystyle = - \int_b^a \sqrt{1+\left(\frac{\mathrm d y}{\mathrm d x}\right)^2} \operatorname d x$

So whether you get a positive or negative result depends on the bounds. There's no guarantee that $a<b$ will give a positive measure.

And similar for the polar coordinate, or parameter vector versions.

Graham Kemp
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    The argument of the square root should be $1 + \left(\frac{dy}{dx}\right)^2$ (not with a $-$ as written); in particular the integrand is always positive, so if $a < b$ the quantity $s$ is always positive. – Travis Willse May 07 '15 at 05:40