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When one formulates the Green Theorem the phrase "curve positively oriented" comes up. After a thorough google search the only description of the above phrase seems to be:

"A curve has positive orientation if a region $R$ is on the left when traveling around the outside of $R$, or on the right when traveling around the inside of $R$. (taken from http://mathworld.wolfram.com/CurveOrientation.html)

This is of course a not rigorous definition as the terms "left,right,outside,inside" are not (to my knowledge) properly defined in the language of mathematics. My question is, what is the rigorous definition of orientation of a curve in $\mathbb{R}^n$?

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2 Answers

up vote 2 down vote accepted

A curve as a "drawn" subset of ${\mathbb R}^2$ has no a priori orientation.

On the other hand, consider a $C^1$-curve $\gamma$ given by a regular parametrization $$\gamma:\quad [a,b]\to{\mathbb R}^2\ ,\qquad t\mapsto z(t)=\bigl(x(t),y(t)\bigr)\ .\qquad(1)$$ Here $z(b)=z(a)$ is allowed, and regular means that $\dot z(t)\ne(0,0)$ for all $t\in[a,b]$. Such a presented $\gamma$ is by definition oriented according to increasing $t$, which means that for any $t\in[a,b]$ the vector $$\dot z(t)=\bigl(\dot x(t),\dot y(t)\bigr)=\lim_{h\to 0+}{z(t+h)-z(t)\over h}$$ is considered as pointing "forward", or in the positive direction.

Any region $R$ in the plane has a boundary $\partial R$, which a priori is only a certain subset of ${\mathbb R}^2$. In many cases (e.g., if $R$ is the unit disk $D$) this boundary set can be presented as a regular closed curve $\gamma$. This curve is positively oriented with respect to $R$, if $R$ is to the left of $\gamma$ (this is not as Wolfram has it).

In terms of formulas this can be expressed as follows: For any $t\in[a,b]$ the vector $\dot z(t)$ is a tangent vector in the positive direction, and $n(t):=\bigl(-\dot y(t),\dot x(t)\bigr)$ is the vector obtained by turning $\dot z(t)$ anticlockwise by $90^\circ$, or "to the left". So $\gamma$ is positively oriented with respect to $R$ if for all $t\in[a,b]$ and sufficiently small $\epsilon>0$ the point $z(t)+\epsilon\ n(t)$ lies in $R$.

When it turns out that the first obtained presentation $(1)$ of $\partial R$ has the wrong orientation with respect to $R$ (i.e., $R$ lies to the right of $\gamma$) then one can easily reverse $\gamma$ by considering the new presentation $$\hat \gamma:\quad [-b,-a]\to{\mathbb R}^2\ ,\qquad t\mapsto z(t)=\bigl(x(-t),y(-t)\bigr)\ .$$

Only if the curve $\gamma=\partial R$ is positively oriented with respect to $R$ Green's formula $$\int_{\partial R} (P\ dx+Q\ dy)=\int_R(Q_x-P_y)\ {\rm d}(x,y)$$ holds.

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When does a curve $\gamma$ have the oposite orientation? –  Nameless Oct 24 '12 at 17:34
    
I suppose you say the same but with $n(t)=(y^{\prime},-x^{\prime}$ for negative orientation. Now how do we prove that a given smooth closed curve with a regular parametrization is either positive or negative oriented? It doesn't seem obvious... –  Nameless Oct 24 '12 at 18:31
    
@Nameless: See my edit concerning reversing $\gamma$. –  Christian Blatter Oct 24 '12 at 19:15
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You can only give a rigorous definition of orientation for a plane curve. Given such a curve, as a continuous function $\gamma\colon[a,b]\to\mathbb{R}^2$ with $\gamma(a)=\gamma(b)$, you can define the index of this curve with respect to any point $c$ not on the curve as follows: Write $$\gamma(t)-c=\lvert\gamma(t)-c\rvert (\cos\theta(t),\sin\theta(t))$$ where $\theta\colon[a,b]\to\mathbb{R}$ is continuous. Such a function $\theta$ exists, and is unique up to an additive constant multiple of $2\pi$. The index of $\gamma$ with respect to $c$ is $$\operatorname{ind}_c(\gamma)=\frac{\theta(b)-\theta(a)}{2\pi}.$$ The curve is positively oriented if the index is $+1$ for points $c$ inside the curve, and negatively oriented if it is $-1$.

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By a plane curve you mean a smooth closed curve on $\mathbb{R}^n$ or exclusively on $\mathbb{R}^2$? –  Nameless Oct 22 '12 at 17:48
    
Exclusively on $\mathbb{R}^2$. Smoothness is not required; continuity is all that's needed. (But the curve should be simple, so Jordan's curve theorem can do its job.) –  Harald Hanche-Olsen Oct 22 '12 at 18:20
    
I see. Thank you for your answer! –  Nameless Oct 22 '12 at 18:21
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