Greens theorem, Curvature and Winding

Greens theorem

Theorem 1 (Greens theorem).

Let $U \subseteq \R^2$ be a bounded connected domain whose boundary is a finite union of connected, piecewise $C^1$ curves. Let $\Gamma$ be the exterior boundary, and $\Gamma_1$, …, $\Gamma_N$ be the interior boundaries. If $F:\bar U \to \R^2$ is $C^1$, then \begin{equation} \int_U \paren[\big]{\partial_1 F_2 - \partial_2 F_1} \, dA = \oint_{\Gamma} F \cdot d\ell + \sum_{i = 1}^N \oint_{\Gamma_i} F \cdot d\ell, \end{equation} where all line integrals above are computed by traversing the exterior boundary counter clockwise, and every interior boundary clockwise.

Remark 2. A common convention is to denote the boundary of $U$ by $\partial U$ and write \begin{equation} \partial U = \Gamma \cup \paren[\big]{ \bigcup_{i = 1}^N \Gamma_i }. \end{equation} Then Theorem 1 becomes \begin{equation}\label{e:greens}\tag{G} \int_U \paren[\big]{\partial_1 F_2 - \partial_2 F_1} \, dA = \oint_{\partial U} F \cdot d\ell, \end{equation} where again the exterior boundary is oriented counter clockwise and the interior boundaries are all oriented clockwise.

Remark 3. In the differential form notation, Greens theorem is stated as \begin{equation} \int_U \paren[\Big]{ \partial_x Q - \partial_y P } \, dA = \int_{\partial U} P \, dx + Q \, dy, \end{equation} $P, Q:\bar U \to \R$ are $C^1$ functions. (We use the same assumptions as before on the domain $U$, and orientations of the line integrals on the boundary.)

Remark 4. Greens theorem is still true if $U$ is not connected; provided you orient the boundaries correctly. The exterior boundary of each connected component needs to be oriented counter clockwise, and all interior boundaries are oriented clockwise.

Example 5 (Failure in unbounded domains). Let $F(x) = (e^{-x_1^2}, 0)$, and $U = \R^2_{+}$. Then \eqref{e:greens} does not hold.

Proposition 6. If $U \subseteq \R^2$ is an unbounded domain and $F \colon \bar U \to \R^2$ is a $C^1$ function such that \begin{equation} \lim_{R \to \infty} \frac{\arclen(\partial U \cap B(0, R))}{R} = 0 \,, \quad\text{and}\quad \lim_{\abs{x} \to \infty} \abs{x}\abs{F(x)} = 0 \,, \end{equation} then Greens theorem (i.e. identity \eqref{e:greens}) holds.

Remark 7. In Example 5 if we instead had $F(x) = (e^{-\abs{x}^2}, 0)$ then Greens theorem does hold.

Example 8. Let $N(x) = \frac{1}{2\pi} \ln \abs{x}$, which is $C^\infty$ on $\R^2 - 0$. Let $U = B(0, 1)$, and $F = \grad^\perp N = (-\partial_2 N, \partial_1 N)$. Clearly $F$ is not defined at $0$; however single points don’t affect area integrals, so let’s ignore that and compute both sides of \eqref{e:greens}. They aren’t equal, showing that Greens theorem may fail even if $F$ is not $C^1$ at even one point in $U$.

Proof of Greens theorem

Lemma 9. Greens theorem is true if $U = (0,1)^2$ .

Proof sketch. The fundamental theorem of calculus gives \begin{equation} \int_U \paren[\big]{ \partial_1 F_2 - \partial_2 F_1 } \, dA = \int_{y = 0}^1 \paren[\big]{ F_2(1, y) - F_2(0, y) } \, dy - \int_{x = 0}^1 \paren[\big]{ F_1(x, 1) - F_1(x, 0) } \, dx \,, \end{equation} and check this is what we need.

Lemma 10. Let $(\Gamma_1, \hat \tau) \subseteq \R^m$ be an oriented curve, $U \subseteq \R^m$ be an open set such that $\bar \Gamma_1 \subseteq U$ and $\varphi \colon \R^m \to \R^n$ is a $C^1$, injective function such that $D \varphi \hat \tau \neq 0$. Let $\Gamma_2 = \varphi(\Gamma_1)$ and suppose $F \colon \Gamma_2 \to \R$ is continuous. Then \begin{equation} \int_{\Gamma_2} F \cdot d\ell = \int_{\Gamma_1} (D \varphi)^T F \circ \varphi \cdot d\ell \,. \end{equation}

Proof sketch. Let $\gamma_1$ parametrize $\Gamma_1$, and note $\varphi \circ \gamma_1$ parametrizes $\Gamma_2$.

Lemma 11. Let $S = (0, 1)^2$. Greens theorem holds on a domain $U$ if there exists a $C^2$ coordinate transformation $\varphi \colon S \to U$, such that $\varphi$ maps $\partial S$ to $\partial U$ and preserves the orientation of the boundaries.

Proof sketch.

  1. Orientation preserving implies $\det D\varphi > 0$ in $S$.
  2. Lemmas 9 and 10 imply \begin{equation} \oint_{\partial U} F \cdot d\ell = \oint_{\partial S} (D \varphi)^T F \circ \varphi \cdot d\ell = \int_S (\partial_1 G_2 - \partial_2 G_1) \, dA, \end{equation} where \begin{equation} G = (D \varphi)^T F \circ \varphi = \sum_{i,j} \partial_i \varphi_j F_j \circ \varphi \, e_i \,. \end{equation}
  3. Now we compute using the chain rule \begin{equation} \partial_1 G_2 - \partial_2 G_1 = \sum_{i, j} \partial_2 \varphi_j \, \partial_i F_j\bigr|_\varphi \, \partial_1 \varphi_i - \partial_1 \varphi_j \, \partial_i F_j\bigr|_\varphi \, \partial_2 \varphi_i = \paren[\big]{ \partial_1 F_2 - \partial_2 F_1 }\circ \varphi \, \det(D \varphi). \end{equation}
  4. Thus, by the change of variable theorem, \begin{equation} \int_S \paren[\big]{\partial_1 G_2 - \partial_2 G_1} \, dA = \int_S \paren[\big]{ \partial_1 F_2 - \partial_2 F_1 }\circ \varphi \, \det(D \varphi) \, dA = \int_U \paren[\big]{ \partial_1 F_2 - \partial_2 F_1} \, dA \,, \end{equation} finishing the proof.

Proof sketch of Theorem 1. Make one or more cuts and divide $U$ into regions where Lemma 11 applies. (Making cuts introduces internal boundaries. The line integrals on these pieces will cancel as each boundary introduced is traversed twice in opposite directions.)

Irrotational fields

Definition 12. Let $U \colon \R^2$ be open. A $C^1$ vector field $F \colon U \to \R^2$ is called irrotational if $\partial_1 F_2 - \partial_2 F_1 = 0$.

Remark 13. In three dimensions, a vector field is irrotational if $\curl F = 0$.

Definition 14 (Simply connected). A region $U \subseteq \R^d$ is called simply connected if every closed curve in $U$ bounds a surface that is completely contained in $U$.

Example 15.

  1. $\R^d$ is simply connected.
  2. $\R^3 - 0$ is simply connected, but $\R^2 - 0$ is not simply connected.
  3. $\R^3 - \set{x_1 = 0}$ is not simply connected.

Proposition 16. Any $C^1$ potential force is irrotational.

Proposition 17. If $U \subseteq \R^2$ is simply connected, and $F \colon U \to \R^2$ is irrotational, then $F$ is a potential force.

Remark 18. If $U$ is not simply connected, there will necessarily exist an irrotational force which is not conservative. The proof of this requires some topology, and we return to it below.

Example 19. Let $U = \R^2 - 0$ (or $\R^2 - B(0, 1)$, if you want a $C^1$ boundary), and $F(x) = x^\perp / \abs{x}^2 = (-x_2, x_1) / |x|^2$. Then $F$ is irrotational but not a potential force.

Computing area

Corollary 20. If $U \subseteq \R^2$ is bounded with a $C^1$ boundary, then \begin{equation} \area(U) = \frac{1}{2} \int_{\partial U} \paren[\big]{ -y \, dx + x \, dy } = \int_{\partial U} -y \, dx = \int_{\partial U} x \, dy \end{equation}

Remark 21. A planimeter is a measuring instrument used to determine the area of an arbitrary two-dimensional shape. The operational principle of the planimeter can be proved using the previous corollary.

Corollary 22 (Surveyor’s formula). Let $P \subseteq \R^2$ be a (not necessarily convex) polygon whose vertices, ordered counter clockwise, are $(x_1, y_1)$, …, $(x_N, y_N)$. Then $$ \area(P) = \frac{(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + \cdots + (x_N y_1 - x_1 y_N)}{2}. $$

Problem 23. Let $a, b > 0$. Find the area enclosed by the ellipse $x^2 / a^2 + y^2 / b^2 = 1$.

Problem 24. Let $\Gamma$ be the curve parametrized by $\gamma(t) = ( a \cos^3 t, a \sin^3 t )$, for $t \in [0, 2\pi]$. Compute the area of the region enclosed by $\Gamma$.

Theorem 25 (Divergence theorem (2D)). Let $U \subseteq \R^2$ be a bounded open set with piecewise $C^1$ boundary. At every point on $\partial U$, let $\hat n$ be the outward pointing unit normal vector. Let $F \colon U \to \R^2$ be a $C^1$ function. Then \begin{equation} \int_U \dv F \, dx = \int_{\partial U} F \cdot \hat n \, \abs{d\ell} \,. \end{equation}

Harmonic functions

Proposition 26 (Greens identity). Let $U \subseteq \R^2$ be a bounded domain whose boundary is a (piecewise) $C^1$ curve, and let $\hat n$ denote the outward pointing normal on $\partial U$. If $f, g: \bar U \to \R$ be $C^2$, show \begin{equation*} \int_U (f \lap g) \, dA = \int_{\partial U} f \grad g \cdot \hat n \, \abs{d\ell} - \int_U (\grad f \cdot \grad g) \, dA. \end{equation*} (Recall $\lap g$ is the Laplacian of $g$ defined by $\lap g = \partial_1^2 g + \partial_2^2 g$.)

Definition 27. A function $u$ is called harmonic if it is $C^2$ and $\lap u = 0$.

Harmonic functions arise in many situations (such as steady states of the heat / wave equations), or the expected reward function for Brownian motion.

Theorem 28 (Mean value property). Let $R > 0$, $a \in \R^2$ and $u \colon \overline{B(a, R)} \to \R$ be a harmonic function. Show that \begin{equation*} u(a) = \frac{1}{2\pi R} \int_{\partial B(a, R)} u \, \abs{d \ell }\,. \end{equation*}

Proof sketch. Let $N(x) = \frac{1}{2\pi} \ln \abs{x - a}$ (this is called the Newton potential). Note $\lap N = 0$ on $\R^2 - 0$. Now for any $\epsilon > 0$, use Greens identity to show that \begin{equation} \int_{\partial B(a, R)} u \cdot \grad N \cdot \hat n \, \abs{d\ell} = \int_{\partial B(a, \epsilon)} u \cdot \grad N \cdot \hat n \, \abs{d\ell} \end{equation} and send $\epsilon \to 0$.

Curvature

Let $\gamma:\R \to \R^d$ be a $C^2$ function, and suppose $\gamma(t)$ represents the position of a particle at time $t$. Let $v(t) = \gamma’(t)$ denote the velocity, and $a(t) = v’(t)$ denote the acceleration.

Problem 29. If $\abs{v(t)} = 1$ for all $t$, then show that $a(t)$ and $v(t)$ are orthogonal vectors.

If a particle moves in a manner that keeps the magnitude of the velocity constant but not the direction, then it will experience acceleration. The magnitude of the acceleration experienced depends on the shape of the path taken. The more curved the path, the higher the acceleration (think about the force you feel when driving around a sharp curve).

Definition 30. If a $C^2$ curve $\Gamma$ is parametrized by the function $\gamma$, then we define the curvature at the point $\gamma(t)$ by \begin{equation} \kappa(\gamma(t)) = \frac{1}{\abs{\gamma’(t)}} \abs[\Bigg]{ \paren[\bigg]{\frac{\gamma’(t)}{ \abs{\gamma’(t)}}}’ } = \frac{1}{\abs{v(t)}} \abs[\Bigg]{ \paren[\bigg]{ \frac{v(t)}{\abs{v(t)}} }’} \,. \end{equation}

Note that if $\abs{\gamma’(t)} = 1$ for all $t$, then the curvature is exactly the magnitude of the acceleration. Also, as $\gamma’$ appears in the denominator, we’ll only consider functions $\gamma$ which always have a non-zero velocity.

Proposition 31. The curvature $\kappa$ is independent of the parametrization. That is if $\beta, \gamma$ are two $C^2$ parametrizations of a $C^2$ curve $\Gamma$, with $\beta(0) = \gamma(0) \in \Gamma$, then $\kappa(\beta(0)) = \kappa(\gamma(0))$.

Remark 32. Since the curvature is independent of the parametrization, given $x \in \Gamma$, one can define the curvature $\kappa(x)$ to be $\kappa(\gamma(0))$ for any $C^2$ parametrization of $\Gamma$ such that $\gamma(0) = x$.

Example 33. If $\Gamma \subseteq \R^2$ is a circle with radius $r$, then $\kappa(x) = 1/r$ for every $x \in \Gamma$.

Problem 34. Compute the curvature of the curve $y^2 = x^2 - 1$ at the point $(x, 1)$.

Problem 35. If $d = 3$, and $\gamma$ is a parametrization of $\Gamma$, let $v = \gamma’$ and $a = \gamma\dprime$. Show that , and hence conclude \begin{equation} \kappa^2 = \frac{\abs{a}^2 \abs{v}^2 - (a \cdot v)^2}{\abs{v}^{6}} \quad\text{and hence}\quad \kappa = \frac{\abs{a \times v}}{\abs{v}^3} \,. \end{equation}

Lemma 36. Let $f:\R \to \R$ be $C^2$, and $\Gamma = \set{ (x, f(x)) \st x \in \R }$ be the graph of $f$. Show that the curvature of $\Gamma$ is given by \begin{equation} \kappa = \frac{\abs{f\dprime}}{(1 + (f’)^2)^{3/2}} \,. \end{equation}

Proposition 37. Let $g:\R^2 \to \R$ be $C^2$, $c \in \R$ and $\Gamma$ be the curve $\set{(x, y) \st g(x, y) = c }$. Suppose further $\grad g \neq 0$ on the curve $\Gamma$. Then the curvature of $\Gamma$ is \begin{equation} \frac{\abs{ \partial_x^2 g (\partial_y g)^2 + \partial_y^2 g (\partial_x g)^2 - 2\partial_x \partial_y g \partial_x g \partial_y g}}{ \abs{\grad g}^3 } \,. \end{equation}

Proof sketch. Let $(a, b) \in \Gamma$, and suppose $\partial_y g(a, b) \neq 0$. Use the implicit function theorem to write $\Gamma$ as the curve $y = f(x)$ near the point $(a, b)$. Now express $f\dprime$ in terms of derivatives of $g$ by implicit differentiation, and use the previous part.

Winding number

Definition 38. Let $\gamma \colon [a, b] \to \R^2 - 0$ be a piecewise $C^1$ function. Define the winding number of $\gamma$ about the origin by \begin{equation} W(\gamma) \defeq \frac{1}{2\pi} \int_a^b \frac{-\gamma_2 \gamma_1’ + \gamma_1 \gamma_2’ }{\abs{\gamma}^2} \, dt \end{equation}

Remark 39. Note for $W(\gamma)$ to be defined we need to ensure that the parametrized curve $\gamma$ doesn’t pass through the origin.

Definition 40. If $(\Gamma, \hat \tau)$ is a (piecewise) $C^1$ oriented curve in $\R^2$ that doesn’t contain the origin then define \begin{equation} W(\Gamma) \defeq \frac{1}{2\pi} \int_\Gamma \frac{x^\perp}{\abs{x}^2} \cdot dx = \frac{1}{2\pi} \int_\Gamma \frac{-x_2 \, dx_1 + x_1 \, dx_2}{\abs{x}^2} = W(\gamma)\,, \end{equation} where $\gamma$ is any parametrization of $(\Gamma, \hat \tau)$.

Example 41. Let $n \in \N$ and $\gamma \colon [0, 1] \to \R^2$ be defined by $\gamma(t) = (\cos(2\pi n t), \sin( 2\pi n t))$. Then $W(\gamma) = n$.

Example 42. Let $a \in \R^2 - 0$, $r < \abs{a}$, $n \in \N$, and $\gamma \colon [0, 1] \to \R^2$ be defined by \begin{equation} \gamma(t) = a + (r \cos(2\pi n t), r \sin( 2\pi n t)) \,. \end{equation} Then $W(\gamma) = 0$.

Lemma 43. Let $U = \R^2 - \set{x_1 \leq 0}$, and $\theta \colon U \to (-\pi , \pi)$ be the angle function (i.e. $\theta(x) = \tan^{-1}(x_2 / x_2)$ for $x_1 > 0$, and extended in the usual way to all of $U$). Let $a, b \in U$ and $\Gamma \subseteq U$ be a curve that goes from $a$ to $b$ (i.e. $\Gamma$ is an oriented curve that starts at $a$ and ends at $b$). Then $2 \pi W(\Gamma) = \theta(b) - \theta(a)$.

Remark 44. This means that for curves that stay in $U$, $2 \pi W(\Gamma)$ is the total angle swept out by the curve. Note, while the function $\theta$ from Lemma 43 is defined on $U$, we compute $\grad \theta = x^\perp / \abs{x}^2$, which is defined on all of $\R^2 - 0$. So it’s tempting to say \begin{equation} W(\Gamma, \tau) = \oint_\Gamma \grad \theta \cdot d\ell = 0 \,, \end{equation} by the fundamental theorem for line integrals. This, however is not valid, as $\theta$ is not defined on all of $\R^2 - 0$, even though our formula for $\grad \theta$ was; so the application of the fundamental theorem is incorrect.

Remark 45. Given that curves in $U$ can’t “wind around” the origin, the total angle is well defined and can be computed using the fundamental theorem. If we look at arbitrary curves in $\R^2 - 0$, they could wind around the origin and we claim $W(\Gamma)$ counts the number of times they wind around the origin.

Theorem 46. If $(\Gamma, \hat \tau)$ is an oriented closed curve in $\R^2 - 0$ then $W(\Gamma) \in \Z$.

Proof sketch of Theorem 46. Let $U_1 = \R^2 - \set{x_1 \leq 0}$ and $U_2 = \R^2 - \set{x_1 \geq 0}$. Parametrize $\Gamma$ by $\gamma \colon [0, 1] \to \R^2$ and divide $[0, 1]$ into smaller intervals $[t_n, t_{n+1}]$ so that on the image of $\gamma$ on each subinterval is contained entirely within $U_1$, or entirely within $U_2$, and $\gamma(t_{n})$ is on the $x$-axis for even $n$ and $y$-axis for odd $n$. Now use Lemma 43 (and it’s analog for the set $U_2$) to show that the winding number of $\gamma$ in this piece is $\pm 1/4$.

Proposition 47. Let $U$ be a bounded domain such that $\partial U$ is a connected $C^1$ curve.

  1. If $0 \not\in \bar U$ then $W(\partial U) = 0$.
  2. If $0 \in U$ and $\partial U$ is oriented counter clockwise, then $W(\partial U) = 1$.

Hint. Find a way to use Greens theorem.

Brower fixed point theorem

Theorem 48 (Brower fixed point theorem). Let $D = \set{ x \in \R^2 \st \abs{x} \leq 1}$. If $F \colon D \to D$ is $C^1$ function then there exists $x \in D$ such that $F(x) = x$.

Remark 49. The Brower fixed point theorem is true even if $F$ is just continuous, and not $C^1$. However, the tools we’ve developed only work with $C^1$ functions.

The proof is based on two ideas:

  1. If you can continuously deform one closed curve into another, without passing through the origin, and by always remaining a closed curve, then the two curves have the same winding number.
  2. If $F$ has no fixed points, we can deform the circle to a point (as above). The circle has winding number $1$, and the point has winding number $0$, so this is a contradiction.

Definition 50 (Free homotopy). Suppose $\gamma_1, \gamma_2 \colon [0, 1] \to \R^2 - 0$ are two $C^1$ functions such that $\gamma_1(0) = \gamma_1(1)$ and $\gamma_2(0) = \gamma_2(1)$. A $C^1$ free homotopy between $\gamma_1$ and $\gamma_2$ is a $C^1$ function $H \colon [0, 1]^2 \to \R^2 - 0$ such that $H(s, 0) = \gamma_1(s)$, $H(s, 1) = \gamma_2(s)$ and $H(0, t) = H(1, t)$ for all $s, t \in [0, 1]$.

Remark 51. The condition $\gamma_i(0) = \gamma_i(1)$ requires both $\gamma_1$ and $\gamma_2$ to be closed (parametrized) curves that don’t pass through the origin. The homotopy $H$ allows you to deform $\gamma_1$ to $\gamma_2$ in a continuous way, so that it is always remains a closed curve that doesn’t pass through the origin.

Lemma 52 (Homotopy lemma). If $\gamma_1$ and $\gamma_2$ are freely homotopic, then $W(\gamma_1) = W(\gamma_2)$.

Proof sketch. Let $\gamma_t(s) = H(s, t)$, and set $w(t) = W(\gamma_t)$. Show that $w$ is continuous and integer valued, and hence constant.

Remark 53. One could define winding number by counting the number of times a curve hits coordinate axis, just as we did in the proof of Theorem 46. This certainly gives us an integer. The advantage of the formula used in Definition 38 is that one can use it to show that $w(t)$ in the above proof is continuous, and hence prove the Lemma 52.

Proof sketch of Theorem 48. Suppose $F$ doesn’t have a fixed point. Given $x \in D$, $F(x) \neq x$ so we can join $x$ and $F(x)$ by a straight line and continue the straight line (on the $x$ side) until it meets $\partial D$. Let $g(x) \in \partial D$ be the point where it meets (there’s an explicit formula for this). Writing out $g$ we can confirm that if $F$ is $C^1$, then $g \colon D \to \partial D$ is also $C^1$. Also, for every $x \in \partial D$, $g(x) = x$ (this uses the fact that $F(x) \in D$).

Now let let $\gamma(s) = (\cos(2\pi s), \sin(2\pi s))$ and define \begin{equation} H(s, t) = g( t \gamma(s) ) \,, \end{equation} and note $H$ gives a free homotopy between $\gamma$ and the curve constant curve $s \mapsto g(0)$ without ever passing through $0$. Since the two curves have different winding numbers this is a contradiction.

Jordan curve theorem

Theorem 54 (Jordan curve theorem). If $\Gamma \subseteq \R^2$ is a closed curve, then $\R^2 - \Gamma$ is the disjoint union of two connected open sets, one bounded one unbounded.

Remark 55. The bounded open set above is called the region enclosed by $\Gamma$. This was originally proved in 1887 by Jordan for continuous curves. Veblen (and several others) claimed there was a gap in his proof, and published a proof 1905 that was correct. Recently, however, Tom Hales revisited Jordan’s original proof, and found it to be correct. Nevertheless, for $C^1$ curves, with winding numbers it is possible to find a simpler proof.

Proof sketch. Suppose first $\Gamma$ is oriented, so we can define line integrals. For $x \in \R^2$, let $\Gamma - x$ be the curve $\Gamma$ shifted by $x$ (i.e. $\Gamma - x = \set{y - x \st y \in \Gamma}$). If $x \not \in \Gamma$, then $\Gamma - x$ doesn’t contain the origin, and so we can define let $F(x) = W(\Gamma - x)$. We claim $U = \set{x \not\in \Gamma \st W(\Gamma - x) = 0}$ is the unbounded component, and $V = \set{x \not\in \Gamma \st W(\Gamma - x) \neq 0}$ is the bounded component. The proof of this is within reach of what we’ve covered, but a bit tricky to get completely correct.

Using the Jordan curve theorem we can prove the existence of irrotational but not potential forces on non-simply connected domains (Remark 18)

Proposition 56. Suppose $U \subseteq \R^2$ is an open connected region, and $\partial U$ is a finite union of piecewise $C^1$ curves. If $U$ is not simply connected, there exists an irrotational force on $U$ which is not a potential force.

Proof sketch. Let $\Gamma \subseteq U$ be a closed curve. By Theorem 54 $\Gamma$ bounds a region $V \subseteq \R^2$. Since $U$ is not simply connected, there must exist some curve $\Gamma$ such that the region it bounds $V \not\subseteq U$. Pick $a \in V - U$, and set $F(x) = \grad^\perp N(x - a) = (-\partial_2 N, \partial_1 N)$, where $N(y) = \frac{1}{2\pi} \ln \abs{y}$ is the Newton potential.

Turning number

Definition 57. Let $\gamma\colon [0, 1] \to \R^2$ be a $C^2$ function such that $\gamma’ \neq 0$, and $\gamma(0) = \gamma(1)$. Let $v = \gamma’$, and $a = v’$, $\hat n = (-v_2, v_1) / \abs{v}$ and define the signed curvature at the point $\gamma(t)$ by \begin{equation} \kappa_s(\gamma(t)) = \frac{1}{\abs{v(t)}} \paren[\bigg]{\frac{v(t)}{\abs{v(t)}}}’ \cdot \hat n \,. \end{equation}

Remark 58. Clearly $\abs{\kappa_s} = \kappa$, where $\kappa$ is the curvature of $\Gamma$. Similar to Proposition 31 one can show that the signed curvature of an oriented curve $(\Gamma, \hat \tau)$ independent of the parametrization.

Definition 59. Given a curve $\Gamma \subseteq \R^2$, define the turning number \begin{equation} T(\Gamma) \defeq \frac{1}{2\pi} \int_\Gamma \kappa_s \cdot \abs{d\ell} \,. \end{equation}

Proposition 60. If $\Gamma$ is a closed $C^2$ curve, then $T(\Gamma) \in \Z$.

Proof sketch. Relate the turning number of $\Gamma$ to the winding number of $\hat \tau$.

Problem 61. What is the winding number and turning number of the curve shown (with $p = 0$).

Hint. The wikipedia page has good intuition (and the same image) 😄.