Crofton's formula
Today I want to present a result that I think deserves to be in every introductory differential geometry course and its consequences.
Recall that a planar curve $\gamma$ is said to be rectifiable if it can be approximated by a piecewise linear curve such that, as the lengths of the approximating chords gets arbitrarily small, the total length of the approximants is bounded above by some real number $L$. The smallest such $L$ is the length length$(\gamma)$ of $\gamma$.
Theorem [Crofton's formula] Let $\gamma$ be a rectifiable plane curve. Let $n_{\gamma}(\theta, r)$ be the number of intersection points between $\gamma$ and the line $l$ pointing in the direction $\theta$ at signed distance $r$ from the origin. Then \[ length(\gamma) =\frac{1}{4} \int \int n_{\gamma}(\theta,r) d\theta dr. \]
We ignore the measure zero set where there are infinitely many intersection points.
Proof We can approximate $\gamma$ be a piecewise linear curve, and it suffices to prove the formula for such curves. Since both sides of the equation are additive, it suffices to prove this for a single line. We now run this process in reverse! Note that the differential/area form $d\theta dr$ is isometry invariant, so there is some constant $C$ such that for any segment of length 1 $\gamma_0$ $\int_N n_{\gamma_0}(l) d\theta dr =C$. By additivity, for the same constant $C$ we have $\int_N n_{\gamma_0}(l) d\theta dr =C$ length$(\gamma)$ for any segment, and hence for any smooth curve $\gamma$. To compute the constant, consider the unit circle centred at the origin. $\blacksquare$
Now that was very slick, but I think Crofton's formula should be more widely known also because it gives nice proofs of a number of results which aren't so easy to prove otherwise.
Lemma Let $\gamma$ and $\Gamma$ be two nested closed convex curves in the plane of lengths $l$ and $L$ respectively. Then $L \geq l$
Proof: A line intersects a convex curve at exactly two points and every line that intersects the inner curve intersects the outer one as well. Hence $n_{\Gamma} \geq n_{\gamma}$, and the result follows from Crofton's formula. $\blacksquare$
Isoperimetric inequality: Let $\gamma$ be a simple closed plane curve of length $L$ that bounds a region $\Omega$ of area $A$. Then $L^2 \geq 4\pi A$ with equality iff $\gamma$ is a circle.
Proof: One can put a metric on the set of closed subsets of Euclidean space. This in fact forms a compact metric space, so there is a curve $\gamma$ maximising area among all curves of fixed length. Assume $\gamma$ is piecewise smooth. If $\gamma$ isn't convex, there is a line between two points $p_1, p_2$ on the curve which passes through the exterior of the curve. Let points $n_1, n_2$ on the curve be such that the line segment $n_1n_2$ is outside $\Omega$. Reflecting the part of the curve between $n_1$ and $n_2$ about this line gives a curve of the same length and larger area, so we may assume $\gamma$ is convex. Parametrise $\gamma$ be arc length $t$ and for a vector $v$ of unit length pointing from a point $p(t)$ on $\gamma$ let $\alpha$ be the angle between $v$ and the positive tangent line of $\gamma$. Let $f(t,\alpha)$ be the length of the segment at angle $v$ from $p(t)$. Let $M$ denote the space of possible choices of $t, \alpha$. For independent copies $M_1, M_2$ with coordinates $(t_1, \alpha_1) $ and $(t_2,\alpha_2)$, we have the following inequality since the integrand is non-negative: \[ 0 \leq \int_{M \times M} (f(t_2,\alpha_2)\sin{\alpha_1} - f(t_1, \alpha_1) \sin{\alpha_2})^2 dt_1 d\alpha_1 dt_2 d\alpha_2 \] By the formula for area in polar coordinates, $\int_0^{\pi} f^2(t, \alpha) d\alpha=2A$ so $\int_M f^2 d\alpha dt=2AL$. Similarly $\int_M \sin^2 d\alpha dt=\frac{\pi L}{2}$. Now consider the space $N$ of oriented lines in the plane. This is characterised by an angle $\theta$ and the signed distance $r$ from the origin as before. Define the map $\lambda: M \to N$ that associates the oriented line with a unit vector. We claim the following:
Let $\omega_M$ denote the area form $\sin{\alpha} d\alpha \wedge dt$. Let $\omega_N$ denote the area form $d\theta \wedge dr$. Then $\lambda^*(\omega_N)=\omega_M$
Proof of claim: Denote by $\psi(t)$ the direction of the positive tangent line to the curve $\gamma$ at point $\gamma(t)$ and let $\gamma_1, \gamma_2$ be the two components of the position vector $\gamma$. By trigonometry one sees that $\theta= \alpha + \psi(t)$ and $r= \gamma \times (\cos{\theta}, \sin{\theta})$. By calculating one sees that $d\theta \wedge dr= (\gamma_1'\sin{\theta}- \gamma_2'\cos{\theta}) d\alpha dt$. But $(\gamma_1',\gamma_2')= (\cos{\psi}, \sin{\psi})$ so by the addition formula for sine one has $(\gamma_1'\sin{\theta}- \gamma_2'\cos{\theta}) =\sin{\alpha} \blacksquare$.
$\int_M f(t, \alpha) \sin{\alpha} d\alpha dt = \int_M f \omega$. Let $h$ be a function on the space of lines $N$ whose value on a line $l$ is the length of its part inside the billiard table. By the previous lemma $\int_M f \omega= \int h dr d\theta$ For a fixed direction $\int h dr$ is the area of the table so $\int_N h dr d\theta = A \int_0^{2\pi} d\theta =2\pi A$. Expanding out the integrand, substituting the values for the integrals and rearranging gives the result. $\blacksquare$
Curves of Constant Width
The next application I want to present concerns so-called curves of constant width.
Let $\gamma$ be a planar closed curve and $v$ be a vector. Let $l_1$ and $l_2$ be distinct lines that are perpendicular to $v$ and tangent to $\gamma$. The width of $\gamma$ in the direction of $v$ is the perpendicular distance between $l_1$ and $l_2$. $\gamma$ is said to be a curve of constant width if its width is the same for all directions.
The easiest example of such a curve is the circle: a circle of radius $r$ has constant width $2r$. However, there are loads of these things. Here's a construction: take a convex polygon $P$ with an odd number of sides, in which each vertex is equidistant to the opposite vertices and closer to all other vertices, e.g. just a regular polygon with an odd number of sides. Replace each side of $P$ by an arc centred at its opposite vertex. This produces what's known as a Releaux polygon, i.e. a curve of constant width made up of circular arcs of constant radius.
Many other constructions, with pretty pictures to accompany them, are available on the Wikipedia page. Even though they are obviously distinct, this requires proof, and one obvious invariant of curves completely fails to distinguish them:
Barbier's theorem: Every curve $\gamma$ of constant width $w$ has perimeter $\pi w$
Proof: Choose an origin inside $\gamma$. Consider the tangent line to $\gamma$ in the direction $\theta$ and let $r(\theta)$ be its distance from the origin. The constant width condition is precisely that $r(\theta)+ r(\theta+ \pi)=w$. By Crofton's formula, \[ length(\gamma) = \frac{1}{4} \int_0^{2\pi}\int_{-r(\theta+\pi}^{r(\theta)}2 dr d\theta = \frac{1}{2}w \int_0^{2\pi}=\pi w \blacksquare \]
As a comment, the Wikipedia page on Buffon's needle suggests an elementary proof of Crofton's formula. Notice however, that the proof also goes through the business of linearity giving the correct expression up to a constant and then determining the constant. I think the proofs are cryptomorphic.
Generalisations
In $R^n$ one can similarly parametrise the set of lines by a direction $\theta$ and a distance from the origin $r$, and the analogous formula \[ area(S) = C_n \int \int n_{\gamma}(\theta,r) d\theta dr. \] holds too, where $C_n$ is $\frac{1}{2 |\text{area of unit ball in}R^{n-1}|}$, holds for any rectifiable codimension 1 hypersurface $S$. The normalisation comes from looking at the case of the unit ball in $R^n$, exactly as in the case of dimension 2. However, the proof method of approximation by hyperplanes given above doesn't work, as shown by the following example:
We approximate a cylinder of radius $r$ and height $l$ by polyhedra such that the area doesn't converge. Fix integer parameters $m$ and $n$. Inscribe a regular $n-$gon in a circle of radius $r$ and construct isosceles triangles of height $\frac{l}{m}$ on each of its edges pointing in the upwards direction perpendicular to the plane of the circle. The vertices of the isosceles triangle form the vertices of another regular $n-$gon. Fill in the triangular gaps between the two $n-$gons Iterate this $m$ times to get $m$ layers of triangles. The area of the polygonal approximation is given by $2mnr \sin{\frac{\pi}{n}}\sqrt{(\frac{l}{m})^2+r^2(1-\cos{\frac{\pi}{n}})^2}$. By choosing $m$ to be a suitable function of $n$, e.g. choosing $m=cn^2$ for different values of $c$, the limit as $n$ tends to infinity converges to arbitrarily large values.
There are further generalisations which go by the name of kinematic formulae, but that's for another time. One can deduce from this generalisation of Crofton's formula the following:
Let $S$ be a convex open set of $R^n$ with compact rectifiable boundary $\partial S$. Let $E(T(S))$ be the expected area of a random orthogonal projection to a hyperplane. Then area$(\partial S) |\text{unit ball in}R^{n-1}|=\int T(S)(\theta) d\theta = E(T(S)) |\text{unit sphere in} R^n|$.
Proof: Apply Crofton's formula to $\partial S$ and integrate over $dr$ then by $d\theta$. The convexity implies that for any direction the integral is exactly twice the area of the projection in that direction. $\blacksquare$
Setting $n=3$ gives the result beautifully animated here.
We end by mentioning a couple of fun results that the reader might enjoy thinking about.
1. A closed curve on the unit sphere which intersects every equator has length at least $2\pi$.
2. Let $Q$ be a unit cube of dimension $n$ and let $H$ be a $k-$dimensional hyperplane passing through its centre. Then the $k-$area of the intersection is at least 1.
Recall that a planar curve $\gamma$ is said to be rectifiable if it can be approximated by a piecewise linear curve such that, as the lengths of the approximating chords gets arbitrarily small, the total length of the approximants is bounded above by some real number $L$. The smallest such $L$ is the length length$(\gamma)$ of $\gamma$.
Theorem [Crofton's formula] Let $\gamma$ be a rectifiable plane curve. Let $n_{\gamma}(\theta, r)$ be the number of intersection points between $\gamma$ and the line $l$ pointing in the direction $\theta$ at signed distance $r$ from the origin. Then \[ length(\gamma) =\frac{1}{4} \int \int n_{\gamma}(\theta,r) d\theta dr. \]
The red points are the intersection points between $\gamma$ and the line, so in this case $n_{\gamma}(\theta, r)=2$.
We ignore the measure zero set where there are infinitely many intersection points.
Proof We can approximate $\gamma$ be a piecewise linear curve, and it suffices to prove the formula for such curves. Since both sides of the equation are additive, it suffices to prove this for a single line. We now run this process in reverse! Note that the differential/area form $d\theta dr$ is isometry invariant, so there is some constant $C$ such that for any segment of length 1 $\gamma_0$ $\int_N n_{\gamma_0}(l) d\theta dr =C$. By additivity, for the same constant $C$ we have $\int_N n_{\gamma_0}(l) d\theta dr =C$ length$(\gamma)$ for any segment, and hence for any smooth curve $\gamma$. To compute the constant, consider the unit circle centred at the origin. $\blacksquare$
Now that was very slick, but I think Crofton's formula should be more widely known also because it gives nice proofs of a number of results which aren't so easy to prove otherwise.
Lemma Let $\gamma$ and $\Gamma$ be two nested closed convex curves in the plane of lengths $l$ and $L$ respectively. Then $L \geq l$
Proof: A line intersects a convex curve at exactly two points and every line that intersects the inner curve intersects the outer one as well. Hence $n_{\Gamma} \geq n_{\gamma}$, and the result follows from Crofton's formula. $\blacksquare$
Isoperimetric inequality: Let $\gamma$ be a simple closed plane curve of length $L$ that bounds a region $\Omega$ of area $A$. Then $L^2 \geq 4\pi A$ with equality iff $\gamma$ is a circle.
Proof: One can put a metric on the set of closed subsets of Euclidean space. This in fact forms a compact metric space, so there is a curve $\gamma$ maximising area among all curves of fixed length. Assume $\gamma$ is piecewise smooth. If $\gamma$ isn't convex, there is a line between two points $p_1, p_2$ on the curve which passes through the exterior of the curve. Let points $n_1, n_2$ on the curve be such that the line segment $n_1n_2$ is outside $\Omega$. Reflecting the part of the curve between $n_1$ and $n_2$ about this line gives a curve of the same length and larger area, so we may assume $\gamma$ is convex. Parametrise $\gamma$ be arc length $t$ and for a vector $v$ of unit length pointing from a point $p(t)$ on $\gamma$ let $\alpha$ be the angle between $v$ and the positive tangent line of $\gamma$. Let $f(t,\alpha)$ be the length of the segment at angle $v$ from $p(t)$. Let $M$ denote the space of possible choices of $t, \alpha$. For independent copies $M_1, M_2$ with coordinates $(t_1, \alpha_1) $ and $(t_2,\alpha_2)$, we have the following inequality since the integrand is non-negative: \[ 0 \leq \int_{M \times M} (f(t_2,\alpha_2)\sin{\alpha_1} - f(t_1, \alpha_1) \sin{\alpha_2})^2 dt_1 d\alpha_1 dt_2 d\alpha_2 \] By the formula for area in polar coordinates, $\int_0^{\pi} f^2(t, \alpha) d\alpha=2A$ so $\int_M f^2 d\alpha dt=2AL$. Similarly $\int_M \sin^2 d\alpha dt=\frac{\pi L}{2}$. Now consider the space $N$ of oriented lines in the plane. This is characterised by an angle $\theta$ and the signed distance $r$ from the origin as before. Define the map $\lambda: M \to N$ that associates the oriented line with a unit vector. We claim the following:
Let $\omega_M$ denote the area form $\sin{\alpha} d\alpha \wedge dt$. Let $\omega_N$ denote the area form $d\theta \wedge dr$. Then $\lambda^*(\omega_N)=\omega_M$
Proof of claim: Denote by $\psi(t)$ the direction of the positive tangent line to the curve $\gamma$ at point $\gamma(t)$ and let $\gamma_1, \gamma_2$ be the two components of the position vector $\gamma$. By trigonometry one sees that $\theta= \alpha + \psi(t)$ and $r= \gamma \times (\cos{\theta}, \sin{\theta})$. By calculating one sees that $d\theta \wedge dr= (\gamma_1'\sin{\theta}- \gamma_2'\cos{\theta}) d\alpha dt$. But $(\gamma_1',\gamma_2')= (\cos{\psi}, \sin{\psi})$ so by the addition formula for sine one has $(\gamma_1'\sin{\theta}- \gamma_2'\cos{\theta}) =\sin{\alpha} \blacksquare$.
$\int_M f(t, \alpha) \sin{\alpha} d\alpha dt = \int_M f \omega$. Let $h$ be a function on the space of lines $N$ whose value on a line $l$ is the length of its part inside the billiard table. By the previous lemma $\int_M f \omega= \int h dr d\theta$ For a fixed direction $\int h dr$ is the area of the table so $\int_N h dr d\theta = A \int_0^{2\pi} d\theta =2\pi A$. Expanding out the integrand, substituting the values for the integrals and rearranging gives the result. $\blacksquare$
Curves of Constant Width
The next application I want to present concerns so-called curves of constant width.
Let $\gamma$ be a planar closed curve and $v$ be a vector. Let $l_1$ and $l_2$ be distinct lines that are perpendicular to $v$ and tangent to $\gamma$. The width of $\gamma$ in the direction of $v$ is the perpendicular distance between $l_1$ and $l_2$. $\gamma$ is said to be a curve of constant width if its width is the same for all directions.
The easiest example of such a curve is the circle: a circle of radius $r$ has constant width $2r$. However, there are loads of these things. Here's a construction: take a convex polygon $P$ with an odd number of sides, in which each vertex is equidistant to the opposite vertices and closer to all other vertices, e.g. just a regular polygon with an odd number of sides. Replace each side of $P$ by an arc centred at its opposite vertex. This produces what's known as a Releaux polygon, i.e. a curve of constant width made up of circular arcs of constant radius.
A Releaux triangle
Many other constructions, with pretty pictures to accompany them, are available on the Wikipedia page. Even though they are obviously distinct, this requires proof, and one obvious invariant of curves completely fails to distinguish them:
Barbier's theorem: Every curve $\gamma$ of constant width $w$ has perimeter $\pi w$
Proof: Choose an origin inside $\gamma$. Consider the tangent line to $\gamma$ in the direction $\theta$ and let $r(\theta)$ be its distance from the origin. The constant width condition is precisely that $r(\theta)+ r(\theta+ \pi)=w$. By Crofton's formula, \[ length(\gamma) = \frac{1}{4} \int_0^{2\pi}\int_{-r(\theta+\pi}^{r(\theta)}2 dr d\theta = \frac{1}{2}w \int_0^{2\pi}=\pi w \blacksquare \]
As a comment, the Wikipedia page on Buffon's needle suggests an elementary proof of Crofton's formula. Notice however, that the proof also goes through the business of linearity giving the correct expression up to a constant and then determining the constant. I think the proofs are cryptomorphic.
Generalisations
In $R^n$ one can similarly parametrise the set of lines by a direction $\theta$ and a distance from the origin $r$, and the analogous formula \[ area(S) = C_n \int \int n_{\gamma}(\theta,r) d\theta dr. \] holds too, where $C_n$ is $\frac{1}{2 |\text{area of unit ball in}R^{n-1}|}$, holds for any rectifiable codimension 1 hypersurface $S$. The normalisation comes from looking at the case of the unit ball in $R^n$, exactly as in the case of dimension 2. However, the proof method of approximation by hyperplanes given above doesn't work, as shown by the following example:
We approximate a cylinder of radius $r$ and height $l$ by polyhedra such that the area doesn't converge. Fix integer parameters $m$ and $n$. Inscribe a regular $n-$gon in a circle of radius $r$ and construct isosceles triangles of height $\frac{l}{m}$ on each of its edges pointing in the upwards direction perpendicular to the plane of the circle. The vertices of the isosceles triangle form the vertices of another regular $n-$gon. Fill in the triangular gaps between the two $n-$gons Iterate this $m$ times to get $m$ layers of triangles. The area of the polygonal approximation is given by $2mnr \sin{\frac{\pi}{n}}\sqrt{(\frac{l}{m})^2+r^2(1-\cos{\frac{\pi}{n}})^2}$. By choosing $m$ to be a suitable function of $n$, e.g. choosing $m=cn^2$ for different values of $c$, the limit as $n$ tends to infinity converges to arbitrarily large values.
There are further generalisations which go by the name of kinematic formulae, but that's for another time. One can deduce from this generalisation of Crofton's formula the following:
Let $S$ be a convex open set of $R^n$ with compact rectifiable boundary $\partial S$. Let $E(T(S))$ be the expected area of a random orthogonal projection to a hyperplane. Then area$(\partial S) |\text{unit ball in}R^{n-1}|=\int T(S)(\theta) d\theta = E(T(S)) |\text{unit sphere in} R^n|$.
Proof: Apply Crofton's formula to $\partial S$ and integrate over $dr$ then by $d\theta$. The convexity implies that for any direction the integral is exactly twice the area of the projection in that direction. $\blacksquare$
Setting $n=3$ gives the result beautifully animated here.
We end by mentioning a couple of fun results that the reader might enjoy thinking about.
1. A closed curve on the unit sphere which intersects every equator has length at least $2\pi$.
2. Let $Q$ be a unit cube of dimension $n$ and let $H$ be a $k-$dimensional hyperplane passing through its centre. Then the $k-$area of the intersection is at least 1.
Comments
Post a Comment