On the Crofton Formula April 11, 2018 The Crofton formula is a classical result in geometry. It Introduction roughly states that the length of a regular curve (by ‘regular’ we mean its length can be measured in some sense) can be given by the measure of all lines that intersects this curve (counting multiplicity). The measure that we put on the set of all straight lines is in fact quite natural. Consider the elementary example of the Buffon’s Needle : Given a needle of length l dropped on a plane ruled with parallel lines t units apart, what is the probability that the needle will cross a line? In the following, we assume l < t , so the needle touches the lines at most once. We can parametrize the position of a needle by x := ‘the distance between the midpoint of the needle and the closest line’ and θ := ‘the acute t angle between the needle and the horizontal line’, with 0 ≤ x ≤ 2 , and 0 ≤ θ ≤ π 2 . We may assume that the position of the needle is uniformly distributed. ‘The needle touches the lines’ translates into: x ≤ l 2 sin θ. So we can calculate the probability as follows: � � ( l/ 2) sin θ π tπ dx dθ = 2 l 4 2 P = tπ. θ =0 x =0 The key observation is to read the formula as a way to calculate l . (Though it was first introduced to give experimental values for π .) Since the needle touches the lines only once, we can delete all lines but one. If we fix the needle on the horizontal line and its midpoint at origin, we can view this experiment as throwing the line randomly and observe if it touches the needle. So now x is ‘the distance between the line and the origin’ and θ 4 is ‘the acute angle between the line and the horizontal line’. The factors tπ are introduced in the framework of probability theory, so we can get rid of 1
them at once. We also want to get rid of the upper bound for the integration in the variable x , so we define: N ( x, θ ) := # { intersection points of the needle and the ( x, θ ) line } So we have the formula: � � ∞ π N ( x, θ ) dx dθ = l 2 2 . θ =0 x =0 Now we can define the parametrization similarly for all ‘unoriented’ lines on the plane: ( x, θ ) ∈ [0 , ∞ ] × [0 , 2 π ) corresponds to an ‘unoriented’ line that has distance d from the origin and ‘points’ at (cos( θ ) , sin( θ )).(On u − v plane, the equation is − sin( θ ) u + cos( θ ) v = x ). dxdθ is the measure we mentioned earlier. Another important observation on the Invariance under rigid motion measure is that it is invariant under rigid motion . A rigid motion is a automorphism of E 2 that preserves its metric and orientation. It is given in general by x → A� T ( ρ,� a ) : � x + � a where A is the rotation matrix with angle ρ . Such T ( ρ,� a ) induces a trans- formation on the set of oriented lines (which is transitive), sending { ( u, v ) : − sin( θ ) u +cos( θ ) v = x } to { ( u, v ) : − sin( θ + ρ ) u +cos( θ + ρ ) v = x − sin( θ + ρ ) a 1 + cos( θ + ρ ) a 2 } So the transformation is: T ( ρ,� a )( x, θ ) = ( x − sin( θ + ρ ) a 1 + cos( θ + ρ ) a 2 , θ + ρ ) It is obvious that the jacobian of this transformation is 1. If we define the � measure of a set S of oriented lines to be µ ( S ) := S dxdθ , then we have proved: µ ( S ) = µ ( T ( ρ,� a ) S ). More generally, for any function defined Proposition on S and a rigid motion T , we have � � f ( x, θ ) dxdθ = f ( T ( x, θ )) dxdθ S T − 1 S Proof Formula for change of variables. It is easy to see that dxdθ is the only measure (up to constant) that is invariant under rigid motion. We are now ready to state and prove our theorem. The main theorem Theorem Given a smooth curve γ of finite length in E 2 . Define N γ ( x, θ ) = # { intersection points of γ and the ( x, θ ) line } 2
then we have: � 2 π � ∞ N γ ( x, θ ) dx dθ = 2 l ( γ ) . θ =0 x =0 where l ( γ ) is the length of γ . Proof First, the calculation earlier shows that the formula holds for seg- ments [ − l/ 2 , l/ 2] (by symmetry). For a segment in a general position, we can use a rigid motion to send it to [ − l/ 2 , l/ 2] and use the proposition above to show that the formula still holds. In general, for a smooth curve γ , we can find a series of polygonal paths γ i to approximate the curve. If lim l ( γ i ) = l ( γ ) and lim N γ i ( x, θ ) = N γ ( x, θ ) in suitable sense, the formula is proved. For technical details, we can choose the γ i so that γ i +1 is a refinement of γ i and each segment in γ i is shorter than 2 − i . Then we have the following: 1. lim l ( γ i ) = l ( γ ) 2. For each fixed ( x, θ ), N γ i ( x, θ ) ≥ N γ i − 1 ( x, θ ) 3. For each fixed ( x, θ ), lim N γ i ( x, θ ) = N γ ( x, θ ) and the result follows. Generalization We give a generalization of the theorem to the situation S 2 . The proof is similar to the case above. Theorem Let γ be a smooth curve on the unit sphere. Then � l ( γ ) = 1 S 2 N γ ( ζ ) dζ 4 where N γ ( ζ ) := # { the intersetion points of γ and the plane perpendicular to ζ } Application We’ll give some application of the formula in this section. An ‘elementary’ proposition Between two convex, closed curves, the inner one is shorter. Proof Denote these two curves by Γ and γ , then we necessarily have N Γ ≥ N γ . A discrete version of Fenchel’s theorem For a polygonal closed path a 0 a 1 . . . a n a 0 in E 3 , we can define its total cur- vature by K := � � ( a i − 1 a i , a i a i +1 ). Then we have: K ≥ 2 π a i − 1 a i Proof Let α i := | a i − 1 a i | . Then � K = l ( � α i α i +1 ) 3
where α i α i +1 denotes the minor arc between α i and α i +1 . Now consider � the spheric polyhegon P determined by α i . We claim that any great circle meets P at least twice. This is true because � | a i − 1 a i | α i = 0. Then apply the spheric version of crofton formula. What’s next? We present some other results in integral geometry. (Crofton) Let D be a bounded domain in E 2 . Let S D ( x, θ ) denote the length of l ( x, θ ) ∩ D . Then: � 2 π � ∞ πA ( D ) = S D ( x, θ ) dx dθ θ =0 x =0 where A ( D ) is the area of D . In the following, we use a more general kinematic density (it means a mea- sure put on a set of geometric objects that is invariant under rigid motion). To parametrize O := { all possible positions of a set C ⊂ R 2 } , we have a map: ( θ, a, b ) → ‘the image of C under the compostion of θ -rotation and ( a, b )-transition’. It is readily checked that ‘ dK := da db dθ ’ is invariant un- der rigid motion. Its importance is showed by the following results. (Poincare) Let C and Γ be piecewise C 1 curves. Let O be all possible posi- tion of Γ in E 2 , then we have � #(Γ ′ ∩ C ) dK = 4 l ( C ) l (Γ) O (Santalo) Let Ω 1 and Ω 2 be two convex domain in E 2 . Let O be all possible position of Ω 2 in E 2 , then we have � χ (Ω ′ 2 ∩ Ω 1 � = ∅ ) dK = 2 π [ A (Ω 1 ) + A (Ω 2 )] + l ( ∂ Ω 1 ) l ( ∂ Ω 2 ) O Remark. With the above two formula, we can give a beautiful proof for the isoperimetric inequality. Lemma Let Ω be a convex domain in E 2 with boundary length L and area A . Then 4 πA ≤ L 2 Proof Fix Ω 1 ∼ = Ω ⊂ E 2 . Let O be the set of all possible position of Ω in E 2 . Define m i := µ ( { Ω ′ ∈ O : #( ∂ Ω ′ ∩ ∂ Ω 1 ) = i } ), the measure of all positions such that the boundaries of two domains meets exactly i times. We use the ‘obvious’ fact that m 1 = 0 to have: � 4 l ( ∂ Ω) 2 = #( ∂ Ω 1 ∩ ∂ Ω ′ ) dK = m 1 + 2 m 2 + 3 m 3 + . . . O = 2 m 2 + 3 m 3 + · · · ≥ 2( m 2 + m 3 + . . . ) � χ (Ω 1 ∩ Ω ′ � = ∅ ) dK = 2(4 πA (Ω) + l ( ∂ Ω) 2 ) = 2 O 4
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