Simple proof of the Cauchy-Crofton formula on the sphere?

Question

Let $\gamma$ be a regular curve on the sphere. In a lecture, the following result was used

$$L(\gamma)=\frac 14 \int_{S^2} \sharp (\gamma \cap \xi ^\perp)d\xi$$

$\xi^\perp$ is the plane with normal $\xi$ going through the origin. $\sharp(\gamma\cap \xi^\perp)$ is the number of points in the intersection of the curve $\gamma$ and the plane $\xi^\perp$.

We're saying we can get the length of $\gamma$ by integrating the function $\xi\mapsto \sharp(\gamma\cap \xi^\perp)$ which counts intersections of $\gamma$ with moving planes. By symmetry this function is even, which means we count every point twice, and I can see why there should be a factor of $\frac 12$. I'm not sure about the $\frac 14$.

I'm having a hard time finding a proof for this result. What is a simple method of proving it?

Answer 1

Consider first a great circle. Each plane meet it in exactly 2 points, and the result follows. For a small arc of length $2\pi \over q$, the result follows because you need $q$ such arcs to cover the circle. The additivity give you the result for arcs of length ${p\over q}\times 2\pi$, and by continuity, for every arc of circle. Now, the result follows by writing a Riemann sum of the rhs as the Crofton formula for a broken geodesic approximating the curve : let $\alpha$ be fixed. Note that for a sufficiently small arc of a $C^1$ curve between two points $a,b$, the number of intersection with a given circle is the same (0 or 1) as its intersection with the geodesic $[a,b]$ if the angle with the direction $\vec { ab}$ is $>\alpha$. Now the total area of part of the sphere made of vector which make an angle $\leq \alpha$ with a given direction is $C\alpha$. If teh curve is nice, you can compute the RHS of Crofton formula, the number $\sharp (\gamma \cup \xi ^\perp)$ is bounded, and the result follows.

Answer 2

For convenience we let $\xi^\perp =\{v\in \mathbb{S}^2| \langle v,x\rangle =0\}$

If $f: \mathbb{S}^2\rightarrow \mathbb{R}$ by $ f(\xi ) =\sharp (\gamma\cap \xi^\perp)$, then note that the restriction $$f|\gamma(t)^\perp \geq 1$$ by $\gamma(t)$ : If $\xi\in \gamma(t)\perp$, then $\xi^\perp $ contains $ \gamma(t)$

We are sufficient to consider the case where $$\gamma(t) = (0,\cos\ t,\sin\ t),\ 0\leq t\leq \theta$$

Then $ f((\pm 1,0,0))=\infty$ and note that $A=\{v\in \mathbb{S}^2| \langle v,\gamma (t)\rangle \geq 0\ {\rm for\ some}\ t\}$ is a union of two lunes whose central angle is $t$.

Note that $f$ has value $1$ on $A-\{(1,0,0),(-1,0,0)\}$ and $f$ has value $0$ on the compliment of $A$. Here $$ \frac{1}{4}\int_{\mathbb{S}^2} f(\xi)d\xi = \frac{1}{4}\int_{A} f(\xi)d\xi = \frac{1}{4} \frac{2t}{2\pi} {\rm area}\ \mathbb{S}^2= t ={\rm length} (\gamma )$$ which completes the proof.