Differentiability of polytope shadow areas

Question

Let $P$ be an opaque convex polyhedron containing the origin in $\mathbb{R}^3$, and let $S$ be an origin-centered sphere strictly containing $P$: $S \supset P$. For a point $x$ on $S$, let $\sigma(x)$ be the area of the shadow of $P$ cast from a light at $x$ onto the plane tangent to $S$ at $-x$:

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My question is:

Q. What is the differentiability class $C^k$ of $\sigma(x): S \to \mathbb{R}$?

I would be surprised if $\sigma$ is a smooth map, $C^\infty$, but it seems to be at least $C^1$...

The question makes sense for a convex polytope in $\mathbb{R}^d$ for $d \ge 2$, with $\sigma(x)$ the $(d{-}1)$-volume of the shadow cast on a $(d{-}1)$ hyperplane.

Answer 1

Let me try to give a computation free (sketch of) proof that the shadow area is not $C^1$. The basic idea is the same as in the answer of Willie Wong: a problem happen when a corner shows up.

Consider a position $x_0$ where some corner $c$ of the polytope is projected right on a facet of the shadow. Then move $x_0$ along a smooth path $x_t$ such that for small negative $t$, $c$ is projected in the interior of the shadow, and for small positive $t$ it is projected "outside the shadow", i.e. it is projected to a vertex of the shadow. Then the area $\sigma(x_t)$ is given by the sum of a smooth function (corresponding to the shadow of the facets already contributing to the shadow at $x_0$) and a function which is zero for $t<0$ (when the facets of $c$ do not contribute to the shadow) and is linear for $t>0$ (when the facets of $c$ start contributing to the shadow). Thus $\sigma$ is not $C^1$ (but it is certainly Lipschitz).

Answer 2

Just doing this very quickly, so maybe computational errors:

In the case $d = 2$, take $P$ as an equilateral triangle centered at the origin inscribed in the unit circle, and $S$ the circle of radius 2, the problem is equivalent to computing the order of tangency of the two functions $$f_1(\theta) = \frac{\sin\theta}{2 - \cos \theta} \quad \text{ and } f_2(\theta) = \frac{\sin(\pi/3 - \theta)}{2 + \cos(\pi/3 -\theta)}$$ at their first positive intersection point.

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  ^{(Image added by J.O'Rourke. They cross at $\theta \approx 0.2709$.)}

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If I did the computations right, this is equivalent to asking for the order of vanishing of the function

$$ 6 \sin \theta + \sqrt{3} - 2 \sqrt{3} \cos\theta $$

at its first positive zero. But using the monotonicity of the $\sin$ and $\cos$ functions it appears that the derivative is strictly positive at the first zero, so that $\sigma$ is not even $C^1$. This suggests that in the $d = 2$ case the answer is Lipschitz + piecewise smooth.