Cauchy’s shadow formula, Geometric Probability

Problem 58.1

Let $K$ be a convex body in $\mathbb{R}^3$ with surface area $S$ . For a direction $\mathbf{d}$ chosen uniformly on the unit sphere, let $A(\mathbf{d})$ be the area of the shadow of $K$ onto a plane perpendicular to $\mathbf{d}$ . What is $\mathbb{E}[A(\mathbf{d})]$ ?

Solution. Write the boundary of $K$ as $\partial K$ . At each point $x \in \partial K$ , let $\mathbf{n}(x)$ be the outward unit normal. For a direction $\mathbf{d}$ , a boundary element at $x$ is sun-facing (visible to a light source at infinity in the direction $-\mathbf{d}$ ) iff $\mathbf{n}(x) \cdot \mathbf{d} > 0$ . Each sun-facing element contributes $\mathbf{n}(x) \cdot \mathbf{d} \, dA(x)$ to the area of the shadow, since its orthogonal projection onto the shadow plane has that size. Hence $A(\mathbf{d}) = \int_{\partial K,\; \mathbf{n} \cdot \mathbf{d} > 0} \mathbf{n}(x) \cdot \mathbf{d} \, dA(x) = \int_{\partial K} \max\!\big( \mathbf{n}(x) \cdot \mathbf{d}, 0 \big) \, dA(x).$

Average over $\mathbf{d}$ uniform on the sphere $S^2$ and exchange the order of integration (Fubini): $\mathbb{E}_{\mathbf{d}}[A(\mathbf{d})] = \int_{\partial K} \mathbb{E}_{\mathbf{d}}[\max(\mathbf{n}(x) \cdot \mathbf{d}, 0)] \, dA(x).$

For a fixed unit vector $\mathbf{n}$ , the quantity $\mathbf{n} \cdot \mathbf{d}$ is the cosine of the angle between $\mathbf{n}$ and $\mathbf{d}$ . For $\mathbf{d}$ uniform on the sphere, this angle $\theta$ has density $\tfrac12 \sin \theta$ on $[0, \pi]$ (the standard spherical Jacobian). Hence $\mathbb{E}_{\mathbf{d}}[\max(\mathbf{n} \cdot \mathbf{d}, 0)] = \int_0^{\pi/2} \cos \theta \cdot \tfrac12 \sin \theta \, d\theta = \tfrac12 \cdot \tfrac12 = \tfrac14.$

Therefore $\boxed{\mathbb{E}[A(\mathbf{d})] = \tfrac14 \cdot \int_{\partial K} dA = \frac{S}{4}.}$ In particular, the unit cube (surface area $S = 6$ ) casts an expected shadow of area $\tfrac{6}{4} = \tfrac32$ . ■

Cauchy’s shadow formula. A unit cube projects onto a plane perpendicular to a direction \mathbf{d} on the unit sphere; the shadow is a hexagon (generically) whose area depends on \mathbf{d} . Averaged uniformly over \mathbf{d} , the expected shadow area is \tfrac14 of the cube’s surface area, which — Cauchy’s shadow formula. A unit cube projects onto a plane perpendicular to a direction $\mathbf{d}$ on the unit sphere; the shadow is a hexagon (generically) whose area depends on $\mathbf{d}$ . Averaged uniformly over $\mathbf{d}$ , the expected shadow area is $\tfrac14$ of the cube’s surface area, which is $\tfrac32$ .

Cauchy’s formula is the three-dimensional analogue of the fact that the average width of a convex plane curve equals its perimeter divided by $\pi$ . In general dimension $d$ , Cauchy’s surface-area formula concerns projections onto $(d-1)$ -dimensional hyperplanes: their average volume is a universal multiple of the body’s surface area. Lower-dimensional projection averages involve the corresponding intrinsic volumes, rather than surface area.

import numpy as np
from scipy.spatial import ConvexHull
# For each uniform random direction d, project the 8 unit-cube vertices onto
# a plane perpendicular to d and compute the area of their convex hull.
N = 10**4
d = np.random.randn(3, N); d /= np.linalg.norm(d, axis=0)
verts = np.array([[x, y, z] for x in [-.5, .5]
                            for y in [-.5, .5]
                            for z in [-.5, .5]]).T      # 3 x 8
areas = np.empty(N)
for i in range(N):
    di = d[:, i]
    h = np.array([1.0, 0, 0]) if abs(di[0]) < 0.9 else np.array([0.0, 1, 0])
    e1 = np.cross(di, h); e1 /= np.linalg.norm(e1)
    e2 = np.cross(di, e1)
    proj = np.column_stack([verts.T @ e1, verts.T @ e2])
    areas[i] = ConvexHull(proj).volume  # 2D "volume" = area
print(f"sim: {areas.mean():.5f}   exact: 1.5")
# sim: 1.50060   exact: 1.5