What is the average width of a given tetrahedron?

Question

I have a tetrahedron with (1, 1, 1), (2, 1, 1), (1, 2, 1), (1, 1, 2) vertex. What is the average width? I don`t know how to start it. I need to find a useful parameterization. Please help me with any ideas.

Answer 1

First, let us recall some definitions to make sure everyone is on the same page.

Given any compact body $\mathcal{B} \subset \mathbb{R}^3$, the support function of $\mathcal{B}$ along a direction $\hat{n} \in S^2$ is defined as $$h_{\mathcal{B}}(\hat{n}) = \max\{ \; \hat{n}\cdot\vec{x} : \vec{x} \in \mathcal{B} \}$$ The expression $b_{\mathcal{B}}(\hat{n}) = h_{\mathcal{B}}(\hat{n}) + h_{\mathcal{B}}(-\hat{n})$ is the width of the image of $\mathcal{B}$ under orthogonal projection onto any line parallel to $\hat{n}$. The mean width of $\mathcal{B}$ (also known as mean breath or mean caliper diameter) is the average of $b_{\mathcal{B}}(\hat{n})$ for $\hat{n}$ over $S^2$.

$$b(\mathcal{B})\;\stackrel{def}{=}\; \frac{1}{4\pi} \int_{S^2} b_{\mathcal{B}}(\hat{n}) d\mu(\hat{n})$$

It is clear $b(\mathcal{B})$ is invariant under translation and rotation of $\mathcal{B}$. To compute the mean width of the tetrahedron at hand (let us call it $\Delta$ ), we can translate it to the standard simplex:

$$\Delta_0 \stackrel{def}{=} \{ (x,y,z) \in \mathbb{R}^3 : x, y, z \ge 0 \land x+y+z \le 1 \}$$ If we parametrize $S^2$ using spherical polar coordinates, $$[0,\pi] \times [-\pi,\pi) \ni (\theta,\phi) \quad\mapsto\quad \hat{n} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta ) \in S^2$$ we have $$\begin{align} b(\Delta) = b(\Delta_0) &= \frac{1}{4\pi} \int_{S^2} \left( h_{\Delta_0}(\hat{n}) + h_{\Delta_0}(-\hat{n})\right) \sin\theta d\theta d\phi\\ &= \frac{1}{2\pi} \int_{S^2} h_{\Delta_0}(\hat{n}) \sin\theta d\theta d\phi\\ &= \frac{1}{2\pi} \int_{S^2} \max\big\{\; 0, \sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\;\big\} \sin\theta d\theta d\phi\\ &= \frac{3}{2\pi} \int_{D} \cos\theta \sin\theta d\theta d\phi \end{align} $$ where $D$ is the region on $S^2$ where $\cos\theta \ge \max\{\; \sin\theta\cos\phi,\sin\theta\sin\phi, 0\;\}$.

Parametrize the upper hemisphere of $S^2$ by $$\mathbb{R}^2 \ni ( u, v ) \quad\mapsto\quad \frac{1}{\sqrt{1+u^2+v^2}} (u, v, 1 ) \in S^2$$ and let $\rho = \sqrt{u^2+v^2} = \tan\theta$, we have

$$\cos\theta = \frac{1}{\sqrt{1+\rho^2}}\quad\text{ and }\quad d\mu(\hat{n}) = \sin\theta d\theta d\phi = \frac{\rho d\rho d\phi}{\sqrt{1+\rho^2}^3}$$ In terms of $u, v$, the region $D$ above corresponds to the condition $u, v \le 1$.
In terms of $\rho,\phi$, this becomes

$$\rho \le \Lambda(\phi) \quad\text{ where }\quad \Lambda(\phi) = \begin{cases} \frac{1}{\cos\phi}, & -\frac{\pi}{2} < \phi \le \frac{\pi}{4}\\ \frac{1}{\sin\phi}, & \frac{\pi}{4} < \phi < \pi\\ \infty & \text{ otherwise } \end{cases}$$

This leads to

$$\begin{align} b(\Delta) &= \frac{3}{2\pi}\int_{u,v \le 1} \frac{\rho d\rho d\phi}{(1+\rho^2)^2} = \frac{3}{2\pi}\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \int_0^{\Lambda(\phi)} \frac{d\rho^2}{(1+\rho^2)^2} d\phi\\ &= \frac{3}{2\pi}\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \left[ 1 - \frac{1}{1 + \Lambda(\phi)^2}\right] d\phi = \frac{3}{2\pi}\left[ \pi - \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\cos^2\phi}{1 + \cos^2\phi} d\phi \right] \end{align} $$ Change variable to $t = \tan\phi$, we get

$$\begin{align} b(\Delta) &= \frac{3}{2\pi}\left[\pi - \int_{-\infty}^1 \frac{dt}{(2+t^2)(1+t^2)}\right] = \frac{3}{2\pi}\left[ \frac{\pi}{4} + \frac{1}{\sqrt{2}}\left(\tan^{-1}\frac{1}{\sqrt{2}} + \frac{\pi}{2}\right) \right]\\ &\approx 1.1131274948822501384717878345303502990597\ldots \end{align} $$

As one can see from above, it is very tedious to compute the mean width by evaluation of its defining integral. There is a short cut for convex polyhedron.

For any compact convex body $\mathcal{B} \subset \mathbb{R}^3$ with smooth enough boundary, there is an interesting formula: $$b(\mathcal{B}) = \frac{1}{2\pi}\int_{\partial \mathcal{B}} H dS\tag{*1}$$ where $H$ and $dS$ is the mean curvature and surface element for the boundary $\partial \mathcal{B}$.

Given a convex polyhedron $P$, we cannot directly apply this formula because $\partial P$ is not smooth enough. To bypass this obstacle, consider the Minkowski sum of $P$ with $\bar{B}(\epsilon)$, a closed ball centered at $0$ with small radius $\epsilon$.

$$P_{\epsilon} \stackrel{def}{=} P + \bar{B}(\epsilon) = \big\{\; \vec{p} + \vec{q}\;:\; \vec{p} \in P, |\vec{q}| \le \epsilon \;\big\}$$

It is clear $\;b(P_{\epsilon})\;$ = $b(P) + 2\epsilon$. Furthermore, the boundary $\partial P_{\epsilon}$ is now smooth enough. Let $K$ be the Gaussian curvature on the boundary. If $V,E,F$ are the number of vertices, edges and faces of $P$, the boundary $\partial P_{\epsilon}$ consists of

• $V$ spherical fragments with radius $\epsilon$, one for each vertex.

  On these spherical fragments, $H = \frac{1}{\epsilon}$ and $K = \frac{1}{\epsilon^2}$. Notice $K = 0$ outside these spherical fragments, Gauss-Bonnet theorem tell us the total area of these fragments is $4\pi \epsilon^2$. As a result, their contribution to RHS of $(*1)$ is $\frac{1}{2\pi}\left(\frac{4\pi\epsilon^2}{\epsilon}\right) = 2\epsilon$.

• $E$ cylindrical fragments with radius $\epsilon$, one for each edge.

  For any edge $e$ of $P$, let $\ell_e$ be its length. Let $\psi_e$ be the angle between the two outward pointing normals of the two faces of $P$ attached to $e$. The cylindrical fragment is the "cartesian product" of a line segment of length $\ell_e$ and a circular arc of length $\theta_e \epsilon$. Since $H = \frac{1}{2\epsilon}$ on such a fragment, each edge contribute $$\frac{1}{2\pi}\left(\frac{(\psi_e \epsilon) \ell_e}{2\epsilon}\right) = \frac{1}{4\pi} \ell_e\psi_e$$ to RHS of $(*1)$.

• $F$ planar polygons, one for each face.

  These planar polygons contributes nothing to RHS of $(*1)$.

Canceling the $2\epsilon$ from both sides of $(*1)$ for $\mathcal{B} = P_\epsilon$, we obtain

$$\bbox[4pt,border:1px solid blue]{ b(P) = \frac{1}{4\pi}\sum_{e \in \text{edges}(P)} \ell_e \psi_e} \tag{*2} $$

Apply these to tetrahedron $\Delta_0$ and notice

• The $4$ outward pointing normals of $\Delta_0$ are $(-1,0,0)$, $(0,-1,0)$, $(0,0,-1)$ and $\frac{1}{\sqrt{3}}(1,1,1)$.

• The $3$ edges attached to $\vec{0}$ has length $1$. For each of these edges, the two faces attached to the edge is perpendicular to each other. ie. $\psi = \frac{\pi}{2}$.

• For the remaining $3$ edges, they have length $\sqrt{2}$. All of them are attached to that face which is an equilateral triangle and having an outward normal pointing in the direction $\frac{1}{\sqrt{3}}(1,1,1)$. This implies the corresponding $\psi = \cos^{-1}\left(-\frac{1}{\sqrt{3}}\right)$.

Combine this, we have

$$\begin{align} b(\Delta) = b(\Delta_0) &= \frac{3}{4\pi}\left[ \frac{\pi}{2} + \sqrt{2}\cos^{-1}\left( -\frac{1}{\sqrt{3}} \right)\right]\\ &\approx 1.1131274948822501384717878345303502990597\dots \end{align} $$ Same answer as before but in a slightly different analytical form.