The n-Category Café

Thanks! A colleague who does the mathematics of materials told me about MacPherson and Srolovitz’s paper a little while back. I was tickled to be able to cite it in a paper on category theory. It’s a spectacular application of the intrinsic volumes.

I’ll re-post their abstract here (with some paragraph breaks inserted). The full paper is behind a paywall (Nature 446 (2007), 1053–1055).

Cellular structures or tessellations are ubiquitous in nature. Metals and ceramics commonly consist of space-filling arrays of single-crystal grains separated by a network of grain boundaries, and foams (froths) are networks of gas-filled bubbles separated by liquid walls. Cellular structures also occur in biological tissue, and in magnetic, ferroelectric and complex fluid contexts.

In many situations, the cell/grain/bubble walls move under the influence of their surface tension (capillarity), with a velocity proportional to their mean curvature. As a result, the cells evolve and the structure coarsens.

Over 50 years ago, von Neumann derived an exact formula for the growth rate of a cell in a two-dimensional cellular structure (using the relation between wall velocity and mean curvature, the fact that three domain walls meet at 120° and basic topology). This forms the basis of modern grain growth theory.

Here we present an exact and much-sought extension of this result into three (and higher) dimensions. The present results may lead to the development of predictive models for capillarity-driven microstructure evolution in a wide range of industrial and commercial processing scenarios—such as the heat treatment of metals, or even controlling the ‘head’ on a pint of beer.

For what it’s worth, a cuboid (in my usage) is something congruent to $[0,a]\times [0,b]\times [0,c]\subseteq \mathbb{R}^3$.

I would probably call that a “rectangular parallelpiped”, a “rectangular prism”, or a “box.” A “cube” to me would be something congruent to $[0,a]^3$. Do you call a rectangle a “squareoid”? (-:

Do you call a rectangle a “squareoid”? (-:

No, but I call a stretched-out version of a group a “groupoid”.

To add plausibility to the British/American thing—I learned to use the term “cuboid” in primary school too. I don’t think I was even as old as ten; maybe seven or eight.

To me, anything of the form $U \times V$ is a “rectangle”. Certainly in place of “cuboid” I would be perfectly happy with “rectangle” or “$n$-dimensional rectangle”.

Thanks for your comments. The results depend on the bricks being able to spin freely. That wasn’t apparent from what I wrote, since in my example it was a ball that was being placed at random—and for a ball, spinning it around makes no difference.

Maybe it would clarify things to put them more formally. First we need to formalize what “probability” means. Let $G$ be the group of Euclidean motions of $\mathbb{R}^3$, generated by translations and rotations. The Haar measure theorem tells us that there’s a measure on $G$, invariant under multiplication $g\cdot-$ by any $g \in G$, and that this measure is unique up to a scalar factor.

This allows us to talk about “random Euclidean motions”. For any $X, \hat{X}, Y \subseteq \mathbb{R}^3$, there’s a well-defined quantity $$Prob(g Y  \text{ meets }  X | g Y  \text{ meets }  \hat{X}).$$ Writing the measure as $\mu$, this is $$\frac{\displaystyle\mu\{g \in G: g Y \cap X \neq \emptyset \}}{\displaystyle\mu\{g \in G: g Y \cap \hat{X} \neq \emptyset \}}.$$ (The choice of normalization doesn’t affect this quantity.)

The kinematic formula tells us what this probability is when $X$, $\hat{X}$ and $Y$ are convex. Here it is. Write $V_0(X)$ for the Euler characteristic of $X$ (which is $0$ if $X$ is empty, and $1$ otherwise). Write $V_1(X)$ for its mean width, $V_2(X)$ for its surface area, and $V_3(X)$ for its volume. Then the probability is $$\frac{\displaystyle c_0 V_0(X) V_3(Y) + c_1 V_1(X) V_2(Y) + c_2 V_2(X) V_1(Y) + c_3 V_3(X) V_0(Y)}{\displaystyle c_0 V_0(\hat{X}) V_3(Y) + c_1 V_1(\hat{X}) V_2(Y) + c_2 V_2(\hat{X}) V_1(Y) + c_3 V_3(\hat{X}) V_0(Y)}$$ where $c_0$, $c_1$, $c_2$ and $c_3$ are constants whose values I could look up (or if I were feeling smart, work out). Though this formula is longish, it’s of a fairly primitive type.

(Professionals might frown and say that some of my definitions of the $V_i$s are out by a constant factor, but seeing as I haven’t specified what the $c_i$s are, it doesn’t matter.)

Incidentally, I’m not aware of places on the web where the basics of this kind of thing are covered. I learned them from a nice little book by Klain and Rota, which doubtless you’ll want to own yourself, but you can “preview” it here.

PS: To answer your questions directly: (1) no, the kinematic theorem doesn’t imply any kind of orientation-independence; (2) yes, specifying the orientations ahead of time does alter the answer (and “mean width” is an intrinsic property of a convex body, not depending on a choice of orientation), (3) yes, choosing to place one brick vertically does change the probability, and (4) yes, the results depend on the bricks being able to spin freely.

If you go down a dimension and think about two straight line segments, things might become clearer. If they’re both constrained to lie horizontally then the probability that they’ll meet is zero. If one is constrained to lie horizontally and the other constrained to lie vertically then the probability is nonzero.