Meissner bodies: A new proof of an old result

Bodies of constant width fascinate mathematicians and artists alike. These elegant shapes roll between two parallel surfaces like ball bearings, yet they can be very far from spherical. These counterintuitive objects, variously called spheroforms, orbiforms, or Gleichdick, depending on who you ask, remain mysterious, despite the intense interest of mathematicians over the last century.

Take, for example, the famous Meissner body (Fig. 1). This shape, first described by the Swiss Mathematician Ernst Meissner around 1911, arises from a bit of a mathematical hack. Meissner rounded three edges of a Relaux tetrahedron—a body of almost constant width—to create a body of precisely constant width. (In this game, close doesn’t count for much.) Despite a nearly one-hundred year investigation, Meissner’s construction still yields the smallest volume body of constant width known. While some mathematicians suspect that Meissner’s body is the smallest volume shape of constant width, a proof of this conjecture remains dreadfully elusive.

Figure 1. The Meissner body rolls like a ball bearing between two parallel planes, but is far from a sphere. Credit: mitrasmit.

In this post, we do not attempt to address that old and apparently difficult conjecture. Rather, we consider another remarkable fact, due to Wilhelm Blaschke in 1915: The surface area of a body of constant width is proportional to its volume. The proof below is relatively elementary, and apparently new. This exposition may be of interest for anyone looking for a deeper understanding of these wonderful shapes.

A precise statement

Let’s start by making things a bit more precise. We say that a convex set $$K\subset \mathbb{R}^3$$ has constant width $$c$$ if every two parallel planes touching the edges of the set are the same distance apart, no matter how the shape is rotated. The surface area and volume of each of these bodies satisfy a remarkable relationship.

Theorem 1. (Blaschke, 1915). Let $$K\subset \mathbb{R}^3$$ be a body of constant width $$c$$. Then the volume $$v(K)$$ and the surface area $$s(K)$$ satisfy

$v(K) = c\frac{1}{2} s(K) – c^3\frac{\pi}{3}.\tag{1}$

The original proof of this result appears in an article [1] that I’ve been unable to locate online, so I can’t compare the proof below to the approach taken by Blaschke. However, a much later article of Santaló provides an analog in arbitrary dimensions that requires more advanced techniques. The proof below, while restricted to three dimensions, extends to readily to higher dimensions, and yields Santaló’s results with relative ease.

Some background

Our proof relies on a few ideas from the theory of convex bodies, in particular, the theory of mixed volumes. The volume $$v(K + \lambda L)$$ of the Minkowski sum of the convex bodies $$K$$ and $$\lambda L$$ is a polynomial in the scale parameter $$\lambda \ge 0$$:

$v(K + \lambda L) = V_0(K,L) + \lambda V_1(K,L) + \lambda^2 V_2(K,L) + \lambda^3 V_3(K,L). \tag{2}$

From this equation, it’s not hard to show that the coefficients $$V_i(K,L)$$, called mixed volumes, satisfy:

1. $$V_{i}(K,L) = V_{3-i}(L,K)$$ for each $$i$$; and
2. $$V(K, L) = V(-K,-L)$$.

If we set $$L= \mathbb{B}_3$$ the 3-ball, then the coefficients in (2) take on special values:

$v(K + \lambda \mathbb{B}_3) = v(K) + s(K) \lambda + 2\pi\bar{w}(K)\lambda^2 + \frac{4 \pi}{3}\lambda^3,\tag{3}$

where $$\bar{w}(K)$$ is the mean width of the convex body $$K$$. (In particular, when $$K$$ has constant width $$c$$, we have $$\bar{w}(K) = c$$.) The result (3) goes by the name Steiner’s formula, after the Swiss mathematician Jakob Steiner who discovered it in 1840. Although Steiner was illiterate until he was 14 and did not attend school until after he turned 18, he managed to make enormous contributions to the theory of analytic geometry during his life.

Finally, we will need another characterization of sets of constant width. A convex set $$K$$ in 3-space has constant width $$c$$ if and only if

$K – K = c \mathbb{B}_3,\tag{4}$

where $$K – K$$ denotes the Minkowski sum of $$K$$ and $$-K$$. For completeness, we prove this relationship at the end of the post.

Proof of Theorem 1.

At this point, you might already sense our strategy: we will compute the volume of $$K + \lambda (-K)$$ using both the mixed volume formula (2) and the Steiner formula (3), and then compare coefficients.

We may assume without loss that the width $$c=1$$, since the general $$c$$ case follows in (1) by homogeneity. Let $$\lambda \in [0,1]$$. Using the mixed volume expansion (2), we find

\begin{align}v(K + \lambda(-K)) &= V_0(K,-K) + 3\lambda V_1(K, -K) + 3\lambda^2 V_2(K, -K) + \lambda^3 V_3(K,-K) \tag{5}\end{align}

On the other hand, we also have $$K + \lambda (-K) = (1-\lambda) K + \lambda \mathbb{B}_3$$ by Equation (4). Applying the Steiner formula (3), we find

$v(K + \lambda(-K)) =(1-\lambda)^3 v(K) +\lambda(1-\lambda)^2 s(K) + \lambda^2 (1-\lambda)2\pi +\lambda^3\frac{4\pi}{3},\tag{6}$

where we’ve used the fact that the mean width $$\bar w (K) = 1$$ because $$K$$ has constant width $$c=1$$. Grouping like powers of $$\lambda$$ in (5) and (6), we arrive at the following system of equations:

\begin{align}V_0(K,-K)&=v(K)\\ 3V_1(K,-K) &= -3 v(K) + s(K) \\ 3V_2(K,-K) &= 3v(K) – 2 s(K) +2\pi\\ V_3(K,-K) &= -v(K) + s(K)  – \frac{2\pi}{3}.\end{align}

The properties of the mixed volumes listed above imply that $$V_0(K,-K) = V_3(K,-K)$$ and $$V_1(K,-K) = V_2(K,-K)$$, so the four equations above reduce to the following two independent statements:

\begin{align}v(k) &= \frac{1}{2}s(K) – \frac{\pi}{3}\tag{7}\\ V_1(K,-K) &= \frac{\pi}{3} – \frac{1}{2} s(K).\end{align}

Equation (7) is precisely Blaschke’s identity (1) for $$c=1$$. The general case follows by homogeneity.

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Appendix: A proof of equation (4)

We make use of properties of the support function $$h_K$$, all of which appear in the linked Wikipedia article. First, it follows from the definition of the support function $$h_K$$ that $$K$$ has constant width $$c$$ if and only if

$h_K(u) + h_K(-u) = c \quad \text{for all} \quad u \in \mathbb{S}^{d-1}.$

Since $$h_K(-u) = h_{-K}(u)$$ and we have the linear relationship  $$h_{K – K} = h_K + h_{-K}$$, we see that $$h_{K -K}(u) = c$$ for all $$u\in \mathbb{S}^{d-1}$$. This latter equation characterizes the scaled  ball $$c\mathbb{B}_d$$, so we conclude that $$K-K = c \mathbb{B}_d$$.

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References

[1] W. Blaschke, Einige Bemerkungen über Kurven und Flächen von konstanter Breite, Ber. Verh. Sächs. Akad. Leipzig, 67 (1915), 290–297.

2 thoughts on “Meissner bodies: A new proof of an old result”

1. A very similar argument shows that symmetric convex bodies $$K=-K\subset \mathbb{R}^d$$ satisfy

$V_i(K, -K) = \binom{d}{i} v(K)$

• Isn’t this obvious? The function V_i is multilinear (if you consider the arguments are repeated) and so Vi(K,-K) must be proportional to the volume v(K)