Whiskey Physics

When I ordered a scotch on the rocks in Korea, I was given this:

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That, as you’re probably wondering, is a single spherical ice cube.  A scotch on the rock.  Now, perfectly spherical ice cube molds are probably much harder to make and handle than an ice tray, so why go to the trouble?  Aesthetics?

Well, no.  (Though the sphere was almost the same diameter as the glass, which was pretty cool.)  The answer, of course, is science.

Let’s talk about ice: ice cools your drink.  How exactly does it do that?  At the most basic level, it’s because the universe really likes things to be the same — nature abhors a vacuum and all that.  In terms of temperature, which can be loosely defined as a measure of the speed of the atoms in an object, this manifests in what we call “heat transfer,” the transfer of kinetic energy from a collection of fast-moving particles (a hot thing) to a collection of slower-moving particles (a less hot thing) until the two objects have roughly the same internal kinetic energy (or, in other words, are about the same temperature).  In my drink example, the part of the slow-moving collection of particles is played by Mr. Ice Sphere, while the role of the fast particles is tonight filled by Mr. Been Sitting on a Shelf Room Temperature Scotch Whisky.

When you place the ice sphere in the glass and poor the scotch over it, heat transfer occurs — it’s not that the ice cools the scotch off, but rather that the ice and the scotch try to get to a common temperature that’s somewhere between the ice’s starting temperature and the scotch’s.  This is an exponential process governed by Fourier’s law, which you need a course in vector calculus to fully appreciate but generally states that the flow of heat between two objects is proportional to the difference in temperature between them and the surface area the heat flows through:

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Initially, the ice is quite cold but the scotch, which has been sitting on a shelf in South Korea in the summertime, is quite tepid — so that temperature difference is very large and dominates the system’s behavior, which tends to cool your drink off very quickly.  Because the equilibrium temperature is much closer to the ice’s starting temperature than the scotch’s and since this step happens relatively rapidly, very little of the ice melts, regardless of the ice’s shape.

But — after the ice has sat in the scotch for a few seconds, that temperature difference is very small, and so heat transfer tends to happen much more slowly.  This is where surface area and the geometry of a sphere come in to play — ideally, no ice would melt while you drink your scotch, keeping all that woody, smokey flavor cold without watering it down, but since the liquid and ice are in contact with the outside air that’s impossible in practice.  Instead, barkeepers can only try to keep the ice intact as long as possible, and that means they must minimize the only thing in Fourier’s law they have any control over: the ice’s surface area.

You’ve probably guessed at this point what shape minimizes surface area for a given volume — the humble sphere.*

It’s exactly that property that makes spherical ice cubes cool (pun most definitely intended).  Since they have a smaller surface area than an equivalent volume of normal ice cubes, they can keep your drink just as cold, but will melt more slowly, letting you savor your scotch and escape the sweltering South Korean summer that much longer.

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* I tried to think of a rigorous proof of this, but couldn’t come up with one… Math-inclined readers, suggestions welcome.  I can only offer the fact that the surface area to volume ratio monotonically decreases for Platonic solids with increasing numbers of faces, and logically it makes sense that a sphere would continue this trend.  Proof by hand-waving.**

** Which is better than the math professor I had who taught me about the Platonic solids and occasionally invoked “proof by intimidation.”

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3 comments

  1. I don’t pretend to know anything about micelles, but that could be more analogous to soap bubbles — they form perfect spheres because that geometry minimizes surface tension. There’s a similar reason why planets are spheres, too, albeit with gravity instead of tension.

  2. One way to approach this problem is variational: Given a region, can I perturb the boundary so as to preserve volume but decrease area? Let’s translate to mathspeak.

    Take a region in R^3. Without loss of generality, it’s convex because taking the convex hull can only increase volume and decrease area. We can then locally write the surface as a graph f(x) in a region R. The area and volume functionals are then

    A = \int_R \sqrt(1+|Df|^2) dx,
    V = \int_R f dx.

    (One may argue that we don’t know f is differentiable, but convex functions are almost everywhere twice differentiable by an awesome theorem due to Alexandrov).

    Now let’s see if perturbing f while keeping the volume the same can decrease area. Take a compactly supported smooth function g on R and add a tiny multiple tg to f, so that area and volume become a function of t. Remember that A is a strictly convex functional, so the minimizer of this problem is unique and must satisfy A'(0)=0.

    The constraint of volume preservation is that V'(0) = 0, i.e.

    \int_R g = 0.

    How does the area change? Taking the derivative of A at t=0, one gets

    A'(0) = \int_R (Df.Dg)/(sqrt(1+|Df|^2)).

    Integrating by parts this becomes

    A'(0) = \int_R gH

    where H is the mean curvature of the surface, div(Df/(sqrt(1+|Df|^2))). (The geometric meaning of mean curvature at a point: write the surface as a graph over this point with derivative 0 at 0, find the maximum and minimum second directional derivatives, and average them, i.e. take the mean of the curvatures! Also known as the Laplacian at this point).

    Together, these constraints give us the Euler-Lagrange equation:

    \int_R g(H-c) = 0

    for all volume-preserving deformations g, where c is a Lagrange multiplier. Thus, a minimizer must have constant mean curvature H = c. (Note: our surface is convex so mean curvature must be positive). Thus, we’ve boiled the problem down to: which closed surfaces have constant positive mean curvature?

    Spheres, of course! This is actually not obvious. The trick is to start with a tangent plane to the surface, move it in a little and reflect what the plane crosses over the plane. Keep on moving the plane until the reflection touches the opposite side for the first time. Rotating, we can see this as two graphs of constant identical mean curvature with one above the other, but touching at a point. But then, roughly, at the touching point the higher surface has higher mean curvature, a contradiction! (Actually, one needs to use a standard PDE result, the strong maximum principle, to conclude this). Thus, the reflection must agree everywhere with the uncrossed part, and the only surface with this reflectional symmetry from all directions is the sphere. Tadaaa!

    P.S. Another idea along the lines of what you suggested is a symmetrization process making our region more spherical while keeping volume the same. Take our region and slice it into thin pancakes, then shift the pancakes so that their centers of mass are all on the same line. This clearly preserves volume and decreases diameter but I’m not sure about area. The rough idea is to first prove that area decreases, then to show that we can find a sequence of symmetrizations converging to the sphere. Doesn’t look easy.

    P.P.S. My opinion is that the variational approach is the most physically meaningful; if a surface with tension is not a sphere, it will seek a perturbation to decrease the tension. The sphere is the only thing where constant-volume perturbations always increase tension.

    P.P.S. great article.

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