Where's the Sphere?

Today, Kate was examining the philosophical concept of balls EXACTLY 1 inch across in mathematics, with pieces of string wrapped around them, which one must inevitably conclude to be exactly π inches long. Luckily, she was kind enough to point out how absurd this concept is to us engineers [pretentiously gestures towards bookshelf full of engineering textbooks]:
Therefore, if I am thinking like a scientist or engineer (possibly using math but not thinking like a mathematician) dealing with only objects I can hold in my hand, I don't think you can point at a piece of string and say "that string is exactly one inch long".
--Kate Nowak

Luckily for you, my endearing reader, I am currently studying the implications of this lack of exactness in the very non-classical Real World™.

If one were to wrap a string around a sphere to measure the circumference of it, ignoring all the difficulties of actually getting the string around a great circle of the sphere, at some point you would need to lay the crossing ends upon the sphere and cut it to get the final length. If it weren't perfectly upon the sphere, the string would be slightly longer than π, and thus, not exact, as per our exacting standards for the exactness of exact measurements. Unfortunately, cutting the string while it is laying upon the sphere requires knowing where the sphere it, which is precluded by the Heisenberg Uncertainty Principle.

Given the root mean square of the certainty of the position and momentum of an object, their product will never be less than a fraction of Planck's constant.
For this example, I'm going to use an aluminum sphere, which generally has a density of 2.70 g/cm3. Since it's an inch in diameter, the radius of the sphere will therefore be 1.27cm. Let's say we also manage to keep the ball pretty well stopped, with the velocity uncertainty of it in the quite admirable order of 10-6 m/s in our super aluminum-ball-holding-string-wrapping-experiment machine, since we can't wrap string around a sphere when we don't even know which way it's going. Calculating the momentum follows:
Δp = mΔv (the unfortunate use of m in this case is mass)
Δp = 2.70 g/cm3 × 1kg ⁄ 1000g × 4 ⁄ 3 π r³ × 10-6 m/s
Δp = 2.317×10-8 kg⋅m ⁄ s

Solving for Δx in the original equation gives our final uncertainty in position of the sphere as:
Δx ≥ 2.277×10-27 meters

Which is pretty dang small. But the question wasn't "A piece of string π inches long to within more orders of magnitude than one could ever dream of measuring," was it? And this is just the best case; you happen to be unlucky, and you could very well be even farther from π.

This is of course ignoring every other source of uncertainty other than where to have the string when you cut it. You could even argue that just the uncertainty of the string would make it impossible to cut it at a specific place, but I'll leave that one as a challenge for the reader.

Popular Posts