Two scaling laws, one theorem

Lecture 1 derives its scaling results three different ways. All three are the same counting argument, and saying so out loud makes the method teachable.

CCBS · EST 5053 · extension mini-essay · on lecture 1 · written by the lecturer as the worked example

1The claim

Lecture 1 presents three scaling results as three separate pieces of cleverness. They are one theorem applied three times, and the theorem is a hundred years old.

Lecture 1 derives three scaling laws and derives each one differently. The blast wave, $R \propto t^{2/5}$, comes from a dimensionless group. Jump height, $h \propto M^{0}$, comes from setting two energies equal and watching $L^{3}$ cancel. Diffusion, $t \approx x^{2}/\kappa$, comes from writing down a partial differential equation and then throwing away the derivatives.

Three tricks. That is how it looks, and it is how I found it as a student, which is why the material felt like a collection of party pieces rather than a method.

The claim of this essay is that all three are the same counting argument, and that the counting argument has a name, a proof and a termination condition. Once you see that, you stop hoping to be clever and start following a procedure.

2Where the scaling arguments came from

The technique is older than the theorem by nearly three centuries, which is why it is usually taught as a knack.

The oldest version in the lecture's ancestry is Galileo's, in the Two New Sciences of 1638.¹ Weight scales as volume and bone strength as cross-section, so a giant of ordinary proportions collapses. He checks it immediately against something anyone can verify: "a small dog could probably carry on his back two or three dogs of his own size; but I believe that a horse could not carry even one of his own size."

That is a scaling argument with no theorem behind it. Galileo could not have told you how many independent conclusions the method was entitled to produce, or when he had exhausted it. Neither could anyone else for the next 276 years. The argument was in constant use and was justified by the fact that it kept working.

Buckingham closed the gap in 1914, in a paper whose title says what it is for: On physically similar systems; illustrations of the use of dimensional equations.²

3The theorem

Two lines of linear algebra, and a number that tells you in advance whether you are done.

Start from something that cannot be argued with: a physical law does not know what units you chose. Rescale your unit of length and every length in the law is multiplied by the same factor. Whatever survives that must be built from combinations in which the factors cancel, which is to say, from dimensionless combinations.

Make it concrete. Suppose your quantities are $Q_{1}, ..., Q_{n}$, with

A product $\Pi = Q_{1}^{p_{1}} \cdot \cdot \cdot Q_{n}^{p_{n}}$ is dimensionless exactly when the exponent of each fundamental unit vanishes:

Three homogeneous linear equations in $n$ unknowns. If they are independent the solution space has dimension $n - 3$, and in general, with $k$ independent fundamental units, $n - k$.

4Both lecture results, in four lines each

The point of the essay. Two of the lecture's three derivations collapse to the same routine.

Both are done at length elsewhere, and the versions below are deliberately shorter than those.⁴ The claim is that the length was never necessary.

Diffusion, which the lecture derives from a PDE

The lecture spends a page on this: Fourier's law, a flux balance on a slab, a limit, then $\partial T/\partial t = \kappa \nabla^{2}T$, then replace the derivatives by their characteristic scales. Here is the same result by counting.

Quantities: a time $t$, a distance $x$, and a diffusivity $\kappa$ with dimensions L²T⁻¹. So $n = 3$. Units appearing: L and T only, so $k = 2$. Therefore $n - k = 1$, and there is exactly one group. Find it: $\kappa t$ has dimensions L², so

Four lines, and no heat equation. The single most useful relation in this course follows from knowing the units of $\kappa$ and nothing else.

The blast wave, which the lecture already does this way

Quantities: energy $E$, time $t$, radius $R$, air density $\rho_{0}$, so $n = 4$. Units: M, L, T, so $k = 3$. One group. Solving (2) with $[E] = ML^{2}T^{-2}$ and $[\rho_{0}] = ML^{-3}$ gives

Same routine, same four lines, and it is the derivation Taylor used to read the Trinity yield off a photograph.³

And the third one, which does not fit

Honesty requires saying that jump height, $h \propto M^{0}$, is not a Π-theorem result. It comes from equating two energies and noticing that $L^{3}$ divides out. That is a conservation argument, and dimensional analysis alone would not give it, because the muscle work density $w$ and the density $\rho$ would have to be on the list and the count would not close.

So the essay's claim needs weakening from "all three" to "two of the three, and the third is a different move". Which is worth stating, because it locates the boundary: the Π theorem handles problems where the physics is entirely in the units, and conservation arguments handle problems where a specific quantity is conserved. Both are cheap. Neither subsumes the other.

5What the theorem cannot do

Three limits, and the first one is the one that bites.

It takes your list as given. The count is arithmetic, but choosing what to count is physics, and there is no formal defence against a wrong list. Include something irrelevant and you get an extra group and lose the sharp answer. Omit something relevant and you get a confidently wrong answer with no warning at all. Taylor's list omits ambient pressure, viscosity and the bomb's own mass, and each omission is a condition on when his answer applies.

It gives the form, never the constant. For a pendulum the theorem gives $T \propto \sqrt{L/g}$ and the true constant is $2\pi \approx 6.3$, which is not of order one. The useful version of the rule is asymmetric: an error in a prefactor costs you a fixed factor, and an error in an exponent compounds with the range you extrapolate over. The theorem gives away the expensive half and keeps the cheap one.

Quantities of the same dimension defeat it. If your list has two lengths, their ratio is already dimensionless and the theorem says nothing about it. That is a real limit, and also a signal: it is telling you that shape matters and units cannot decide it.

6So what?

The practical payoff is a ten-second test you can run before deciding whether a problem is worth a week.

The reason to know the theorem is not elegance. It is that $n - k$ is computable in ten seconds and tells you which of three situations you are in, per Figure 1. If it is 1, stop and write down the answer. If it is 2, you have a universal curve and one measurement will pin it down, which is a research plan. If it is 3 or more, units have given you what they can and the real work starts now.

Compare that with the alternative, which is what I did as a student: stare at the problem hoping to notice a combination. Sometimes I did. When I did not, I had no way to tell whether I had missed something or whether there was nothing to find. The theorem converts a hope into a procedure with a termination condition, and that is what makes it teachable.

The thing I would go and check next, and did not have time for here: whether the same count can be run on a chemical reaction network, where the fundamental units include concentration and the quantities include rate constants. If it can, then the reaction-order geometry of lecture 11 might have a dimensional-analysis reading, and that would be a nicer entry point to it than the one currently planned.

†Sources, and what I checked

An extension mini-essay should say what it actually read, and at what depth. This section is part of the format.