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Phys 239 Quantitative Physics Lecture 5: Buckingham Pi
Dimensional Analysis
We have messed around a bit with mixing and matching units in the previous lecture in the context of electromagnetic units to come up with propagation speed and impedance of a coaxial cable, and also in the exercise of forming Planck units.
Now we turn to a more rigorous approach for solving problems dimensionally. This means that we should be able to get a functional solution up to some numerical constant. It’s not a free ride: we have to supply a fair bit of intuition and limiting-case expectations. But it can get us out of logjams when we don’t know how to approach problems at first.
Buckingham Pi Theorem
Named after Edgar Buckingham and presented in Physical Review, 4 , 345 in 1914, the Pi theorem (named for a product, Π, rather than the stuff you eat) lays out a systematic way by which to establish appropriate scaling relations for physical problems.
The Central Idea
Units are arbitrary. Inches, millimeters, light years, all describe the same physical concept of length, with simple (some might say meaningless) conversion factors between them. Getting the physics right should not depend on unit systems. Any physical relation can be expressed as a series of terms, all moved to one side of the equation to produce zero on the right hand side:
Y 1 + Y 2 + Y 3 +... + YN = 0.
If we change some internal unit, like inches to centimeters, for instance, then we incur a factor of 2.54 to some rational power in each term, depending on how many powers of length appear in each term. In fact, since it is meaningless to add dimensionally inhomogeneous quantities, each Yj term must have the same units and thus the same power of a particular unit like centimeters.
Then what’s to keep us from making our physical relation dimensionless by dividing through by one of the Yj? For instance:
1 +
Y 2
Y 1
Y 3
Y 1
YN
Y 1
The simple power of this manipulation is that now one may change units at will without changing the equation, since the equation is now dimensionless. Inches, light years—who cares! More generally, we can cast any physical equation as f (vi) = 0, (1)
which is a dimensionless function of n variables vi, i = 1,... , n. Variables may include things like physical constants, important parameters in the problem, and—very importantly—a variable representing the actual quantity being sought (like a drag force, a tidal displacement, etc.).
The Buckingham Approach
Nothing profound yet, really. Buckingham suggested writing Eq. 1 in the following way:
f (Π 1 , Π 2 , Π 3 ,... , Πn−r ) = 0, (2)
where the n variables, vi, employing r distinct fundamental units (length, time, mass, charge) have been re-cast into dimensionless “Π” variables of the form
Πk = (v 1 )αk^1 (v 2 )αk^2 · · · (vn)αkn^ , (3)
with the αki as rational powers. It is the fact that these new variables are products of all the others that earns the Π designation. Importantly, the αki powers are chosen so that each Πk is dimensionless and thus invariant to changes of units (equivalently, scaling any of the r fundamental dimensions).
There are n − r independent Π constructions, up to arbitrary powers of each Πk. For instance, some function
f (Πk) is not fundamentally distinct from a (different) function g(Π
(^83) k ), achieving the same physical end.
We can usefully change the form of Eq. 2 to a different function of all but one of the Πk, pulling out Π 1 (or any other) in the following way: Π 1 = g(Π 2 ,... , Πn−r , m), (4)
where the index, m, is a subtle deal: an index to keep track of the possibility for multiple roots/solutions. One way to put this is that if we vary only Π 1 in Eq. 2 while keeping all other Πk fixed, we must find at least one root (which is what allows us to adopt the form in Eq. 4 in the first place), but possibly more—in which case the index m is used to label multiple solutions. We will ultimately see an example of this for drag in viscous vs. inertial regimes.
That’s it. Really? Did we actually accomplish anything? Well, no. Just rearrangments after coercing our original problem into dimensionless variables via Eq. 3. If anything, the primary usefulness is recognizing that there are n − r such combinations, and a formulaic way to write the equation. But we really need to see examples to make this come to life. And we’ll see that we still have to breathe physical intuition into the problem to make it puff up.
Example 1: Kepler’s Third Law
Starting gently on something we already know (and how to get at by comparing centripetal acceleration to gravity), let’s employ the method to arrive at Kepler’s third law. This is a relationship between period and orbital radius (semi-major axis). Hey! We already have two of the variables we need: T and a. We’ll expect it to depend on the mass of the central body, M , and also gravity, as embodied in the gravitational constant, G. Time to make a table!
i vi units notes 1 T s include variable for answer 2 a m obvious/sought dependency 3 M kg central body mass 4 G m
3 kg·s^2 gravity, yeah? units via^ GM/r
(^2) = m s^2
So n = 4 and r = 3 (s, m, kg), leaving only n − r = 1 unique combination of variables for a single Π value. Three of the units are simple, so we build them into something having the same units as G—namely, a^3 /M T 2. Divide G by this composite to get our single Π variable:
GM T 2
a^3
→ f
GM T 2
a^3
The Buckingham bus stops here. Now we have to apply physical intuition and limiting cases to make it real. Obviously, the function is more than just a multiple or power of Π 1 , or we end up with a trivial/nonsense case. Furthermore, there is no reason/call to apply any power at all. But we should apply an additive constant, K, to avoid the trivial solution:
GM T 2 a^3
− K = 0,
Had we used M⊕ as a variable instead of g, we would have ended up with the rather symmetric (insightful?) construction:
∆h ∼ R⊕ Mm M⊕
R⊕
a
Had we done this, we would have seen that we would not need or want G, as getting rid of g in favor of M⊕ leaves no other variables with a time dimension to play against G. We would have had n = 5, r = 2, and three Π variables essentially exactly following the simple/direct ratios in the expression above.
Synopsis
The Buckingham Pi theorem is not quite magic. It’s just a way to organize a mess of relevant variables, helping us focus on physical reasoning. But in the end, it can’t build the physics for us.
