Dimensional Analysis & Buckingham Pi

Dimensional analysis uses the fact that valid physics equations must be dimensionally consistent. Learn base dimensions (M, L, T, Θ, I), dimensional homogeneity, checking and deriving formulas, the Buckingham Pi theorem step by step, the famous dimensionless groups (Re, Fr, Ma, Nu, Pr) and how experiments are planned with far fewer tests.

Before running a single experiment, you can often find the form of a law just from units. The pendulum period can only depend on √(L/g); the drag on a sphere must be ρV²D² times some function of Reynolds number.

The Buckingham Pi theorem makes this systematic: n variables built from k base dimensions can be combined into n − k dimensionless groups. An experiment that seemed to need thousands of runs collapses to one curve.

An analogy

Dimensional analysis is like balancing a chemical equation: atoms (dimensions) must balance on both sides. That alone tells you a lot about what reaction (formula) is possible.

Checking an equation

Replace every symbol by its dimensions. s = ut + ½at²: [L] = [LT⁻¹][T] + [LT⁻²][T²] = L ✓. If terms disagree, the equation is wrong. Arguments of exp, sin, log must be dimensionless.

Buckingham Pi — the recipe

A systematic method for any problem:

Physical meaning of groups

Most named groups are ratios of competing effects: Re = inertia/viscosity decides laminar vs turbulent; Fr = inertia/gravity decides wave making; Bi = internal/external heat resistance decides whether a body heats uniformly.

Limits and pitfalls

The theorem gives the form, never the function F — that needs experiments or theory. Missing a relevant variable gives wrong groups; including irrelevant ones creates extra groups that experiments show do not matter. Non-dimensionalising governing equations (not just variables) reveals which groups multiply which terms, guiding asymptotic simplifications.

Formulas

Worked example: Pendulum period

Given: T depends on L, g, m

Find: Form of T.

Answer: T = C√(L/g); mass is irrelevant (C = 2π → 2.006 s for L = 1 m)

Worked example: Drag on a sphere

Given: F depends on ρ, V, D, μ

Find: Dimensionless form.

Answer: C_D = F/(½ρV²A) = φ(Re)

Worked example: Reynolds number check

Given: Water ν = 1×10⁻⁶ m²/s, V = 2 m/s, D = 0.05 m

Find: Re and regime.

Answer: Re = 1×10⁵ → turbulent

Finding Pi groups with a null-space computation

import numpy as np

# Pipe flow: dp, rho, V, D, mu, L   (rows: M, L, T exponents)
names = ["dp", "rho", "V", "D", "mu", "L"]
A = np.array([[ 1,  1,  0, 0,  1, 0],     # M
              [-1, -3,  1, 1, -1, 1],     # L
              [-2,  0, -1, 0, -1, 0]])    # T
print("rank =", np.linalg.matrix_rank(A), "-> Pi groups =", len(names) - np.linalg.matrix_rank(A))

# Check candidate groups: exponent vectors must satisfy A @ e = 0
groups = {"dp/(rho V^2)": [1, -1, -2, 0, 0, 0],
          "rho V D/mu":   [0, 1, 1, 1, -1, 0],
          "L/D":          [0, 0, 0, -1, 0, 1]}
for g, e in groups.items():
    print(f"{g:14s} dimensions ->", A @ np.array(e))
# rank = 3 -> Pi groups = 3
# dp/(rho V^2)   dimensions -> [0 0 0]
# rho V D/mu     dimensions -> [0 0 0]
# L/D            dimensions -> [0 0 0]

The dimensional matrix's null space contains every dimensionless combination. So Δp/(ρV²) = F(Re, L/D) — the Darcy/Moody form, found without any fluid mechanics.

Common mistakes

Applications

Key takeaways

Going further