Fluid mechanics Foundations and kinematics of flow Theorem

Buckingham π theorem

Builds on Dimensionless numbers in fluid mechanics Rank of a matrix Null space Dimension Basis

Statement

Let $q_1, \dots, q_n$ be positive physical quantities whose dimensions are expressed in $k$ base dimensions. Let $D$ be the $k \times n$ dimension matrix: its column $j$ lists the exponents of the base dimensions in $[q_j]$ . Let $r$ be the rank of $D$ .

The exponent vectors $a \in \Q^n$ for which $q_1^{a_1} \cdots q_n^{a_n}$ is dimensionless are the elements of the null space of $D$ , which has dimension $n - r$ . A basis $a^{(1)}, \dots, a^{(n-r)}$ of it gives dimensionless products $\pi_i = q_1^{a^{(i)}_1} \cdots q_n^{a^{(i)}_n}$ , and every dimensionless power product of the $q_j$ is a product of real powers of $\pi_1, \dots, \pi_{n-r}$ .
Let $f(q_1, \dots, q_n) = 0$ be a dimensionally homogeneous physical law. This means that it stays true when the units of the base dimensions are changed by any positive factors. Then for some function $F$ of $n - r$ variables, $f(q_1, \dots, q_n) = 0$ holds if and only if $F(\pi_1, \dots, \pi_{n-r}) = 0$ .

Remarks

Part 1 is linear algebra: the dimension of a power product is $D a$ , and rank–nullity gives the dimension $n - r$ of the null space. For part 2, choose $r$ of the quantities with independent dimensions. A change of units sets these all equal to $1$ , and then the other quantities become dimensionless products, which are the $\pi_i$ . Example: the period $T$ of a pendulum of length $\ell$ and mass $m$ under gravity $g$ involves $n = 4$ quantities and the $k = 3$ dimensions mass, length and time, with $r = 3$ . So there is one product $\pi_1 = T \sqrt{g / \ell}$ , and the law reads $F(\pi_1) = 0$ , that is $T = C \sqrt{\ell / g}$ for a constant $C$ ; the mass drops out. The theorem does not give $F$ or the constant $C$ , and it gives nothing if a relevant quantity is left out of the list.

Going deeper

For a student

The theorem is the formal basis of dimensional analysis and of the dimensionless numbers of fluid mechanics. Take the drag force on a sphere in a viscous fluid. The quantities force, density, speed, diameter and viscosity give $5 - 3 = 2$ dimensionless groups, the drag coefficient and the Reynolds number. So the drag coefficient is a function of the Reynolds number alone. This is what makes model tests possible: matching the dimensionless groups of a model and of a full-scale object makes their dimensionless behaviour equal. The choice of the basis of the null space is not unique, so different authors may use different but equivalent sets of $\pi$ groups.

Reading path

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Basis concept
Dimension concept
Dimensionless numbers in fluid mechanics concept
Null space concept
Rank of a matrix concept
Buckingham π theorem statement

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Monin–Obukhov similarity theory

Tag	`4Q9PB2WPAY`
URI	`m2://4Q9PB2WPAY`
Kind	statement, theorem
Writing	W0 stub
MSC	76M55, 00A71
Status	This copy of the site has no build data. Run `m2 status 4Q9PB2WPAY`.
About	Dimensionless numbers in fluid mechanics
Uses	Dimensionless numbers in fluid mechanics, Rank of a matrix, Null space, Dimension, Basis
Same as	Wikidata Q999783 Wikipedia Buckingham π theorem
Changed	3 November 2026

Buckingham π theorem

Statement

Remarks

Going deeper

For a student

Reading path

Where this leads

Further reading

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