Dimensional Analysis with Linear Algebra

Exploration of Buckingham’s Pi Theorem

Author

Jasper Day

Source

Buckingham’s Pi theorem states that any relations between natural quantities can be expressed in an equivalent form using Pi groups, dimensionless quantities formed between those quantities.

1 Assumptions:

The following assumptions must hold:

  1. \textit{u}, our quantity of interest, must equal some function f\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right), that is, \textit{n} measurable quantities expressed as independent variables & parameters x_{i}. It is further assumed that the equation

u = f\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right) is dimensionally homogeneous.

  1. The quantities \{u, x_{1}, x_{2}, x_{3}, \ldots, x_{n}\} are measured in terms of \text{m} fundamental dimensions \{ L_{1}, L_{2}, L_{3}, \ldots, L_{n} \}

  2. If \text{W} is any quantity of \{ u, x_{1}, \ldots, x_{n}\}, then

\left[W\right] = L_{1}^{p_{1}} \cdot L_{2}^{p_{2}} \cdot \ldots \cdot L_{m}^{p_{m}}

Then we can create \textbf{P} = \begin{bmatrix}p_{1} \\ p_{2} \\ \vdots \\ p_{m} \\\end{bmatrix}, the dimension vector of W.

This gives us the m\times n dimension matrix

\textbf{A} = \begin{bmatrix} \textbf{P}_{1} | \textbf{P}_{2} | \cdots | \textbf{P}_{n} \\ \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} & \cdots & p_{1n} \\ p_{21} & p_{22} & \cdots & p_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ p_{m1} & p_{m2} & \cdots & p_{mn} \\ \end{bmatrix}

2 Conclusions of the Buckingham Pi Theorem

  1. The relation u = f\left(x_{1}, x_{2}, \ldots, x_{n} \right) can be expressed in terms of dimensionless quantities.

  2. The number of dimensionless quantities is

k + 1 = n + 1 - \texttt{rank}\left(A\right)

(The reason for k + 1 is that we pull out the original quantity u from the matrix \textbf{A}. Otherwise this term would not appear.)

  1. Since \textbf{A} has \texttt{rank}\left(A\right) = n - k, there are k linearly independent solutions of \textbf{Az} = \textbf{0} denoted as z^{1}, z^{2}, \ldots, z^{k}.

Let \textbf{a}, an m-column vector, be the dimension vector of u, and let \textbf{y}, an n-column vector, be a solution of

\textbf{Ay} = -\textbf{a}

Then the relation u = f\left(x_{1}, x_{2}, \ldots, x_{n} \right) simplifies to g\left(\Pi_{1}, \Pi_{2}, \ldots, \Pi_{k} \right).

There is one \Pi group for each linearly indepenent set of \textbf{Az} = \textbf{0}, plus one \Pi group for u. The parameters in each pi group are raised to the respective row of z\prime.

3 Why it Works:

Recall that the nullspace of a matrix \textbf{A} is the space of all vectors \textbf{z} for which \textbf{Az} = \textbf{0}. The multiplication \textbf{Az} is a linear combinations of the columns of \textbf{A}:

\textbf{Az} = \left[ z_{1}\textbf{P}_{1} | z_{2}\textbf{P}_{2} | \ldots | z_{n}\textbf{P}_{n} \right]

This linear combination of the columns of \textbf{A} is the same thing that you get when you raise each of the parameters x_{n} to the respective element of \textbf{z}:

\left[x_{i}^{z_{i}}\right] = \left[W\right]^{z_{i}} = \left(L_{1}^{p_{1}} \cdot L_{2}^{p_{2}} \cdot \ldots \cdot L_{m}^{p_{m}}\right)^{z_{i}} = L_{1}^{p_{1}z_{i}} \cdot L_{2}^{p_{2}z_{i}} \cdot \ldots \cdot L_{m}^{p_{m}z_{i}}

Which corresponds to column i of \textbf{Az}. Finally, since \textbf{z} is in the nullspace of \textbf{A}, the sum of the powers on each of the base units L will be 0, resulting in an overall dimensionless quantity.