Buckingham Pi Theorem

Suppose we had a physical equation relating $n$ quantities $f(q_1,\dots ,q_n)=0$. Suppose the $n$ quantities only involve $k$ independent physical dimensions. Then the equation can be restated as

$$F({\Pi }_1,\dots ,{\Pi }_p)=0$$

for some $p=n-k$ parameters ${\Pi }_1,\dots ,{\Pi }_p$. We can express the dimension of each physical quantity $q_i$ as a product of the base dimensions ${\mathsf{B}}_1,\dots ,{\mathsf{B}}_{\ell }$, such that

$$[q_i]={\mathsf{B}}_1^{m_{1i}}\cdots {\mathsf{B}}_{\ell }^{m_{\ell i}}$$

Then, we take the logarithm of the dimensions

$$\log [q_i]=m_{1i}\log {\mathsf{B}}_1+\cdots +m_{\ell i}\log {\mathsf{B}}_{\ell }$$

giving us a change of basis formula

$$\left(\log [q_1]\cdots \log [q_n]\right)=\left(\log {\mathsf{B}}_1\cdots \log {\mathsf{B}}_{\ell }\right)\underset{\mathbf{M}}{\underbrace{\left(\begin{array}{ccc}{m}_{11}& \cdots & m_{1n}\\ ⋮& \ddots & ⋮\\ m_{\ell 1}& \cdots & m_{\ell n}\end{array}\right)}}$$

where $\mathbf{M}$ is the $\ell \times n$ dimension matrix. The physical quantities involve only $k$ independent dimensions $⟺$ the rank of $\mathbf{M}$ is $k$. Using the Rank-Nullity Theorem, the null space of $\mathbf{M}$ has dimension $p=n-k$.

Solving the equation $\mathbf{M}{a}_j=0$ gives us the basis for the null space, size $p$. Indeed,

$$a_1=\left(\begin{array}{c}{a}_{11}\\ ⋮\\ a_{1n}\end{array}\right),\dots ,a_p=\left(\begin{array}{c}{a}_{p1}\\ ⋮\\ a_{pn}\end{array}\right)$$

Then

$$\left(\log [q_1]\cdots \log [q_n]\right)a_j=0$$

That is,

$${\Pi }_j:=C_j{q}_1^{a_{1j}}\cdots q_n^{a_{nj}}$$

are $p$ dimensionless variables. This theorem mathematically proves how many parameters actually govern a physical system so that we can perform a test or a simulation. This allows us to reliably scale models (like a small car in a wind tunnel) to predict real-world performance.

Example 1 (Atomic Bomb)

We can use the Buckingham Pi Theorem to find the dimensionless parameters governing the blast radius of an atomic bomb. First, we’ll identify the physical quantities involved.

Physical Quantity	Symbol	Dimension
Radius of the fireball	$[r]$	${\mathsf{M}}^0{\mathsf{L}}^1{\mathsf{T}}^0$
Density of surrounding air	$[\rho ]$	${\mathsf{M}}^1{\mathsf{L}}^{-3}{\mathsf{T}}^0$
Energy released by the bomb	$[E]$	${\mathsf{M}}^1{\mathsf{L}}^2{\mathsf{T}}^{-2}$
Time since the ignition	$[t]$	${\mathsf{M}}^0{\mathsf{L}}^0{\mathsf{T}}^1$

We get the dimension matrix

$$\mathbf{M}=\left(\begin{array}{cccc}0& 1& 1& 0\\ 1& -3& 2& 0\\ 0& 0& -2& 1\end{array}\right)$$

Because of the rectangular shape, we have a one dimensional null space. In particular, it is spanned by

$$a=\left(\begin{array}{c}-5\\ -1\\ 1\\ 2\end{array}\right)$$

This gives us the dimensionless variable

$$\Pi =r^{-5}{\rho }^{-1}{E}^1{t}^2=\frac{E{t}^2}{\rho r^5}$$

Since we only have dimensionless parameter, the physical law governing the explosion must take the form $F(\Pi )=0$, which mathematically implies $\Pi =\text{constant}$. Let this constant be $C$. So,

$$\frac{E{t}^2}{\rho r^5}=C$$

On smaller experiments, $C\approx 1.033$. Given (from the photograph),

• $t=0.006$ seconds
• $r=80$ meters
• ${\rho }_{\text{TestSite}}=1.1$ kg/m${}^3$

then the energy released by the bomb is approximately

$$\begin{array}{rl}E& =C\rho r^5{t}^{-2}\\ & \approx 10^{14}\text{ joules}\\ & \approx 24\text{ kilotons of TNT}\end{array}$$

With more frames, G.I. Taylor got $22$ kilotons of TNT using more frames. The ground truth was $20$ kilotons of TNT.

The original paper is Taylor’s “The Formation of a Blast Wave by a Very Intense Explosion”.

Example 2 (Drag of a Car)

A moving car will experience aerodynamic drag. We can reasonably postulate there exists a function relating the following $5$ physical quantities:

Physical Quantity	Symbol	Dimension
Car’s length scale	$[L]$	${\mathsf{M}}^0{\mathsf{L}}^1{\mathsf{T}}^0$
Car’s speed	$[v]$	${\mathsf{M}}^0{\mathsf{L}}^1{\mathsf{T}}^{-1}$
Air density	$[\rho ]$	${\mathsf{M}}^1{\mathsf{L}}^{-3}{\mathsf{T}}^0$
Air viscosity	$[\mu ]$	${\mathsf{M}}^1{\mathsf{L}}^{-1}{\mathsf{T}}^{-1}$
Drag force	$[F]$	${\mathsf{M}}^1{\mathsf{L}}^1{\mathsf{T}}^{-2}$

We get the dimension matrix

$$\mathbf{M}=\left(\begin{array}{ccccc}0& 0& 1& 1& 1\\ 1& 1& -3& -1& 1\\ 0& -1& 0& -1& -2\end{array}\right)$$

The null space of $\mathbf{M}$ is two dimensional, spanned by

$$a_1=\left(\begin{array}{c}1\\ 1\\ 1\\ -1\\ 0\end{array}\right)a_2=\left(\begin{array}{c}-2\\ -2\\ -1\\ 0\\ 1\end{array}\right)$$

such that

$$\begin{array}{rl}{\Pi }_1& =L^1{v}^1{\rho }^1{\mu }^{-1}=\frac{\rho vL}{\mu }\\ {\Pi }_2& =L^{-2}{v}^{-2}{\rho }^{-1}{F}^1=\frac{F}{\rho L^2{v}^2}\end{array}$$

By the Buckingham Pi Theorem, the physical law governing the drag force must take the form $F({\Pi }_1,{\Pi }_2)=0$, which mathematically implies ${\Pi }_2=\Phi ({\Pi }_1)$ for some function $\Phi$. Note that ${\Pi }_1$ is the Reynolds number. In other words, if we wanted to set up a wind tunnel so simulate this system, we’d need to ensure

$${\text{Re}}_{\text{WindTunnel}}={\text{Re}}_{\text{RealWorld}}$$

such that the drag force in the wind tunnel is a scaled version of the drag force in the real world. In particular,

$${\left(\frac{F}{\rho L^2{v}^2}\right)}_{\text{WindTunnel}}={\left(\frac{F}{\rho L^2{v}^2}\right)}_{\text{RealWorld}}$$

From this equality, we can measure the force on the small model in the wind tunnel and use it to accurately calculate the force the full-sized car will experience in the real world.