The Buckingham Pi Theorem

January 20, 2012 in Uncategorized with 1 Comment

1. Introduction

In this post I outline the Buckingham \pi Theorem which shows how to use dimensional analysis to compute answers to seemingly intractable physical problems. For instance, in 1950 Geoffrey Taylor used the theorem to work out the energy payload released by the 1945 Trinity test atomic explosion in New Mexico simply by looking at slow motion video records. My main source for this post is Bluman and Kumei (1989).

Frame from slow motion footage of the Trinity nuclear test with a distance measurement showing the blast radius as well as a time measure showing seconds elapsed since denotation.

2. Basic Framework

The Buckingham \pi Theorem concerns physical problems with the following form: There is a variable of interest, y, which is some unknown function of N different physical quantities x_1, x_2, \ldots, x_N.

(1)   \begin{align*} y &= f(x_1, x_2, \ldots, x_N) \end{align*}

Each of these physical quantities is composed of measurements in only M \leq (N-1) fundamental dimensions labeled c_1, c_2, \ldots, c_M. Thus, I can define a dimension operator which gives the dimensions of an arbitrary variable w \in \{y, x_1, x_2, \ldots, x_N\} and write its output as:

(2)   \begin{align*} \mathtt{dim}[z] &= \prod_{m=1}^{M} c_m^{a_m} \end{align*}

So, for example, if z is measuring pressure on the surface of a table, I could write \mathtt{dim}[z] = \mathtt{lb}/\mathtt{in}^2 where c_1 = \mathtt{lb}, c_2 = \mathtt{in}, a_1 = 1 and a_2 = -2.

Definition (Dimensionless Quantity):
An arbitrary variable w \in \{y, x_1, x_2, \ldots, x_N\} is dimensionless if:

(3)   \begin{align*} \mathtt{dim}[z] &= 1 \end{align*}

3. Main Result

Now, I show how to reformulate this problem and apply linear algebra to the dimensional exponents to derive a characterization of the solution to f(x_1,x_2,\ldots,x_N) - y = 0 as a function of dimensionless quantities. First, I define A as an M \times N matrix of dimensional exponents for x_1,x_2,\ldots,x_N and B as the M \times 1 vector of dimensional exponents for y:

(4)   \begin{align*} A &= \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,N} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,N} \\ \vdots & \vdots & \ddots & \vdots \\ a_{M,1} & a_{M,2} & \cdots & a_{M,N} \end{bmatrix}, \quad B = \begin{bmatrix} b_1 & b_2 & \cdots & b_M \end{bmatrix}^{\top} \end{align*}

Next, we know that if A has full rank, then there will N - M = K solutions to the system of equations 0 = AU. Define U as this N \times K matrix of solutions.

(5)   \begin{align*} U &= \begin{bmatrix} u_{1,1} & u_{1,2} & \cdots & u_{1,K} \\ u_{2,1} & u_{2,2} & \cdots & u_{2,K} \\ \vdots & \vdots & \ddots & \vdots \\ u_{N,1} & u_{N,2} & \cdots & u_{N,K} \end{bmatrix} \end{align*}

Finally, define V as the solution to system of equations 0 = AV + B:

(6)   \begin{align*} V &= \begin{bmatrix} v_1 & v_2 & \cdots & v_N \end{bmatrix}^{\top} \end{align*}

With these objects in hand, I can now state the Buckingham \pi Theorem.

Proposition (Buckingham \pi):
A physical system y = f(x_1,x_2,\ldots,x_N) with M fundamental dimensions can be restated as:

(7)   \begin{align*} y &= \frac{g(\pi_1,\pi_2,\cdots,\pi_K)}{x_1^{u_1} \cdot x_2^{u_2} \cdots x_N^{u_N}}, \end{align*}

where K = N - M, g(\cdot) is an unknown function and \{\pi_1,\pi_2,\cdots,\pi_K\} are dimensionless parameters constructed from the physical parameters \{x_1,x_2,\ldots,x_N\} using equations of the form below:

(8)   \begin{align*} \pi_k &= x_1^{v_1} \cdot x_2^{v_2} \cdots x_N^{v_N}, \end{align*}

4. An Example

I now give an example of how to employ this theorem by working out the nuclear payload example from the introduction. For more information on this example, take a look at this blog post. Suppose that an atomic blast has a shock wave radius of \delta = f(\epsilon,\tau,\rho,\phi) with variables:

  1. \epsilon: Energery released by the explosion,
  2. \tau: Time elapsed since the explosion took place,
  3. \rho: Initial density, and
  4. \phi: Initial pressure.

For this problem N = 4 and M=3 with fundamental dimensions of length l, mass m, and time t yielding the dimensional matrix A written below:

(9)   \begin{align*} A &= \begin{bmatrix} 2 & 0 & -3 & -1 \\ 1 & 0 & 1 & 1 \\ -2 & 1 & 0 & -2 \end{bmatrix} \end{align*}

The energy (force) released by the bomb is the amount of mass accelerating through a unit square on the surface of the blast wave. Since K = N - M = 1, we have only 1 dimensionless constant which can be computed as the solution to the system of equations 0 = AU:

(10)   \begin{align*} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} &= \begin{bmatrix} 2 & 0 & -3 & -1 \\ 1 & 0 & 1 & 1 \\ -2 & 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} - \alpha \cdot 2 \\ \alpha \cdot 6 \\ - \alpha \cdot 3 \\ \alpha \cdot 5 \end{bmatrix} \end{align*}

Setting the scalar free parameter \alpha = 1, I can write \pi_1 as:

(11)   \begin{align*} \pi_1 &= \epsilon^{-2} \cdot \tau^{6} \cdot \rho^{-3} \cdot \phi^{5} \end{align*}

The dimension of the shock wave radius \delta is in units of distance yielding the B dimension matrix below:

(12)   \begin{align*} B &= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^{\top} \end{align*}

The system of 3 equations 0 = AV + B has 4 unknowns:

(13)   \begin{align*} 0 &= 1 + 2 \cdot v_1 - 3 \cdot v_3 - v_4 \\ 0 &= v_1 + v_3 + v_4 \\ 0 &= -2 \cdot v_1 + v_2 - 2 \cdot v_4 \end{align*}

Thus, the vector V will thus be defined up to a single free parameter \hat{\alpha} \geq 0:

(14)   \begin{align*} V &= \begin{bmatrix} (8 \cdot \hat{\alpha} - 1)/5 \\ 2 \cdot (3 \cdot \hat{\alpha} - 1) / 5 \\ (1 - 3 \cdot \hat{\alpha}) / 5 \\ \hat{\alpha} \end{bmatrix}^{\top} \end{align*}

Using the formula \pi = \delta/\left( \epsilon^{v_1} \cdot \tau^{v_2} \cdot \rho^{v_3} \cdot \phi^{v_4} \right) and tuning \hat{\alpha} = 0, I can compute \pi as:

(15)   \begin{align*} \pi &= \delta \cdot \left[ \frac{\epsilon \cdot \tau^2}{\rho} \right]^{-1/5} \end{align*}

Combining all of these results yields a formulation for \delta in terms of a constant (\epsilon \cdot \tau^2/\rho)^{1/5} and an unknown function of the dimensionless quantity g(\pi_1):

(16)   \begin{align*} \delta &= \left[ \frac{\epsilon \cdot \tau^2}{\rho} \right]^{1/5} \cdot g \left( \pi_1 \right) \end{align*}

Taylor expanding around g(0) where g(0) \neq 0 yields a formulation where \delta \propto t^{2/5} with scaling constant c given by:

(17)   \begin{align*} c &= \left( \frac{E}{\rho_0} \right)^{1/5} \cdot g(0) \end{align*}

Setting g(0) = 1, the \log \times \log plot of \delta vs \tau yielded an accurate fit where the predicted values fall on the solid line and the (declassified) empirically observed values are denoted by +‘s in the plot below:

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