Buckingham π theorem

In engineering, applied mathematics, and physics, the Buckingham π theorem is a key theorem in dimensional analysis. It is a formalization of Rayleigh's method of dimensional analysis. Loosely, the theorem states that if there is a physically meaningful equation involving a certain number n of physical variables, then the original equation can be rewritten in terms of a set of p = n − k  dimensionless parameters π₁, π₂, ..., π_p constructed from the original variables. (Here k is the number of physical dimensions involved; it is obtained as the rank of a particular matrix.)

The theorem can be seen as a scheme for nondimensionalization because it provides a method for computing sets of dimensionless parameters from the given variables, even if the form of the equation is still unknown.

Historical information

Although named for Edgar Buckingham, the π theorem was first proved by French mathematician J. Bertrand in 1878. Bertrand considered only special cases of problems from electrodynamics and heat conduction, but his article contains in distinct terms all the basic ideas of the modern proof of the theorem and clear indication of its utility for modelling physical phenomena. The technique of using the theorem (“the method of dimensions”) became widely known due to the works of Rayleigh (the first application of the π theorem in the general case to the dependence of pressure drop in a pipe upon governing parameters probably dates back to 1892, a heuristic proof with the use of series expansion, to 1894).

Formal generalization of the π theorem for the case of arbitrarily many quantities was given first by A. Vaschy in 1892, then in 1911—apparently independently—by both A. Federman, and D. Riabouchinsky, and again in 1914 by Buckingham. It was Buckingham's article that introduced the use of the symbol "π_i" for the dimensionless variables (or parameters), and this is the source for the theorem's name.

Statement

More formally, the number of dimensionless terms that can be formed, p, is equal to the nullity of the dimensional matrix, and k is the rank. For the purposes of the experimenter, different systems that share the same description in terms of these dimensionless numbers are equivalent.

In mathematical terms, if we have a physically meaningful equation such as

$f(q_1,q_2,\ldots,q_n)=0\,$

where the q_i are the n physical variables, and they are expressed in terms of k independent physical units, then the above equation can be restated as

$F(\pi_1,\pi_2,\ldots,\pi_p)=0\,$

where the π_i are dimensionless parameters constructed from the q_i by p = n − k dimensionless equations —the so-called Pi groups— of the form

$\pi_i=q_1^{a_1}\,q_2^{a_2}\cdots q_n^{a_n} \,$

where the exponents a_i are rational numbers (they can always be taken to be integers: just raise it to a power to clear denominators).

Significance

The Buckingham π theorem provides a method for computing sets of dimensionless parameters from the given variables, even if the form of the equation is still unknown. However, the choice of dimensionless parameters is not unique: Buckingham's theorem only provides a way of generating sets of dimensionless parameters, and will not choose the most 'physically meaningful'.

Two systems for which these parameters coincide are called similar (as with similar triangles, they differ only in scale); they are equivalent for the purposes of the equation, and the experimentalist who wants to determine the form of the equation can choose the most convenient one. To find out relation between the number of variables and fundamental dimensions Buckingham's theorem is most important.

Proof

Outline

It will be assumed that the space of fundamental and derived physical units forms a vector space over the rational numbers, with the fundamental units as basis vectors, and with multiplication of physical units as the "vector addition" operation, and raising to powers as the "scalar multiplication" operation: represent a dimensional variable as the set of exponents needed for the fundamental units (with a power of zero if the particular fundamental unit is not present). For instance, the gravitational constant g has units of $\ell/t^2=\ell^1t^{-2}$ (distance over time squared), so it is represented as the vector $(1,-2)$ with respect to the basis of fundamental units (distance,time).

Making the physical units match across sets of physical equations can then be regarded as imposing linear constraints in the physical unit vector space.

Formal proof

Given a system of n dimensional variables (physical variables), in k (physical) dimensions, write the dimensional matrix M, whose rows are the dimensions and whose columns are the variables: the (i, j)th entry is the power of the ith unit in the jth variable. The matrix can be interpreted as taking in a combination of the dimensional quantities and giving out the dimensions of this product. So

$M\begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix}$

is the units of

$q_1^{a_1}\,q_2^{a_2}\cdots q_n^{a_n}. \,$

A dimensionless variable is a combination whose units are all zero (hence, dimensionless), which is equivalent to the kernel of this matrix; a dimensionless variable is a linear relation between units of dimensional variables.

By the rank-nullity theorem, a system of n vectors in k dimensions (where all dimensions are necessary) satisfies a (p = n − k)-dimensional space of relations. Any choice of basis will have p elements, which are the dimensionless variables.

The dimensionless variables can always be taken to be integer combinations of the dimensional variables (by clearing denominators). There is mathematically no natural choice of dimensionless variables; some choices of dimensionless variables are more physically meaningful, and these are what are ideally used.

Examples

Speed

This example is elementary, but demonstrates the general procedure: Suppose a car is driving at 100 km/hour; how long does it take to go 200 km?

This question has two fundamental physical units: time t and length $\ell$, and three dimensional variables: distance D, time taken T, and velocity V. Thus there are 3 − 2 = 1 dimensionless quantity. The units of the dimensional quantities are:

$D \sim \ell,\ T \sim t,\ V \sim \ell/t.$

The dimensional matrix is:

$M=\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & -1 \end{bmatrix}$

The rows correspond to the dimensions $\ell$, and t, and the columns to the dimensional variables D, T, V. For instance, the 3rd column (1, −1), states that the V (velocity) variable has units of $\ell^1 t^{-1} = \ell/t$.

For a dimensionless constant $\pi=D^{a_1}T^{a_2}V^{a_3}$ we are looking for a vector $a=[a_1,a_2,a_3]$ such that the matrix product of M on a yields the zero vector [0,0]. In linear algebra, this vector is known as the kernel of the dimensional matrix, and it spans the nullspace of the dimensional matrix, which in this particular case is one-dimensional. The dimensional matrix as written above is in reduced row echelon form, so one can read off a kernel vector within a multiplicative constant:

$a=\begin{bmatrix}-1\\ 1 \\ 1\end{bmatrix}.$

If the dimensional matrix were not already reduced, one could perform Gauss–Jordan elimination on the dimensional matrix to more easily determine the kernel. It follows that the dimensionless constant may be written:

$\pi = D^{-1}T^1V^1 = TV/D$

or, in dimensional terms:

$\pi \sim (\ell)^{-1}(t)^1(\ell/t)^1 \sim 1$

Since the kernel is only defined to within a multiplicative constant, if the above dimensionless constant is raised to any arbitrary power, it will yield another equivalent dimensionless constant.

Dimensional analysis has thus provided a general equation relating the three physical variables:

$f(\pi)=0 \,$

which may be written:

$T=\frac{C D}{V}$

where C is one of a set of constants, such that $C=f^{-1}(0)$. The actual relationship between the three variables is simply $D=VT$ so that the actual dimensionless equation ($f(\pi)=0$) is written:

$f(\pi)=\pi-1 = VT/D - 1 = 0 \,$

In other words, there is only one value of C and it is unity. The fact that there is only a single value of C and that it is equal to unity is a level of detail not provided by the technique of dimensional analysis.

File:Pendel PT.svg

The simple pendulum

We wish to determine the period T  of small oscillations in a simple pendulum. It will be assumed that it is a function of the length L , the mass M , and the acceleration due to gravity on the surface of the Earth g, which has dimensions of length divided by time squared. The model is of the form

$f(T,M,L,g) = 0.\,$

(Note that it is written as a relation, not as a function: T isn't written here as a function of M, L, and g.)

There are 3 fundamental physical dimensions in this equation: time t, mass m, and length l, and 4 dimensional variables, T, M, L, and g. Thus we need only 4 − 3 = 1 dimensionless parameter, denoted π, and the model can be re-expressed as

$f(\pi) = 0 \,$

where π is given by

$\pi =T^{a_1}M^{a_2}L^{a_3}g^{a_4} \,$

for some values of a₁, ..., a₄.

The dimensions of the dimensional quantities are:

$T = t, M = m, L = \ell, g = \ell/t^2. \,$

The dimensional matrix is:

$M=\begin{bmatrix} 1 & 0 & 0 & -2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{bmatrix}$

(The rows correspond to the dimensions t, m, and l, and the columns to the dimensional variables T, M, L and g. For instance, the 4th column, (−2, 0, 1), states that the g variable has dimensions of $t^{-2}m^0 \ell^1$.)

We are looking for a kernel vector a = [a₁, a₂, a₃, a₄] such that the matrix product of M on a yields the zero vector [0,0,0]. The dimensional matrix as written above is in reduced row echelon form, so one can read off a kernel vector within a multiplicative constant:

$a=\begin{bmatrix}2\\ 0 \\ -1 \\ 1\end{bmatrix}.$

Were it not already reduced, one could perform Gauss–Jordan elimination on the dimensional matrix to more easily determine the kernel. It follows that the dimensionless constant may be written:

$\begin{align}\pi &= T^2M^0L^{-1}g^1\\ &= gT^2/L\end{align}$

In fundamental terms:

$\pi=(t)^2(m)^0(\ell)^{-1}(\ell/t^2)^1 = 1 \,$

which is dimensionless. Since the kernel is only defined to within a multiplicative constant, if the above dimensionless constant is raised to any arbitrary power, it will yield another equivalent dimensionless constant.

This example is easy because three of the dimensional quantities are fundamental units, so the last (g) is a combination of the previous. Note that if a₂ were non-zero there would be no way to cancel the M value—therefore a₂ must be zero. Dimensional analysis has allowed us to conclude that the period of the pendulum is not a function of its mass. (In the 3D space of powers of mass, time, and distance, we can say that the vector for mass is linearly independent from the vectors for the three other variables. Up to a scaling factor, $\vec g - 2 \vec T - \vec L$ is the only nontrivial way to construct a vector of a dimensionless parameter.)

The model can now be expressed as:

$f(gT^2/L) = 0.\$

Assuming the zeroes of f  are discrete, we can say gT²/L = C_n  where C_n  is the nth zero. If there is only one zero, then gT²/L = C . It requires more physical insight or an experiment to show that there is indeed only one zero and that the constant is in fact given by C = 4π².

For large oscillations of a pendulum, the analysis is complicated by an additional dimensionless parameter, the maximum swing angle. The above analysis is a good approximation as the angle approaches zero.

See also

• Blast wave
• Dimensional analysis
• Dimensionless quantity
• Natural units
• Nondimensionalization
• Similitude (model)
• Rayleigh's method of dimensional analysis

References

Exposition

• Mike Sheppard, 2007 Systematic Search for Expressions of Dimensionless Constants using the NIST database of Physical Constants

External links

• Some reviews and original sources on the history of pi theorem and the theory of similarity (in Russian)