Buckingham π theorem

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The Buckingham π theorem is a key theorem in dimensional analysis. The theorem loosely states that if we have a physically meaningful equation involving a certain number, n, of physical variables, and these variables are expressible in terms of k  independent fundamental physical quantities, then the original expression is equivalent to an equation involving a set of p = nk  dimensionless variables constructed from the original variables. This 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'.

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[edit] 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 which 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 qi  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 πi are dimensionless parameters constructed from the qi  by p = nk  equations of the form

\pi_i=q_1^{m_1}\,q_2^{m_2}\cdots q_n^{m_n}

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

The use of the πi as the dimensionless parameters was introduced by Edgar Buckingham in his original 1914 paper on the subject from which the theorem draws its name.

[edit] 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.

[edit] Proof

[edit] Outline

Begin by considering the space of fundamental and derived physical units as 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.

[edit] Formal proof

(The examples make this clearer.)

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}m_1\\ \vdots \\ m_n\end{bmatrix}

is the units of

q_1^{m_1}\,q_2^{m_2}\cdots q_n^{m_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.

[edit] Examples

[edit] Speed

This example is elementary, but demonstrates the general procedure.

Suppose a car is driving at 100 km/hour; how long does it take it to go 200 km?

This question has 2 fundamental physical units: time t and length l, and 3 dimensional variables: distance D, time taken T, and velocity V. Thus there is 3-2=1 dimensionless quantity.

The units of the dimensional quantities are

D = \ell,\  T = t,\  V = \ell/t.

The dimensional matrix is:

\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 .)

This is in reduced row echelon form, so one can read off that the kernel is generated by

\begin{bmatrix}-1\\ 1 \\ 1\end{bmatrix}.

(Were it not already reduced, one could perform Gauss-Jordan elimination on the dimensional matrix.)

Thus

\begin{align}\pi &= D^{-1}T^1V^1\\
                        &= TV/D\end{align}

(or some power thereof).

In dimensions:

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

is dimensionless.

Admittedly, here the relation is simply D = VT (or rather DVT = 0), so VT / D = 1 is dimensionless, and the dimensionless equation (f(π) = 0) is

VT / D − 1 = 0

which can be solved for time to

T = \frac{D}{V}.

However, the above dimensional analysis does not require any physical understanding, and is useful in less familiar situations.

[edit] 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 units 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 here written as a function of M, L, and g.)

There are 3 fundamental physical units 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(π) = 0

where π is given by

\pi =T^{m_1}M^{m_2}L^{m_3}g^{m_4}

for some values of m1, ..., m4.

The units of the dimensional quantities are:

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

The dimensional matrix is:

\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 units of t^{-2}m^0 \ell^1.)

This is in reduced row echelon form, so one can read off that the kernel is generated by

\begin{bmatrix}2\\ 0 \\ -1 \\ 1\end{bmatrix}.

(Were it not already reduced, one could perform Gauss-Jordan elimination on the dimensional matrix.)

Thus

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

(or some power thereof).

In dimensions:

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

is dimensionless.

This example is easy because 3 of the dimensional quantities are fundamental units, so the last (g) is a combination of the previous.

Note that if m2 were non-zero there would be no way to cancel the M value—therefore m2 must be zero. Dimensional analysis has allowed us to conclude that the period of the pendulum is not a function of its mass.

The model can now be expressed as

f(gT2 / L) = 0.

[edit] Solving

The above is how far dimensional analysis takes up; solving the equation requires further analysis or experimentation.

Assuming the zeroes of f  are discrete, we can say gT²/L = Kn  where Kn  is the nth zero. If there is only one zero, then gT²/L = K . 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 K = 4π² .

[edit] Generalizations

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 in the limit that this angle is zero.

[edit] The atomic bomb

In 1941, Sir Geoffrey I. Taylor used dimensional analysis to estimate the energy released in an atomic bomb explosion (Taylor, 1950a,b). The first atomic bomb was detonated near Alamogordo, New Mexico on July 16, 1945. In 1947, movies of the explosion were declassified, allowing Sir Geoffrey to complete the analysis and estimate the energy released in the explosion, even though the energy release was still classified. The actual energy released was later declassified and its value was remarkably close to Taylor's estimate.

Taylor supposed that the description of the process was adequately described by five physical quantities: the time t  since the detonation, the energy E  which is released at a single point in space at detonation, the radius R  of the shock wave at time t , the atmospheric pressure p  and the ambient density ρ. There are only three fundamental physical units in this equation: mass, time, and length. Thus we need only 5 − 3 = 2 dimensionless parameters, which can be found to be

\pi_0=R\,\left(\frac{\rho}{Et^2}\right)^{1/5}

and

\pi_1=p\,\left(\frac{t^6}{E^2\rho^3}\right)^{1/5}.

The process can now be described by an equation of the form

f(\pi_0,\pi_1)=0,\,

or, equivalently

R=\left(\frac{Et^2}{\rho}\right)^{1/5}g(\pi_1),

where g1) is some function of π1. The energy in the explosion is expected to be huge, so that for times of the order of a second after the explosion, we can estimate π1 to be approximately zero, and experiments using light explosives can be conducted to determine that g(0) is on the order of unity so that

R\approx\left(\frac{Et^2}{\rho}\right)^{1/5}.

This is Taylor's equation which, once he knew the radius of the explosion as a function of the time, allowed him to calculate the energy of the explosion. (Wan, 1989)

[edit] See also

[edit] References

[edit] Exposition

  • Kline, Stephen J. (1986). Similitude and Approximation Theory. Springer-Verlag, New York. ISBN 0387165185. 
  • Wan, Frederic Y.M. (1989). Mathematical Models and their Analysis. Harper & Row Publishers, New York. ISBN 0060469021. 

[edit] Original sources

  • Buckingham, E. (1915). "The principle of similitude". Nature 96: 396-397. 
  • Buckingham, E. (1915). "Model experiments and the forms of empirical equations". Trans. A.S.M.E 37: 263-296. 
  • Taylor, Sir G. (1950). "The Formation of a Blast Wave by a Very Intense Explosion. I. Theoretical Discussion". Proc. Roy. Soc. A 201: 159-174. 
  • Taylor, Sir G. (1950). "The Formation of a Blast Wave by a Very Intense Explosion. II. The Atomic Explosion of 1945". Proc. Roy. Soc. A 201: 175-186. 

[edit] External links