# 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 = nk dimensionless parameters π1, π2, ..., πp constructed from the original variables. (Here k is the number of physical dimensions involved; it is obtained as the rank of a particular matrix.)

Edgar Buckingham circa 1886

The theorem provides a method for computing sets of dimensionless parameters from the given variables, or nondimensionalization, even if the form of the equation is still unknown.

The Buckingham π theorem indicates that validity of the laws of physics does not depend on a specific unit system. A statement of this theorem is that any physical law can be expressed as an identity involving only dimensionless combinations (ratios or products) of the variables linked by the law (for example, pressure and volume are linked by Boyle's law – they are inversely proportional). If the dimensionless combinations' values changed with the systems of units, then the equation would not be an identity, and the theorem would not hold.

## History

Although named for Edgar Buckingham, the π theorem was first proved by French mathematician Joseph Bertrand[1] 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 clearly indicates the theorem's 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[note 1] to the dependence of pressure drop in a pipe upon governing parameters probably dates back to 1892,[2] a heuristic proof with the use of series expansions, to 1894.[3]

Formal generalization of the π theorem for the case of arbitrarily many quantities was given first by A. Vaschy in 1892,[4] then in 1911—apparently independently—by both A. Federman[5] and D. Riabouchinsky,[6] and again in 1914 by Buckingham.[7] It was Buckingham's article that introduced the use of the symbol "${\displaystyle \pi _{i}}$ " for the dimensionless variables (or parameters), and this is the source of the theorem's name.

## Statement

More formally, the number ${\displaystyle p}$  of dimensionless terms that can be formed is equal to the nullity of the dimensional matrix, and ${\displaystyle k}$  is the rank. For experimental purposes, 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

${\displaystyle f(q_{1},q_{2},\ldots ,q_{n})=0,}$

where ${\displaystyle q_{1},\ldots ,q_{n}}$  are the ${\displaystyle n}$  independent physical variables, and they are expressed in terms of ${\displaystyle k}$  independent physical units, then the above equation can be restated as
${\displaystyle F(\pi _{1},\pi _{2},\ldots ,\pi _{p})=0,}$

where ${\displaystyle \pi _{1},\ldots ,\pi _{p}}$  are dimensionless parameters constructed from the ${\displaystyle q_{i}}$  by ${\displaystyle p=n-k}$  dimensionless equations — the so-called Pi groups — of the form
${\displaystyle \pi _{i}=q_{1}^{a_{1}}\,q_{2}^{a_{2}}\cdots q_{n}^{a_{n}},}$

where the exponents ${\displaystyle a_{i}}$  are rational numbers. (They can always be taken to be integers by redefining ${\displaystyle \pi _{i}}$  as being raised to a power that clears all denominators.)

## Significance

The Buckingham π theorem provides a method for computing sets of dimensionless parameters from given variables, even if the form of the equation remains unknown. However, the choice of dimensionless parameters is not unique; Buckingham's theorem only provides a way of generating sets of dimensionless parameters and does not indicate 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. Most importantly, Buckingham's theorem describes the relation between the number of variables and fundamental dimensions.

## 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 standard gravity ${\displaystyle g}$  has units of ${\displaystyle D/T^{2}=D^{1}T^{-2}}$  (distance over time squared), so it is represented as the vector ${\displaystyle (1,-2)}$  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-units vector space.

### Formal proof

Given a system of ${\displaystyle n}$  dimensional variables ${\displaystyle q_{1},\ldots ,q_{n}}$  (with physical dimensions) in ${\displaystyle k}$  fundamental (basis) dimensions, the dimensional matrix is the ${\displaystyle k\times n}$  matrix ${\displaystyle M}$  whose ${\displaystyle k}$  rows are the fundamental dimensions and whose ${\displaystyle n}$  columns are the dimensions of the variables: the ${\displaystyle (i,j)}$ th entry (where ${\displaystyle 1\leq i\leq k}$  and ${\displaystyle 1\leq j\leq n}$ ) is the power of the ${\displaystyle i}$ th fundamental dimension in the ${\displaystyle j}$ th variable. The matrix can be interpreted as taking in a combination of the dimensions of the variable quantities and giving out the dimensions of this product in fundamental dimensions. So the ${\displaystyle k\times 1}$  (column) vector that results from the multiplication

${\displaystyle M{\begin{bmatrix}a_{1}\\\vdots \\a_{n}\end{bmatrix}}}$

consists of the units of
${\displaystyle q_{1}^{a_{1}}\,q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}}$

in terms of the ${\displaystyle k}$  fundamental independent (basis) units.[note 2]

A dimensionless variable is a quantity that has all of its fundamental dimensions raised to the zeroth power (the zero vector of the vector space over the fundamental dimensions). The dimensionless variables are exactly the vectors that belong to the kernel ${\displaystyle \ker M}$  of this matrix.[note 2]

By the rank–nullity theorem, a system of ${\displaystyle n}$  vectors (matrix columns) in ${\displaystyle k}$  linearly independent dimensions (the rank of the matrix is the number of fundamental dimensions) leaves a nullity ${\displaystyle p=\dim(\ker M)}$  satisfying ${\displaystyle p=n-k,}$  where the nullity is the number of extraneous dimensions which may be chosen to be dimensionless.

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.

The International System of Units defines ${\displaystyle k=7}$  base units, which are the ampere, kelvin, second, metre, kilogram, candela and mole. It is sometimes advantageous to introduce additional base units and techniques to refine the technique of dimensional analysis. (See orientational analysis and reference.[8])

## Examples

### Speed

This example is elementary but serves to demonstrate the procedure.

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

This question considers ${\displaystyle n=3}$  dimensioned variables: distance ${\displaystyle d,}$  time ${\displaystyle t,}$  and speed ${\displaystyle v,}$  and we are seeking some law of the form ${\displaystyle t=\operatorname {Duration} (v,d).}$  These variables admit a basis of ${\displaystyle k=2}$  dimensions: time dimension ${\displaystyle T}$  and distance dimension ${\displaystyle D.}$  Thus there is ${\displaystyle p=n-k=3-2=1}$  dimensionless quantity.

The dimensional matrix is

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

in which the rows correspond to the basis dimensions ${\displaystyle D}$  and ${\displaystyle T,}$  and the columns to the considered dimensions ${\displaystyle D,T,{\text{ and }}V,}$  where the latter stands for the speed dimension. The elements of the matrix correspond to the powers to which the respective dimensions are to be raised. For instance, the third column ${\displaystyle (1,-1),}$  states that ${\displaystyle V=D^{0}T^{0}V^{1},}$  represented by the column vector ${\displaystyle \mathbf {v} =[0,0,1],}$  is expressible in terms of the basis dimensions as ${\displaystyle V=D^{1}T^{-1}=D/T,}$  since ${\displaystyle M\mathbf {v} =[1,-1].}$

For a dimensionless constant ${\displaystyle \pi =D^{a_{1}}T^{a_{2}}V^{a_{3}},}$  we are looking for vectors ${\displaystyle \mathbf {a} =[a_{1},a_{2},a_{3}]}$  such that the matrix-vector product ${\displaystyle M\mathbf {a} }$  equals the zero vector ${\displaystyle [0,0].}$  In linear algebra, the set of vectors with this property is known as the kernel (or nullspace) of (the linear map represented by) the dimensional matrix. In this particular case its kernel is one-dimensional. The dimensional matrix as written above is in reduced row echelon form, so one can read off a non-zero kernel vector to within a multiplicative constant:

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

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, replacing the dimensions by the corresponding dimensioned variables, may be written:

${\displaystyle \pi =d^{-1}t^{1}v^{1}=tv/d.}$

Since the kernel is only defined to within a multiplicative constant, the above dimensionless constant raised to any arbitrary power yields another (equivalent) dimensionless constant.

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

${\displaystyle F(\pi )=0,}$

or, letting ${\displaystyle C}$  denote a zero of function ${\displaystyle F,}$
${\displaystyle \pi =C,}$

which can be written in the desired form (which recall was ${\displaystyle t=\operatorname {Duration} (v,d)}$ ) as
${\displaystyle t=C{\frac {d}{v}}.}$

The actual relationship between the three variables is simply ${\displaystyle d=vt.}$  In other words, in this case ${\displaystyle F}$  has one physically relevant root, and it is unity. The fact that only a single value of ${\displaystyle C}$  will do and that it is equal to 1 is not revealed by the technique of dimensional analysis.

### The simple pendulum

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

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

(Note that it is written as a relation, not as a function: ${\displaystyle T}$  is not written here as a function of ${\displaystyle M,L,{\text{ and }}g.}$ )

There are ${\displaystyle k=3}$  fundamental physical dimensions in this equation: time ${\displaystyle t,}$  mass ${\displaystyle m,}$  and length ${\displaystyle \ell ,}$  and ${\displaystyle n=4}$  dimensional variables, ${\displaystyle T,M,L,{\text{ and }}g.}$  Thus we need only ${\displaystyle p=n-k=4-3=1}$  dimensionless parameter, denoted by ${\displaystyle \pi ,}$  and the model can be re-expressed as

${\displaystyle F(\pi )=0,}$

where ${\displaystyle \pi }$  is given by
${\displaystyle \pi =T^{a_{1}}M^{a_{2}}L^{a_{3}}g^{a_{4}}}$

for some values of ${\displaystyle a_{1},a_{2},a_{3},a_{4}.}$

The dimensions of the dimensional quantities are:

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

The dimensional matrix is:

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

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

We are looking for a kernel vector ${\displaystyle a=\left[a_{1},a_{2},a_{3},a_{4}\right]}$  such that the matrix product of ${\displaystyle \mathbf {M} }$  on ${\displaystyle a}$  yields the zero vector ${\displaystyle [0,0,0].}$  The dimensional matrix as written above is in reduced row echelon form, so one can read off a kernel vector within a multiplicative constant:

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

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:

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

In fundamental terms:
${\displaystyle \pi =(t)^{2}(m)^{0}(\ell )^{-1}\left(\ell /t^{2}\right)^{1}=1,}$

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.

In this example, three of the four dimensional quantities are fundamental units, so the last (which is ${\displaystyle g}$ ) must be a combination of the previous. Note that if ${\displaystyle a_{2}}$  (the coefficient of ${\displaystyle M}$ ) had been non-zero then there would be no way to cancel the ${\displaystyle M}$  value; therefore ${\displaystyle a_{2}}$  must be zero. Dimensional analysis has allowed us to conclude that the period of the pendulum is not a function of its mass ${\displaystyle M.}$  (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, ${\displaystyle {\vec {g}}+2{\vec {T}}-{\vec {L}}}$  is the only nontrivial way to construct a vector of a dimensionless parameter.)

The model can now be expressed as:

${\displaystyle F\left(gT^{2}/L\right)=0.}$

Assuming the zeroes of ${\displaystyle F}$  are discrete and that they are labelled ${\displaystyle C_{1},C_{2},\ldots ,}$  then this implies that ${\displaystyle gT^{2}/L=C_{i}}$  for some zero ${\displaystyle C_{i}}$  of the function ${\displaystyle F.}$  If there is only one zero, call it ${\displaystyle C,}$  then ${\displaystyle gT^{2}/L=C.}$  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 ${\displaystyle C=4\pi ^{2}.}$

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.

### Other examples

An example of dimensional analysis can be found for the case of the mechanics of a thin, solid and parallel-sided rotating disc. There are five variables involved which reduce to two non-dimensional groups. The relationship between these can be determined by numerical experiment using, for example, the finite element method.[9]

The theorem has also been used in fields other than physics, for instance in sport sciences.[10]

## References

### Notes

1. ^ When in applying the π–theorem there arises an arbitrary function of dimensionless numbers.
2. ^ a b If these basis units are ${\displaystyle b_{1},\ldots ,b_{k}}$  and if ${\displaystyle q_{i}=m_{1i}b_{1}+\cdots +m_{ki}b_{k}}$  for every ${\displaystyle 1\leq i\leq n}$  then ${\displaystyle M={\begin{bmatrix}m_{11}&\cdots &m_{1i}&\cdots &m_{1n}\\\vdots &&\vdots &&\vdots \\m_{k1}&\cdots &m_{ki}&\cdots &m_{kn}\\\end{bmatrix}}}$  so that for instance, the units of ${\displaystyle q_{1}}$  in terms of these basis units are ${\displaystyle M\left(\left[1\ 0\ \cdots \ 0\right]^{\operatorname {T} }\right)={\begin{bmatrix}m_{11}\\\vdots \\m_{k1}\\\end{bmatrix}}.}$  For a concrete example, suppose that the ${\displaystyle k=2}$  fundamental units are meters ${\displaystyle b_{1}=m}$  and seconds ${\displaystyle b_{2}=s,}$  and that there are ${\displaystyle n=3}$  dimensional variables: ${\displaystyle q_{1}=m/s^{2},q_{2}=1/m,q_{3}=s/m.}$  By definition of vector addition and scalar multiplication in ${\displaystyle V,}$
${\displaystyle q_{1}=ms^{-2}=1m+(-2)s,\quad q_{2}=m^{-1}=(-1)m+0s,\quad q_{3}=m^{-1}s=(-1)m+1s}$

so that ${\displaystyle M={\begin{bmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\\end{bmatrix}}={\begin{bmatrix}1&-1&-1\\-2&0&1\\\end{bmatrix}}.}$  By definition, the dimensionless units are those of the form ${\displaystyle m^{0}s^{0},}$  which are exactly the vectors in ${\displaystyle \ker M=\operatorname {span} \left\{[1,-1,2]^{\operatorname {T} }\right\}=\left\{\left(q_{1}-q_{2}+2q_{3}\right)^{s}:s\in \mathbb {Q} \right\}.}$  This can be verified by a direct computation:
${\displaystyle q_{1}-q_{2}+2q_{3}=\left(ms^{-2}\right)^{1}+\left(m^{-1}\right)^{-1}+\left(sm^{-1}\right)^{2}=m^{1}s^{-2}+m^{1}+m^{-2}s^{2}=m^{1+1+(-2)}s^{-2+0+2}=m^{0}s^{0},}$

which is indeed dimensionless. Consequently, if some physical law states that ${\displaystyle q_{1},q_{2},q_{3}}$  are necessarily related by a (presumably unknown) equation of the form ${\displaystyle f\left(q_{1},q_{2},q_{3}\right)=0}$  for some (unknown) function ${\displaystyle f}$  with ${\displaystyle \operatorname {domain} f\subseteq \mathbb {R} ^{3}}$  (that is, the tuple ${\displaystyle \left(q_{1},q_{2},q_{3}\right)}$  is necessarily a zero of ${\displaystyle f}$ ), then there exists some (also unknown) function ${\displaystyle F:\mathbb {R} ^{1}\to \mathbb {R} }$  that depends on only ${\displaystyle p=3-2=1}$  variable, the dimensionless variable ${\displaystyle \pi _{1}:=q_{1}-q_{2}+2q_{3}=q_{1}q_{3}^{2}/q_{2}}$  (or any non-zero rational power ${\displaystyle {\hat {\pi }}_{1}:=\pi _{1}^{s}}$  of ${\displaystyle \pi _{1},}$  where ${\displaystyle 0\neq s\in \mathbb {Q} }$ ), such that ${\displaystyle F\left(\pi _{1}\right)=0}$  holds (if ${\displaystyle {\hat {\pi }}_{1}:=\pi _{1}^{s}}$  is used instead of ${\displaystyle \pi _{1}}$  then ${\displaystyle F}$  can be replaced with ${\displaystyle {\hat {F}}(x):=F\left(x^{1/s}\right)}$  and once again ${\displaystyle {\hat {F}}\left({\hat {\pi }}_{1}\right)=0}$  holds). Thus in terms of the original variables, ${\displaystyle F\left(q_{1}q_{3}^{2}/q_{2}\right)=0}$  must hold (alternatively, if using ${\displaystyle {\hat {\pi }}_{1}:=\pi _{1}^{1/2}={\sqrt {\pi _{1}}}}$  for instance, then ${\displaystyle {\hat {F}}\left({\sqrt {q_{1}q_{3}^{2}/q_{2}}}\right)=0}$  must hold). In other words, the Buckingham π theorem implies that ${\displaystyle q_{1}q_{3}^{2}/q_{2}\in F^{-1}(0),}$  so that if it happens to be the case that this ${\displaystyle F}$  has exactly one zero, call it ${\displaystyle C,}$  then the equation ${\displaystyle q_{1}q_{3}^{2}/q_{2}=C}$  will necessarily hold (the theorem does not give information about what the exact value of the constant ${\displaystyle C}$  will be, nor does it guarantee that ${\displaystyle F}$  has exactly one zero).

### Citations

1. ^ Bertrand, J. (1878). "Sur l'homogénéité dans les formules de physique". Comptes Rendus. 86 (15): 916–920.
2. ^ Rayleigh (1892). "On the question of the stability of the flow of liquids". Philosophical Magazine. 34 (206): 59–70. doi:10.1080/14786449208620167.
3. ^ Strutt, John William (1896). The Theory of Sound. Vol. II (2nd ed.). Macmillan.
4. ^ Quotes from Vaschy's article with his statement of the pi–theorem can be found in: Macagno, E. O. (1971). "Historico-critical review of dimensional analysis". Journal of the Franklin Institute. 292 (6): 391–402. doi:10.1016/0016-0032(71)90160-8.
5. ^ Федерман, А. (1911). "О некоторых общих методах интегрирования уравнений с частными производными первого порядка". Известия Санкт-Петербургского политехнического института императора Петра Великого. Отдел техники, естествознания и математики. 16 (1): 97–155. (Federman A., On some general methods of integration of first-order partial differential equations, Proceedings of the Saint-Petersburg polytechnic institute. Section of technics, natural science, and mathematics)
6. ^ Riabouchinsky, D. (1911). "Мéthode des variables de dimension zéro et son application en aérodynamique". L'Aérophile: 407–408.
7. ^
8. ^ Schlick, R.; Le Sergent, T. (2006). "Checking SCADE Models for Correct Usage of Physical Units". Computer Safety, Reliability, and Security. Lecture Notes in Computer Science. Berlin: Springer. 4166: 358–371. doi:10.1007/11875567_27. ISBN 978-3-540-45762-6.
9. ^ Ramsay, Angus. "Dimensional Analysis and Numerical Experiments for a Rotating Disc". Ramsay Maunder Associates. Retrieved 15 April 2017.
10. ^ Blondeau, J. (2020). "The influence of field size, goal size and number of players on the average number of goals scored per game in variants of football and hockey: the Pi-theorem applied to team sports". Journal of Quantitative Analysis in Sports. 17 (2): 145–154. doi:10.1515/jqas-2020-0009. S2CID 224929098.