Buckingham π theorem

Suppose we have quantities ${\displaystyle q_{1},q_{2},\dots ,q_{n}}$, where the units of ${\displaystyle q_{i}}$ contain length raised to the power ${\displaystyle c_{i}}$. If we originally measure length in meters but later switch to centimeters, then the numerical value of ${\displaystyle q_{i}}$ would be rescaled by a factor of ${\displaystyle 100^{c_{i}}}$. Any physically meaningful law should be invariant under an arbitrary rescaling of every fundamental unit; this is the fact that the pi theorem hinges on.

Given a system of ${\displaystyle n}$ dimensional variables ${\displaystyle q_{1},\ldots ,q_{n}}$ in ${\displaystyle \ell }$ fundamental (basis) dimensions, the dimensional matrix is the ${\displaystyle \ell \times n}$ matrix ${\displaystyle M}$ whose ${\displaystyle \ell }$ rows correspond to the fundamental dimensions and whose ${\displaystyle n}$ columns are the dimensions of the variables: the ${\displaystyle (i,j)}$th entry (where ${\displaystyle 1\leq i\leq \ell }$ 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 variable quantities and giving out the dimensions of the combination in terms of the fundamental dimensions. So the ${\displaystyle \ell \times 1}$ (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 ${\displaystyle \ell }$ fundamental independent (basis) units.[note 3]

If we rescale the ${\displaystyle i}$th fundamental unit by a factor of ${\displaystyle \alpha _{i}}$, then ${\displaystyle q_{j}}$ gets rescaled by ${\displaystyle \alpha _{1}^{-m_{1j}}\,\alpha _{2}^{-m_{2j}}\cdots \alpha _{\ell }^{-m_{\ell j}}}$, where ${\displaystyle m_{ij}}$ is the ${\displaystyle (i,j)}$th entry of the dimensional matrix. In order to convert this into a linear algebra problem, we take logarithms (the base is irrelevant), yielding

$${\displaystyle {\begin{bmatrix}\log {q_{1}}\\\vdots \\\log {q_{n}}\end{bmatrix}}\mapsto {\begin{bmatrix}\log {q_{1}}\\\vdots \\\log {q_{n}}\end{bmatrix}}-M^{\operatorname {T} }{\begin{bmatrix}\log {\alpha _{1}}\\\vdots \\\log {\alpha _{\ell }}\end{bmatrix}},}$$

which is an action of ${\displaystyle \mathbb {R} ^{\ell }}$ on ${\displaystyle \mathbb {R} ^{n}}$. We define a physical law to be an arbitrary function ${\displaystyle f\colon (\mathbb {R} ^{+})^{n}\to \mathbb {R} }$ such that ${\displaystyle (q_{1},q_{2},\dots ,q_{n})}$ is a permissible set of values for the physical system when ${\displaystyle f(q_{1},q_{2},\dots ,q_{n})=0}$. We further require ${\displaystyle f}$ to be invariant under this action. Hence it descends to a function ${\displaystyle F\colon \mathbb {R} ^{n}/\operatorname {im} {M^{\operatorname {T} }}\to \mathbb {R} }$. All that remains is to exhibit an isomorphism between ${\displaystyle \mathbb {R} ^{n}/\operatorname {im} {M^{\operatorname {T} }}}$ and ${\displaystyle \mathbb {R} ^{p}}$, the (log) space of pi groups ${\displaystyle (\log {\pi _{1}},\log {\pi _{2}},\dots ,\log {\pi _{p}})}$.

We construct an ${\displaystyle n\times p}$ matrix ${\displaystyle K}$ whose columns are a basis for ${\displaystyle \ker {M}}$. It tells us how to embed ${\displaystyle \mathbb {R} ^{p}}$ into ${\displaystyle \mathbb {R} ^{n}}$ as the kernel of ${\displaystyle M}$. That is, we have an exact sequence

${\displaystyle 0\to \mathbb {R} ^{p}\xrightarrow {\ K\ } \mathbb {R} ^{n}\xrightarrow {\ M\ } \mathbb {R} ^{\ell }.}$

Taking tranposes yields another exact sequence

${\displaystyle \mathbb {R} ^{\ell }\xrightarrow {\ M^{\operatorname {T} }\ } \mathbb {R} ^{n}\xrightarrow {\ K^{\operatorname {T} }\ } \mathbb {R} ^{p}\to 0.}$

The first isomorphism theorem produces the desired isomorphism, which sends the coset ${\displaystyle v+M^{\operatorname {T} }\mathbb {R} ^{\ell }}$ to ${\displaystyle K^{\operatorname {T} }v}$. This corresponds to rewriting the tuple ${\displaystyle (\log q_{1},\log q_{2},\dots ,\log q_{n})}$ into the pi groups ${\displaystyle (\log \pi _{1},\log \pi _{2},\dots ,\log \pi _{p})}$ coming from the columns of ${\displaystyle K}$.

The International System of Units defines seven 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])

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 ${\displaystyle t=\operatorname {Duration} (v,d).}$ Any two of these variables are dimensionally independent, but the three taken together are not. 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 ${\displaystyle L}$ and ${\displaystyle T,}$ and the columns to the considered dimensions ${\displaystyle L,T,{\text{ and }}V,}$ 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 ${\displaystyle V=L^{0}T^{0}V^{1},}$ represented by the column vector ${\displaystyle \mathbf {v} =[0,0,1],}$ is expressible in terms of the basis dimensions as ${\displaystyle V=L^{1}T^{-1}=L/T,}$ since ${\displaystyle M\mathbf {v} =[1,-1].}$

For a dimensionless constant ${\displaystyle \pi =L^{a_{1}}T^{a_{2}}V^{a_{3}},}$ 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 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.

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 ${\displaystyle M,L,{\text{ and }}g.}$)

Period, mass, and length are dimensionally independent, but acceleration can be expressed in terms of time and length, which means the four variables taken together are not dimensionally independent. 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:

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

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.

To demonstrate the application of the π theorem, consider the power consumption of a stirrer with a given shape. The power, P, in dimensions [M · L²/T³], is a function of the density, ρ [M/L³], and the viscosity of the fluid to be stirred, μ [M/(L · T)], as well as the size of the stirrer given by its diameter, D [L], and the angular speed of the stirrer, n [1/T]. Therefore, we have a total of n = 5 variables representing our example. Those n = 5 variables are built up from k = 3 independent dimensions, e.g., length: L (SI units: m), time: T (s), and mass: M (kg).

According to the π-theorem, the n = 5 variables can be reduced by the k = 3 dimensions to form p = n − k = 5 − 3 = 2 independent dimensionless numbers. Usually, these quantities are chosen as ${\textstyle \mathrm {Re} ={\frac {\rho nD^{2}}{\mu }}}$, commonly named the Reynolds number which describes the fluid flow regime, and ${\textstyle N_{\mathrm {p} }={\frac {P}{\rho n^{3}D^{5}}}}$, the power number, which is the dimensionless description of the stirrer.

Note that the two dimensionless quantities are not unique and depend on which of the n = 5 variables are chosen as the k = 3 dimensionally independent basis variables, which, in this example, appear in both dimensionless quantities. The Reynolds number and power number fall from the above analysis if ${\textstyle \rho }$, n, and D are chosen to be the basis variables. If, instead, ${\textstyle \mu }$, n, and D are selected, the Reynolds number is recovered while the second dimensionless quantity becomes ${\textstyle N_{\mathrm {Rep} }={\frac {P}{\mu D^{3}n^{2}}}}$. We note that ${\textstyle N_{\mathrm {Rep} }}$ is the product of the Reynolds number and the power number.

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 sports science.[10]