Buckingham π theorem

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 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.

Given a system of n dimensional variables (with physical dimensions) in k fundamental (basis) dimensions, write the dimensional matrix M, whose rows are the fundamental dimensions and whose columns are the dimensions of the variables: the (i, j)th entry is the power of the ith fundamental dimension in the jth 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

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

is the units of

${\displaystyle q_{1}^{a_{1}}\,q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}.}$

A dimensionless variable is a quantity with fundamental dimensions raised to the zeroth power (the zero vector of the vector space over the fundamental dimensions), which is equivalent to the kernel of this matrix.

By the rank–nullity theorem, a system of n vectors (matrix columns) in k linearly independent dimensions (the rank of the matrix is the number of fundamental dimensions) leaves a nullity, p, satisfying (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 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 [9])

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 three dimensioned variables: distance d, time t, and velocity v, and we are seeking some law of the form t = Duration(v, d) . These variables admit a basis of two dimensions: time dimension T and distance dimension D. Thus there is 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 D and T, and the columns to the considered dimensions D, T, and V, where the latter stands for the velocity dimension. The elements of the matrix correspond to the powers to which the respective dimensions are to be raised. For instance, the third column (1, −1), states that V = D⁰T⁰V¹, 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 Ma equals the zero vector [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 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 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 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

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

(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 ${\displaystyle t}$, mass ${\displaystyle m}$, and length ${\displaystyle \ell }$, 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

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

where π is given by

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

for some values of a₁, ..., a₄.

The dimensions of the dimensional quantities are:

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

The dimensional matrix is:

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

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

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:

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

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:

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

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, ${\displaystyle {\vec {g}}+2{\vec {T}}-{\vec {L}}}$ is the only nontrivial way to construct a vector of a dimensionless parameter.)

The model can now be expressed as:

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

Assuming the zeroes of f are discrete, we can say gT²/L = C_n, where C_n is the nth zero of the function f. 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.

Drinks cooled with small ice cubes cool faster than drinks cooled with the same mass of larger ice cubes. The common explanation for this phenomenon is that smaller cubes have greater surface area, and this greater area causes greater heat conduction and therefore faster cooling. If this explanation were correct, then it would imply that the rate of cooling should be proportional to ${\displaystyle 1/L}$, where ${\displaystyle L}$ is the length of the cube edges, and thus the time for the drink to cool should be proportional to ${\displaystyle L}$. In fact, dimensional analysis shows this common explanation to be incorrect, and gives the surprising result that the time to cool the drink is proportional to ${\displaystyle L^{2}}$.

The important dimensional quantities are the length scale of the cubes ${\displaystyle L}$ (dimension ${\displaystyle \ell }$), the time ${\displaystyle T}$ (dimension ${\displaystyle t}$), the temperature ${\displaystyle \Theta }$ (dimension ${\displaystyle \theta }$), the thermal conductivity ${\displaystyle \kappa }$ (dimensions ${\displaystyle \ell ^{1}t^{-3}\theta ^{-1}m^{1}}$), and the volumetric heat capacity ${\displaystyle s}$ (dimensions ${\displaystyle \ell ^{-1}t^{-2}\theta ^{-1}m^{1}}$). The dimensional matrix is:

$${\displaystyle M={\begin{bmatrix}1&0&0&1&-1\\0&1&0&-3&-2\\0&0&1&-1&-1\\0&0&0&1&1\end{bmatrix}}.}$$

The nullspace of M is 1-dimensional, and the kernel is spanned by the vector

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

and therefore ${\displaystyle \pi =L^{2}T^{-1}\kappa ^{-1}s^{1}}$. (Note that the temperature ${\displaystyle \Theta }$ does not appear in the dimensionless group.) Therefore, the cooling time of the drink is solved by an implicit function

$${\displaystyle f(L^{2}T^{-1}\kappa ^{-1}s^{1})=0}$$

that is, when the argument of the function ${\displaystyle L^{2}T^{-1}\kappa ^{-1}s^{1}}$ is some constant c. Therefore, the drink cooling time is ${\displaystyle T=cL^{2}\kappa ^{-1}s^{1}}$, so that the cooling time is proportional to the length scale of the ice cube squared, not just the length scale.

Why is the cooling time proportional to ${\displaystyle L^{2}}$ and not ${\displaystyle L}$? While it's true that the total ice cube surface area is proportional to ${\displaystyle 1/L}$, that accounts for only one factor in the increased cooling rate. Because the cubes are smaller, the thermal gradients are larger, again by a factor of ${\displaystyle 1/L}$, which increases the cooling rate by a factor of ${\displaystyle 1/L}$. So there are two effects that increase the cooling rate, each by a factor of ${\displaystyle 1/L}$, and the conventional explanation accounts for only one of these.

A simple 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.[10]

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