Reducing Non-Dimensionless Product to Dimensionless Form

## K-transformation

Let x1, x2, …, xn denote independent variables that represent magnitudes of physical quantities.

Geometrically, we may consider x1, x2, …, xn as coordinates in a space S.

Also, let A, B and C be any positive constants and a, b and c be any integer such that the variables K1, K2, …, Kn are given by

Ki = Aai Bbi Cci     ∀ i = 1, 2, …, n
Then, any independent variable xi may be considered to be
xi = Ki xi'     ∀ i = 1, 2, …, n
That is, a point transformation that carries the point xi' to the point xi in space S. This is called K-transformation.

K-space is the set of all points that can be derived from a given point xi' by K-transformation.
The resultant of two K-transformations is also a K-transformation. And, the set of all K-transformations is algebraically a group.
If the magnitudes of a physical quantity xi are restricted to have positive values, then the space S will consist entirely of points whose coordinates are positive.

### Four Lemmas

A K-space is generated by any one of its points.
Let the magnitudes of physical quantities xi' i = 1, …, n represent points that by K-transformation generate the K-space.

Let xi'' be another point in the K-space such that

xi'' = Ki' xi'
where Ki' = (A')ai (B')bi (C')ci

If the point generated by xi' is

xi = Ki xi'
where Ki = Aai Bbi Cci and the point xi also lies in the K-space, then
xi = Ki (xi''/Ki')
xi = K̄i xi''
where K̄i ≜ (Ki/Ki'), thus xi = K̄i xi'' tells us that the point xi in K-space is derived from the point xi'' by a K-transformation.

Therefore, any point xi derived from point xi' by a K-transformation may also be derived from the point xi'' by a K-transformation provided that the point xi'' lies in the same K-space as the point xi'.∎

The space S is completely partitioned by nonoverlapping K-spaces.
Since by definition
any point in the space S generates a K-space
and from the preceeding theorem that
a K-space is generated by any one of its points
therefore, the space S is partitioned into nonoverlapping K-spaces.∎

A dimensionally homogeneous dimensionless function, R = f(x1, x2, …, xn) is constant in any K-space.
Since π is dimensionless
K = Aa Bb Cc = A0 B0 C0 = 1
where A, B and C are any positive constants with respective dimensional exponents a, b and c.

Also, π is dimensionally homogeneous

Kπ = f(k1x1, k1x2, …, knxn)
substituting K = 1 we get
π = f(k1x1, k1x2, …, knxn)
Thus, the point (x1, x2, …, xn) generates π in K-space and π is a constant.∎

Since dimensionless products of x's are constants througout a K-space, if π1, π2, …, πp is a complete set of dimensionless products of the x's, then to each K-space of the space S there is a corresponding single set of values of the π's. This is restated in the next theorem.
If π1, π2, …, πp is a complete set of dimensionless products of the x's, there corresponds to each set of values of the π's a single K-space of the space S.
Let {π1', π2', …, πp'} be a complete set of constant values of π's and let xi' and xi'' be points (i = 1, …, n) of the space S that corresponds to the values of the elements in the set. Thus,
πh' = (x1')k1h (x2')k2h … (xn')knh
πh' = (x1'')k1h (x2'')k2h … (xn'')knh
or
(x1')k1h (x2')k2h … (xn')knh = (x1'')k1h (x2'')k2h … (xn'')knh
Since the magnitudes of physical quantities x's are assumed to have only positive values, taking logarithm of both sides yields
k1h log(x1') + k2h log(x2') + … + knh log(xn') = k1h log(x1'') + k2h log(x2'') + … + knh log(xn'')
or
k1h [log(x1') − log(x1'')] + k2h [log(x2') − log(x2'')] + … + knh [log(xn') − log(xn'')] = 0
or
k1h log(x1'/x1'') + k2h log(x2'/x2'') + … + knh log(xn'/xn'') = 0
Substituting the definition ri ≜ log(xi'/xi'') we get
r1 k1h + r2 k2h + … + rn knh = 0     for all h = 1, 2, …, p
Recall that {π1', π2', …, πp'} is a complete set of dimensionless products and from the theorem(ibid. 6.) that
exponents k1, k2, …, kn
of a complete set of dimensionless products of the independent variables x1, x2, …, xn
are a fundamental system of solutions of
a1k1 + a2k2 + … + ankn = 0
b1k1 + b2k2 + … + bnkn = 0
c1k1 + c2k2 + … + cnkn = 0
Thus, the system
r1 k1h + r2 k2h + … + rn knh = 0     for all h = 1, 2, …, p
will have the solution ki1, ki2, …, kip and the coefficients r1, r2, …, rn are linearly related to the coefficients of the fundamental system, that is, there exists α, β and γ such that
ri = log(xi'/xi'') = αai + βbi + γci     for all h = 1, 2, …, p
For logarithm of base 10
log10(xi'/xi'') = αai + βbi + γci
xi' = xi'' 10ai + βbi + γci)
Take A = 10α, B = 10β, C = 10γ, then
xi' = xi'' Aai Bbi Cci
xi' = Ki xi''
where KiAai Bbi Cci

Therefore, points xi' and xi'' belong in the same K-space.∎

## Buckingham's Theorem

If an equation is dimensionally homogeneous, it can be reduced to a relationship among a complete set of dimensionless products.
Let y = f(x1, x2, …, xn) be a dimensionally homogeneous equation. Then, we have the theorem(ibid. 7.) that tells us that there exists a product of powers of x's for such an equation. In other words, the dimensionally homogeneous equation y = f(x1, x2, …, xn) can be expressed as
π = F(x1, x2, …, xn)
where π is dimensionless.

Furthermore because of lemma

If π1, π2, …, πp is a complete set of dimensionless products of the x's, there corresponds to each set of values of the π's a single K-space of the space S.
we know that each set of values of π1, π2, …, πp corresponds to a single K-space.

And from lemma

A dimensionally homogeneous dimensionless function, R = f(x1, x2, …, xn) is constant in any K-space.
each K-space will correspond to a single value of π

Thus, each set of values of π1, π2, …, πp will correspond to a single value of π. That is, π is a single-valued function of π1, π2, …, πp.

Therefore, an arbitrary dimensionally homogeneous equation y = f(x1, x2, …, xn) is reduced to the form π = F(x1, x2, …, xn). Alternatively, an equation that relates dimensionless products is dimensionally homogeneous.∎

Since(ibid. 6.)

The number of products in a complete set of dimensionless products of the independent variables x1, x2, …, xn is nr, in which r is the rank of the dimensional matrix of the variables.
for the complete set of dimensionless products π1, π2, …, πp
p = nr

### Systematic steps for deriving a complete set of dimensionless products

The illustration is made using the example by Langhaar (1951d). Imagine that the investigation involves a unknown function f which is dependent a collection of variables and/or parameters: P, Q, R, S, T, U, V and all of them can be derived from three base dimensions M, L, T. Thus, the problem f(P, Q, R, S, T, U, V) is a MLT-dimensional system.

The relationship of all of the independent variables/parameters of the f to the three base dimensions can be summarized as

P Q R S T U V
M 2 −1 3 0 0 −2 1
L 1 0 −1 0 2 1 2
T 0 1 0 3 1 −1 2
This is the dimensional matrix. Hence, Step-1: Get the Dimensional Matrix is accomplished.

From the earlier discussions we know that the dimensionless products about to be derived will be of the form

π = Pk1Qk2Rk3Sk4Tk5Uk6Vk7
The entries of each row of the dimensional matrix can be taken as the coefficients of the k's giving us the system of homogeneous equations
2k1k2 + 3k3 + 0k4 + 0k5 − 2k6 + k7 = 0
k1 + 0k2k3 + 0k4 + 2k5 + k6 + 2k7 = 0
0k1 + k2 + 0k3 + 3k4 + k5k6 + 2k7 = 0
Thus, performing Step-2: Solve the Homogeneous Equation returns
k5 = −11k1 + 9k2 − 9k3 + 15k4
k6 = 5k1 − 4k2 + 5k3 − 6k4
k7 = 8k1 − 7k2 + 7k3 − 12k4
These are the expressions for k5, k6 and k7. What about k1, k2, k3 and k4?

To get the solution we start from k1 = 1 and set the rest to 0 and then set k2 and rest to 0 and so on until k4 = 1 as follows

Set: k1 = 1, rest to 0     Then: k5 = −11, k6 = 5, k7 = 8
Set: k2 = 1, rest to 0     Then: k5 = 9, k6 = −4, k7 = −7
Set: k3 = 1, rest to 0     Then: k5 = −9, k6 = 5, k7 = 7
Set: k4 = 1, rest to 0     Then: k5 = 15, k6 = −6, k7 = −12
Putting this in tabular form we get
k1 k2 k3 k4 k5 k6 k7
1 0 0 0 −11 5 8
0 1 0 0 9 −4 −7
0 0 1 0 −9 5 7
0 0 0 1 15 −6 −12
This is the solution matrix. Hence, Step-3: Get the Solution Matrix is accomplished.
k1 k2 k3 k4 k5 k6 k7
P Q R S T U V
π1 1 0 0 0 −11 5 8
π2 0 1 0 0 9 −4 −7
π3 0 0 1 0 −9 5 7
π4 0 0 0 1 15 −6 −12
Finally, for Step-4: Get the Dimensionless Products we have
π1 = PT−11U5V8
π2 = QT9U−4V−7
π3 = RT−9U5V7
π4 = ST15U−6V−12 diman© is my in-house Clojure based software for dimensional analysis. It is based on seven base dimensions: [M], [L], [T], [K], [A], [mol] and [cd] representing the quantities mass, length, time, thermodynamic temperature, electric current, amount of substance and luminous instensity respectively. Therefore, diman© is based on this seven dimensional system.

diman© is capable of performing dimensional consistency checks and derive dimensionless products. The program aims at simplifying the tedious computational steps particularly while deriving dimensionless products.

### Accomplishing the steps in diman©

For a given problem before on can get the results of a complete set of dimensionless products the user must perform some minimum initialization steps.

Setting up the dimensional formulae of all the independent variables of the unknown function f
``` (def formula_of_manifold_eqn   [{:quantity "term-p", :dimension "[M^(2)*L^(1)]"}   {:quantity "term-q", :dimension "[M^(-1)*T^(1)]"}   {:quantity "term-r", :dimension "[M^(3)*L^(-1)]"}   {:quantity "term-s", :dimension "[T^(3)]"}   {:quantity "term-t", :dimension "[L^(2)*T^(1)]"}   {:quantity "term-u", :dimension "[M^(-2)*L^(1)*T^(-1)]"}   {:quantity "term-v", :dimension "[M^(1)*L^(2)*T^(2)]"}]) ```

Since these are derived dimensional formulae, it must be placed temporarily inside the `standard_formula` entity. This is done with
`(update-sformula formula_of_manifold_eqn)`

Finally, to call the dimensions for respective independent variable of f we define
``` (def varpars   [{:symbol "P", :quantity "term-p"}   {:symbol "Q", :quantity "term-q"}   {:symbol "R", :quantity "term-r"}   {:symbol "S", :quantity "term-s"}   {:symbol "T", :quantity "term-t"}   {:symbol "U", :quantity "term-u"}   {:symbol "V", :quantity "term-v"}]) ```

Steps-1, 2 and 3 in one code

The processes for generating the dimensional matrix, solving the homogeneous equation and determining the solution matrix can be achieved in one code block as shown
``` (def solution_matrix (get-solution-matrix                          (solve (get-augmented-matrix                                      (generate-dimmat varpars))))) ```
The solution matrix is therefore
``` => (view-matrix solution_matrix) [1 0 0 0 -11N 5N 8N] [0 1 0 0 9N -4N -7N] [0 0 1 0 -9N 5N 7N] [0 0 0 1 15N -6N -12N] Size -> 4 x 7 ```

For the final Step-4: Get the Dimensionless Products we use `get-dimensionless-products`. Thus,
``` => (pprint (get-dimensionless-products solution_matrix varpars))   [{:symbol "pi0", :expression "P^(1)*T^(-11)*U^(5)*V^(8)"}   {:symbol "pi1", :expression "Q^(1)*T^(9)*U^(-4)*V^(-7)"}   {:symbol "pi2", :expression "R^(1)*T^(-9)*U^(5)*V^(7)"}   {:symbol "pi3", :expression "S^(1)*T^(15)*U^(-6)*V^(-12)"}] ```

Therefore, diman© saves the analyst from labouring in computational tasks but at the same time provides the ability to follow each step of the derivation process.

Bibliography
• Anton, H. (1977a). Elementary Linear Algebra (2nd ed.). John Wiley & Sons, Inc.
As of 2021 this book is in its 11th edition but I consider the earlier editions particular those published in the 1970s to be an excellent pedagogical text for a learner interested in the concepts of Linear Algebra. Calculus is not a prerequisite for this text.

• Anton, H. (1977b). Vector Spaces. In Elementary Linear Algebra (2nd ed., pp. 121–204). John Wiley & Sons, Inc.
The concepts of vector space discussed here(ibid. 4.) were based on this chapter.

Bureau International des Poids et Mesures is the organization whose mission is to provide standards on matters related to measurement science. diman© is a software for doing dimensional analysis; It can do consistency checks and derive dimensionless products. Diman(c) is based on the Internation System of Units (SI), base units.

• Buckingham, E. (1914). On Physically Similar Systems; Illustrations of the Use of Dimensional Equations. Phys. Rev., 4(4), 345–376. https://doi.org/10.1103/PhysRev.4.345
This is the paper where Buckingham illustrates the possibility of reducing a given dimensionally homogeneous equation into a relationship among the complete set of dimensionless products of the equation. This has come to be known as Buckingham's Theorem.

• Jerrard, H. G., & McNeill, D. B. (1992). Dictionary of Scientific Units: Including dimensionless numbers and scales (6th ed.). Chapman & Hall.
This is a good resource for someone doing science let alone dimensional analysis.

• Langhaar, H. L. (1951a). Algebraic Theory of Dimensional Analysis. In Dimensional Analysis and Theory of Models (pp. 47–59). John Wiley & Sons, Inc.
Most of the materials for this lecture "Theory of Dimensionless Products" were influenced by this chapter.

• Langhaar, H. L. (1951b). General Remarks on Dimensional Analysis. In Dimensional Analysis and Theory of Models (pp. 14–16). John Wiley & Sons, Inc.
This section provides the definition of dimensional analysis.

• Langhaar, H. L. (1951c). Principles and Illustrations of Dimensional Analysis. In Dimensional Analysis and Theory of Models (pp. 13–28). John Wiley & Sons, Inc.
This preliminary chapter provides an overall view of why one might want to consider using dimensionless products. The section "General Remarks on Dimensional Analysis" is part of this chapter.

• Langhaar, H. L. (1951d). Systematic Calculation of Dimensionless Products. In Dimensional Analysis and Theory of Models (pp. 29-46). John Wiley & Sons, Inc.
This chapter provides the steps for deriving dimensionless products. It has two examples. This chapter and the chapter on algebraic theory (Langhaar, 1951a) were the foundation for how diman© can derive dimensionless products.

• McNish, A. G. (1957, April 1). Dimensions units and standards. Physics Today, 10(4), 19. https://doi.org/10.1063/1.3060330
This is from a talk at the National Bureau of Standards, 1956. The talk addresses various fundamental questions like What is a dimension? What is the purpose of a dimensional system? Why should there be at least seven base/elemental dimensions? How should we choose the minimum number of quantities and hence the "absolute" units to build a consistent system of units for the given experimentally derived equations?

• Preussner, G. M. (2018, May 24). Dimensional Analysis in Programming Languages. Personal Homepage. https://gmpreussner.com/research/dimensional-analysis-in-programming-languages
This article provides a fairly exaustive landscape of the software implementation — in terms of available packages that deal with dimensions or dimensional analysis and in some cases at the level of some high-level programming language — with regards to dimensional analysis. From this paper one notices that the notion of "dimensional analysis" used for most softwares are often not defined or poorly defined and hence vague and not necessarily alligned to its mathematical definition. Consequently, out of more than fifty or so dimensional analysis related softwares mentioned in the article only a very few addressed dimensionless product.