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
Kbar_i = (A/A')^a_i (B/B')^b_i (C/C')^c_i
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 Logo
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-solved-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