Theory of Dimensionless Products: Non-Dimensionless Products

Non-Dimensionless Products

Existence of a non-dimensionless product

Imagine the dimensional matrix

$example of dimensional matrix with fundamental dimensions M, L, and T$ . such that, the constants a, b, and c are not all zero. Under what conditions does there exist a product of the form y = x₁^k₁ ⋅ x₂^k₁ ⋅ … ⋅ x_n^k_n?

First condition

If y is not dimensionless, a product of the form y = x₁^k₁ ⋅ x₂^k₁ ⋅ … ⋅ x_n^k_n exists if, and only if, the dimensional matrix of the variables x₁, x₂, …, x_n has the same rank as the dimensional matrix of the variables y, x₁, x₂, …, x_n.

Suppose, the exponents k₁, k₂, …, k_n exists such that they are a solution of the linear system of equations a₁k₁ + a₂k₂ + … + a_nk_n = a
b₁k₁ + b₂k₂ + … + b_nk_n = b
⋮
g₁k₁ + g₂k₂ + … + g_nk_n = g
where a, b, …, g are the dimensional exponents.

For the dimensional matrix of coefficients of k's

$dimensional matrix of coefficients$ let its rank be r.

And, let R be the rank of the above dimensional matrix augmented by the column a, b, …, g

$dimensional matrix of coefficients augmented with constants a to g$ . But, theorem for consistency of a system of linear equations tells us that for the above system of equations to be consistent the ranks must be equal, r = R.

Because the system is consistent and k_i's are a solution, the product y = x₁^k₁ ⋅ x₂^k₁ ⋅ … ⋅ x_n^k_n must therefore exist.∎

Before one jumps into the topic of consistency and inconsistency one must familiarize with two other terminologies, system of linear equations and solution of the system.

Consider the equation

4x₁ − x₂ + 3x₃ = −1. One will notice that this is a linear equation in the variables x₁, x₂, and x₃.

Take another equation which is given by

3x₁ + x₂ + 9x₃ = −4. This is also not only a linear equation but equation in the variables x₁, x₂, and x₃.

Since both equations above are linear in the variables x₁, x₂, and x₃, values of these variables that satisfy one equation must also satisfy the other equation. Therefore, they can be lumped together as

4x₁ − x₂ + 3x₃ = −1
3x₁ + x₂ + 9x₃ = −4 where, the values x₁ = 1, x₂ = 2, and x₃ = −1 statisfy both equations.

Generalizing this, we define a system of linear equation as

A finite set of linear equations in the variables x₁, x₂, …, x_n.

Whose solution is defined as

A sequence of numbers s₁, s₂, …, s_n such that, x₁ = s₁, x₂ = s₂, …, x_n = s_n is a solution of every equation in the system.

Furthermore, there may exist another sequence of numbers t₁, t₂, …, t_n which also solves every equation in the system. Therefore, a solution set is

The set of all the solutions of the equation.

Not all systems of linear equations have solutions

Consider the system given by x₁ + x₂ = 4
2x₁ + 2x₂ = 6. Reducing the second equation by dividing by ½ we see that the above system is the same as x₁ + x₂ = 4
x₁ + x₂ = 3. In other words, the equations in the system contradict each other and they do not have a solution.

We then define a system to be inconsistent as

A system of equations that has no solution.

And, consistent defined as

A system of equations that has at least one solution.

Recall that rank^{(ibid. 5.)} is

The dimension of the row and column space of a matrix A is called the rank of A.

Furthermore

A system of linear equations Ax = b is consistent if and only if b is in the column space of A.

Let a system of linear equations be Ax = b such that it is equivalent to $[a11 a12 ... a1n; a21 a22 ... a2n; ... ; am1 am2 ... amn][x1; x2; ...; xn]^T = [b1; b2; ...; bm]^T$ This can be further written as $[a11*x1 + a12*x2 + ... + a1n*xn; a21*x1 + a22*x2 + ... + a2n*xn; ... ; am1*x1 + am2*x2 + ... + amn*xn] = [b1; b2; ...; bm]^T$ or $x1 [a11; a21; ... ; am1] + x2 [a12; a22; ... ; am2] + ... + xn [a1n; a2n; ... ; amn] = [b1; b2; ...; bm]^T$ Therefore b is a linear combination of column vectors of matrix A.

From definition^{(ibid. 5.)} we know that for the possibility of a linear combination of vectors at least one set of the scalars x₁, x₂, …, x_n must exist. It follows that if x representing the scalars exist then b must be a linear combination of vectors. Above shows that this is a linear combination of column vectors of A.

Therefore, the system is consistent only if b is a linear combination of the column vectors and consequently b is in the column space.∎

A system of linear equations Ax = b is consistent if and only if the rank of its augmented matrix is equal to the rank of A.

Consider the consistent linear system Ax = b such that A is a m × n matrix, x a n × 1 vector and b a m × 1 column vector. [A b] is the augmented matrix.

From the lemma^{(ibid. 5)}

Nonzero row vectors in a row-echelon form of a matrix A form a basis for the row space of A. we know that the number of the resulting nonzero vectors gives us the dimension of the column or row space of A and consequently the rank of matrix A.

From the lemma

Ax = b is consistent if and only if b is in the column space of A and since our system is consitent, the only unique column vector b between matrices A and [A b] must be in the column space of A. Therefore, the number of nonzero vectors following row operations on the A and [A b] will be equal.

Therefore, for a consistent system Ax = b, matrices A and [A b] will have equal ranks.∎

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Second condition

If y = ƒ (x₁, x₂, …, x_n) is a dimensionally homogeneous equation, and if y is not dimensionless, there exists a product of powers of the x's that has the same dimension as y.

This theorem depends on the theorem for the first condition (see proof).

Hypothesis: y = ƒ (x₁, x₂, …, x_n) is dimensionally homogeneous and y = x₁^k₁ ⋅ x₂^k₁ ⋅ … ⋅ x_n^k_n does not exist.

Consider the dimensional matrix of the variables y, x₁, x₂, …, x_n involving three fundamental units of dimensions [M], [L], and [T] given below

$example of dimensional matrix with fundamental dimensions M, L, and T$ . where the constants a, b, and c are not all zero and rank of this dimensional matrix is R.

The hypothesis tells us that a product of the form y = x₁^k₁ ⋅ x₂^k₁ ⋅ … ⋅ x_n^k_n does not exist. If r is the rank of the dimensional matrix of the independent variable x₁, x₂, …, x_n and we already know that the rank of the dimensional matrix of the dependent and independent variables is R then, according to the theorem for the first condition

r ≠ R. Consequently, the definition of rank of a matrix tells us that the rank of the dimensional matrix with the first column deleted must be less than R. Thus, r < R. Case: R = 3 Therefore, its determinant $determinant of matrix of order R = 3$ is not equal to zero.

Using Cramer's rule for cofactor expansion

$Cofactor C11 = b1c2 - b2c1$
$Cofactor C21 = a2c1 - a1c2$
$Cofactor C31 = a1b2 - a2b1$

Δ = aC₁₁ + bC₂₁ + cC₃₁ ≠ 0. If the first column is deleted and replaced by any other column from the parent dimensional matrix $dimensional matrix of order R = 3 with first column deleted and replaced by any column from the parent dimensional matrix$ For i = 1 or i = 2 the determinant is automatically equal to zero. For instance, for i = 1 $dimensional matrix of order R = 3 replaced by the column vector [a1 b1 c1]$

Δ = a₁(b₁c₂ − b₂c₁) + b₁(a₂c₁ − a₁c₂) + c₁(a₁b₂ − a₂b₁)
= a₁b₁c₂ −a₁b₂c₁ +a₂b₁c₁ −a₁b₁c₂ +a₁b₂c₁ −a₂b₁c₁
= 0.                                                                                            And, for i = 2 $dimensional matrix of order R = 3 replaced by the column vector [a2 b2 c2]$

Δ = a₂(b₁c₂ − b₂c₁) + b₂(a₂c₁ − a₁c₂) + c₂(a₁b₂ − a₂b₁)
= a₂b₁c₂ −a₂b₂c₁ +a₂b₂c₁ −a₁b₂c₂ +a₁b₂c₂ −a₂b₁c₂
= 0.                                                                                            But, for i ≥ 3 the determinant is not automatically equal to zero. For i = 3 $dimensional matrix of order R = 3 replaced by the column vector [a3 b3 c3]$

Δ = a₃(b₁c₂ − b₂c₁) + b₃(a₂c₁ − a₁c₂) + c₃(a₁b₂ − a₂b₁)
= a₃b₁c₂ − a₃b₂c₁ + a₂b₃c₁ − a₁b₃c₂ + a₁b₂c₃ − a₂b₁c₃ Based on the definition of rank of a matrix we know that for our R = 3 $determinant of matrix of order R = 3 is not equal to zero$ And, from the theorem for the first condition we know that r < R. Therefore, $rank of the dimensional matrix of order R = 3 replaced by the column vector [a3 b3 c3] must be less than R$ hence, $determinant of matrix of order R = 3 replaced by the column vector [a3 b3 c3] must be equal to zero$ Extending this we can state that for i ≥ 3 the determinants are all equal to zero.

Let,

C₁₁ → α, C₂₁ → β, C₃₁ → γ then our above results for determinants is summarized as Δ = aα + bβ + cγ ≠ 0,
a_iα + b_iβ + c_iγ = 0, for all i = 1, 2, …, n.

Since, our hypothesis claim that y = ƒ (x₁, x₂, …, x_n) is dimensionally homogeneous then,

K ⋅ ƒ (x₁, x₂, …, x_n) = ƒ (K₁x₁, K₂x₂, …, K_nx_n) where, K = A^a ⋅ B^b ⋅ C^c
K_i = A^a_i ⋅ B^b_i ⋅ C^c_i for all i = 1, 2, …, n.

If G is some arbitrary positive constant such that

A = G^α, B = G^β, and C = G^γ. Substituting these into the above K's we get, K = A^a ⋅ B^b ⋅ C^c
= G^αa ⋅ G^βb ⋅ G^γc
= G^{(aα + bβ + cγ)}
K_i = G^{(a_iα +
b_iβ + c_iγ)}
= G⁰ = 1, for all i = 1, 2, …, n. Because K_i = 1 and K = G^{(aα + bβ + cγ)} is an arbitrary constant K ⋅ ƒ (x₁, x₂, …, x_n) = ƒ (K₁x₁, K₂x₂, …, K_nx_n) becomes, K y = ƒ (x₁, x₂, …, x_n). This means that any arbitrary positive constant multiplied by y will be returned by the operator ƒ working on the independent variables. In other words, for this equation there is no correspondence from the independent variables x₁, x₂, …, x_n to the dependent variable y. That is, ƒ (x₁, x₂, …, x_n) cannot be a function. This is contradictory to the hypothesis that y = ƒ (x₁, x₂, …, x_n) is dimensionally homogeneous and hence by definition y is a function.

We can therefore state that for a rank of dimensional matrix of three if y = ƒ (x₁, x₂, …, x_n) is dimensionally homogeneous, then there exist a dimensionless product.

Following the above steps one can prove the above statement for other ranks.∎

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Next:

Buckingham's Theorem (p:8) ➽

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