Dimensional Homogeneity on Some Arithmetic Operations

## Dimensional homogeneity of a sum

Recall that dimensional homogeneity(ibid. 2.) is defined mathematically by the theorem

The function ƒ (x1, x2, …, xn) is dimensionally homogeneous if, and only if,
Kƒ (x1, x2, …, xn) = ƒ (K1x1, K2x2, …, Knxn)
is an identity in the variables (x1, x2, …, xn, A, B, C, D, E, F, G) where
yK = yAaBbCcDdEeFfGg
x1K1 = x1Aa1Bb1Cc1Dd1Ee1Ff1Gg1
x2K2 = x2Aa2Bb2Cc2Dd2Ee2Ff2Gg2

xnKn = xnAanBbnCcnDdnEenFfnGgn.
Then
The (sum) function ƒ (x1, x2, …, xn) = x1 + x2 + … + xn is dimensionally homogeneous if, and only if, in
K ⋅ (x1 + x2 + … + xn) = K1x1 + K1x2 + … + Knxn
all terms in the sum have the same dimensions as the sum
K = K1 = K2 = … = Kn.

Let y = ƒ (x1, x2, …, xn) be such that ƒ is the sum of xi's,

ƒ (x1, x2, …, xn) = x1 + x2 + … + xn.
Substituting this into the theorem for dimensional homogeneity given by
Kƒ (x1, x2, …, xn) = ƒ (K1x1, K2x2, …, Knxn)
we get,
K ⋅ (x1 + x2 + … + xn) = K1x1 + K2x2 + … + Knxn
Kx1 + Kx2 + … + Kxn = K1x1 + K2x2 + … + Knxn.
Because this is an identity in the xi's
K = K1 = K2 + … = Kn.∎
Since, we know that for the coordinates of magnitude y, x1, x2, …, xn their corresponding units are given by K, K1, K2, …, Kn respectively, the identity of all the K's means all the terms have the same dimension. This can be shown as follows.

We know that,

K = AaBbCcDdEeFfGg
Ki = AaiBbiCciDdiEeiFfiGgi    for all i = 1, 2, …, n.
Because the same A to G are shared among the K's, from K = K1 = K2 + … = Kn we can see that for all i = 1, 2, …, n
a = ai
b = bi
⋮
g = gi.∎
Since, these small letters represent powers to their respective unit of magnitude the identity proves that all the terms have the same dimension. Hence this is a necessary condition.

## Dimensional homogeneity of a product

The (product) function
y = ƒ (x1, x2, …, xn)
= x1k1x2k1 ⋅ … ⋅ xnkn
is dimensionally homogeneous if, and only if, the exponents k1, k2, …, kn are a solution of the linear equations
a = a1k1 + a2k2 + … + ankn
b = b1k1 + b2k2 + … + bnkn

g = g1k1 + g2k2 + … + gnkn
where a, b, …, g are the dimensional exponents.
Note that this is a necessary condition to
K = K1k1 K2k2Knkn.

Because the expression
y = x1k1x2k1 ⋅ … ⋅ xnkn
plays an important part in dimensional analysis this expression is called products.

Let y = ƒ (x1, x2, …, xn) be such that ƒ is the product of xi's,

ƒ (x1, x2, …, xn) = x1k1x2k1 ⋅ … ⋅ xnkn.
Substituting this into the theorem for dimensional homogeneity given by
Kƒ (x1, x2, …, xn) = ƒ (K1x1, K2x2, …, Knxn)
we get,
K ⋅ (x1k1x2k1 ⋅ … ⋅ xnkn) = (K1x1)k1 ⋅ (K2x2)k2 ⋅ … ⋅ (Knxn)kn
K ⋅ (x1k1x2k1 ⋅ … ⋅ xnkn) = K1k1 K2k2Knkn ⋅ (x1k1x2k1 ⋅ … ⋅ xnkn) .
Because this is an identity
K = K1k1 K2k2Knkn.
We know that,
K = AaBbCcDdEeFfGg
Ki = AaiBbiCciDdiEeiFfiGgi    for all i = 1, 2, …, n.
Let us say that the function y = x1k1x2k1 ⋅ … ⋅ xnkn involves three dimensions. Then,
K = K1k1 K2k2Knkn
AaBbCc = (Aa1Bb1Cc1)k1 ⋅ (Aa2Bb2Cc2)k2 ⋅ … ⋅ (AanBbnCcn)kn
AaBbCc = A(a1k1 + a2k2 + … + ankn)B(b1k1 + b2k2 + … + bnkn)C(c1k1 + c2k2 + … + cnkn).
Therefore,
a = a1k1 + a2k2 + … + ankn
b = b1k1 + b2k2 + … + bnkn
c = c1k1 + c2k2 + … + cnkn
which in matrix form is where the exponents k1, k2, …, kn are a solution of the linear equations.∎