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,
K ⋅ ƒ (x1, x2, …, xn) = ƒ (K1x1, K2x2, …, Knxn)
is an identity in the variables (x1, x2, …, xn, A, B, C, D, E, F, G) where
y ⋅ K = y ⋅ Aa ⋅ Bb ⋅ Cc ⋅ Dd ⋅ Ee ⋅ Ff ⋅ Gg         
x1 ⋅ K1 = x1 ⋅ Aa1 ⋅ Bb1 ⋅ Cc1 ⋅ Dd1 ⋅ Ee1 ⋅ Ff1 ⋅ Gg1
x2 ⋅ K2 = x2 ⋅ Aa2 ⋅ Bb2 ⋅ Cc2 ⋅ Dd2 ⋅ Ee2 ⋅ Ff2 ⋅ Gg2
⋮
xn ⋅ Kn = xn ⋅ Aan ⋅ Bbn ⋅ Ccn ⋅ Ddn ⋅ Een ⋅ Ffn ⋅ Ggn.
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
K ⋅ ƒ (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 = Aa ⋅ Bb ⋅ Cc ⋅ Dd ⋅ Ee ⋅ Ff ⋅ Gg                                            
Ki = Aai ⋅ Bbi ⋅ Cci ⋅ Ddi ⋅ Eei ⋅ Ffi ⋅ Ggi    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.

❶

Dimensional homogeneity of a product

The (product) function
y = ƒ (x1, x2, …, xn)           
= x1k1 ⋅ x2k1 ⋅ … ⋅ 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 K2k2 … Knkn.

Because the expression
y = x1k1 ⋅ x2k1 ⋅ … ⋅ 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) = x1k1 ⋅ x2k1 ⋅ … ⋅ xnkn.
Substituting this into the theorem for dimensional homogeneity given by
K ⋅ ƒ (x1, x2, …, xn) = ƒ (K1x1, K2x2, …, Knxn)
we get,
K ⋅ (x1k1 ⋅ x2k1 ⋅ … ⋅ xnkn) = (K1x1)k1 ⋅ (K2x2)k2 ⋅ … ⋅ (Knxn)kn                
K ⋅ (x1k1 ⋅ x2k1 ⋅ … ⋅ xnkn) = K1k1 K2k2 … Knkn ⋅ (x1k1 ⋅ x2k1 ⋅ … ⋅ xnkn) .
Because this is an identity
K = K1k1 K2k2 … Knkn.
We know that,
K = Aa ⋅ Bb ⋅ Cc ⋅ Dd ⋅ Ee ⋅ Ff ⋅ Gg                                            
Ki = Aai ⋅ Bbi ⋅ Cci ⋅ Ddi ⋅ Eei ⋅ Ffi ⋅ Ggi    for all i = 1, 2, …, n.
Let us say that the function y = x1k1 ⋅ x2k1 ⋅ … ⋅ xnkn involves three dimensions. Then,
K = K1k1 K2k2 … Knkn                                                                                   
Aa ⋅ Bb ⋅ Cc = (Aa1 ⋅ Bb1 ⋅ Cc1)k1 ⋅ (Aa2 ⋅ Bb2 ⋅ Cc2)k2 ⋅ … ⋅ (Aan ⋅ Bbn ⋅ Ccn)kn                      
Aa ⋅ Bb ⋅ Cc = 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
dimensional matrix of coefficients times vector of exponents = vector of dimensional exponents
where the exponents k1, k2, …, kn are a solution of the linear equations.∎

❷

Next:

On Vector Space (p:4) ➽

✪