Principle of Homogeneity of Dimensions - Param Himalaya

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. 

This principle is helpful because

1. To convert a physical quantity from one system of units to another : 

This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, 

nu = constant

$n_{1} u_{1} = n_{2} u_{2}$

Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are $M_{1}$, $L_{1}$ and $T_{1}$ and the other system are $M_{2}$ , $L_{2}$ and $T_{2}$ respectively, then we can write

$n_{1}[M_{1}^{a} \ L_{1}^{b} \ T_{1}^{c}]=n_{2}[M_{2}^{a} \ L_{2}^{b} \ T_{2}^{c}]$

2. The magnitude of physical quantities may be added together or subtracted from one another only if they have the same dimensions.

For Example, in the physical expression $v^{2} = u^{2} + 2as$, the dimensions of $v^{2}$, $u^{2}$ and 2as are the same and equal to $[L^{2}T^{-2}]$.

3. Any physical quantity can be multiplied or divided with the same or different dimensions.

As an example 1 kg and 2 m cannot be added or subtracted but they can be multiplied or divided.

W = mgh

4. To check the dimensional correctness of a given physical equation :

Example : The equation $\frac{1}{2}mv^{2} = mgh$ can be checked by using this method as follows :

Solution : Dimensional formula for

$\frac{1}{2}mv^{2} = [M] [LT^{-1}]^{2} = [ML^{2}T^{-2}]$

Dimensional formula for 

$mgh = [M][LT^{-2}][L] =[ML^{2}T^{-2}]$

Hence

$[ML^{2}T^{-2}] = [ML^{2}T^{-2}]$

$L.H.S = R.H.S$

Both sides are dimensionally the same, hence the equation $\frac{1}{2}mv^{2}$ =mgh is dimensionally correct. 

5. To establish the relation among various physical quantities: 

Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows. 

Let time period T depend upon mass m of the bob, length l of the pendulum and acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.

Solution : 

$$T \propto m^{a}l^{b}g^{c}$$

$$T = k. m^{a}l^{b}g^{c}$$

Here k is the dimensionless constant 

Rewriting the above equation with dimensions. 

$$[T^{1}] = [M^{a} ] [L^{b} ] [LT^{-2} ]^{c}$$

$$[M^{0}L^{0}T^{1}] = [M^{a}L^{b+c}T^{-2c}]$$

Comparing the powers of M, L and T on both sides, 

a =0

b+c = 0

-2c = 1

Solving for a,b and c

a=0, b = 1/2  and c= - 1/2

$$T \propto m^{a}l^{b}g^{c}$$

$$T = k.m^{0}l^{l/2}g^{-l/2}$$

$$T = k \left ( \frac{l} {g} \right )^{1/2}$$

$$T = k {\sqrt{\frac{l} {g} }}$$

Experimentally $k = 2 \pi$, hence 

$$T = 2 \pi {\sqrt{\frac{l} {g} }}$$