Uses and Limitations of Dimensional Analysis

Uses of dimensional analysis

The method of dimensional analysis is used to

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

2.    check the dimensional correctness of a given equation.

3.      establish a relationship between different physical quantities in an equation.

(i) To convert a physical quantity from one system of units to another

Given the value of G in cgs system is 6.67 x 10^âˆ'8dyne cm² g^âˆ'2. Calculate its value in SI units.

In cgs system

G_cgs     = 6.67 x 10^âˆ'8       

M₁ = 1g

L₁      = 1 cm

T₁      = 1s

In SI system

G = ?

M₂ = 1 kg

L₂      = 1m

T₂      = 1s

The dimensional formula for gravitational constant is  [M^-1L³T^-2]

In cgs system, dimensional formula for G is [M₁^xL₁^yT₁^z]

In SI system, dimensional formula for G is [M₂^xL₂^yT₂^z]

Here x = âˆ'1, y = 3, z = âˆ'2

[M₂^xL₂^yT₂^z] = G_cgs[M₁^xL₁^yT₁^z]

G = G_cgs[M₁/m₂]^x[M₁/m₂]^y[M₁/m₂]^z

= 6.67 x 10^âˆ'8 [1g/1kg]^-1[1cm/1m]³[1s/1s]^-2

= 6.67 x 10^âˆ'11

In SI units,

G = 6.67 x 10^âˆ'11 N m² kg^âˆ'2

(ii) To check the dimensional correctness of a given equation

Let us take the equation of motion

s = ut + (1/2)at²

Applying dimensions on both sides

[L] = [LT^âˆ'1] [T] + [LT^âˆ'2] [T²]

(1/2 is a constant having no dimension)

[L] = [L] + [L]

As the dimensions on both sides are the same, the equation is dimensionally correct.

(iii) To establish a relationship between the physical quantities in an equation

Let us find an expression for the time period T of a simple pendulum.  The time period T may depend upon (i) mass m of the bob (ii) length l of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended.

(i.e) T α m^x l^y g^z

T = k m^x l^y g^z

where k is a dimensionless constant of propotionality. Rewriting equation (1) with dimensions,

[T¹] = [M^x] [L ^y] [LT^âˆ'2]^z

[T¹] = [M^x L ^{y + z} T^âˆ'2z]

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

x = 0, y + z = 0 and âˆ'2z = 1

Solving for x, y and z, x = 0, y = 1/2 and z =-1/2

From equation (1), T = k m^o l ^1/2 g^âˆ'1/2

T =k root(l/g)

Experimentally the value of k is determined to be 2Ï€.

T=2Ï€ root(l/g)

Limitations of Dimensional Analysis

(i)                The value of dimensionless constants cannot be determined by this method.

(ii)             This method cannot be applied to equations involving exponential and trigonometric functions.

(iii)           It cannot be applied to an equation involving more than three physical quantities.

(iv)            It can check only whether a physical relation is dimensionally correct or not. It cannot tell whether the relation is absolutely correct or not. For example applying this technique s =ut + (1/4) at² is dimensionally correct whereas the correct relation is s = ut + (1/2) at²