# Dimensional Formula , Dimensional Equation , Principle of Homogeneity

## Dimension :

Dimension of a physical quantity is defined as the power to which the fundamental units have to be raised to represent the derived unit of that quantity.

Dimensional Formula of Some Physical Quantities:

 Quantity                       Dimensions Quantity                    Dimensions Acceleration                         $LT^{-2}$ Angular acceleration          T-2 Angular frequency/ speed   T-1 Angular Momentum       $M L^2 T^{-1}$ Angular velocity              T-1 Area                                    L2 Displacement                   L Energy                      $M L^2 T^{-2}$ (Total /Kinetic /potential/ Internal) Force                             $MLT^{-2}$ Frequency                    T–1 Gravitational Field strength   LT–2 Gravitational potential      L2T–2 Length                        L Mass                           M Density                     ML–3 Momentum            $M L T^{-1}$ Power                      $ML^2 T^{-3}$ Pressure                 $ML^{-1}T^{-2}$ Rotational Inertia     ML2 Time                       T Torque                 $ML^2T^{-2}$ Velocity                 LT–1 Heat                       $ML^2T^{-2}$ Capacitance      $M^{-1}L^{-2}T^4 I^2$ Charge                        IT Conductivity      $M^{-1}L^{-3}T^3I^2$ Current                      IT° Current Density      L–2T°I1 Electric dipole moment     LIT Electric field Strength       MLT–3I–1 Electric Flux              $ML^3T^{-3}I^{-1}$ Electric Potential      $ML^2T^{-3}I^{-1}$ Electromotive force  $ML^2T^{-3}I^{-1}$ Inductance        $ML^2T{-2}I{-2}$ Magnetic dipole moment     L2T0I Magnetic field Strength     MT0I1 Magnetic Flux             $M L^2 T^{-2}I^{-1}$ Magnetic Induction     $M T^{-2}I^{-1}$ Permeability            $M L T^{-2} I^{-2}$ Permittivity            $M^{-1}L^{-3} T^4 I^2$ Resistance               $M L^2 T^{-3}I^{-2}$ Resistivity              $M L^3 T^{-3}I^{-2}$ Voltage                 $M L^2 T^{-3}I^{-1}$ Volume                  L3 Wavelength         L Work /Energy      $M L^2 T^{-2}$

## Dimensional Equation :

Whenever the dimension of a Physical quantity is equated with its dimensional formula , we get a dimensional Equation .

## Principle of Homogeneity :

According to this , we can multiply Physical quantity with same or different dimensional formula , however in addition & subtraction only like quantities can be added or subtracted .
e.g. If P + Q ⇒ P & Q both represent same Physical quantity .

Solved Example :  Calculate the dimensional formula of Energy from the equation $E = \frac{1}{2}mv^2$

Solution : Dimensionally , E = mass × (velocity)2

Since 1/2 is a number and has no dimension .

$\large [E] = M \times (\frac{L}{T})^2$

$\large [E] = M L^2 T^{-2}$

Solved Example : The kinetic Energy of a particle moving along elliptical trajectory is given by $K = \alpha s^2$ ; Where s is the distance travelled by the particle . Determine dimension of α .

Solution : $\large K = \alpha s^2$

$\Large \alpha = \frac{K}{s^2}$

$\large [\alpha] = \frac{M L^2 T^{-2}}{L^2}$

$\large [\alpha] = M L^0 T^{-2}$

Solved Example : The distance covered by the particle is given by $\large x = a + bt + ct^2 + dt^3$ . Find the dimension of a , b , c & d .

Solution : According to principle of homogeneity , all the five quantities should have same dimension . Since [x] = [L]

[a] = [L]

[bt] = [L] ⇒ [b] = [LT-1]

[ct2] = [L] ⇒ [c] = [LT-2]

[dt3] ⇒ [L] = [d] = [LT-3]

## What are the Uses of Dimensional equations ?

(i) Conversion of one system of units into another.

(ii) Checking the accuracy of various formulae.

(iii) Derivation of formula

#### (i) Conversion of one system of units into another.

It is based on fact that ;

Numerical value × Unit = Constant

So on changing unit the numerical value also gets changed . If n1 and n2 are the numerical values of a given physical quantity and u1 & u2 be the units respectively in two different system of units , then

n1 u1 = n2 u2

$\large n_2 = n_1 [\frac{M_1}{M_2}]^a [\frac{L_1}{L_2}]^b [\frac{T_1}{T_2}]^c$

Solved Example : Convert 1 N into dyne.

Solution: Dimensional formula of force is

F = [M1L1T-2]

Now we have to convert M.K.S system into C.G.S system

M = 1kg    L = 1m    T = 1s (M.K.S)

M = 1g    L = 1cm    T = 1s (C.G.S)

F = M1L1T-2

F (SI) = [1 kg] [1m] [s-2]

= [103g] [102 cm] [s-2]

= 105 dyne

Solved Example : Check the accuracy of the relation

$\displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

where ν is the frequency, l is length , T is tension and m is mass per unit length of the string.

Solution : The given relation is

$\displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

Writing the dimensions on either side, we get

LHS = ν = [T-1] = [M0L0T-1]

RHS = $\displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

$\displaystyle = \frac{1}{L}\sqrt{\frac{MLT^-2}{ML^-1}}$

= [T-1]

As LHS = RHS

Dimensionally the formula is correct

Solved Example :    If force, length and time would have been the fundamental units what would have been the dimensional formula for mass ?

Solution:Let   M = K FaLbTc

[M1L0T0] =  [MLT-2]a [Lb] Tc

[M1L0T0]= [MaL(a+b)T(-2a+c)]

a = 1 , a + b = 0     &  – 2a + c = 0

=> a = 1 , b = – 1 ,  c = 2

Dimensional formula for mass : [FL1T]

Solved Example : In the equation y = A sin(ωt – kx) obtain the dimensional formula of ω and k. Given x is distance and t is time.

Solution:     The given equation is

y = A sin(ωt – kx)

The argument of a trigonometrical function i.e. angle is dimensionless

i.e.   ωt = θ

$\displaystyle \omega = \frac{\theta}{t} = \frac{1}{T}$

= [T-1]

= [M0L0T-1]

Also      kx = θ

$\displaystyle k = \frac{\theta}{x} = \frac{1}{L}$

= [ L-1 ]

= [M0L-1T0]

## Limitations of Dimensional Analysis:

1. Dimension does not depend on the magnitude of the quantity involved. Therefore, a dimensionally correct equation need not be actually correct.

e.g. : dimension of 1/T and 2π/T are same

2. Dimensional method cannot be used to derive relations other than those involving products of physical parameters.

e.g. : y = a cos(ωt − kx) can not be derived using this method.

3. This method cannot be applied to derive formula if in mechanics a physical quantity depends on more than three physical quantities (mass, length, time). e.g. : $\displaystyle T = 2\pi \sqrt{\frac{I}{m g L}}$ cannot be derived by using dimensions .

Exercise :   “The dimension of torque is equal to the dimension of work”. Yet, the two quantities are different. Explain.