Dimensional Analysis

Most of the physical things are measurable in this world. The system developed by humans to measure these things is called the measuring system. Every measurement has two parts, a number (n) and a unit(u). The unit describes the number, what this number is and what it signifies. For example, 46 cm, here 46 is the number and cm is the unit. Without units, it’s impossible to describe the quantity.  

In the physical world, there are all types of quantities to measure. As small as the size of an atom, to as large as the distance between the planets. It also becomes necessary to convert them from one unit to another. This conversion is called unit analysis or dimensional analysis. The unit analysis is just another form of proportional reasoning. In it, measurement is multiplied by some known proportion and gives a result having a different unit. To put it in general, it is a method to multiply or divide a number by a known ratio to find another unit.

Types of Units

1. Fundamental units: Quantities that are not derived from any other quantity are called fundamental quantities. Units used to measure these quantities are called fundamental units. e.g. Fundamental quantities are Length, Mass, time, Electric current, temperature, intensity, the quantity of substance. Ampere, Kelvin, Candela, mole. Units measuring these quantities are called fundamental units.
2. Derived units: Quantities that are derived from other quantities are called derived quantities. The units which are used to measure these quantities are called derived units. e.g. Force, acceleration, pressure, energy, power are derived quantities.  

Dimensions and dimensional formula

Every quantity needs to be expressed in a single unit. For that, all fundamental units involved in that quantity are raised to certain powers. These powers are called dimensions.

All the fundamental units raised to certain power are put together into an expression. This expression is called the dimensional formula of that quantity.

For ex – velocity is represented as v = L¹T^-1. Here 1 and -1 are called the dimensions and L¹T^-1 is the dimensional formula.  

Dimensional Analysis

If we need to check the validity of an equation, then dimensional analysis comes to the rescue. Dimensional analysis is also called factor label method or unit factor method as conversion factors are used to get the same units. If you want to check whether a given equation is correct or not, you can compute the dimensions on both sides (LHS and RHS), if both dimensions are equal, then the equation is correct, otherwise, it’s wrong.

For example –  

If you forget whether  

1. Speed = distance*time

2. Speed = distance/time

Dimensional analysis on 1

LHS = LT-1

RHS = LT

LHS != RHS  

Hence its wrong

Dimensional analysis on 2

LHS = LT-1

RHS = L/T = LT-1

LHS = RHS

Hence its right.

Homogeneity Principle of Dimensional Analysis

If each term on both sides in an equation has the same dimensional formula, then the equation is dimensionally correct. This is the principle of homogeneity. It is helpful because it helps in converting units from one system to another.  

For example –  

We have a physics equation,  

S = ut + 1/2at²

LHS is distance,  

So dimension of LHS = L¹M⁰T⁰

RHS,

In dimensional frame,

= [u][t] + [a][t]²

= [LT^-1][T] + [LT^-2][T]²

= [L] + [L]

Both the terms of RHS are equal to LHS, therefore this equation is dimensionally correct and the homogeneity principle of dimensional analysis holds.  

Applications of Dimensional Analysis

One of the most important aspects of measurement is dimensional analysis, and it has various applications such as,

1. It is used to check the correctness of an equation or any relation by using the principle of homogeneity. If dimensions on both sides are equal then the equation is also dimensionally correct.

2. It is also used to convert the units from one system to another.

3. They also represent the nature of physical quantity.

4. Dimensional analysis is also used to derive formulas.

Limitations of Dimensional Analysis

There are also many limitations of dimensional analysis. Few of them are

1. It doesn’t provide any information about dimensional constants.

2. Trigonometric, exponential, and logarithmic functions cannot be derived from dimensional analysis.

3. Using dimensional analysis, we cannot find whether a quantity is a scalar or vector.

Dimensional and dimensionless variables

Variables that have dimensions and do not have fixed values are called dimensional variables. Example – velocity, acceleration, work, power etc.

Variables that have no fixed values and do not have dimensions are called dimensionless variables. Example – specific gravity, friction, Poisson’s ratio.

There are dimensionless quantities with units – Angular displacement – radian, Joule’s constant – joule/calorie

Also dimensionless quantities without units – Pure numbers – π, e, sin θ, cos θ.

Variables that have dimensions and also the fixed values are called dimensional constants. Example – Gravitational Constant(G), Universal gas constant(R), speed of light in vacuum(c).

Deducing the relation among physical quantities using dimensional analysis

Lets derive the formula for centripetal force, F, acting on an object moving in a uniform circle.

We know that, centripetal force acting on a particle depends on its mass(m), velocity with which its moving(v) and the radius of the circle(r).

 F = m^av^br^c

 Writing the dimensions of all quantities

 [MLT-2] = M^a[LT^-1]^b L^c

 On simplifying,  

 [MLT^-2] = M^a[L^b+cT^-b]

 Using the principle of homogeneity,

 a = 1,  

 b + c = 1

 b = 2

 On solving, we get

 a = 1, b = 2, c = -1

Therefore the centripetal force, F is

F= k (m*v²)/r

Sample Problems 

Problem 1: Determine the dimensional formula of kinetic energy.

Solution:

The formula for the Kinetic energy is 1/2 mv²

The dimensional formula for velocity can be calculated as,

 = [M]

   We know that,

   Velocity = distance/time

   Dimensional formula of velocity = [LT^-1]

   Therefore, kinetic energy = [M^1L2T-2]

Problem 2: You remember πr² and 2πr as formulas of area and circumference of circle, but you are unable to recall which formula is for area and which for circumference. Find using dimensional analysis.

Solution:

   Formula 1 – πr² = [π].[r]²

   As π is a constant, so it is dimensionless

   So, =  1. [L]²

   Radius has the dimension of length, therefore formula 1 has dimensional formula of area.

Area = πr²

   Formula 2 – 2πr = 2.[π].[r]

   As π is a constant, so it is dimensionless

   So, =  2.1. [L] = dimension of length

   Radius has the dimension of length, therefore formula 2 has dimensional formula of circumference.

Circumference = 2πr

Problem 3: Is v = at dimensionally correct?

Solution:

Here, LHS: velocity, v 

v = [LT^-1]

And, RHS = at i.e. acceleration × time

Acceleration = [LT^-2]

Time = [T]

Therefore, at = [LT^-1]

Hence,

LHS = RHS

and the given equation is dimensionally correct.

Problem 5: What are dimensional constants?

Solution:

Those Constant which have dimension are called dimensional constant. For example: Planck’s constant, Joule’s constant etc.

Problem 5: What are dimensional variables?

Solution:

The physical quantities that have dimension but do not have a fixed value. For Example: velocity, displacement etc.