Dimensional Analysis

Most physical quantities are measurable and require a measurement system. Each measurement consists of a numerical value and a unit, which gives it meaning. For example, in 46 cm, 46 is the number and cm is the unit. Physics deals with minuscule to very large quantities, so unit conversion is often needed. This process is called dimensional analysis, which helps convert units using known relations.

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, and 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 that are used to measure these quantities are called derived units. e.g., Force, acceleration, pressure, energy, and power are derived quantities.  

Dimensions and Dimensional Formula

Every physical quantity is expressed in terms of fundamental units. In dimensional analysis, these fundamental units are raised to certain powers, and these powers are called dimensions.

The expression formed by combining all fundamental units with their respective powers is called the dimensional formula of a physical quantity.

Example: Velocity is expressed as: v = LT^{-1}

Here, the powers of L and T (1 and −1) represent the dimensions, and (LT^{-1}) is the dimensional formula of velocity.

Unit Factor Method

If we need to check the validity of an equation, then dimensional analysis comes to the rescue. Dimensional analysis is also called the factor label method or the 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

Physical quantities can be classified based on whether they have dimensions or not.

Dimensional Variables

Dimensional variables are physical quantities that have dimensions and do not have fixed values. These quantities can vary from one situation to another. Examples: velocity, acceleration, work, power.

Dimensionless Variables

Dimensionless variables are physical quantities that do not have dimensions. They are generally pure ratios and have no fixed units. Examples: specific gravity, coefficient of friction, Poisson’s ratio. Some dimensionless quantities may still have units for representation, such as angular displacement (radian) and Joule’s constant (joule/calorie). Others have no units at all, such as pure numbers like π, e, sinθ, and cosθ.

Dimensional Constants

Dimensional constants are physical quantities that have fixed values and also possess dimensions. Examples: gravitational constant (G), universal gas constant (R), speed of light in vacuum (c).

Derivation of Relation Using Dimensional Analysis (Centripetal Force)

Let the formula for centripetal force (F), acting on an object moving in a uniform circular path, be derived. The centripetal force depends on mass (m), velocity (v), and radius (r).

Assume: F = m^a v^b r^c

Writing dimensions of each quantity:

[MLT^-2] = M^a (LT^-1)^b L^c

= M^a L^b+c T^-b

Using the principle of homogeneity:

• (a = 1)
• (b = 2)
• (c = -1)

Hence, the required relation is: F = k \frac{mv^2}{r}

where (k) is a constant of proportionality.

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 4: 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.