Dimensional Analysis - Meaning, Examples, FAQs

Principle of Homogeneity:

The principle of homogeneity of dimensions says that “ In any physical mathematical equation the dimensions of each term appearing in the equation are the same on each side of that equation”. This is called the principle of homogeneity.

What is meant by Dimension?

In physics, any physical quantity can be expressed in terms of fundamental units and representation of a physical quantity in terms of fundamental units is called the dimension of that physical quantity.

Following are the symbols for fundamental units used in Dimensional Analysis.

What is Dimensional Analysis?

When we represent each physical quantity of a mathematical equation in its dimensional form then analysis of dimensions to determine whether a given equation is correct or not dimensionally is known as dimensional analysis.

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Application of Dimensional Analysis:

The most widely uses of dimensional analysis are mentioned as:

1. Dimensional analysis is used to check the validity of a physical equation.
2. Dimensional analysis is used to determine the dimensions of any unknown variable’s dimension in a given physical equation.
3. Dimensional analysis is used to convert units from one system to another.

Examples of Dimensional Analysis:

When we analyse the physical equation by using their dimensions such as Distance, velocity and Time relation.

We know that Dimension of physical quantity Velocity is [LT^-1] while Dimensions of quantity Time is [T] and from relation, we know that, Distance=Velocity×time so we can find the dimension of quantity distance by multiplying the dimensions of Velocity and Time and we get, Dimension of Distance as LT^-1T=[L] Hence, this is a simple example of dimensional analysis showing a valid physical equation can be checked by using dimensional analysis.

Check the correctness of the equation f=mv²/r

We will use the dimensional analysis and principle of homogeneity which can be used as If the dimension of quantity ‘f’ represent force and dimension of quantity mv²/r where m represents mass, v represents velocity and r represents radius are the same then the given equation will be correct dimensionally.

The force which has a dimension of mass×acceleration so, dimension of ‘f’ can be written as M[LT^-2] or Dimension of force f=[MLT^-2]

Now, the dimension of radius which is simply the distance will be Dimension of r=[L] and for mass Dimension of mass m=[M] and for velocity Dimension of velocity v=[LT^-1] Now, on putting these dimensions on the right-handed part of the equation which is mv²/r we get, [M][LT^-1]²/[L] on solving we get, [MLT^-2]

Hence, the dimension of quantity f is Dimension of force f=[MLT^-2] and the dimension of quantity mv²/r is, [MLT^-2] which are the same, so from the principle of homogeneity this physical equation has dimensions the same on both sides so, this is a correct equation.

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Check the correctness of the equation v²-u²=2aS

We will use the dimensional analysis and principle of homogeneity which can be used as If the dimension of quantity v²-u² where v, u represents velocity and has a dimension of [LT^-1] and the dimension of quantity 2aS where a represent acceleration and has a dimension of [LT^-2] and S represent distance which has a dimension of [L] is the same, then the equation will be correct dimensionally. Now, using these dimensions let us find the dimension of quantity v²-u² as [LT^-1]²=[L²T^-2] since, u and v are both velocities so their difference is also a velocity.

Now, let us find the dimension of quantity 2aS as, 2 is a constant which is dimensionless and multiplication of a and S will have the dimension of LT^-2L=[L²T^-2] so, we see that both parts have the same dimension of [L²T^-2] so, according to the principle of homogeneity, both parts have the same dimension which shows, given equation is correct.

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Dimensional analysis of S=ut+0.5at²

We will do the Dimensional analysis of S=ut+1/2at² by using the principle of homogeneity to check the correctness of the following equation S=ut+1/2at² where S represent distance having a dimension of [L] which is on the left side of the equation. Now coming to the right side, we have u which is velocity having the dimension of [LT^-1] and t is time having the dimension of [T] so the net dimension of the product of velocity u and time t will be LT^-1T=[L] and similarly another part of right side equation is 0.5at² where a is acceleration having the dimension of [LT^-2] and the dimension of the square of t is [T²] so the net dimension of term 0.5at² will be [L]. Hence, the net right-sided equation has a dimension of L+L=[L] since the addition of two dimensions is the same dimension. Hence, the left side and right side of the equation has the same dimension of [L] so, by the principle of homogeneity, the equation S=ut+0.5at² is correct.

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