Dimensional Analysis

PhysicsUnits and measurements

Introduction

Dimensions, in very simple words, are to study the nature of any object mathematically. To check how tall a person is, or to see how big the circle is, we have been using dimensions with the magnitude of course for quantifying the objects around us in dayto- day life. Dimensional analysis just refers to the relationship of the different physical quantities with the units of measurement. In this article, we shall discuss what exactly it means, where it comes from and what are its various applications that we see around us at present.

What is Dimensional Analysis?

Dimensional analysis is an analysis done to understand the relationship between the different physical quantities and the base quantities. The base quantities as given by the International System of Quantities are- length, mass, time, current, amount of a substance, and luminous intensity.

The basic way of performing the dimensional analysis requires us to represent the parameters of any measurable quantity in terms of these 6 units. For example, the force can be represented as mass(kg) times acceleration, with the SI unit called Newton. But, the acceleration in the formula can be broken down into length times the time squared($\mathrm{ms^{-2}}$).

Hence, this gives us the units of force finally all represented in the base units ($\mathrm{kg.m.s^{-2}}$)i.e mass, length, and time. This example gives us a rough overview of dimensional analysis. It can be represented using the square brackets and M, L, and T as −

$$\mathrm{F=[MLT^{-2}]}$$

The principle of Homogeneity

The principle of Homogeneity is the most basic rule of dimensional analysis. For our reference to understand this principle, let’s look at the following terms and their meanings −

Commensurable quantities: The physical quantities that have the same dimension Incommensurable quantities: The physical quantities that have different dimensions The principle of Homogeneity states that no two or more incommensurable quantities can be added, equaled, subtracted, or even compared.

There is a very famous analogous phrase for explaining this principle- “apples and oranges”. It is often used to talk about the differences between the two objects and how they are not to be physically compared

This principle implies that for an expression or an equation to be physically meaningful, addition, subtraction, and comparison of just commensurable quantities have to be done. One cannot add the mass of the apple to the length of the banana. But the mass of the apple and mass of the banana can be added (this will safely give us total mass) Whereas, the mass of the apple divided by the length of the banana is fine and well accepted. In very simple words, units of both sides of the equation should have the same unit/dimension.

Applications of Dimensional Analysis

Dimensional analysis is widely used to solve and analyze real-life problems involving quantities. Major applications of the dimensional analysis can be listed as follows

• It can be used to check the correctness of any formula or equation. As the Principle of Homogeneity restricts us to have the same units on both sides of the equation, it is a great way of checking if any complex equation is valid or not.

• It can be used to find out units of any unknown physical constant: Again, while validating and knowing the Principle of Homogeneity, we can find out the unknown quantity in an equation.

• It is used to find out the unit of the physical quantity in some other system of units.

• It can also be used to derive formulas and new relations for various complex quantities.

Solved Examples

Q1. Using dimensional analysis, check if the formula $\mathrm{F=\frac{mv^{2}}{r^{2}}}$ is correct or not?

Ans. The dimensional formula of force is $\mathrm{MLT^{-2}}$

$$\mathrm{[MLT^{-2}]=\frac{[M][LT^{-1}]^{2}}{[L]^{2}}}$$

$$\mathrm{=[ML^{2-2}T^{-2}]}$$

$$\mathrm{=[MT^{-2}]}$$

RHS$\mathrm{\neq}$LHS, i.e the dimensions on both sides of the formula are unequal. Hence, due to the principle of homogeneity, the formula is incorrect.

Q2. Find the dimensions of energy?

Ans. Energy is defined as the capacity of doing work. For our convenience let’s consider potential energy, the formula for which is given by $\mathrm{m\times g\times h}$, where ‘m’ is mass, ‘g’ is the acceleration due to gravity, and h is height.

Hence,

$$\mathrm{E=[M][LT^{-2}][L]}$$

$$\mathrm{E=[ML^{2}T^{-2}]}$$

Q3. Check the correctness of the equation of motion: v=u+at, where all the variables hold their usual meanings.

Ans. Given that $\mathrm{v=u+at}$

RHS: $\mathrm{v=[LT^{-1}]}$

LHS: $\mathrm{u+at=[LT^{-1}]+[LT^{-2}][T]}$

$$\mathrm{[LT^{-1}]=[LT^{-1}]+[LT^{-2+1}]}$$

$$\mathrm{=[LT^{-1}]=[LT^{-1}]+[LT^{-1}]}$$

In LHS we see two quantities of the same dimensions being added, which is very much permissible as they will add up to give a quantity of the same dimension. Further, it matches the dimension at RHS. Hence, the equation is valid and correct.

Q4. In some theories, the time period T is given to be dependent on pressure P, density $\mathrm{\rho }$, and on energy E. Using dimensional analysis, derive the formula indicating the relationship between T, P, E and $\mathrm{\rho }$.

Ans. According to the question, we have

$$\mathrm{T=CP^{a}\rho ^{b}E^{c}}$$

where C is a dimensionless constant

Putting dimensions in this equation,

$$\mathrm{[T]=C[ML^{-1}T^{-2}]^{a}[ML^{-3}]^{b}[ML^{2}T^{-2}]^{c}}$$

$$\mathrm{[T]=C[M^{+b+c}L^{-a-3b+2c}T^{-2a-2c}]}$$

Now, comparing the powers of the base units from both sides

$$\mathrm{0=a+b+c}$$

$$\mathrm{1=-2a-2c}$$

$$\mathrm{0=-a-3b+2c}$$

On solving the above linear equations, we get

$$\mathrm{a=-\frac{5}{6};b=1;c=\frac{1}{3}}$$

i.e The formula becomes $\mathrm{T=C\frac{\rho ^{1/2} E^{1/3}}{P^{5/6}}}$, where C is a dimensionless constant.

Limitations

Although the dimensional analysis is extremely helpful in so many domains to simply the difficult technical problems into manageable ones, we still have some limitations of this analysis which can be listed as follows

• It doesn’t give us any information about the value of the physical constants.

• While using dimensional analysis to derive the equations, the method fails to give us an idea about the expressions involving the trigonometric, logarithmic, and exponential terms.

• The dimensional analysis doesn’t tell us if a physical quantity is a scalar or a vector.

• It cannot be used to derive the expressions that involve a quantity that depends on more than three physical quantities.

Conclusion

Dimensional analysis helps in developing relationships between the units of different physical quantities with the base quantities. It may be used to keep a check of the physical formulas and equations: the two sides of any equation must have the same dimensions. Further, it serves as an amazing guide in deriving equations of a physical system much more easily instead of using any rigorous derivations.

FAQs

Q1. What do you mean by a dimensionless quantity? Give some examples.

Ans. The quantities which have no physical dimension and have the corresponding SI unit as 1 are known as the dimensionless quantities. Example: refractive index, Reynolds number, relative permeability.

Q2. What do you mean by a dimensional constant. Give some examples.

Ans. The constants that have fixed dimensions and have a fixed value are known as dimensional constants. They have some corresponding SI units too. Example: Planck's constant and gravitational constant.

Q3. What is the other name for dimensional analysis?

Ans. Dimensional analysis is also known as the unit factor method or factor label method.

Q4. Find the dimensional formula for entropy?

Ans. Entropy is given by

$$\mathrm{\Delta S=\frac{\Delta Q}{T}}$$

Here the Q heat has the dimensions of thermal energy, which is given by $\mathrm{\left [ ML^{2}T^{-2} \right ]}$. So the dimensional formula for entropy is given by

$$\mathrm{S=\frac{\left [ ML^{2}T^{-2} \right ]}{K}=[ML^{2}T^{-2}K^{-1}]}$$

Q5. Is the division and multiplying of the two physical quantities with different dimensions allowed by the Principle of Homogeneity? Why?

Ans. Yes. Both multiplication and taking ratios of incommensurable quantities are allowed. It will give us another unit that can match the RHS of the equation and hence satisfy the relation.