Dimensional analysis is the practice of checking relations amongst physical quantities by identifying their dimensions and units of measurement. However, dimensional analysis is possible only if the dimensions of various terms on either side of the equation are the same. This rule is known as the principle of homogeneity of dimensions. The principle is based on the fact that two quantities of the same dimension only can be added, subtracted, or compared.

Also, dimensional analysis is an amazing tool for checking whether or not equations are dimensionally correct. It is also possible to use dimensional analysis to generate plausible equations if we know the quantities involved. Quantification of the size and shape of things can be done using dimensional analysis. The mathematical study of the nature of objects is possible today, thanks to dimensional analysis.

We have already learned to express most physical quantities in terms of basic dimensions. We will now learn about dimensional analysis and its applications with the help of fundamental quantities like mass, length, time, etc.

Principle of Homogeneity of Dimensions

The equations depicting physical situations must have the same dimensions. This principle is based on the fact that only two quantities of the same dimension can be added, subtracted, or compared.

Principle of Homogeneity states that “dimensions of each of the terms of a dimensional equation on both sides should be the same.”

An equation of the form $x=a+b+c+\dots .$ is dimensionally correct if and only if the variables $x$, $a$, $b$, $c$ all have the same dimensions. This principle may be applied to differential equations and integral equations, as well as to algebraic equations. It should not be assumed, however, that an empirical equation is necessarily dimensionally homogeneous.

Study Everything About Stokes Law Here

Example:

1) $F=\frac{m{v}^2}{r^2}$

By substituting dimension of the physical quantities in the above relation:

$[ML{T}^{–2}]=[M][L{T}^{–1}{]}^2/[L{]}^2$

i.e., $[ML{T}^{–2}]=[M{T}^{–2}]$

As in the above equation dimensions of both sides are not the same; this formula is not correct dimensionally, so it can never be physically accurate.

Note:

• 1. If $[M{]}^a[L{]}^b[T{]}^c=[M{]}^x[L{]}^y[T{]}^z$ then from the principle of homogeneity, we have $a=x,b=y,c=z$.
• 2. A dimensionally correct equation may not be physically correct, but a physically correct equation is always dimensionally correct.

Dimensional Analysis as Factor Label Method

Units can be converted from one system to the other. The method used for this is called factor label method, unit factor method, or dimensional analysis.

In this method, a unit can be converted from one system to another by using a conversion factor that describes the relationship between units. It is based on the fact that the ratio of each fundamental quantity in one unit with their equivalent quantity in another unit is equal to one.

For example,
1) How many minutes are there in $3$ hours?
Solution:
$1$ hour $=$ $60$ minutes
$3$ hour $=$ $3\times 60$ minutes $=$ $180$ minutes
(Here the conversion factor is $60$ minutes)\

We use conversion factors accordingly so that the answer comes in the desired unit and biased results are avoided. In this way, each fundamental quantity like mass, length, and time is converted into another desired unit system using the conversion factor.

Applications of Dimensional Analysis

Dimensional analysis is used to solve problems in real-life physics. We make use of dimensional analysis for five prominent reasons:

1. Finding unit of a physical quantity in a given system of units
2. Finding dimensions of physical constant or coefficients
3. Converting a physical quantity from one system to the other
4. Checking the dimensional correctness of a given physical relation
5. Tool for research to derive new relations

Let us try to understand each application of Dimensional analysis in detail-

1. Finding Unit of a Physical Quantity in a Given System of Units

James Clerk Maxwell and Jenkin invented the dimensional formula in the early 1860’s for unit conversion, and the modern concept of dimension started in 1863 with Maxwell. He synthesized earlier formulations by Fourier, Weber, and Gauss.

To write the formula of a physical quantity, we find its dimensions using dimensional analysis. Now in the dimensional formula replacing $M$, $L$ and $T$ by the fundamental units of the required system, we get the unit of the physical quantity. However, sometimes to this unit, we further assign a specific name, e.g., $\text{Work = Force}\times \text{Displacement}$.

So $[W]=[ML{T}^{–2}]\times [L]=[M{L}^2{T}^{–2}]$

So its units in C.G.S. system will be $\text{gc}{\text{m}}^2/{\text{s}}^2$ which is called $\mathrm{e}\mathrm{r}\mathrm{g}$ while in M.K.S. system will be $\text{kg}{\text{m}}^2/{\text{s}}^2$ which is called $\mathrm{j}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{e}$.

2. Finding Dimensions of Physical Constant or Coefficients

Dimensions of a physical quantity are unique and special. We write an equation by putting a proportional constant calculated by substituting the dimensional formulae of all other physical quantities.

• (i) Gravitational constant: According to Newton’s law of gravitation $F=G\frac{m_1{m}_2}{r^2}$ or $G=\frac{F{r}^2}{m_1{m}_2}$
• Substituting the dimensions of all physical quantities $[G]=\frac{[ML{T}^{–2}][L^2]}{[M][M]}=[M^{–1}{L}^3{T}^{–2}]$
• (ii) Planck constant: According to Planck $E=hv$ or $h=\frac{E}{\nu }$
• Substituting the dimensions of all physical quantities $[h]=\frac{[M{L}^2{T}^{–2}]}{[T^{–1}]}=[M{L}^2{T}^{–1}]$
• (iii) Coefficient of viscosity: According to Poiseuille’s formula $\frac{dV}{dt}=\frac{\pi p{r}^4}{8\eta l}$ or $\eta =\frac{\pi p{r}^4}{8l(dV/dt)}$
• Substituting the dimensions of all physical quantities $[\eta ]=\frac{[M{L}^{–1}{T}^{–2}][L^4]}{[L][L^3/T]}=[M{L}^{–1}{T}^{–1}]$.

3. Converting a Physical Quantity from One System to the Other

The measure of a physical quantity is $nu=$ constant

If a physical quantity $X$ has the dimensional formula $[M^a{L}^b{T}^c]$ and if (derived) units of that physical quantity in two systems are $[M_1^a{L}_1^b{T}_1^c]$ and $[M_2^a{L}_2^b{T}_2^c]$ respectively and $n_1,n_2$ are the numerical values in the two systems respectively, then $n_1[u_1]=n_2[u_2]$.

$\Rightarrow n_1[M_1^a{L}_1^b{T}_1^c]=n_2[M_2^a{L}_2^b{T}_2^c]$

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

Where,

$M_1,L_1$ and $T_1=$ basic units of mass, length, and time in the first (known) system and

$M_2,L_2$ and $T_2=$ basic units of mass, length, and time in the second (unknown) system

Example:

1. Conversion of Newton into Dyne.

Newton is the S.I. unit of force and has the dimensional formula $[ML{T}^{–2}]$.

So $1\text{N}=1\frac{\text{kg}{\text{m}}^2}{{\text{s}}^2}$

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

$=1{\left[\frac{\text{kg}}{\text{gm}}\right]}^1{\left[\frac{\text{m}}{\text{cm}}\right]}^1{\left[\frac{\text{sec}}{\text{sec}}\right]}^{–2}$

$=1{\left[\frac{{10}^3\text{gm}}{\text{gm}}\right]}^1{\left[\frac{{10}^2\text{cm}}{\text{cm}}\right]}^1{\left[\frac{\text{sec}}{\text{sec}}\right]}^{–2}$

$={10}^5$

$\therefore 1\text{N}={10}^5\text{Dyne}$

2. Conversion of gravitational constant ($G$) from C.G.S. to M.K.S. system

The value of $G$ in C.G.S. system is $6.67\times {10}^{–8}$ C.G.S. units while its dimensional formula is $[M^{–1}{L}^3{T}^{–2}]$

So $G=6.67\times {10}^{–8}\frac{\text{c}{\text{m}}^{\text{3}}}{\text{g}{\text{s}}^{\text{2}}}$

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

$=6.67\times {10}^{–8}{\left[\frac{\text{gm}}{\text{kg}}\right]}^{–1}{\left[\frac{\text{cm}}{\text{m}}\right]}^3{\left[\frac{\text{sec}}{\text{sec}}\right]}^{–2}$

$=6.67\times {10}^{–8}{\left[\frac{\text{gm}}{\text{1}{\text{0}}^{\text{3}}\text{gm}}\right]}^{–1}{\left[\frac{\text{cm}}{\text{1}{\text{0}}^{\text{2}}\text{cm}}\right]}^3{\left[\frac{\text{sec}}{\text{sec}}\right]}^{–2}=6.67\times {10}^{–11}$

$\therefore G=6.67\times {10}^{-11}$ M.K.S. units.

4. Checking the Dimensional Correctness of a Given Physical Relation

This application is based on the principle of homogeneity of dimensions. According to this rule, only those terms can be added or subtracted with the same dimensions.

If $X=A\pm (BC{)}^2\pm \sqrt{DEF}$,

Then according to the principle of homogeneity $[X]=[A]=[(BC{)}^2]=[\sqrt{DEF}]$

If the dimensions of each term on both sides of an equation are the same, then the equation is dimensionally correct.

Example:

1. $s=ut–(1/2)a{t}^2$

By substituting respective dimensions of the physical quantities in the above equation –

$[L]=[L{T}^{–1}][T]–[L{T}^{–2}][T^2]$

i.e., $[L]=[L]–[L]$

As in the above equation, the dimensions of each term on both sides of an equation are the same; the equation is dimensionally correct. However, from equations of motion, we know that $s=ut–(1/2)a{t}^2$.

5. Tool for Research to Derive New Relations

Suppose one knows the dependency of a physical quantity on other quantities which is of the product type. In that case, using the dimensional analysis method, a relation between the quantities can be derived.

Example:

i. Time-period of a Simple Pendulum

Let the Time-period of a simple pendulum be a function of the mass of the bob $m$, effective length $l$, acceleration due to gravity $g$, then assuming the equation to be the product of power function of $m,l$ and $g$.

i.e., $T=K{m}^x{l}^y{g}^z$, where $K=$ dimensionless constant

If the above relation is dimensionally correct, then by substituting the dimensions of quantities –

$[T]=[M{]}^x[L{]}^y[L{T}^{–2}{]}^z$

Or $[M^0{L}^0{T}^0]=[M^x{L}^{y+z}{T}^{–2z}]$

Equating the exponents of similar quantities $x=0,y=\frac{1}{2}$ and $z=\frac{1}{2}$

So the required physical relation becomes $T=K\sqrt{\frac{l}{g}}$

ii. Stokes’ Law

When a small sphere of radius $r$ moves at a constant terminal speed $v$ through a fluid, the viscous force $F$ is experimentally dependent on the radius $r$, the velocity of the sphere $v$, and the fluid’s viscosity $\eta$.

So $F=f(\eta ,r,v)$

If the function is a product of power functions of $\eta$, $r$ and $v$, then $F=K{\eta }^x{r}^y{v}^z$; where $K$ is a dimensionless constant

If the above relation is dimensionally correct

$[ML{T}^{–2}]=[M{L}^{–1}{T}^{–1}{]}^x[L{]}^y[L{T}^{–1}{]}^z$

or $[ML{T}^{–2}]=[M^x{L}^{–x+y+z}{T}^{–x–z}]$

Equating the exponents of similar quantities $x=1;–x+y+z=1$ and $–x–z=–2$

Solving these for $x$, $y$ and $z$, we get $x=y=z=1$

So equation becomes $F=K\eta rv$

On experimental grounds $K=6\pi$; so $F=6\pi \eta rv$

This equation is called the Stokes’ law.

Solved Examples on Dimensional Analysis

Problem 1. The equation $\left(P+\frac{a}{V^2}\right)\left(V–b\right)=\text{constant}$. The unit of the constant $a$ is
(a) $\text{Dyne}\times {\text{cm}}^{\text{5}}$
(b) $\text{Dyne}\times {\text{cm}}^{\text{4}}$
(c) $\text{Dyne}\times {\text{cm}}^{\text{3}}$
(d) $\text{Dyne}\times {\text{cm}}^{\text{2}}$
Solution: (b) According to the principle of dimensional homogeneity $[P]=\left[\frac{a}{V^2}\right]$
$\Rightarrow [a]=[P][V^2]=[M{L}^{–1}{T}^{–2}][L^6]$
$=[M{L}^5{T}^{–2}]$
or unit of $a=\text{gm}\times {\text{cm}}^5\times {\text{s}}^{–2}=\text{Dyne}\times {\text{cm}}^4$.

Problem 2. If $x=at+b{t}^2$, where $x=$ distance travelled by the body in kilometres and $t=$ time in seconds, then the unit of the constant $b$ is
(a) $\text{km/s}$
(b) $\text{km – s}$
(c) ${\text{km/s}}^2$
(d) $\text{km – }{\text{s}}^2$
Solution: (c) From the principle of homogeneity of dimensions,
$[x]=[b{t}^2]\Rightarrow [b]=\left[\frac{x}{t^2}\right]$
$\therefore$ Unit of constant $b=\text{km/}{\text{s}}^2$.

Problem 3. A force $F$ is given by $F=at+b{t}^2,$ where $t$ is time. What are the dimensions of $a$ and $b$
(a) $ML{T}^{–3}$ and $M{L}^2{T}^{–4}$
(b) $ML{T}^{–3}$ and $ML{T}^{–4}$
(c) $ML{T}^{–1}$ and $ML{T}^0$
(d) $ML{T}^{–4}$ and $M{L}^2{T}^1$
Solution: (b) From the principle of dimensional homogeneity $[F]=[at]$
$\therefore [a]=\left[\frac{F}{t}\right]=\left[\frac{ML{T}^{–2}}{T}\right]$
$=[ML{T}^{–3}]$
Similarly $[F]=[b{t}^2]$
$\therefore [b]=\left[\frac{F}{t^2}\right]=\left[\frac{ML{T}^{–2}}{T^2}\right]=[ML{T}^{–4}]$

Problem 4. The dimensions of a physical quantity $X$ in the equation Force $=\frac{X}{\text{Density}}$ is given by
(a) $M^1{L}^4{T}^{–2}$
(b) $M^2{L}^{-2}{T}^{–1}$
(c) $M^2{L}^{-2}{T}^{–2}$
(d) $M^1{L}^{-2}{T}^{–1}$
Solution: (c) $[X]=[\text{Force}]\times [\text{Density}]$
$=[ML{T}^{–2}]\times [M{L}^{–3}]=[M^2{L}^{–2}{T}^{–2}]$.

Problem 5. A physical quantity is measured, and its value is found to be $nu$ where $n=$ numerical value and $u=$ unit. Then which of the following relations is true
(a) $n\propto u^2$
(b) $n\propto u$
(c) $n\propto \sqrt{u}$
(d) $n\propto \frac{1}{u}$
Solution: (d) We know $P=nu=$ constant
$\therefore n_1{u}_1=n_2{u}_2$ or $n\propto \frac{1}{u}$.

Problem 6. In the C.G.S. system, the magnitude of the force is $100\mathrm{d}\mathrm{y}\mathrm{n}\mathrm{e}\mathrm{s}$. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is
(a) $0.036$
(b) $0.36$
(c) $3.6$
(d) $36$
Solution: (c) $n_1=100$, $M_1=\text{g},L_1=\text{cm,}{T}_1=\sec ,$ and $M_2=\text{kg},L_2=\text{meter,}{T}_2=\text{minute,}x=1,y=1,z=–2$
By substituting these values in the following conversion formula $n_2=n_1{\left[\frac{M_1}{M_2}\right]}^x{\left[\frac{L_1}{L_2}\right]}^y{\left[\frac{T_1}{T_2}\right]}^2$
$n_2=100{\left[\frac{\text{gm}}{\text{kg}}\right]}^1{\left[\frac{\text{cm}}{\text{meter}}\right]}^1{\left[\frac{\sec }{\text{minute}}\right]}^{–2}$
$n_2=100{\left[\frac{\text{gm}}{{10}^3\text{gm}}\right]}^1{\left[\frac{\text{cm}}{{10}^2\text{cm}}\right]}^1{\left[\frac{\sec }{60\sec }\right]}^{–2}=3.6$.

Problem 7. If velocity $v$ acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,A$ and $F$ would be
(a) $F{A}^{–1}v$
(b) $F{v}^3{A}^{–2}$
(c) $F{v}^2{A}^{–1}$
(d) $F^2{v}^2{A}^{–1}$
Solution: (b) Given, $v=$ velocity $=[L{T}^{–1}]$, $A=$ Acceleration $=[L{T}^{–2}]$, $F=$ force $=[ML{T}^{–2}]$
By substituting the dimension of each quantity, we can check the accuracy of the formula
[Angular momentum] $=F{v}^3{A}^{–2}$
$[M{L}^2{T}^{–1}]=[ML{T}^{–2}][L{T}^{–1}{]}^3[L{T}^{–2}{]}^{–2}=[M{L}^2{T}^{–1}]$
L.H.S. = R.H.S., i.e., the above formula is Correct.

Problem 8. If velocity $V$, force $F$ and energy $E$ are taken as fundamental units, then the dimensional formula for mass will be
(a) $V^{–2}{F}^{0}E$
(b) $V^{0}F{E}^2$
(c) $V{F}^{–2}{E}^0$
(d) $V^{–2}{F}^{0}E$
Solution: (d) Let $M=V^a{F}^b{E}^c$
Putting dimensions of each quantity on both sides $[M]=[L{T}^{–1}{]}^a[ML{T}^{–2}{]}^b[M{L}^2{T}^{–2}{]}^c$
Equating powers of dimensions, We have $b+c=1,a+b+2c=0$ and $–a–2b–2c=0$
Solving these equations, $a=–2,b=0$ and $c=1$
So $M=[V^{–2}{F}^{0}E]$.

Limitations of Dimensional Analysis

Although dimensional analysis is very useful, it has many limitations, viz.

• 1. The uniqueness of physical quantities doesn’t depend on dimensional formula, as many physical quantities have the same dimensions. For example, if the dimensional formula of a physical quantity is $[M{L}^2{T}^{–2}]$, the quantity may be work or energy or torque.
• 2. Numerical constant $K$ having no dimensions such as $1$ or $2\pi$ cannot be deduced by the dimensional methods.
• 3. The method of dimensions can be used only when the equation is in product form. For example,
• $s=ut+\left(\frac{1}{2}\right)a{t}^2$ or $y=a\sin \omega t$
• The above equations can’t be derived by using dimensional analysis because they have addition and sine terms. However, the dimensional correctness of these equations can be checked.
• 4. The method of dimensions cannot be applied to derive a formula in which a physical quantity depends on more than three physical quantities. However, we can check the correctness of the given equation dimensionally. For example, the formula $T=2\pi \sqrt{\frac{1}{mgl}}$ cannot be derived by dimensional analysis, but its dimensional correctness can be checked.
• 5. Suppose a physical quantity depends on three variables, of which two have the same dimensions. In that case, the formula connecting the variable can’t be derived by dimensional analysis, e.g., the formula for the frequency of a tuning fork $f=(d/L^2)v$ cannot be derived by the theory of dimensions but can be checked.

Summary

1. Dimensional analysis is an amazing tool to check relations among physical quantities by identifying their dimensions.
2. The principle of homogeneity is based on the fact that two quantities of the same dimension only can be added, subtracted, or compared.
3. Dimensional analysis can be used to generate plausible equations if we know the type of quantities involved.

FAQs on Dimensional Analysis

Q.1. What is dimensional analysis?
Ans: Dimensional analysis is the practice of checking relations amongst physical quantities by identifying their dimensions.

Q.2. What are the basic rules of dimensional analysis?
Ans: Dimensional analysis is possible only if the dimensions of various terms on either side of the equation are the same. This rule is known as the principle of homogeneity of dimensions. The principle is based on the fact that two quantities of the same dimension only can be added, subtracted, or compared.

Q.3. State the principle of homogeneity of dimensions.
Ans: The principle of homogeneity states that “Only those terms can be added or subtracted which have same dimensions.”

Q.4. What is the main goal of dimensional analysis?
Ans: The main idea in the dimensional analysis is to create a conversion ratio (unit factor) that has the units you want in the numerator and the units you already have in the denominator.

Q.5. What are the applications of dimensional analysis?
Ans: We make use of dimensional analysis for five prominent reasons:
1. Finding unit of a physical quantity in a given system of units
2. Finding dimensions of physical constant or coefficients
3. Converting a physical quantity from one system to the other
4. Checking the dimensional correctness of a given physical relation
5. Tool for research to derive new relations

Q.6. What are the limitations of dimensional analysis?
Ans: The limitations of dimensional analysis are-
1. It is impractical for the correlation of more than three parameters.
2. Data obtained from a large number of experiments may be undetermined.
3. Subtraction and addition of parameters cannot be reflected in dimensional analysis.
4. Dimensional analysis cannot confirm the validity of a relationship of the physical quantities.
5. The dimensional analysis cannot determine the nature of the unknown physical quantities.

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