Dimensions of Physical quantities and Dimensional Analysis: The nature of any physical quantity is determined by its dimensions. We have basically seven fundamental units, and the rest of the units are derived using those seven fundamental units. These units are known as derived units, and they can be expressed in terms of some combination of those seven fundamental units. Therefore, these seven quantities are known as dimensions. Dimensions of any physical quantities are denoted using \(‘[]’\) square brackets.

From this article, you will come to know more about how to derive dimensional formulae for different physical quantities; then, we will look at a list of different dimensional formulae and, finally, what are they used for and how to use them.

Deriving a Dimensional Formula

Whenever the unit of any physical quantity is expressed in terms of the seven fundamental units, then that formula is known as its dimensional formula. The power or the exponents to which the units of base quantities (or the units of fundamental quantities) are raised to represent that quantity are known as dimensions. Detailed analysis of these dimensional formulae can help one understand more about that physical quantity.

For example, the dimensional formula for length can be written as \(\left[ {\rm{L}} \right]\), mass can be written as \(\left[ {\rm{M}} \right]\), time can be written as \(\left[ {\rm{T}} \right]\), form current \(\left[ {\rm{A}} \right]\) and form temperature \(\left[ {\rm{K}} \right]\). Using these, other-dimensional formulae can be derived.

Learn About Dimensional Analysis and its Applications

Let us say we want to know about the dimensional formula for area. The basic formula for area is length times the breadth.

\({\rm{Area}} = {\rm{length}} \times {\rm{breadth}}\)

The dimensional formula for length and breadth will be the same because both are the physical quantities relating to length and measured in metres. Hence, we can write for the dimensional formula for the area as,

\([{\rm{Area}}] = [{\rm{L}}] \times [{\rm{L}}]\)

\([{\rm{Area}}] = [{{\rm{L}}^2}]\)

Therefore, the dimensional formula of the area is \([{{\rm{L}}^2}]\). Similarly, volume is length times breadth times height. Therefore,

\({\rm{Volume}} = {\rm{length}} \times {\rm{breadth}} \times {\rm{height}}\)

\([{\rm{Volume}}] = [{\rm{L}}] \times [{\rm{L}}] \times [{\rm{L}}]\)

\(\therefore \,[{\rm{Volume}}] = [{{\rm{L}}^3}]\)

Let us try to find the dimensional formula for density. Basically, density is the ratio of mass over volume.

\({\rm{Density}} = \frac{{{\rm{ Mass }}}}{{{\rm{ Volume }}}}\)

\([{\rm{Density}}] = \frac{{[{\rm{M}}]}}{{\left[ {{{\rm{L}}^3}} \right]}}\)

\([\rho ] = \left[ {{\rm{M}}{{\rm{L}}^{ – 3}}} \right]\)

OR

\([\rho ] = \left[ {{\rm{M}}{{\rm{L}}^{ – 3}}\;{{\rm{T}}^0}} \right]\)

Here, time raised to zero will represent that this physical quantity (density) does not have any dimensions of time. So, this is how we can derive the dimensional formula for any unknown physical quantity.

Dimensional Formulae of Some Physical Quantities

Physical quantity	General formula	\(SI\) unit	Dimensional formula
Area	\(A = l \times b\)	\({{\rm{m}}^2}\)	\(\left[ {{{\rm{L}}^2}} \right]\)
Volume	\(V = l \times b \times h\)	\({{\rm{m}}^3}\)	\(\left[ {{{\rm{L}}^3}} \right]\)
Density	\(\rho = \frac{m}{V}\)	\({\rm{kg}}/{{\rm{m}}^3}\)	\(\left[ {{\rm{M}}{{\rm{L}}^{ – 3}}} \right]\)
Electric charge	\(q = i \times t\)	\({\rm{As}}\) (Coulomb)	\([{\rm{AT}}]\)
Speed	\(v = \frac{d}{t}\)	\({\rm{m}}/{\rm{s}}\)	\(\left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right]\)
Acceleration	\(a = \frac{v}{t}\)	\({\rm{m}}/{{\rm{s}}^2}\)	\(\left[ {{\rm{L}}{{\rm{T}}^{ – 2}}} \right]\)
Linear momentum	\(p = mv\)	\({\rm{kg}}\,{\rm{m}}/{\rm{s}}\)	\(\left[ {{\rm{ML}}{{\rm{T}}^{ – 1}}} \right]\)
Anqular momentum	\(L = r \times p\)	\({\rm{kg}}\,{{\rm{m}}^2}/{\rm{s}}\)	\(\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 1}}} \right]\)
Force	\(F = ma\)	\({\mathop{\rm kg}\nolimits} \frac{{\rm{m}}}{{{{\rm{s}}^2}}}\) (Newton)	\(\left[ {{\rm{ML}}{{\rm{T}}^{ – 2}}} \right]\)
Torque	\(\tau = r \times F\)	\({\rm{Nm}}\)	\({\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 2}}} \right]}\)
Work	\(W = Fd\)	\({\rm{Nm}}\)(Joule)	\({\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 2}}} \right]}\)
Eneray	\(\Delta E=W\)	\({\rm{J}}\)	\({\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 2}}} \right]}\)
Power	\(P = \frac{W}{t}\)	\(\frac{{\rm{J}}}{{\rm{s}}}\) (Watt)	\({\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 3}}} \right]}\)
Frequency	\(f = \frac{1}{T}\)	\({{\rm{s}}^{ – 1}}\) (Hertz)	\(\left[ {{{\rm{T}}^{ – 1}}} \right]\)
Electric field	\(E = \frac{F}{q}\)	\({\rm{N/C}}\)	\(\left[ {{\rm{ML}}{{\rm{T}}^{ – 3}}\;{{\rm{A}}^{ – 1}}} \right]\)
Electric potential	\(V = \frac{W}{q}\)	\({\rm{J/C}}\)	\(\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 3}}\;{{\rm{A}}^{ – 1}}} \right]\)

These are just a few of the basic physical quantities. There are many more, and we will even learn how to determine the dimensional formula of any unknown physical quantity.

Dimensional Analysis

Can you add length and mass? It is obviously no. Why? Because both are different, and both have a different dimensional formula. We can’t add or subtract physical quantities with different dimensional formulas. But this length and mass example was a simple one. Often students make a mistake in such things which involves physical quantities with unknown dimensions. Or often do they forget the general formula.

By thoroughly understanding dimensional formulae and by using dimensional analysis, all the problems related to understanding the formula are resolved. As mentioned earlier, the dimensional formula helps us to understand the nature of any physical quantity. For example, we mentioned earlier about density that it doesn’t have any dimensions of time. This is its nature. It should only include mass and volume and nothing else. And this nature was clearly visible from the dimensional formula. Hence, this is how we can understand its nature because a dimensional formula will break any equation into its fundamental quantities.

In the further part of this article, we will learn some important applications of dimensional analysis.

Checking Dimensional Validity

In the previous point, we discussed that two physical quantities could not be added or subtracted unless they have the same dimensional formula. Similarly, we can check the dimensional validity of any equation based on its dimensional formula. But there is one limitation that although the equation may be dimensionally valid, it is not necessary that it is the correct equation.

Follow these simple steps to check dimensional validity. An equation is dimensionally valid if,

1. Dimensional formula on both the side of equals to sign \((=)\) must be same.
2. Dimensional formula on both the side of plus and minus signs \({\rm{( + ,}}\,{\rm{ – )}}\) must be the same.
3. Any numerical constant included in the equation will be considered as a dimensionless constant.

Let us understand this by a simple example:

\(s = ut + a{t^2}\)

Here, \(s\) is the displacement of a particle in motion
\(u\) is the initial speed of the particle
\(t\) is the time of its motion
\(a\) is the linear acceleration of the same particle

Taking dimensional formula for all the quantities:

\([{\rm{L}}] = \left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right][{\rm{T}}] + \left[ {{\rm{L}}{{\rm{T}}^{ – 2}}} \right]\left[ {{{\rm{T}}^2}} \right]\)

\(\therefore \,[{\rm{L}}] = [{\rm{L}}] + [{\rm{L}}]\)

Since all the three-dimensional formulas here are the same, this formula is said to be dimensionally valid. But, if you might have studied this formula in kinematics, then the actual equation is:

\(s = ut + \frac{1}{2}a{t^2}\)

But because dimensions are not considered for constants, we can’t be sure whether the given equation is \(100\% \) correct or not. We can just check whether if it is dimensionally valid or not.

Let us see one more example:

\(v – u = 2as\)

Here, \(v\) is the final speed of an object in motion
\(u\) is its initial speed
\(a\) is its acceleration
and \(s\) is its displacement

Looking at the dimensional formula on \({\rm{LHS}}\) and \({\rm{RHS}}\)

\({\rm{LHS}} = \left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right] + \left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right] = \left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right]\)

\({\rm{RHS}} = \left[ {{\rm{L}}{{\rm{T}}^{ – 2}}} \right][{\rm{L}}] = \left[ {{{\rm{L}}^2}\;{{\rm{T}}^{ – 2}}} \right]\)

Here, we can clearly see that \({\rm{LHS}} \ne {\rm{RHS}}\), which means that the dimensional formula is invalid. If the dimensional formula is invalid, then the formula is also completely incorrect. But we can write the same thing vice versa.

Determining Dimensional Formula of an Unknown Quantity

Suppose there is an unknown quantity or an unknown constant in the equation, which is also dimensionally valid. Then, we can determine its dimensional formula using the dimensional formula for other quantities.

Let us understand this by a simple example:

\(F = \frac{{G{m_1}{m_2}}}{{{r^2}}}\)

Here, \(F\) is the gravitational force between two masses
\({{m_1}}\) and \({{m_2}}\) are the two masses
\(r\) is the distance between both the masses
and \(G\) is the universal gravitational constant

So, if we want to find the dimensional formula of gravitational constant, we can just make it the subject of the equation

\(G = \frac{{F{r^2}}}{{{m_1}{m_2}}}\)

Substituting with dimensional formulae,

\([G] = \frac{{\left[ {{\rm{ML}}{{\rm{T}}^{ – 2}}} \right]\left[ {{{\rm{L}}^2}} \right]}}{{\left[ {{{\rm{M}}^2}} \right]}}\)

\(\therefore \,[G] = \left[ {{{\rm{M}}^{ – 1}}\;{{\rm{L}}^3}\;{{\rm{T}}^{ – 2}}} \right]\)

Hence, the dimensional formula for universal gravitational constant is \(\left[ {{{\rm{M}}^{ – 1}}\;{{\rm{L}}^3}\;{{\rm{T}}^{ – 2}}} \right]\)

Let’s see another example:

\(F = 6\pi r\eta v\)

Here, \(F\) is the viscous force
\(r\) is the radius of the sphere
\(v\) is the velocity of the sphere
\(\eta \) is the coefficient of viscosity

We need to find the dimensional formula of the coefficient of viscosity. Making it the subject of the equation:

\(\eta = \frac{F}{{6\pi rv}}\)

Substituting, respective dimensional formulae,

\([\eta ] = \frac{{\left[ {{\rm{ML}}{{\rm{T}}^{ – 2}}} \right]}}{{[{\rm{L}}]\left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right]}}\)

\(\therefore \,[\eta ] = \left[ {{\rm{M}}{{\rm{L}}^{ – 1}}\;{{\rm{T}}^{ – 1}}} \right]\)

Hence, the dimensional formula for the coefficient of viscosity is \(\left[ {{\rm{M}}{{\rm{L}}^{ – 1}}\;{{\rm{T}}^{ – 1}}} \right]\).

So, this is how we can determine the dimensional formula for any unknown quantity using dimensional analysis.

Deducing Relation Between Physical Quantities

Let us say we don’t know the formula for a physical quantity. We can’t make an exact equation, but at least we can make a dimensionally correct equation that will somehow at least help us understand the relationship between those physical quantities.

Let us understand this by an example

A question says that the speed of travelling waves in a string \((v)\) depends on the tension in the string \((T)\) and its linear mass density \((\mu )\). Find the relation between these quantities.

We will understand the whole process step by step. Since we don’t know about the power to be raised, we will simply assume it to be \(x\) and \(y\)

\(v = k{T^x}{\mu ^y}\)

Here, \(k\) is a dimensionless constant. Taking dimensional formula on both the sides

\(\left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right] = {\left[ {{\rm{ML}}{{\rm{T}}^{ – 2}}} \right]^x}{\left[ {{\rm{M}}{{\rm{L}}^{ – 1}}} \right]^y}\)

\(\left[ {{\rm{L}}{{\rm{T}}^{ – 1}}} \right] = \left[ {{{\rm{M}}^{x + y}}\;{{\rm{L}}^{x – y}}\;{{\rm{T}}^{ – 2x}}} \right]\)

Comparing dimensions on both sides:

\(x + y = 0\)

\(x – y = 1\)

\(-2x = -1\)

Solving the equations, we get:

\(x = \frac{1}{2}\) or \(y = – \frac{1}{2}\)

Therefore, the relation will become:

\(v = k\sqrt {\frac{T}{\mu }} \)

And the dimensionless constants are obtained using experimental results

Taking another example

The heat energy \((Q)\) lost in a resistor depends on the current flowing through the resistor \((i)\), the value of resistance \((R)\), and time \((t)\). Find the relation between all the mentioned quantities.

\(Q = k{i^x}{R^y}{t^z}\)

Taking dimensional formula:

\(\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 2}}} \right] = {[{\rm{A}}]^x}{\left[ {{\rm{M}}{{\rm{L}}^2}\;{{\rm{T}}^{ – 3}}\;{{\rm{A}}^{ – 2}}} \right]^y}{[\;{\rm{T}}]^z}\)

\(\left[ {{\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ – 2}}{{\rm{A}}^0}} \right] = \left[ {{{\rm{M}}^y}{{\rm{L}}^{2y}}{{\rm{T}}^{ – 3y + z}}{{\rm{A}}^{x – 2y}}} \right]\)

Comparing dimensions on both sides,

\(y = 1\)
\(2y = 2\)
\( – 3y + z = – 2\)
\( x – 2y = 0\)

Solving all the equations:

\(x = 2,\,y = 2\) and \(z = 1\)

Therefore, the relation will become,

\(Q = k{i^2}Rt\)

Summary

From this article, we came to know what dimensional and dimensional formulas its importance is and how to apply dimensional analysis for any equation to find the validity of the equation, dimension of any unknown quantity, and relation between physical quantities. Every equation is incomplete without verifying its dimensional validity. During derivation, if there is a small mistake, the equation can come out to be incorrect, and to avoid this problem, we can even match the dimensions of physical quantities.

Frequently Asked Questions

Q.1. What are the dimensions of physical quantity?
Ans: Dimensions are the power raised to the fundamental quantity in a dimensional formula. They are very important to know about the nature of the quantity.

Q.2. What are the applications and limitations of dimensional analysis?
Ans: Using dimensional analysis, we can find the relation between physical quantities, study nature in terms of fundamental quantities and find the dimensional formula of an unknown quantity. Although any equation is dimensionally valid, the equation may or may not be correct.

Q.3. How do you find dimensional formula?
Ans: One can find a dimensional formula by the combination of physical quantities in a general formula or by using its units.

Q.4. What is the dimensional formula of distance?
Ans: Since distance includes length, its dimensional formula is \(\left[ {{{\rm{M}}^0}\;{{\rm{L}}^1}\;{{\rm{T}}^0}} \right]\)

Q.5. What is the use of dimensional formula?
Ans: It is used to verify the correctness of the equation. If the equation is correct, then it should be dimensionally correct as well.

Now you are provided with all the necessary information on the dimensions of physical quantities and dimensional analysis and we hope this detailed article is helpful to you. If you have any queries regarding this article, please ping us through the comment section below and we will get back to you as soon as possible.