## What is called Dimension of a quantity?

*All the derived physical quantities can be expressed in terms of some combination of symbolic representation of the seven fundamental or base quantities. These symbols are called dimensions of fundamental quantities.*

Thus we have seven dimensions for seven numbers of fundamental quantities. These seven dimensions of seven fundamental quantities are called as the seven dimensions of the world. These are denoted with square brackets [ ] as given below.

- Dimension of length = [ L ]
- Dimension of mass = [ M ]
- Dimension of time = [ T ]
- Dimension of electric current = [ I ]
- Dimension of thermodynamic temperature = [ K ]
- Dimension of luminous intensity = [ \text {cd} ]
- Dimension of amount of substance = [ \text {mol} ]

The dimension of a *physical quantity* is expressed as the powers or exponents of dimensions to which the fundamental quantities must be raised to represent that quantity completely.

Out of the seven dimensions of *fundamental quantities*, three dimensions are sufficient and normally used to express the dimensions of most of the derived quantities. These are –

- Dimension of length = [ L ]
- Dimension of mass = [ M ]
- Dimension of time = [ T ]

**EXAMPLE –**

*Density* of a substance is defined as the mass per unit volume of that substance. Therefore –

\text {Density} = \frac { \text {Mass} }{ \text {Volume} }

Or, \quad \text {Density} = \frac { \text {Mass} }{ \text {Length} \times \text {Breadth} \times \text {Height} }

Putting the dimensions of these quantities, we will get the dimension of density. Therefore, dimension of density will be –

= \frac { [ M ] }{ [ L ] \times [ L ] \times [ L ] }

= \frac { [ M ] }{ [ L ]^3 }

Therefore, dimension of density will be expressed as –

= [ M^1 \ L^{ -3 } \ T^0 ]

Hence, the dimensions of density are 1 in mass, 3 in length and 0 in time.

## Dimensional Analysis

*Dimensional analysis is defined as the method of studying a physical phenomenon on the basis of dimensions of fundamental quantities involved in that phenomenon.*

Following are the three main uses of dimensional analysis –

- To convert a
*physical quantity*from one system of unit to another system. - To check the correctness of a given physical relation.
- To derive a relationship between different physical quantities.

### Conversion of System of Units

By use of dimensional analysis, it is possible to convert a physical quantity from one system of units to another. It is based on the fact that the magnitude of the physical quantity remains the same whatever be the system of measurement of units.

Let, ( u_1 ) and ( u_2 ) are the units of measurement of a physical quantity ( Q ) and ( n_1 ) and ( n_2 ) are the corresponding numerical values of measurement. Then –

Q = n_1 u_1 = n_2 u_2

Also, let ( M_1, \ L_1, \ T_1 ) and ( M_2, \ L_2, \ T_2 ) are the sizes of fundamental units of *mass*, length and time in system 1 and system 2 respectively.

If, the dimensional formula of quantity ( Q ) is expressed as [ M^a \ L^b \ T^c ] , then –

- u_1 = [ M_1^a \ L_1^b \ T_1^c ]
- u_2 = [ M_2^a \ L_2^b \ T_2^c ]

Therefore, \quad n_1 [ M_1^a \ L_1^b \ T_1^c ] = n_2 [ M_2^a \ L_2^b \ T_2^c ]

Or, \quad 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

This equation can be used to find the numerical value of ( n_2 ) in the new system of units.

#### EXAMPLE –

J ( Joule ) is the SI unit of *energy* and ( erg ) is the CGS unit of energy. Dimensional formula of energy is [ M^1 \ L^2 \ T^{-2} ] .

Comparing it with the general expression of dimension of a physical quantity [ M^a \ L^b \ T^c ] , we will get –

a = 1, \quad b = 2, \quad c = - 2

Values of quantities in CGS and SI systems are –

In SI system. | In CGS system. |

M_1 = 1 \ kg = 1000 \ g | M_2 = 1 \ g |

L_1 = 1 \ m = 100 \ cm | L_2 = 1 \ cm |

T_1 = 1 \ s | T_2 = 1 \ s |

n_1 = 1 \ J (joule) | n_2 = ? (erg) |

By dimensional analysis, we have –

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

Therefore, \quad n_2 = 1 \left [ \frac { 1000 }{ 1 } \right ]^1 \left [ \frac { 100 }{ 1 } \right ]^2 \left [ \frac { 1 }{ 1 } \right ]^{-2}

Or, \quad n_2 = 1 \times 10^3 \times 10^4 = 10^7

Thus, \quad 1 \ \text {joule} = 10^7 \ \text {erg}

### Checking Correctness of Physical Relations

To check the dimensional correctness of a physical relation we use the principle of homogeneity of dimensions.

#### Principle of Homogeneity of Dimensions

* According to this principle, a physical equation will be dimensional correct if the dimensions of all the terms occurring on both sides of the equation are the same. *

This principle is based on the fact that only the physical quantities of the same kind can be added, subtracted or compared. Thus, velocity can be added to velocity but not to force.

**EXAMPLE –**

Let us check the dimensional *accuracy* of Newton’s formula for *laws of motion* –

s = ut + \left ( \frac {1}{2} \right ) at^2

Dimensions of different terms used in this relation are –

- Distance traveled ( s ) = [ L^1 ]
- Initial velocity ( u ) = [ L^1 \ T^{-1} ]
- Time ( t ) = [ T^1 ]
- Acceleration ( a ) = [ L^1 \ T^{-2} ]
- Constant term \left ( \frac {1}{2} \right ) = [ M^0 \ L^0 \ T^0 ] = 1

Therefore, dimension for LHS of the expression is –

LHS = [ L^1 ]

And, dimension for RHS of the expression will be –

RHS = ut + \left ( \frac {1}{2} \right ) at^2

= [ L^1 \ T^{-1} ] \times [ T^1 ] + 1 \times [ L^1 \ T^{-2} ] \times [ T^1 ]^2

= [ L^{1} \ T^{( -1+1 )} ] + [ L^1 \ T^{( -2 + 2 )} ]

= [ L^1 ]

Thus we see that all the terms on both sides of the equation are having same dimensions. So, the given relation is dimensional correct.

### Deducing Relation among Physical Quantities

If we know the various factors on which a physical quantity depends, then by the use of principle of homogeneity of dimensions, we can derive an expression for that physical quantity.

**EXAMPLE –**

Let, us derive an expression for the *centripetal force* ( F ) acting on a particle of mass ( m ) moving with velocity ( v ) in a circular path of radius ( r ) .

We know that, the centripetal force depends upon the following quantities –

- Mass of particle ( m )
- Velocity of particle ( v )
- Radius ( r ) of circular path.

Let, the centripetal force is expressed by a relation as –

F \propto m^a \ v^b \ r^c

Then \quad F = K \ m^a \ v^b \ r^c …….. (1)

Where, ( K ) is a dimensionless constant called as **constant of proportionality. **

Writing the dimensions of various quantities in equation (1), we will get –

[ M^1 \ L^1 \ T^{-2} ] = 1 \ [ M ]^a \ [ L T^{-1} ]^b \ [ L ]^c

= [ M^a \ L^{(b + c)} \ T^{-b} ]

Comparing the dimensions of similar quantities on both sides, we get –

a = 1 \quad ( b + c ) = 1 \quad - b = - 2

By solving above relations, we get –

a = 1 \quad b = 2 \quad c = - 1

Putting these values in equation (1) we get –

F = K \ m \ v^2 \ r^{-1}

= K \left ( \frac {mv^2}{r} \right )

This is the required expression for the centripetal force.

## Limitations of Dimensional Analysis

Method of dimensional analysis in deducing relation among physical quantities has some limitations described as follows –

- The method does not give any information about the dimensionless
( K ) .**constant of proportionality** - Method of dimensional analysis sometime fails when a physical quantity depends on more than three physical quantities.
- It fails when a physical quantity viz. \left [ s = ut + \left ( \frac {1}{2} \right ) at^2 \right ] is the sum or difference of two or more quantities.
- It fails to derive relationships which involve trigonometric, logarithmic or exponential functions.
- Sometimes, it is difficult to identify the factors on which the physical quantity depends.
- The method becomes more complicated when universal dimensional constants like ( G ) and ( R ) etc. are involved.