Dimensional Analysis


Dimension

Dimension is the power to which the fundamental unit (mass, length and time) has to be raised to get the unit of any physical quantity. Units of mass, length and time are represented by bracketed capital letter; $[M]$, $[L]$ and $[T]$ respectively.
For example, \[\text{Volume}=\text{length}×\text{breadth}×\text{height}\] \[\text{Volume}=[L]×[L]×[L]=[L^3]\] Thus, volume possesses three dimensions in length. To measure volume, units of mass and time are not used, So, we can represent it as $[M^0L^3T^0]$.For velocity, \[\text{Velocity}=\frac{\text{Displacement}}{\text{Time}}\] \[\text{Velocity}=\frac{[L]}{[T]}\] \[\text{Velocity}=\left[LT^{-1}\right]\] \[\text{Velocity}=\left[M^0LT^{-1}\right]\] Thus, velocity consists of zero dimension in mass, one dimension in length and minus one dimension in time.
The expression obtained above for velocity is the dimensional formula of the velocity.
Thus, dimensional formula of a physical equation is defined as the expression containing suitable powers of fundamental units (mass, length and time) such that it expresses the dimension of that physical quantity.
If we denote velocity by $[v]$, then, \[\left[v\right]=\left[M^0LT^{-1}\right]\] This equation is known as dimensional equation of velocity.
Thus, the equation obtained when a physical quantity is equated with its dimensional formula is known as dimensional equation.

Types of Physical Quantities on the basis of Dimensional Analysis

On the basis of dimensional analysis, there are four types of physical quantities;

Dimensional variables

These physical quantities do not have fixed value but have dimensions. Examples: velocity, acceleration, force, pressure, work, power, etc.

Dimensionless variables

These physical quantities neither have fixed value nor have dimensions. Examples: angle, efficiency, strain, etc.
Angle does not have dimensions because, \[\text{Angle} = \frac{\text{Arc}}{\text{Radius}} = \frac{[L]}{[L]} = 1\] Efficiency does not have dimensions because, \[\text{Efficiency}=\frac{\text{Output Work}}{\text{Input Work}}\] Both output and input work have same unit. So, efficiency does not have dimension.
Strain does not have dimensions because, \[\text{Strain}=\frac{\text{Change in dimension of body}}{\text{Original dimension of the body}}\] Both numerator and denominator have same unit. So, strain does not have dimension.

Dimensional constants

These physical quantities have both fixed value as well as dimensions. Examples: gravitational constant $(G)$, Planck’s constant $(h)$, permittivity of free space $(ε_0)$, etc.

Dimensionless constants

These physical quantities have fixed value but do not have dimensions. Examples: $π$, numbers $(1,2,3,4,…)$, trigonometric quantities, etc.

Some Physical Quantities with their S.I. Units and Dimensional Formula

Physical Quantity

S.I. unit

Dimensional Formula

Acceleration

ms-2

[M0LT-2]

Angular Acceleration

rad s-2

[MoL0T-2]

Angular Momentum

kg m-2 s-1

[ML2T-1]

Angular Velocity

rad s-1

[MoL0T-1]

Area

m2

[MoL2T0]

Coefficient of Elasticity

N m-2

[ML-1T-2]

Coefficient of Viscosity

kg m-1 s-1 [Pa s or Decapoise]

[ML-1T-1]

Density

kg m-3

[ML-3T0]

Electric Field

kg m s-3 A-1 [Vm-1 or NC-1]

[MLT-3A-1]

Energy

kg m2 s-2 [Joule (J)]

[ML2T-2]

Force

kg m s-2 [Newton (N)]

[MLT-2]

Frequency

s-1 [Hertz (Hz)]

[MoL0T-1]

Gravitational Constant

N m2 kg-2

[M-1L3T-2]

Heat Capacity

J K-1

[ML2T-2K-1]

Impulse

N s

[MLT-1]

Linear Momentum

kg m s-1

[MLT-1]

Moment of Inertia

kg m2

[ML2T0]

Planck’s Constant

J s

[ML2T-1]

Power

J s-1 [Watt (W)]

[ML2T-3]

Power of lens

m-1  [Diaptor (D)]

[MoL-1T0]

Pressure

N m-2 [Pascal (Pa)]

[ML-1T-2]

Pressure Gradient

Kg m-2 s-2

[ML-2T-2]

Radius of Gyration

m

[M0LT0]

Stress

N m-2

[ML-1T-2]

Surface Energy

J m-2

[ML0T-2]

Surface Tension

N m-1

[ML0T-2]

Torque

N m

[ML2T-2]

Velocity

ms-1

[M0LT-1]

Velocity Gradient

s-1

[MoL0T-1]

Voltage

Kg m2 s-3 A-1 [Volt]

[ML2T-3A-1]

Volume

m3

[M0L3T0]

Work

kg m2 s-2 [Joule (J)]

[ML2T-2]

 


To check the correctness of physical equation

The principle of homogeneity is used to check the correctness of a physical equation. Principle of homogeneity states that
If a physical equation is dimensionally correct, then the dimension of each quantities on right hand side must be equal to the dimension of left hand side. 
i.e. if $A + B = C + D$ is a dimensionally correct equation then the dimension of $A$, $B$, $C$ and $D$ are equal.
For example: Checking the correctness of $PV=RT$.

Here, Dimensional formula of: \[\text{Pressure (P)}=\left[ML^{-1}T^{-2}\right]\] \[\text{Volume (V)}=\left[L^3\right]\] \[\therefore \text{PV}=\left[ML^{-1}T^{-2}\right] \left[L^3\right]=\left[ML^2T^{-2}\right]\] \[\text{Universal Gas Constant (R)}=\left[ML^2T^{-2}K^{-1}\right]\] \[\text{Temperature (T)}=\left[K\right]\] \[\text{RT}=\left[ML^2T^{-2}K^{-1}\right] \left[K\right]\] \[=\left[ML^2T^{-2}\right]\] Since the dimension of $PV$ is equal to the dimension of $RT$. The equation is dimensionally correct.
Checking the correctness of $v^2=u^2+2as$.

Here, Dimension of \[\text{LHS}=\left[ LT^{-1} \right]^2=\left[ L^2T^{-2} \right]\] \[\text{First term of RHS}=\left[ LT^{-1} \right]^2\] \[=\left[ L^2T^{-2} \right]\] \[\text{Second term of RHS}=\left[ LT^{-2} \right]\left[L\right]\] \[=\left[ L^2T^{-2} \right]\] Since, dimension of $LHS$ is equal to the dimension of each terms of $RHS$, the physical relation is dimensionally correct.

Note- All physical equations are dimensionally correct but every dimensionally correct equations are not necessarily physically correct equations. For example, Suppose an equation, \[s=ut+at^2\] This equation is dimensionally correct equation but it is not physically correct.


To determine the dimension of unknown quantity in a physical quantity

Dimensional method can also be used to determine the dimension of a constant in a physical equation. For example, To determine the dimensional formula of Gravitational Constant $(G)$, we have, \[F=\frac{GMm}{R^2}\] \[G=\frac{FR^2}{Mm}\] \[=\frac{\left[MLT^{-2}\right] \left[L^2\right]}{\left[M\right]\left[M\right]}=\left[M^{-1}L^3T^{-2}\right]\]
Again, for the dimension of Specific Heat Capacity $(S)$, we have, \[Q=mSdT\] \[S=\frac{Q}{mdT}\] \[=\frac{\left[ML^2T^{-2}\right]} {\left[M\right]\left[K\right]}=\left[L^2T^{-2}K^{-4}\right]\]


To convert one type of unit into another

The value of a physical quantity $X$ is given by, \[X=nu\] where,
$u$ = unit of the measurement
$n$ = number of times the unit is contained in $X$
Here, for a given quantity, $nu$ is constant. Hence, for two systems of a physical quantity, \[n_1u_1=n_2u_2\] Now- \[u_1=\left[M_1^xL_1^yT_1^z\right]\] \[u_2=\left[M_2^xL_2^yT_2^z\right]\] where, $x$, $y$ and $z$ represent the dimensions of the physical quantity. \[n_1\left[M_1^xL_1^yT_1^z\right]=n_2\left[M_2^xL_2^yT_2^z\right]\] \[n_2=n_1\left[\frac{M_1}{M_2}\right]^x \left[\frac{L_1}{L_2}\right]^y \left[\frac{T_1}{T_2}\right]^z\] This formula can be used to convert one system of unit into another.
Converting $5\text{ms}^{-1}$ into CGS unit by dimensional method \[\text{Dimension of velocity} = \left[M^0LT^{-1}\right]\] Thus, $x=0$, $y=1$ and $z=-1$. In SI, \[M_1=1 \text{ kg} = 1000\text{ g}\] \[L_1=1\text{ m}=100\text{ cm}\] \[T_1=1 \text{ sec}\] \[n_1=5\] In CGS, \[M_2=1 \text{ g}\] \[L_2=1 \text{ cm}\] \[T_2=1 \text{ sec}\] \[n_2=?\] we have, \[n_2=n_1\left[\frac{M_1}{M_2}\right]^x \left[\frac{L_1}{L_2}\right]^y \left[\frac{T_1}{T_2}\right]^z\] \[=5\left[\frac{1000}{1}\right]^0 \left[\frac{100}{1}\right]^1 \left[\frac{1}{1}\right]^{-1}\] \[=5*100\] \[=500\] \[5\text{ m/s}=500 \text{ cm/s}\]
Converting $1 \text{ joule}$ into \text{ erg} \[\text{Dimension of work} = \left[ML^2T^{-2}\right]\] Thus, $x=1$, $y=2$ and $z=-2$. In SI, \[M_1=1 \text{ kg} = 1000\text{ g}\] \[L_1=1\text{ m}=100\text{ cm}\] \[T_1=1 sec\] \[n_1=1\] In CGS, \[M_2=1 \text{ g}\] \[L_2=1 \text{ cm}\] \[T_2=1 \text{ sec}\] \[n_2=?\] we have, \[n_2=n_1\left[\frac{M_1}{M_2}\right]^x \left[\frac{L_1}{L_2}\right]^y \left[\frac{T_1}{T_2}\right]^z\] \[=1\left[\frac{1000}{1}\right]^1 \left[\frac{100}{1}\right]^2 \left[\frac{1}{1}\right]^{-2}\] \[=1000*100^2\] \[=10^7\] \[1 \text{ joule} = 10^7 \text{ erg}\]


​To derive the relation between different physical quantities

Principle of homogeneity is used to determine the relationship between different physical quantities. We have to find all the possible factors on which the physical quantity may depend. By using the dimensions of the factors, we find out the relation between the quantities. Let’s find out the relation between force $(F)$, pressure $(P)$ and area $(A)$. Let \[P ∝F^x \] \[P ∝A^y \] combining above relations, \[P ∝F^xA^y\] \[P=kF^xA^y\text{ ____(1)}\] where, $k$ = proportionality constant. We have to find $x$ and $y$. Writing dimensional equation of each terms of $(1)$, \[\left[ML^{-1}T^{-2}\right]=\left[MLT^{-2}\right]^x\left[L^2\right]^y\] \[\left[ML^{-1}T^{-2}\right]=\left[M^xL^xT^{-2x}\right]\left[L^{2y}\right]\] \[\left[ML^{-1}T^{-2}\right]=\left[M^xL^{x+2y}T^{-2x}\right]\] equating, we get, \[x=1\] and, \[x+2y=-1\] \[1+2y=-1\] \[y=-1\] Putting the values of $x$ and $y$ in $(1)$, \[P=kF^1A^{-1}\] \[P=k\frac{F}{A}\] The value of $k$ is found to be $1$. \[P=\frac{F}{A}\] This is the relation between pressure, force and area.
​Determining the relation between time period of a simple pendulum $(T)$, length of the simple pendulum $(l)$ and acceleration due to gravity $(g)$: Let \[T ∝l^x \] \[T ∝ g^y \] combining above relations, \[T ∝l^xg^y\] \[T=kl^xg^y\text{ ____(1)}\] where, $k$ = proportionality constant. Writing dimensional equation of each terms of $(1)$, \[\left[T\right]=\left[L\right]^x \left[LT^{-2}\right]^y\] \[\left[T\right]=\left[L^{x+y}T^{-2y}\right]\] equating, we get, \[-2y=1\] \[y=-\frac{1}{2}\] and, \[x+y=0\] \[x=\frac{1}{2}\] putting the values of $x$ annd $y$ in $(1)$, \[T=kl^{\frac{1}{2}}g^{-\frac{1}{2}}\] \[T=k\sqrt{\frac{l}{g}}\] Experimentally, the value of $k$ is found to be $2π$. \[T=2π\sqrt{\frac{l}{g}}\] This is the required relation.


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