Dimensions and Dimensional Analysis

 

Base Units

According to international system of units, known as SI units, seven units are introduced as base units. They are as follows:

Base Unit	Name	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric Current	Ampere	A
Temperature	Kelvin	K
Amount of Substance	mole	mol
Luminosity	candela	cd

 

 

Derived Units

The units derived from base units are called derived units.

E.g.
Force = mass x acceleration
N = kg x ms^-2
Therefore, the derived units of force is kgms^-2.
N = kgms^-2

Derived Quantity	Name	Symbol
Pressure	Pascal	Pa = kgm-1s-2
Power	Watt	W = kgm2s-3
Resistance	Ohm	Ω = kgm2s-3A-2
Magnetic Flux Density	Tesla	T = kgs-2A-1
Frequency	Hertz	Hz = s-1

 

Food for thought

The units of power is Watt, W, yet that of force is kgms^-2. How do you account for the use of uppercase and lowercase letters?

 

Dimensions

The way a physical unit is related to base units indicates the dimensions of the physical unit in question.

E.g.1
Area = length x width
The dimensions of area is denoted as follows:
[Area] = L²
E.g.2
Volume = length x width x height
The dimensions of volume is denoted as follows:
[Volume] = L³
E.g.3
Density = mass/volume
The dimensions of density is denoted as follows:
[Density] = kgL^-3
E.g.4
Force = mass x acceleration
The dimensions of force is denoted as follows:
[Force] = MLT^-2

Dimensions of Base Units

The dimensions of the base units are as follows:

Base Unit	Dimension
Length	L
Mass	M
Time	T
Electric Current	I
Temperature	θ
Amount of Substance	N
Luminosity	J

 

Dimensions of Derived Units

The dimensions of some derived units are as follows:

Derived Units	Dimensions
[Current]	A
[Energy]	ML2T-2
[Power]	ML2T-3
[Voltage]	ML2T-3A-1
[Resistance]	ML2T-3A-2
[Coefficient of Viscosity]	ML-1T-1

 

Dimensional Homogeneity

If an equation is correct, the dimensions on both sides must be the same - or consistent.

E.g.
v = u + at
LHS: LT^-1
RHS: LT^-1 + LT^-2T = 2LT^-1
The numerical constants are ignored in dimensions, as they do not have dimensions.
LHS = RHS
The formula is dimensionally homogeneous.

The following formula is not dimensionally homogeneous

E.g.
Force = m x v
LHS: MLT^-2
RHS: MLT^-1 = MLT^-1
LHS ≠ RHS
The formula is not dimensionally homogeneous; therefore, it is not correct.

Food for thought

Check whether the formula, F = t/3mdl, is dimensionally homogeneous. F, m, d, t and l are force, mass, distance, time and length respectively.

Dimensional Analysis - deriving formulae

Since physical relationship must be dimensionally homogeneous to be correct, we can use the very fact to derive formulae, having observed how the physical quantities are related.

E.g.1
Newton's Second Law says that the force exerted on an object is directly proportional to the rate of change of momentum.
F ∝ mv/t
F = (mv/t)^x → 1
[F] = [(MLT^-1/T)^x]
MLT^-2 = M^xL^xT^-2x
By making powers of each dimension on both sides equal,
T → -2 = -2x
x = 1
From 1 → F = mv/t
F = ma

E.g.2
The voltage across a resistor was found out to be proportional to the current and resistance.
V ∝ IR
V = k I^xR^y, where k is a dimensionless constant and x and y are to be found by dimensional analysis.
[V] = [I^x][R^y]
M¹L²A^-1T^-3 = (A)^x(ML²A^-2T^-3)^y
M¹L²A^-1T^-3 = A^x-2yM^yL^2yA^-2yT^-3y
By making the powers of each dimension on both sides equal,
M → 1 = y
A → -1 = x - 2y = x - 2 → x = 1
RHS: A¹ (ML²A^-2T^-3) = IR
V = IR

E.g.3

The time period of a pendulum is thought to be proportional to the mass of the bob, length and the acceleration due to gravity.

 

 

period ∝ m l g
period = k m^x l^y g^z → 1, where k is a dimensionless constant.
[period] = [m^x l^y g^z]
T = M^x L^y (LT^-2)^z
M⁰L⁰T¹ = M^x L^y+zT^-2z
By making the powers of each dimension on both sides equal,
M → x = 0
T → 1 = -2z → z = -1/2 L → 0 = y + z → y = 1/2
From 1,
T = k √l/g
k = 2π √(l/g)

E.g.4
Stokes' Law says that the viscous force(drag force) acts on a sphere is proportional to the radius, coefficient of viscosity and the velocity.
F ∝ r η v
F = k r^x η^y v^z → 1,
[F] = k[r^x η^y v^z]
MLT^-2 = L^x (ML^-1T^-1)^y (LT^-1)^z
M¹L¹T^-2 = M^y L^x-y+z T^-2y-z
By making the powers of each dimension on both sides equal,
M → 1 = y
T → -2 = -y - z
-1 - z = -2
z = 1
L → 1 = x - y + z
x -1 + 1 = 1
x = 1
From 1,
F = k rηv
k = 6π F = 6πηrv

E.g.5
Albert Einstein found out that the rest energy, E, of mass, m, is proportional to both the mass and the speed of light.
E ∝ mc
E = k m^xc^y → 1
[E] = [m^xc^y]
ML²T^-2 = M^x(LT^-1)^y
ML²T^-2 = M^x L^y T^-y
By making the powers of each dimension on both sides equal,
M → 1 = x
L → 2 = y
T → -2 = -y
y = 2
Since k=1, from 1,
E = mc²

E.g.6
The kinetic energy, KE, of an object is related to its velocity and mass. Derive the formula.
KE ∝ m v
kE = k m^xv^y → 1
[kE] =[m^x] [v^y]
ML²T^-2 = M^x (LT^-1)^y
M¹L²T^-2 = M^x L^yT^-y
By making the powers of each dimension on both sides equal,
M → 1 = x
T → -2 = -y → y = 2
From 1,
KE = k mv²
Since k= 1/2
KE = 1/2 mv²

E.g.7
The centripetal force, F_c, of an object that moves at a velocity v in a circle of radius r, are related Derive the formula.
F_c ∝ m v r
F_c = k m^xv^y r^z → 1
[F_c] =[m^x] [v^y] [r^z]
MLT^-2 = M^x (LT^-1)^y L^z
M¹L¹T^-2 = M^x L^y+zT^-y
By making the powers of each dimension on both sides equal,
M → 1 = x
T → -2 = -y → y = 2
L → 1 = y + z
1 = 2 + z → z = -1
From 1
F_c = k mv² / r
k = 1;
F_c = mv² / r

Deriving the dimensions of a physical quantity

Dimensions of known quantities help us find the dimensions of unknown physical quantities in a formula.

E.g
According to Coulomb's Law, F = Q₁Q₂ / d², where F, Q and d are force, charge and the distance respectively.
Let's find out the dimensions of charge from the above.
[F] = [Q₁Q₂] / [d²]
MLT^-2 = [Q²]/L² - since the dimensions of charge is the same
M¹L¹T^-2 = [Q²]/L²
[Q²] = M¹L¹T^-2 L²
[Q²] = ML³T^-2

Question: Derive the dimensions of ρ in V = √(T/ρ), where V and T are speed of sound along a wire under tension, T.

Deriving the dimensions of a constant

Some constants in physical formulae have dimensions. For example, G, the gravitational constant, in F = G M₁M₂ / d², has dimensions. They can be found as follows:

E.g.
F = G M₁M₂ / d²
G = F d² / M₁M₂
[G] = [F] [d²] / [M₁M₂]
[G] = MLT^-2 L² / M²
[G] = M^-1 L³ T^-2

Question: Derive the dimensions of the Plank's Constant, h, in E = hf, where E and f are energy and frequency of a photon respectively.

Practice Questions

1. The potential energy of a falling apple depends on the mass, height and acceleration due to gravity. Derive a formula to connect them.
2. The pressure at the bottom of a lake depends on the depth, density of water and acceleration due to gravity. Derive a formula for pressure.