Prerequisites
Before studying dimensional analysis, you should understand:
- SI Units — Base and derived units
- Dimensions — Dimensional formulas of quantities
What is Dimensional Analysis?
Dimensional analysis is a powerful mathematical tool that uses the dimensions of physical quantities to:
- Check if equations are dimensionally correct
- Derive relationships between physical quantities
- Convert units from one system to another
Applications of Dimensional Analysis
1. Checking Dimensional Correctness
Principle: A physical equation is dimensionally correct if all terms have the same dimensions.
You can only add/subtract quantities with identical dimensions.
Examples
Given: $s = ut + \frac{1}{2}at^2$
LHS: [s] = [L]
RHS Term 1: [ut] = [LT⁻¹][T] = [L] ✓
RHS Term 2: [½at²] = [LT⁻²][T²] = [L] ✓
All terms = [L]
Conclusion: ✓ Dimensionally correct!
LHS: [v²] = [LT⁻¹]² = [L²T⁻²]
RHS:
- [u²] = [L²T⁻²] ✓
- [2as] = [LT⁻²][L] = [L²T⁻²] ✓
Conclusion: ✓ Dimensionally correct!
Wrong equation: $s = ut + at$ (missing the square on time)
Check:
- [ut] = [L]
- [at] = [LT⁻²][T] = [LT⁻¹] ≠ [L] ✗
This equation is dimensionally INCORRECT!
2. Deriving Relationships (Most Important for JEE)
When a physical quantity depends on several other quantities, we can find the relationship using dimensional analysis.
Standard Method:
- Identify the quantity to derive and the quantities it depends on
- Assume a power-law relationship: $Q = k \cdot A^a B^b C^c ...$ (where k is dimensionless constant)
- Write dimensions of all quantities
- Equate powers of M, L, T on both sides
- Solve the system of equations
- Write the final relationship
Problem: The time period T of a simple pendulum depends on:
- Length of pendulum (l)
- Mass of bob (m)
- Acceleration due to gravity (g)
Find the relationship.
Solution:
Step 1: Assume $T = k \cdot l^a m^b g^c$
Step 2: Write dimensions:
- [T] = [T]
- [l] = [L]
- [m] = [M]
- [g] = [LT⁻²]
Step 3: Substitute:
$$[T] = [L]^a [M]^b [LT^{-2}]^c$$ $$[M^0 L^0 T^1] = [M^b L^{a+c} T^{-2c}]$$Step 4: Equate powers:
- For M: 0 = b → b = 0
- For T: 1 = -2c → c = -1/2
- For L: 0 = a + c → a = -c → a = 1/2
Step 5: Final relation:
$$\boxed{T = k\sqrt{\frac{l}{g}}}$$The actual formula is $T = 2\pi\sqrt{l/g}$, where k = 2π.
Key Insight: Time period is independent of mass! Dimensional analysis revealed this.
Interactive Demo: Visualize Dimensional Analysis
Explore how to derive physical relationships using dimensional analysis.
Dimensional analysis cannot determine:
- Dimensionless constants (like 2π above)
- Whether quantities appear in numerator or denominator (we get √(l/g) or √(g/l) — need physics!)
- The exact form when exponential, logarithmic, or trigonometric functions are involved
More Derivation Examples
Problem: The viscous force F on a sphere moving through a fluid depends on:
- Radius of sphere (r)
- Velocity of sphere (v)
- Coefficient of viscosity (η), where [η] = [ML⁻¹T⁻¹]
Derive the relation.
Solution:
Assume: $F = k \cdot r^a v^b \eta^c$
Dimensions:
$$[MLT^{-2}] = [L]^a [LT^{-1}]^b [ML^{-1}T^{-1}]^c$$ $$[MLT^{-2}] = [M^c L^{a+b-c} T^{-b-c}]$$Equating powers:
- M: 1 = c → c = 1
- T: -2 = -b - c = -b - 1 → b = 1
- L: 1 = a + b - c = a + 1 - 1 → a = 1
The actual Stokes’ law: $F = 6\pi\eta rv$ (k = 6π)
Problem: The energy E of a photon depends on its frequency ν. If E = hν, find dimensions of Planck’s constant h.
Solution:
$$E = h\nu$$ $$h = \frac{E}{\nu}$$ $$[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]}$$ $$\boxed{[h] = [ML^2T^{-1}]}$$This is angular momentum dimension! (Indeed, h/2π is the quantum of angular momentum)
3. Converting Units Between Systems
Dimensional analysis helps convert quantities from one unit system to another.
Formula for conversion:
$$n_2 u_2 = n_1 u_1 \left(\frac{M_1}{M_2}\right)^a \left(\frac{L_1}{L_2}\right)^b \left(\frac{T_1}{T_2}\right)^c$$where:
- n₁, u₁ = numerical value and unit in system 1
- n₂, u₂ = numerical value and unit in system 2
- a, b, c = powers in dimensional formula [MᵃLᵇTᶜ]
Problem: Convert 1 dyne to newtons.
Solution:
Force: [F] = [MLT⁻²] → a = 1, b = 1, c = -2
CGS system (system 1):
- M₁ = 1 gram
- L₁ = 1 cm
- T₁ = 1 s
SI system (system 2):
- M₂ = 1 kg = 1000 g
- L₂ = 1 m = 100 cm
- T₂ = 1 s
Limitations of Dimensional Analysis
Cannot find dimensionless constants
- We got $T = k\sqrt{l/g}$ but not k = 2π
Cannot distinguish between similar quantities
- Work and torque both have [ML²T⁻²]
- Cannot tell which is which
Cannot handle addition/subtraction
- Equation like $F = ma + mg$ cannot be derived
Cannot work with transcendental functions
- $e^x$, $\sin x$, $\log x$ must have dimensionless arguments
Cannot determine exact form
- Could be $F \propto v$ or $F \propto v²$ if both work dimensionally
May give wrong numerator/denominator
- Could get $\sqrt{g/l}$ instead of $\sqrt{l/g}$ — need physics!
Special Techniques
Technique 1: Dimensionless Products
When deriving formulas, look for dimensionless combinations.
For a sphere moving in fluid, dimensionless number:
$$\text{Reynolds number} = \frac{\rho v d}{\eta}$$where ρ = density, v = velocity, d = diameter, η = viscosity
Check: $[\rho vd/\eta] = \frac{[ML^{-3}][LT^{-1}][L]}{[ML^{-1}T^{-1}]} = [M^0L^0T^0] = 1$ ✓
Technique 2: Matching Unknown Dimensions
If equation contains unknown constant with unknown dimensions:
Given: $F = Av + Bv^3$, find dimensions of A and B.
Solution:
From homogeneity, both terms must equal [F] = [MLT⁻²]
Term 1: $[A][v] = [MLT^{-2}]$
$$[A] = \frac{[MLT^{-2}]}{[LT^{-1}]} = \boxed{[MT^{-1}]}$$Term 2: $[B][v^3] = [MLT^{-2}]$
$$[B] = \frac{[MLT^{-2}]}{[L^3T^{-3}]} = \boxed{[ML^{-2}T^1]}$$Practice Problems
Level 1: JEE Main Basics
Check if the equation $v = u + at$ is dimensionally correct.
Solution:
- [v] = [LT⁻¹]
- [u] = [LT⁻¹]
- [at] = [LT⁻²][T] = [LT⁻¹]
All terms have [LT⁻¹]. ✓ Correct!
The velocity v of sound in a gas depends on pressure P and density ρ. Find the relation.
Solution:
Let $v = k P^a \rho^b$
Dimensions:
- [v] = [LT⁻¹]
- [P] = [ML⁻¹T⁻²]
- [ρ] = [ML⁻³]
Equating:
- M: 0 = a + b → b = -a
- T: -1 = -2a → a = 1/2
- L: 1 = -a - 3b = -1/2 + 3/2 ✓
Therefore: b = -1/2
$$\boxed{v = k\sqrt{\frac{P}{\rho}}}$$(Actual: $v = \sqrt{\gamma P/\rho}$ where γ is specific heat ratio)
Level 2: JEE Main/Advanced
The centripetal force F on a body of mass m moving in a circle depends on mass m, velocity v, and radius r. Derive the formula.
Solution:
Let $F = k m^a v^b r^c$
$$[MLT^{-2}] = [M]^a [LT^{-1}]^b [L]^c$$ $$[MLT^{-2}] = [M^a L^{b+c} T^{-b}]$$Equating:
- M: 1 = a → a = 1
- T: -2 = -b → b = 2
- L: 1 = b + c = 2 + c → c = -1
(k = 1, so $F = mv^2/r$)
If force F, length L, and time T are fundamental units, express mass in terms of F, L, T.
Solution:
From $F = ma$:
$$m = \frac{F}{a} = \frac{F}{L/T^2} = \frac{FT^2}{L}$$ $$\boxed{[m] = [F T^2 L^{-1}]}$$In new system: mass has dimensions F¹L⁻¹T²
Level 3: JEE Advanced
The period of oscillation T of a drop of liquid depends on:
- Radius r
- Density ρ
- Surface tension S (dimensions [MT⁻²])
Find the relationship.
Solution:
Let $T = k r^a \rho^b S^c$
$$[T] = [L]^a [ML^{-3}]^b [MT^{-2}]^c$$ $$[T] = [M^{b+c} L^{a-3b} T^{-2c}]$$Equating:
- M: 0 = b + c → b = -c
- T: 1 = -2c → c = -1/2
- Therefore: b = 1/2
- L: 0 = a - 3b = a - 3/2 → a = 3/2
(Actual formula: $T = 2\pi\sqrt{r^3\rho/S}$)
In the equation $P = \frac{a}{V} - \frac{b}{V^2}$, where P is pressure and V is volume, find dimensions of a and b.
Solution:
From homogeneity, both terms must equal [P] = [ML⁻¹T⁻²]
For term 1:
$$\frac{[a]}{[V]} = [P]$$ $$[a] = [P][V] = [ML^{-1}T^{-2}][L^3]$$ $$\boxed{[a] = [ML^2T^{-2}]}$$(dimension of energy!)
For term 2:
$$\frac{[b]}{[V^2]} = [P]$$ $$[b] = [P][V^2] = [ML^{-1}T^{-2}][L^6]$$ $$\boxed{[b] = [ML^5T^{-2}]}$$Strategy for JEE Problems
For checking equations:
- Find dimensions of each term separately
- Check if all terms match
- Special attention to arguments of sin, cos, log, exp (must be dimensionless)
For deriving formulas:
- List all relevant quantities
- Assume power-law relationship
- Write dimensional formula for each
- Set up equations for M, L, T
- Solve systematically
- Write final relation (with “k” for unknown constant)
For conversions:
- Identify dimensional formula [MᵃLᵇTᶜ]
- Note base unit ratios
- Apply conversion formula
- Calculate carefully!
Summary Table
| Application | What It Can Do | What It Cannot Do |
|---|---|---|
| Check equations | Verify dimensional correctness | Prove equation is physically correct |
| Derive formulas | Find power-law relationships | Find numerical constants |
| Unit conversion | Convert between systems | — |
| Find dimensions | Determine dimensions of constants | Distinguish similar quantities |
Related Topics
Within Units and Measurements
- SI Units — Foundation of dimensional analysis
- Dimensions — Dimensional formulas needed here
- Errors — Error propagation uses dimensions
- Significant Figures — Precision in dimensional calculations
Connected Chapters
- Kinematics — Deriving kinematic formulas
- Newton’s Laws — Checking force equations
- Gravitation — Deriving orbital formulas
- Oscillations — Period of pendulum, springs
- Fluid Mechanics — Viscosity, surface tension formulas
Math Connections
- Algebra — Solving simultaneous equations
- Logarithms — Dimensionless arguments