What is Dimensional Analysis?

Dimensional analysis is the process of using the dimensions of physical quantities to:

  • Derive relationships
  • Check correctness of equations
  • Convert units

This method relies on the principle of homogeneity, which states:

Every term in a physical equation must have the same dimensions.

Applications of Dimensional Analysis

1. Checking the dimensional consistency of equations

  • Example: Is s=ut+12at2s = ut + \frac{1}{2}at^2 dimensionally correct?

Breakdown:

  • ss: [L][L]
  • utut: [LT−1][T]=[L][L T^{-1}] [T] = [L]
  • at2at^2: [LT−2][T2]=[L][L T^{-2}] [T^2] = [L] ✅ All terms are of dimension [L][L] → equation is correct.

2. Deriving relationships between physical quantities (up to a constant)

Assume that quantity depends on variables, then equate dimensions.

Example: Find the formula for time period TT of a simple pendulum: Let T∝lagbT \propto l^a g^b [T]=[L]a[LT−2]b=[La+bT−2b][T] = [L]^a [L T^{-2}]^b = [L^{a+b} T^{-2b}] Equating powers: a+b=0a + b = 0, −2b=1-2b = 1 → b=−12b = -\frac{1}{2}, a=12a = \frac{1}{2} So, T=kl/gT = k \sqrt{l/g} (k is dimensionless constant)

3. Conversion between unit systems

Use: n1[U1]=n2[U2]n_1 [U_1] = n_2 [U_2]

Dimensional analysis cannot provide numerical constants or trigonometric functions.

Example: Dimensional Derivation

Derive the expression for speed vv of a wave on a string. Assume: v∝Taμbv \propto T^a \mu^b where TT = tension, μ\mu = mass per unit length.

Dimensions:

  • vv: [LT−1][L T^{-1}]
  • TT: [MLT−2][M L T^{-2}], μ\mu: [ML−1][M L^{-1}]

Let: [LT−1]=[MLT−2]a[ML−1]b=[Ma+bLa−bT−2a][L T^{-1}] = [M L T^{-2}]^a [M L^{-1}]^b = [M^{a+b} L^{a-b} T^{-2a}] Equating powers: a+b=0a + b = 0, a−b=1a - b = 1, −2a=−1-2a = -1 → a=12a = \frac{1}{2}, b=−12b = -\frac{1}{2}

So, v=kTμv = k \sqrt{\frac{T}{\mu}}