Chapter 5: Dimensional Formulas and Dimensional Analysis

5.1 What Are Dimensions?

The dimensions of a physical quantity are the powers to which the fundamental (base) quantities must be raised in order to represent that physical quantity. Since SI recognises seven base quantities, the dimensional formula of any physical quantity is written using the symbols M (mass), L (length), T (time), A (electric current), K (thermodynamic temperature), mol or N (amount of substance), and cd (luminous intensity), each raised to an appropriate power. In most mechanics and heat problems, only M, L, and T are needed. For example, the dimensional formula of speed, which is distance divided by time, is [M⁰L¹T⁻¹], usually written simply as [LT⁻¹]. The dimensional formula of force, which is mass times acceleration, is [M¹L¹T⁻²], usually written [MLT⁻²]. Note that dimensional formulas describe the nature of a quantity, not its size — they tell us that force is "mass times length divided by time-squared" regardless of whether it is measured in newtons, dynes, or pounds-force.

Quantities that have the same dimensional formula are said to be dimensionally similar, even if they represent physically different concepts — for instance, work, energy, and torque are all dimensionally [ML²T⁻²], even though torque is a vector-like quantity distinct from the scalar quantities work and energy. Conversely, quantities such as angle, strain, refractive index, and specific gravity have no dimensions at all (their dimensional formula is [M⁰L⁰T⁰], i.e., they are dimensionless), because they are ratios of two similar quantities whose units cancel out.

5.2 Dimensional Formulas of Important Physical Quantities

5.3 Principle of Homogeneity of Dimensions

The principle of homogeneity of dimensions states that a physical equation can be correct only if the dimensions on both sides of the equation are the same, and every term that is added or subtracted within the equation must also have identical dimensions. This is the foundation of dimensional analysis: since only quantities of the same dimension can be meaningfully added, subtracted, or set equal to each other, checking dimensions provides a powerful, quick way to test whether a proposed formula could possibly be correct.

5.4 Uses of Dimensional Analysis

• To check the correctness (dimensional consistency) of a physical equation or formula, by verifying that both sides have the same dimensional formula.
• To convert a physical quantity's value from one system of units to another (for example, from CGS to SI), using the relation between the numerical value and the units in both systems.
• To derive a plausible relationship between physical quantities, when the quantities on which a given quantity depends are known but the exact form of the relation is not (this method is limited to relations involving simple products of powers, and cannot fix dimensionless constants).
• To express a derived unit in terms of the base units and to identify the dependence of a quantity on the fundamental quantities.

5.5 Worked Examples

Example 1: Checking the correctness of an equation

Verify the equation for the time period of a simple pendulum, T = 2π√(l/g), using dimensional analysis, where l is length and g is acceleration due to gravity.

The dimension of l is [L] and the dimension of g is [LT⁻²]. So the dimension of l/g is [L]/[LT⁻²] = [T²], and the dimension of √(l/g) is [T]. Since 2π is a dimensionless constant, the right-hand side has dimension [T], which matches the dimension of time period T on the left-hand side, [T]. Hence the equation is dimensionally correct (though dimensional analysis alone cannot verify that the numerical constant 2π itself is correct — it can only confirm that the equation is dimensionally consistent).

Example 2: Deriving the dimensional formula of a derived quantity

Find the dimensional formula of the coefficient of viscosity η, given by Stokes' law F = 6πηrv, where F is the viscous force, r is the radius of a sphere, and v is its velocity through the fluid.

Rearranging, η = F/(6πrv). Dimensionally, [F] = [MLT⁻²], [r] = [L], [v] = [LT⁻¹]. So [η] = [MLT⁻²] / ([L][LT⁻¹]) = [MLT⁻²] / [L²T⁻¹] = [ML⁻¹T⁻¹]. This confirms the dimensional formula for the coefficient of viscosity listed in the table above.

Example 3: Converting a quantity between two systems of units

Convert a force of 1 newton into dynes (the CGS unit of force). We know force has dimensions [MLT⁻²]. Using n₁u₁ = n₂u₂ where n is the numerical value and u the unit: 1 N = 1 kg × 1 m × 1 s⁻² . Since 1 kg = 1000 g and 1 m = 100 cm, we get 1 N = 1000 g × 100 cm × 1 s⁻² = 10⁵ g·cm·s⁻² = 10⁵ dyne. Therefore, 1 newton = 10⁵ dyne, a conversion frequently tested directly in exams.

5.6 Limitations of Dimensional Analysis

• It cannot determine dimensionless constants (such as 2π, 1/2, or other pure numbers) that appear in a formula; these must be found by experiment or by detailed derivation.
• It cannot be used for equations involving trigonometric, logarithmic, or exponential functions, since the arguments of such functions must themselves be dimensionless, and dimensional analysis gives no information about them.
• It fails when a physical quantity depends on more than three independent variables (in an M-L-T system), because there are then not enough equations (only three, one each for M, L, and T) to solve for all the unknown powers.
• It cannot distinguish between physical quantities that have the same dimensional formula but are conceptually different (for example, it cannot distinguish work from torque, since both are [ML²T⁻²]).
• It cannot be applied if a physical quantity is the sum or difference of two or more terms with different physical origins that happen to be combined in a way not reducible to a single power relationship — it only checks consistency, and does not itself derive the complete correct equation from scratch, including numerical coefficients.