Dimensional analysis — how to derive formulas and check equations

Question

How do we use dimensional analysis to check whether an equation is correct, and to derive the form of a physical formula?

Solution — Step by Step

Every valid physics equation must be dimensionally homogeneous — both sides must have the same dimensions. The seven base dimensions in SI are:

Mass [ $M$ ], Length [ $L$ ], Time [ $T$ ], Current [ $A$ ], Temperature [ $K$ ], Amount [ $mol$ ], Luminosity [ $cd$ ]

For mechanics, we mainly use $M$ , $L$ , and $T$ .

Is $v^2 = u^2 + 2as$ dimensionally correct?

LHS: $v^2 = [LT^{-1}]^2 = [L^2T^{-2}]$
RHS first term: $u^2 = [L^2T^{-2}]$
RHS second term: $2as = [LT^{-2}][L] = [L^2T^{-2}]$

All terms have dimensions $[L^2T^{-2}]$ . The equation is dimensionally correct.

Now check $v = u + at^2$ (deliberately wrong):

$v = [LT^{-1}]$
$at^2 = [LT^{-2}][T^2] = [L]$

Dimensions do not match ( $LT^{-1} \neq L$ ), so the equation is wrong.

Suppose we know the time period $T$ of a simple pendulum depends on mass $m$ , length $L$ , and acceleration due to gravity $g$ . Find the formula.

Assume: $T = k \cdot m^a \cdot L^b \cdot g^c$ (where $k$ is a dimensionless constant)

Dimensions: $[T] = [M]^a [L]^b [LT^{-2}]^c = [M^a L^{b+c} T^{-2c}]$

Matching both sides:

$M$ : $a = 0$ (time period does not depend on mass)
$T$ : $1 = -2c$ , so $c = -1/2$
$L$ : $0 = b + c$ , so $b = 1/2$

T = k \sqrt{\frac{L}{g}}

Dimensional analysis gives us $k = 2\pi$ cannot be determined — only the form of the equation.

graph TD
    A[Dimensional Analysis] --> B{Purpose?}
    B -->|Check equation| C[Find dimensions of each term]
    C --> D{All terms same dimension?}
    D -->|Yes| E[Equation may be correct]
    D -->|No| F[Equation is definitely wrong]
    B -->|Derive formula| G[Assume power law dependence]
    G --> H[Match dimensions on both sides]
    H --> I[Solve for powers]
    I --> J["Get formula form (not constants)"]

Why This Works

Dimensional analysis works because physical laws must be independent of the choice of units. If an equation holds in SI units, it must hold in CGS units too — and this is only possible if both sides have the same dimensions.

Limitations:

Cannot determine dimensionless constants ( $2\pi$ , $1/2$ , etc.)
Cannot distinguish between quantities with the same dimensions (work and torque both have $[ML^2T^{-2}]$ )
Cannot derive equations involving sums of multiple terms with different functional forms (like $v = u + at$ )

Alternative Method

Some useful dimensional formulas to memorise for quick checking:

Quantity	Dimensional Formula
Force	$[MLT^{-2}]$
Energy/Work	$[ML^2T^{-2}]$
Power	$[ML^2T^{-3}]$
Pressure	$[ML^{-1}T^{-2}]$
Momentum	$[MLT^{-1}]$
Angular momentum	$[ML^2T^{-1}]$
Gravitational constant $G$	$[M^{-1}L^3T^{-2}]$
Planck’s constant $h$	$[ML^2T^{-1}]$

For JEE Main, a common question type is “which of the following has the same dimensions as X?” This requires knowing the dimensional formula of each option. Practice matching — angular momentum and Planck’s constant have the same dimensions ( $[ML^2T^{-1}]$ ), and this appears almost every year.

Common Mistake

A dimensionally correct equation is not necessarily physically correct. For example, $v = u + 2at$ is dimensionally correct (all terms have $[LT^{-1}]$ ) but physically wrong (the correct equation is $v = u + at$ ). Dimensional analysis can only tell us if an equation is definitely wrong (dimensions do not match) — it cannot confirm that an equation is right.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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