Dimensional analysis — how to derive formulas and check equations

Question

How do you use dimensional analysis to (a) check the correctness of a formula, (b) derive a new formula, and (c) convert units between systems? What are the limitations of this method?

(JEE Main 2024 asked a derivation using dimensions; CBSE 11 boards test the checking method)

Solution — Step by Step

Every valid physics equation must have the same dimensions on both sides. Dimensions are expressed in terms of the 7 fundamental quantities: $[M], [L], [T], [A], [K], [mol], [cd]$ .

For mechanics, we mostly use $[M], [L], [T]$ . For example: force has dimensions $[MLT^{-2}]$ , energy has $[ML^2T^{-2}]$ .

To check if $v = u + at$ is dimensionally correct:

LHS: $[v] = [LT^{-1}]$
RHS: $[u] = [LT^{-1}]$ , $[at] = [LT^{-2}][T] = [LT^{-1}]$

Both terms on the RHS have the same dimensions as the LHS. The equation is dimensionally consistent.

Suppose the time period of a simple pendulum depends on length ( $l$ ), mass ( $m$ ), and gravitational acceleration ( $g$ ).

Let $T = k \cdot l^a \cdot m^b \cdot g^c$

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

Comparing: $b = 0$ , $a + c = 0$ , $-2c = 1$ → $c = -1/2$ , $a = 1/2$

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

The dimensionless constant $k = 2\pi$ cannot be determined by this method.

Cannot determine dimensionless constants ( $2\pi$ , $1/2$ , etc.)
Cannot distinguish between quantities with the same dimensions (work and torque both have $[ML^2T^{-2}]$ )
Does not work if the relation involves exponential, logarithmic, or trigonometric functions
Can only be used when the relationship involves power law dependence

flowchart TD
    A["Dimensional Analysis"] --> B["Check formula:<br/>LHS dimensions = RHS dimensions?"]
    A --> C["Derive formula:<br/>Assume power law dependence"]
    A --> D["Convert units:<br/>Ratio of fundamental units"]
    B --> E["If dimensions match: may be correct<br/>If not: definitely wrong"]
    C --> F["Compare exponents of M, L, T<br/>Solve for powers"]
    F --> G["Cannot find dimensionless constants"]
    style G fill:#ff6b6b,stroke:#333

Why This Works

Dimensional analysis works because the laws of physics are independent of the units we choose. An equation that is true in SI must also be true in CGS — and this invariance places strict constraints on how quantities can combine.

If an equation has different dimensions on its two sides, it is definitely wrong (you are adding apples and oranges). If dimensions match, the equation may be correct — but dimensional consistency is necessary, not sufficient.

Alternative Method

For unit conversion using dimensions: $1 \text{ J} = 1 \text{ kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$ . In CGS: $= 1000 \text{ g} \cdot (100 \text{ cm})^2 \cdot \text{s}^{-2} = 10^7 \text{ erg}$ . The conversion factor comes from replacing each fundamental unit. This is faster than memorising conversion factors.

Common Mistake

Students assume a dimensionally correct equation is necessarily physically correct. Dimensional analysis only checks necessary consistency, not sufficiency. For example, $F = mv$ has correct dimensions of $[MLT^{-2}]$ on both sides… wait, it does not: $[mv] = [MLT^{-1}]$ . But $F = ma$ and $F = mv^2/r$ both have the same dimensions — dimensional analysis cannot distinguish between them. The correct formula depends on the physics, not just the dimensions.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next