Dimensional Analysis — Checking Physics with Units

Dimensions are not units

A dimension is the kind of quantity; a unit is the scale you measure it with. Length is a dimension — meters, miles, and light-years are units of it. Dimensional analysis ignores the scale and watches the kind.

The base dimensions of mechanics are mass $[M]$, length $[L]$, and time $[T]$. (Temperature $[\Theta]$ and electric current $[I]$ join the set when heat and circuits enter.) Everything else is built from these: velocity is $[L]/[T]$, force is $[M][L]/[T]^2$, energy is $[M][L]^2/[T]^2$.

Rule 1: only add like dimensions

You can add $3\ \mathrm{m} + 2\ \mathrm{m}$, but $3\ \mathrm{m} + 2\ \mathrm{s}$ is meaningless — and dimensional analysis enforces this ruthlessly. Every term added or subtracted in an equation must carry identical dimensions.

This single rule kills a huge class of mistakes. If your derived expression for a distance contains a term with dimensions of $[L]/[T]$, something went wrong upstream — no calculator needed to know it.

Rule 2: both sides must match

Check $v^2 = u^2 + 2as$. Each term: $[L]^2/[T]^2$ — velocity squared on the left, velocity squared and $[L][L]/[T]^2 = [L]^2/[T]^2$ on the right. Balanced.

Now try a corrupted memory, $s = vt^2$: left is $[L]$, right is $[L][T]$. Mismatched — the formula is wrong, and you caught it in five seconds. During exams this is the cheapest error detector ever invented.

Deriving the pendulum period from nothing

Suppose you forget the pendulum formula but suspect the period $T$ depends on length $L$, gravity $g$, and mass $m$: $T \propto L^a g^b m^c$. Match dimensions:

$$[T] = [L]^a \left(\frac{[L]}{[T]^2}\right)^b [M]^c$$

Mass gives $c = 0$ — mass drops out entirely, which is why a heavy bob and a light bob swing together. Time gives $b = -1/2$, length gives $a = 1/2$. The full formula is the proportionality below times a dimensionless constant (which turns out to be $2\pi$):

What dimensions cannot tell you

Dimensional analysis is blind to dimensionless constants: it can never produce the $2\pi$ in the pendulum formula or the $1/2$ in $\tfrac{1}{2}mv^2$. It also cannot distinguish functions with the same dimensions — $\sin(\omega t)$ and $\cos(\omega t)$ look identical to it.

One more restriction with teeth: the argument of any transcendental function — $\sin$, $\exp$, $\log$ — must be dimensionless. If you ever write $e^{3\ \mathrm{meters}}$, stop: the exponent must be a pure number.

Scaling laws: the real superpower

Because $T \propto \sqrt{L/g}$, doubling a pendulum's length multiplies its period by $\sqrt{2}$, and quadrupling the length doubles it — no formula memorized, just the scaling. Engineers use this constantly: model ships, wind tunnels, and crash tests all rely on dimensional scaling to predict big behavior from small experiments.

Whenever a problem asks 'what happens if this doubles?', reach for dimensions before algebra. The scaling exponent usually falls out in a line or two.

A worked check: pressure

Is $P = F/A$ dimensionally sane? Force is $[M][L]/[T]^2$, area is $[L]^2$, so $F/A$ gives $[M]/[L][T]^2$ — mass per length per time squared. That is exactly the dimension of pressure (and the reason the pascal is $\mathrm{kg/(m\cdot s^2)}$).

Try it on every formula you meet this week. Within days the check becomes reflex, and reflexes save exams.

Variables and units

Common mistakes

Real-world applications

• Students check every derived formula in seconds during exams.

• Engineers scale model tests to full size — ships, aircraft, crash tests — via dimensional similarity.

• Physicists guessing a new formula's form start by matching dimensions.

• The Reynolds number and other dimensionless groups, built purely from dimensions, predict fluid behavior across scales.