Dimensional Analysis

Dimensions tell you the kind of physical quantity (length, time, mass, …) independent of units. Dimensional analysis is a fast, powerful tool for checking equations, guessing formulas, and converting units.

Concept

We denote dimensions with capital letters inside brackets:

• Length: $[L]$
• Mass: $[M]$
• Time: $[T]$
• Current: $[A]$
• Temperature: $[\Theta]$

Any mechanical quantity $Q$ has dimensions $[Q] = M^a L^b T^c$ for some exponents.

The Three Main Uses

1. Check Equation Consistency

Both sides of any physical equation must have the same dimensions.

For $v = u + at$:

• $[v] = LT^{-1}$
• $[u] = LT^{-1}$
• $[at] = (LT^{-2})(T) = LT^{-1}$ ✓

2. Derive Formulas (up to a constant)

If $T$ (period of simple pendulum) depends on $l$, $g$, $m$, we write $T = k\, l^a g^b m^c.$ Equate dimensions:

• LHS: $T$.
• RHS: $L^a (LT^{-2})^b M^c = M^c L^{a+b} T^{-2b}$.

So $c = 0$, $-2b = 1 \Rightarrow b = -1/2$, $a + b = 0 \Rightarrow a = 1/2$. Therefore $T = k\sqrt{l/g}.$ The dimensionless constant $k = 2\pi$ must come from a full derivation.

3. Convert Units

A quantity is invariant; only its numerical value changes between unit systems.

$n_1 [\text{Unit}_1] = n_2 [\text{Unit}_2]$

So if $1\,\text{erg} = 10^{-7}\,\text{J}$, then $5 \times 10^7$ erg $= 5\,\text{J}$.

Worked Example

Q: The velocity $v$ of a wave on a stretched string depends on tension $T$ (force) and linear mass density $\mu$ (mass per length). Derive $v$ dimensionally.

Solution: Write $v = k\, T^a \mu^b$.

Dimensions:

• $[v] = LT^{-1}$
• $[T] = MLT^{-2}$
• $[\mu] = ML^{-1}$

So $LT^{-1} = (MLT^{-2})^a (ML^{-1})^b = M^{a+b}\, L^{a-b}\, T^{-2a}$.

Solving: $-2a = -1 \Rightarrow a = 1/2$; $a + b = 0 \Rightarrow b = -1/2$; check: $a - b = 1$ ✓.

$v = k\sqrt{T/\mu}.$ The exact derivation gives $k = 1$, so $v = \sqrt{T/\mu}$.

Common Confusions

• Constants like $\pi$, $2$, $1/2$ are dimensionless — dimensional analysis cannot recover them.
• Trigonometric, exponential, and logarithmic arguments must be dimensionless: $\sin(\omega t)$ requires $[\omega t] = 1$, so $[\omega] = T^{-1}$.
• Dimensional consistency is necessary but not sufficient: $E = mv$ is dimensionally wrong, but even $E = mv^2$ is dimensionally right despite missing the $1/2$.

Key Takeaways

• Use dimensional analysis to (i) check equations, (ii) guess formulas up to a constant, (iii) convert units.
• Powers, exponentials, and trig arguments must be dimensionless.
• Numerical constants and pure functions cannot be found dimensionally — full derivation is needed.