Dimensional analysis — checking equations, deriving relations: Worked examples

The scenario matrix

Think of each row as a cell — a distinct trap or trick the topic can present. We will hit every one.

Cell	Case class	What is special / dangerous
A	Straight check of a given equation	passes → "not disproved" only
B	Failed check (spot the broken term)	a decisive verdict
C	Derive a relation, 3 unknowns = 3 equations	fully solvable
D	Under-determined derivation (unknowns > 3)	method breaks — must show WHY
E	Zero / degenerate exponent (a power comes out $0$ )	a variable drops out
F	Dimensionless argument of $sin / ln / e^{x}$	forces a hidden constraint
G	Unit conversion via dimensions (sign & power of each base)	exponents steer the factors
H	Real-world word problem	translate story → dimensions
I	Exam twist — find dimension of an unknown constant	isolate the symbol

Nine cells, nine worked examples. Each is labelled with the cell it fills.

Cell A — a plain check that passes

Cell B — a check that fails (decisive)

Cell C — derive a relation, 3 unknowns, 3 equations

Worked example How does the speed

$v$ of a wave on a string depend on the tension $F$ , the mass-per-length $μ$ , and possibly the length $ℓ$ ? Forecast: do you expect $ℓ$ to survive, or drop out? Bet now — the method will tell us.

Setup. Guess a product of powers: $v = k F^{a} μ^{b} ℓ^{c}, k dimensionless .$ Dimensions: $[v] = L T^{- 1}$ , $[F] = M L T^{- 2}$ , $[μ] = M L^{- 1}$ (mass per length), $[ℓ] = L$ .

Step 1 — expand the right side. $L T^{- 1} = (M L T^{- 2})^{a} (M L^{- 1})^{b} (L)^{c} = M^{a + b} L^{a - b + c} T^{- 2 a} .$ Why this step? Collect each base separately — each of $M, L, T$ is an independent axis, so their exponents must match one axis at a time.

Step 2 — match powers of each base.

$M : a + b = 0$
$L : a - b + c = 1$
$T : - 2 a = - 1 \Rightarrow a = \frac{1}{2}$

Step 3 — solve. From $M$ : $b = - a = - \frac{1}{2}$ . Substitute into $L$ : $\frac{1}{2} - (- \frac{1}{2}) + c = 1 \Rightarrow 1 + c = 1 \Rightarrow c = 0$ .

Step 4 — assemble. $v = k F^{1/2} μ^{- 1/2} ℓ^{0} = k \frac{F}{μ} .$

Verify: $[F / μ] = \frac{M L T ^{- 2}}{M L ^{- 1}} = L^{2} T^{- 2} = L T^{- 1} = [v]$ ✅. And $c = 0$ says length drops out — your forecast is now settled. (True value $k = 1$ .)

Cell D — under-determined: more unknowns than equations

Worked example The drag force

$F$ on a sphere might depend on speed $v$ , radius $r$ , fluid density $ρ$ , and viscosity $η$ . Try to derive it. Forecast: we have four unknown powers but only three base dimensions. Do you think one clean answer will pop out? Guess yes/no.

Setup. $F = k v^{a} r^{b} ρ^{c} η^{d}$ with $[F] = M L T^{- 2}$ , $[v] = L T^{- 1}$ , $[r] = L$ , $[ρ] = M L^{- 3}$ , $[η] = M L^{- 1} T^{- 1}$ .

Step 1 — match powers. $M L T^{- 2} = (L T^{- 1})^{a} (L)^{b} (M L^{- 3})^{c} (M L^{- 1} T^{- 1})^{d} .$

$M : c + d = 1$
$L : a + b - 3 c - d = 1$
$T : - a - d = - 2 \Rightarrow a + d = 2$

Why this step? Same recipe as Cell C — but count the unknowns.

Step 2 — count. Four unknowns $a, b, c, d$ ; only three equations. One free parameter remains — say we leave $d$ undetermined.

Step 3 — express in terms of $d$ . From the equations: $a = 2 - d$ , $c = 1 - d$ , and $b = 1 - a + 3 c + d = 1 - (2 - d) + 3 (1 - d) + d = 2 - d$ . $F = k v^{2 - d} r^{2 - d} ρ^{1 - d} η^{d} = k (ρ v^{2} r^{2}) (\frac{η}{ρ v r})^{d} .$

Verify / lesson: the leftover group $\frac{η}{ρ v r}$ is dimensionless (it is $1/ Reynolds number$ ). Dimensional analysis cannot fix $d$ — that needs experiment or the Buckingham Pi theorem, which formalises exactly this "one free dimensionless group per extra variable." This is the boundary the parent note warned about.

Cell E — a power comes out zero (a variable disappears)

Cell F — dimensionless argument of a function

Worked example A pulse travels as

$y = A sin (k x - ω t)$ . Find the dimensions of $k$ and $ω$ , and confirm the argument is pure. Forecast: $x$ is a length and $t$ is a time. What must $k$ and $ω$ carry so that $k x$ and $ω t$ are pure numbers? Guess.

Step 1 — the governing rule. The argument of $sin$ must be dimensionless (parent corollary). So each piece inside must be $[M^{0} L^{0} T^{0}]$ : $[k x] = 1 and [ω t] = 1.$ Why this step? $sin$ is a sum of powers of its argument; you cannot add a length to its cube — the argument has to be a pure number.

Step 2 — solve for $k$ . $[k] [L] = 1 \Rightarrow [k] = L^{- 1}$ (per metre — a wave number).

Step 3 — solve for $ω$ . $[ω] [T] = 1 \Rightarrow [ω] = T^{- 1}$ (per second — an angular frequency).

Step 4 — dimension of the amplitude $A$ . Since $sin (\cdot)$ is dimensionless and $y$ is a displacement, $[A] = [y] = L$ .

Verify: $[k x - ω t] = (L^{- 1}) (L) - (T^{- 1}) (T) = 1 - 1$ , both pure numbers — subtraction is legal (same kind). ✅ Notice homogeneity even inside the bracket: both terms must be dimensionless, not just their difference.

Cell G — unit conversion where each power steers the factor

Worked example Express

$1 joule$ in ergs (CGS energy unit, $g\cdotcm^{2} \cdots^{- 2}$ ). Forecast: energy is $[M L^{2} T^{- 2}]$ . Mass scales by $1 0^{3}$ , length by $1 0^{2}$ — but length is squared. What total power of ten? Guess before Step 3.

Step 1 — dimension drives the recipe. $[E] = M^{1} L^{2} T^{- 2}$ . Each base unit is replaced by (new-per-old factor) raised to its power in the dimension. $1 J = (1 kg)^{1} (1 m)^{2} (1 s)^{- 2} .$ Why this step? The exponents $1, 2, - 2$ are copied straight from $[E]$ ; that is the whole point of tracking dimensions.

Step 2 — substitute the conversion factors. $1 kg = 1 0^{3} g, 1 m = 1 0^{2} cm, 1 s = 1 s .$ $1 J = (1 0^{3} g)^{1} (1 0^{2} cm)^{2} (1 s)^{- 2} .$

Step 3 — combine the powers of ten. $1 0^{3} \times 1 0^{2 \cdot 2} = 1 0^{3} \times 1 0^{4} = 1 0^{7} .$ $1 J = 1 0^{7} g\cdotcm^{2} \cdots^{- 2} = 1 0^{7} erg .$

Verify: the exponent of length being $2$ is what turned $1 0^{2}$ into $1 0^{4}$ — exactly the trap the forecast flagged. If you had forgotten the square you would get $1 0^{5}$ , wrong by $100 \times$ . See Units and the SI system for the general scaling rule.

Cell H — a real-world word problem

Worked example A raindrop of mass

$m$ falling through air reaches a terminal speed where drag balances gravity. The drag force has magnitude $F = c v^{2}$ for some constant $c$ with $[c] = M L^{- 1}$ . Estimate how terminal speed $v_{t}$ scales with $m$ and $g$ . Forecast: heavier drops fall faster or slower? Stronger gravity — faster or slower? Guess both trends.

Step 1 — translate the physics into a balance. At terminal speed the net force is zero, so weight = drag: $m g = c v_{t}^{2} .$ Why this step? "Terminal" means no more acceleration — the two forces are equal in size, so their dimensions must already match ( $[m g] = M L T^{- 2}$ , $[c v^{2}] = (M L^{- 1}) (L T^{- 1})^{2} = M L T^{- 2}$ ✅ — homogeneity confirms the setup before we solve).

Step 2 — solve for $v_{t}$ . $v_{t} = \frac{m g}{c} .$

Step 3 — read the scaling. $v_{t} \propto m$ and $v_{t} \propto g$ . Bigger mass → faster; stronger gravity → faster. Compare with your forecast.

Verify: $[m g / c] = \frac{M \cdot L T ^{- 2}}{M L ^{- 1}} = L^{2} T^{- 2} = L T^{- 1} = [v]$ ✅. See Equations of motion (kinematics) for what happens before terminal speed is reached.

Cell I — exam twist: find the dimension of an unknown constant

Worked example In the van der Waals equation for a real gas, one term reads

$\frac{a}{V ^{2}}$ and it is added to a pressure $P$ . Given $V$ is a volume, find $[a]$ . Forecast: $a$ divided by volume-squared must be a pressure. So $a$ carries pressure × volume². What dimension is that? Guess.

Step 1 — use homogeneity on the sum. Since $\frac{a}{V ^{2}}$ is added to $P$ , it must share $P$ 's dimension: $[\frac{a}{V ^{2}}] = [P] = M L^{- 1} T^{- 2} .$ Why this step? You may only add like to like — that single rule pins the whole answer.

Step 2 — isolate $[a]$ . Multiply both sides by $[V^{2}] = (L^{3})^{2} = L^{6}$ : $[a] = [P] [V^{2}] = (M L^{- 1} T^{- 2}) (L^{6}) = M L^{5} T^{- 2} .$

Verify: put it back — $\frac{[ a ]}{[ V ^{2} ]} = \frac{M L ^{5} T ^{- 2}}{L ^{6}} = M L^{- 1} T^{- 2} = [P]$ ✅. The trick for every "find the dimension of the constant" twist: whatever the constant sits inside must combine to the dimension of the term it lives beside.

Figures — the geometry of two derivations

The wave-on-a-string result (Cell C) and the exponent-matching machine (Cell C/D) are worth seeing as pictures.

Figure — Dimensional analysis — checking equations, deriving relations

The figure above shows the three base "axes" ( $M$ , $L$ , $T$ ) as bookkeeping columns: derivation means balancing each column separately.

The second figure plots how terminal speed (Cell H) grows as $m$ — a curve that starts steep and flattens, the visual signature of a square root.

Recap — one line per cell

Recall Which move solves which cell?

Check that passes (A) ::: compute both sides, conclude "not disproved" only. Check that fails (B) ::: a single mismatched exponent proves the equation wrong. Solvable derivation (C) ::: 3 base equations, 3 unknown powers → unique law up to constant $k$ . Under-determined (D) ::: unknowns > 3 leaves a dimensionless group free; needs experiment / Buckingham Pi. Zero exponent (E) ::: a base with no carrier forces that power to $0$ ; the variable drops out. Dimensionless argument (F) ::: inside $sin / ln / e^{x}$ every piece must be a pure number, fixing $[k], [ω]$ . Unit conversion (G) ::: scale each base factor raised to its exponent in the dimension. Word problem (H) ::: translate the physics to a force/energy balance, then match powers. Unknown constant (I) ::: the constant's dimension is whatever makes its term match its neighbour.