Dimensional analysis — checking equations, deriving relations: Visual walkthrough

Step 1 — Meet the pendulum and name every part

WHAT. A pendulum is a small heavy ball (the bob) hanging from a string, pulled sideways and let go so it swings. We give names to the three things it is made of:

$ℓ$ (the letter "ell") — the length of the string, measured in metres. Its dimension is $[L]$ .
$m$ — the mass of the bob (how much stuff it contains), in kilograms. Its dimension is $[M]$ .
$g$ — the strength of gravity: how fast falling things speed up. Its dimension is $[L T^{- 2}]$ (a length gained per second, every second).

The thing we want is $T$ — the period, the time for one full swing over and back. Its dimension is $[T]$ (a time).

WHY these three ingredients? Because those are the only knobs on the apparatus a person could turn. You can change the string length, swap the ball, or do the experiment on the Moon (different $g$ ). Everything else (colour, string thickness) plausibly has nothing to do with timing, so we bet the period is built from $ℓ$ , $m$ , $g$ alone.

PICTURE. The swing and its labelled parts.

Figure — Dimensional analysis — checking equations, deriving relations

Step 2 — Make a guess with unknown powers

WHAT. We guess that the period is a product of these ingredients, each raised to some unknown power:

$T = k \cdot [L]^{a} ℓ^{a} \cdot [M]^{b} m^{b} \cdot [L T^{- 2}]^{c} g^{c}$

Reading the symbols left to right:

$T$ — the period we are hunting for.
$k$ — a plain number with no dimension (like $2$ or $π$ ). It carries no units, so homogeneity can never see it. We accept from the start that this method cannot find $k$ .
$a, b, c$ — unknown powers we must solve for. If $a = \frac{1}{2}$ that means $ℓ$ ; if $b = 0$ that means mass drops out entirely.

WHY a product of powers, and not a sum? Because a sum like " $ℓ + g$ " would force us to add a length to an acceleration — different dimensions, forbidden by homogeneity. A product never mixes kinds; it just multiplies them. So the only dimensionally safe shape for a one-term guess is a product of powers. That is exactly why the derivation trick works on single-term laws and fails on sums (see the parent's limitations).

PICTURE. The guess drawn as a machine: three dials $a, b, c$ we can turn until the output kind matches $[T]$ .

Step 3 — Turn the guess into pure dimensions

WHAT. We now replace every quantity by its kind — its dimension — because homogeneity is a statement about kinds, not about numbers. Substituting $[T] = T$ , $[ℓ] = L$ , $[m] = M$ , $[g] = L T^{- 2}$ :

$T^{1} = L^{a} M^{b} (L T^{- 2})^{c}$

WHY drop $k$ ? Because $k$ is dimensionless — its "kind" is $M^{0} L^{0} T^{0}$ , an invisible factor of $1$ in a dimension equation. It contributes nothing here (and is exactly why it stays hidden forever).

Next we expand the bracket. The power $c$ hits both letters inside $(L T^{- 2})^{c}$ :

$(L T^{- 2})^{c} = L^{c} T^{- 2 c}$

so the whole right side becomes

$T^{1} = M^{b} L^{a + c} T^{- 2 c}$

Term by term on the right:

$M^{b}$ — all the mass, gathered onto one $M$ .
$L^{a + c}$ — length comes from two sources: the string ( $a$ ) and the length hidden inside gravity ( $c$ ). They add.
$T^{- 2 c}$ — the only time on the right comes from gravity's "per second squared", so its power is $- 2 c$ .

PICTURE. Three separate bins — an $M$ bin, an $L$ bin, a $T$ bin — with the powers dropped into each.

Step 4 — Match each base dimension separately

WHAT. The left side is $M^{0} L^{0} T^{1}$ (period is pure time: no mass, no length, one time). The right side is $M^{b} L^{a + c} T^{- 2 c}$ . For two products of the same three letters to be equal, the powers of each letter must match one-to-one:

M : L : T : b = 0 a + c = 0 - 2 c = 1

Reading each line as a sentence:

$b = 0$ — the mass power on the right ( $b$ ) must equal the mass power on the left ( $0$ ).
$a + c = 0$ — the total length power on the right must equal the left's length power ( $0$ ).
$- 2 c = 1$ — the time power on the right must equal the left's ( $1$ ).

WHY this is the whole engine. One equation fell out for free per base dimension. We started with three unknowns $a, b, c$ and homogeneity handed us exactly three equations — a perfectly determined system. (If we'd guessed four ingredients we'd still get only three equations: underdetermined, and the trick would stall. That is limitation #2 in the parent.)

PICTURE. The left–right "balance scale", one pan per base dimension, showing which powers must equal which.

Step 5 — Solve the three tiny equations

WHAT. Solve them in the easy order:

$b = 0$ — done immediately.
$- 2 c = 1 \Rightarrow c = - \frac{1}{2}$ — divide both sides by $- 2$ .
$a + c = 0 \Rightarrow a = - c = + \frac{1}{2}$ — move $c$ across.

So $a = \frac{1}{2}$ , $b = 0$ , $c = - \frac{1}{2}$ .

WHY $b = 0$ is the headline. Nothing on the right except $m$ carried any mass, and the left carried none — so the only way to balance the $M$ -axis is to use zero mass. The mathematics forces the pendulum period to ignore the bob's weight. Drop a bowling ball or a marble on the same string: same swing time.

WHY $a$ and $c$ come out opposite. The length axis had to cancel to zero, and the time axis fixed $c = - \frac{1}{2}$ ; length conservation ( $a + c = 0$ ) then mirrors it into $a = + \frac{1}{2}$ . Longer string → bigger $ℓ$ → slower swing; stronger gravity → bigger $g$ in the denominator → faster swing.

PICTURE. The three dials from Step 2 now locked to their solved values, with a swing sketch showing "longer = slower, stronger gravity = faster."

Step 6 — Assemble the law

WHAT. Put the solved powers back into the guess $T = k ℓ^{a} m^{b} g^{c}$ :

$T = k ℓ^{1/2} m^{0} g^{- 1/2}$

Now simplify each factor:

$ℓ^{1/2} = ℓ$ — square-root of length in the numerator.
$m^{0} = 1$ — mass vanishes, as promised.
$g^{- 1/2} = 1/ g$ — square-root of gravity in the denominator.

$T = k \frac{ℓ}{g}$

WHY we can't get $k$ . The boxed law is correct in form, but $k$ is a bare number invisible to dimensions. Full physics (or one experiment) reveals $k = 2 π$ , giving the textbook $T = 2 π ℓ / g$ . Dimensional analysis got us 90% of the way with zero physics.

PICTURE. The final formula with each piece annotated and its effect on the swing.

Step 7 — The degenerate and limiting cases (never skip these)

WHAT. Test the formula at its extremes to be sure it never breaks:

$ℓ \to 0$ (no string): $T = k 0/ g = 0$ . A pendulum with no length has no swing — instant. ✅ Sensible.
$g \to 0$ (deep space, weightless): $ℓ / g \to \infty$ , so $T \to \infty$ . With no gravity there is no restoring pull, so it never comes back — an infinite period. ✅ Exactly right, and the $g$ in the denominator is what produces this blow-up.
$g$ large (a heavy planet): bigger denominator ⇒ smaller $T$ ⇒ faster swing. ✅
$m$ anything: $m^{0} = 1$ regardless, so a feather-bob and a lead-bob on equal strings tie. ✅ This is the mass-independence, re-confirmed.

WHY show these. A formula must survive its edge cases or it is not trustworthy. Each limit here matches physical common sense and points back to which exponent ( $a = + \frac{1}{2}$ vs $c = - \frac{1}{2}$ ) is responsible — length in the top, gravity in the bottom.

PICTURE. Two mini-panels: the $ℓ \to 0$ collapse (period shrinks to zero) and the $g \to 0$ runaway (period shoots to infinity).

The one-picture summary

The whole journey on one board: guess → dimensions → three-axis matching → solve → assemble → check limits.

Recall Feynman: the walkthrough in plain words

I wanted to know how long a swing takes to go over and back. I looked at the swing and saw only three dials I could turn: the string's length, the ball's weight, and how hard gravity pulls. So I guessed the answer is those three multiplied together, each raised to some mystery power — a product, never a sum, because you're not allowed to add a length to a gravity. Then I stripped away all the numbers and kept only the kinds: length-kind, weight-kind, time-kind. The rule "you can only equate matching kinds" split into three little demands — one for weight, one for length, one for time. The weight demand said "use zero weight," so the ball's mass simply doesn't matter — heavy or light, same swing. The time demand pinned the gravity power to minus-a-half, and length had to mirror it to plus-a-half. Snapping the pieces back together gave: swing time grows like the square-root of length and shrinks like the square-root of gravity. The only thing this trick can't tell me is a plain multiplying number out front — that turned out to be two-pi, which you learn from an experiment, not from bookkeeping.

Connections

Dimensional analysis — checking equations, deriving relations — the parent; this page is its central result drawn out in full.
Buckingham Pi theorem — the general theory behind "count the base dimensions to count your equations."
Equations of motion (kinematics) — where the same homogeneity check catches wrong formulas.
Units and the SI system — the units that put actual numbers on $ℓ$ , $g$ , $T$ .
Vectors — components and addition — another place where "add only like kinds" quietly governs the algebra.