Deriving relations using dimensions

You already know that dimensions can check a formula — plug in $[M]$ , $[L]$ , $[T]$ on both sides and see if they match. The far more powerful trick is to run that check backwards: if you don't yet know the formula, but you have a good guess about which quantities it depends on, dimensional analysis can hand you the shape of the relation before you've done a single experiment. This is the second prescribed use of dimensional analysis in ISC Class 11 Physical World and Measurement, sitting right after the principle of homogeneity. It is a genuine problem-solving tool — physicists use exactly this trick to sanity-check new formulas in unfamiliar situations — but it comes with sharp limits that are just as much a part of the syllabus as the method itself. This article rebuilds the method as a systematic derivation, re-solves the pendulum and Stokes'-law examples step by step, and drills the traps that examiners love to test.

The core insight

If a quantity $Q$ is assumed to depend as a pure power product on quantities $A,B,C$ , then equating $Q$ 's dimensional formula to $A^aB^bC^c$ 's and comparing powers of $M$ , $L$ , $T$ gives up to three equations that pin down the exponents — but never the dimensionless constant.

The core idea

The method has three moving parts: a guess, a comparison, and a solve.

The guess. You assume, on physical grounds, that the target quantity $Q$ depends only on quantities $A$ , $B$ , $C$ , combined as a pure product of powers with an unknown dimensionless constant $k$ :

Q = k\,A^{a}B^{b}C^{c}

where: $Q$ is the quantity you want a relation for, $A$ , $B$ , $C$ are the quantities it is assumed to depend on (each with a known dimensional formula), $a$ , $b$ , $c$ are unknown real-number exponents to be found, and $k$ is a dimensionless constant that dimensional analysis can never supply.

The comparison. Because both sides of a physically meaningful equation must have identical dimensions (the principle of homogeneity), you replace every symbol by its dimensional formula $[M^{x}L^{y}T^{z}]$ and multiply out the right-hand side. This turns the algebraic guess into a purely dimensional equation.

The solve. Matching the powers of $M$ , $L$ , and $T$ independently on both sides gives (at most) three simultaneous linear equations in the unknowns $a$ , $b$ , $c$ . Solving them gives the exponents — and therefore the form of the relation, up to the missing constant $k$ . The diagram below lays out this pipeline as one flow, which is worth memorising as a fixed procedure rather than reinventing it each time.

Figure

A three step flow from guessed power law to dimensional equation to three exponent equations solved for a comma b comma c

The pipeline also exposes exactly why the method has hard limits. Only three independent equations ever exist — one each from $M$ , $L$ , $T$ — so it works cleanly only when there are exactly three unknown exponents. It only recognises product-power relationships, so it is blind to sums, differences, and the arguments of trigonometric, exponential, or logarithmic functions, since none of those change the dimensional formula in a way the method can decode. And because $k$ never appears in the dimensional equation at all, no amount of algebra will ever recover it — that number has to come from experiment or a full mechanical derivation.

Derivation ASetting up and solving a dimensional equation

Goal: given that the time period $T$ of a simple pendulum is assumed to depend only on its length $L$ , the acceleration due to gravity $g$ , and the bob's mass $m$ , use dimensional analysis to find how $T$ depends on $L$ , $g$ , and $m$ , recovering the exponents in $T = k\,L^{a}g^{b}m^{c}$ .

Step 1. Write down the assumed relation with unknown exponents: $T = k\,L^{a}g^{b}m^{c}$ , where $k$ is dimensionless.

Step 2. List the dimensional formula of every quantity involved:

[T]=[M^0L^0T^1],\quad [L]=[M^0L^1T^0],\quad [g]=[M^0L^1T^{-2}],\quad [m]=[M^1L^0T^0]

Step 3. Substitute these into the assumed relation to build the dimensional equation, then multiply out the right-hand side by adding exponents:

[M^0L^0T^1] = [M^0L^1T^0]^{a}[M^0L^1T^{-2}]^{b}[M^1L^0T^0]^{c} = [M^{c}L^{a+b}T^{-2b}]

Step 4. Equate the powers of $M$ , $L$ , $T$ separately on both sides, since the two sides can only be equal if every power matches:

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

Step 5. Solve the three equations in order. The $T$ equation gives $b=-\tfrac12$ directly. Substituting into the $L$ equation gives $a=-b=\tfrac12$ . The $M$ equation gives $c=0$ immediately, telling us $T$ does not depend on the bob's mass at all.

Step 6. Substitute the exponents back into the assumed relation and simplify:

T = k\,L^{1/2}g^{-1/2}m^{0} = k\sqrt{\dfrac{L}{g}}

This is as far as dimensional analysis can take you — it correctly predicts the mass-independence and the $\sqrt{L/g}$ shape, but it cannot tell you that $k=2\pi$ ; that fact only comes out of the full SHM derivation covered later in Oscillations and Waves.

Worked example

Problem. The viscous drag force $F$ on a small sphere moving through a fluid is assumed to depend on the coefficient of viscosity $\eta$ (dimensional formula $[ML^{-1}T^{-1}]$ ), the sphere's radius $r$ , and its speed $v$ , as $F = k\,\eta^{a}r^{b}v^{c}$ . Find $a$ , $b$ , $c$ , and identify the resulting law.

Step 1. Write the dimensional formulae of all four quantities: $[F]=[MLT^{-2}]$ , $[\eta]=[ML^{-1}T^{-1}]$ , $[r]=[L]$ , $[v]=[LT^{-1}]$ .

Step 2. Build the dimensional equation by substituting and multiplying out the right side:

[MLT^{-2}] = [ML^{-1}T^{-1}]^{a}[L]^{b}[LT^{-1}]^{c} = [M^{a}L^{-a+b+c}T^{-a-c}]

Step 3. Match powers of $M$ , $L$ , $T$ :

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

Step 4. Solve in sequence. From $M$ : $a=1$ . Substituting into the $T$ equation, $-2=-1-c$ , so $c=1$ . Substituting both into the $L$ equation, $1=-1+b+1$ , so $b=1$ .

Step 5. Substitute back to get the relation:

F = k\,\eta\,r\,v

This matches the structure of Stokes' law, $F = 6\pi\eta rv$ — dimensional analysis correctly finds that drag scales linearly with viscosity, radius, and speed, but the numerical constant $k = 6\pi$ is supplied only by the full fluid-dynamics derivation, never by this method.

Competitive stretch

A fast sanity check that avoids resolving simultaneous equations from scratch: count your unknown exponents against your available equations. Both examples above assumed exactly three independent variables, giving exactly three equations from $M$ , $L$ , $T$ — a clean, unique solve. If a problem instead gives you four assumed variables, you have three equations for four unknowns, and the exponents cannot be pinned down uniquely unless you are told (or can argue) that one exponent has a specific value, such as a known dependence being linear.

Common traps

Assuming $k=1$ once the exponents are found — dimensional analysis never determines the dimensionless constant, so it must always be left as an unknown $k$ or supplied from experiment/full derivation.
Listing an incomplete or wrong set of dependent variables — the whole method silently assumes the guessed list is exactly right; a missing or extra variable produces exponents that balance the dimensions but do not describe real physics.
Trying to force the method onto a relation that is a sum or difference of terms (like $s = ut + \tfrac12at^2$ ) — dimensional analysis only handles pure products of powers, not additive combinations.
Skipping the final check — always re-substitute the solved exponents into the dimensional equation from Step 3/Step 2 and confirm $M$ , $L$ , $T$ balance on both sides before trusting the answer.

Ask the tutor: want to try deriving the exponents for another relation yourself, like the frequency of a stretched string or the drag force on a falling raindrop?

If dimensional analysis cannot find k, why bother using it at all?

Because the exponents are usually the hard part. Knowing that $T \propto \sqrt{L/g}$ or that $F \propto \eta r v$ tells you how the quantity scales — double the length and the period grows by $\sqrt2$ , for instance — which is exactly the kind of prediction you can check against data or use in further reasoning, well before you have done the full mechanical derivation needed for the constant.

How do I know which variables to include when I set up the guess?

There is no formula for this — it comes from physical reasoning about what the quantity could plausibly depend on, informed by the situation described. A pendulum's period could plausibly depend on length, gravity, and mass, so all three go into the guess; something clearly irrelevant, like the colour of the bob, is left out. If your guess is wrong, the exponents you compute may balance dimensionally yet still not match reality, which is exactly why the method only gives a plausible relation.

What do I do if I have four unknown variables but only three equations?

You cannot solve uniquely with dimensional analysis alone. Either you drop or justify removing one variable on physical grounds, or you need extra information from outside dimensional analysis, such as a known exponent value or an experimental result, to close the gap. The syllabus explicitly flags "more than three independent variables" as one of the situations where this method fails.

Deriving relations using dimensions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads