Dimensional formulae

Strip a physical quantity of its numerical value and its unit label, and what remains is its dimensional formula — a compact statement of exactly how it depends on mass, length and time. Force is always $[MLT^{-2}]$ whether you measure it in newtons, dynes or pounds-force; the formula is a property of the quantity itself, not of the system of units chosen to report it. This single idea — that every mechanical quantity can be written as $[M^aL^bT^c]$ — turns out to be one of the most useful bookkeeping tools in physics. It lets you check whether an equation could possibly be correct before you trust a single number it produces, and it lets you build a formula for an unfamiliar quantity from first principles, just by tracking how the powers of $M$ , $L$ and $T$ combine. This section builds that skill from the ground up, using the same worked numbers as your first pass through the topic.

The core insight

Every mechanical quantity's dependence on mass, length and time can be written as $[M^aL^bT^c]$ , derived directly from its defining equation, and the principle of homogeneity — that every term in a valid physical equation must share the same dimensional formula — is what makes this bookkeeping useful for checking and deriving relations.

The core idea

Why three base quantities are enough for mechanics. Every mechanical quantity — velocity, force, energy, pressure — is built out of mass, length and time combined by multiplication, division and integer powers. No mechanical formula ever needs a fourth ingredient, so tracking just three exponents captures everything.

The general form. For a quantity $Q$ , the dimensional formula is written

[Q] = [M^{a}L^{b}T^{c}]

where: $a$ , $b$ , $c$ are the powers (usually small integers) to which mass, length and time are raised, and the square brackets mean "the dimensional formula of," not a numerical value — $[Q]$ throws away every number and unit choice and keeps only the exponent pattern.

Reading it off the defining equation. You never guess a dimensional formula; you derive it from how the quantity is defined. Velocity is displacement over time, so $[v] = [L]/[T] = [LT^{-1}]$ (mass does not appear, so its power is $0$ : $[M^0LT^{-1}]$ ). Acceleration is velocity over time, giving $[a] = [LT^{-2}]$ . Force is mass times acceleration, $[F] = [M][LT^{-2}] = [MLT^{-2}]$ — the single most-used dimensional formula in the whole subject. From force you can chain onward: work is force times displacement, $[W] = [MLT^{-2}][L] = [ML^2T^{-2}]$ , and pressure is force over area, $[P] = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$ .

Dimensionless quantities. Angles, strain, refractive index and the coefficient of friction are pure ratios of two quantities with the same dimensions, so everything cancels and their dimensional formula is $[M^0L^0T^0]$ . They carry no unit dependence on $M$ , $L$ , $T$ at all — this is a common trap, covered below.

The principle of homogeneity. In any dimensionally correct equation, every term — on either side, and every term that is added or subtracted — must carry the identical dimensional formula. You can equate or combine two lengths, but you can never meaningfully add a length to a time; the equation would be dimensionally nonsensical even if someone wrote it down. This one principle underlies every later use of dimensional analysis: checking a derived formula, building an unfamiliar relation from scratch, and converting a measurement from one system of units to another.

Figure

A balance scale with mass length and time blocks stacked on the left pan and the single symbol F equals M L T to the power minus two on the right pan, showing that a physical quantity dimensional formula is just a bookkeeping combination of the three base quantities

Derivation ADimensional formula of kinetic energy

The goal is to derive the dimensional formula of kinetic energy, $K = \tfrac12 mv^2$ , from its defining equation alone — showing, along the way, that a purely numerical factor such as $\tfrac12$ never contributes to a dimensional formula, and that the result must match the dimensional formula of work, since energy is a single physical kind of quantity regardless of the form it takes.

Step 1. Identify which parts of $K = \tfrac12 mv^2$ carry dimensions and which do not. The factor $\tfrac12$ is a pure number — it has no mass, length or time dependence, so it contributes $[M^0L^0T^0]$ and can simply be dropped when writing the dimensional formula.

Step 2. Write the dimensional formulae of the two physical factors that remain: mass and velocity. Mass is a base quantity, $[m] = [M]$ . Velocity is displacement over time, so $[v] = [LT^{-1}]$ .

Step 3. Combine according to the structure of the defining equation, $K \propto m v^2$ , remembering that squaring a quantity squares its dimensional formula too:

[K] = [M]\,[LT^{-1}]^2 = [M][L^2T^{-2}] = [ML^2T^{-2}]

Step 4. Check the result against a known formula. Work is force times displacement, $[W] = [MLT^{-2}][L] = [ML^2T^{-2}]$ — identical to $[K]$ . This is not a coincidence: kinetic energy, potential energy and work are all the same physical kind of quantity, so the principle of homogeneity requires them to share one dimensional formula.

Worked example

Problem. Newton's law of gravitation states $F = \dfrac{Gm_1m_2}{r^2}$ . Find the dimensional formula of the universal gravitational constant $G$ .

Step 1. Rearrange the defining equation to isolate $G$ : $G = \dfrac{Fr^2}{m_1m_2}$ .

Step 2. Write the known dimensional formulae of every quantity on the right: $[F] = [MLT^{-2}]$ , $[r^2] = [L^2]$ , and $[m_1] = [m_2] = [M]$ .

Step 3. Substitute and combine exponents by treating $[M]$ , $[L]$ , $[T]$ as algebraic symbols:

[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]

So $G$ has dimensional formula $\boxed{[M^{-1}L^3T^{-2}]}$ , consistent with its SI unit $\text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ — the negative power of $M$ is a useful sanity check, since $G$ appears in the denominator alongside two masses but the force and distance-squared in the numerator dominate.

Competitive stretch

The same method extends to angular momentum, $L_{\text{ang}} = mvr$ : with $[m]=[M]$ , $[v]=[LT^{-1}]$ , $[r]=[L]$ , you get $[L_{\text{ang}}] = [M][LT^{-1}][L] = [ML^2T^{-1}]$ — one power of $[L]$ higher than momentum's $[MLT^{-1}]$ , exactly what you'd expect from multiplying momentum by a radius.

Common traps

Assigning a dimensional formula to angles, strain, or refractive index — dimensionless quantities are $[M^0L^0T^0]$ , not "no formula at all."
Slipping up on exponent arithmetic when combining formulae — add and subtract powers of $[M]$ , $[L]$ , $[T]$ separately and carefully, the way you would with any algebraic exponents.
Assuming that two quantities sharing a dimensional formula must be the same physical quantity — work and torque are both $[ML^2T^{-2}]$ , yet one is a scalar and the other a vector.
Letting a numerical constant like $\tfrac12$ , $\pi$ , or $4\pi$ leak into a dimensional formula — such constants are always dimensionless and never appear in $[M^aL^bT^c]$ .

Ask the tutor: want to practise deriving the dimensional formula of torque, impulse, or surface tension step by step? Ask the AI tutor to work through a few more with you.

If work and torque have the same dimensional formula, are they the same quantity?

No. Sharing a dimensional formula only means the two quantities are built from mass, length and time in the same combination — it says nothing about the physical nature of the quantity. Work, $[ML^2T^{-2}]$ , is a scalar produced by force acting through a displacement in the direction of the force. Torque, also $[ML^2T^{-2}]$ , is a vector produced by a force acting at a perpendicular distance from a pivot. Dimensional formulae group quantities by "how big" their unit dependence is, not by what they physically represent, so always check the defining equation, not just the exponents, before treating two quantities as identical.

Why does the numerical factor in a formula, like the half in kinetic energy, never show up in the dimensional formula?

Because a pure number carries no dependence on mass, length or time at all — its dimensional formula is $[M^0L^0T^0]$ , which multiplies anything else by $1$ and leaves it unchanged. This is true of $\tfrac12$ , $\pi$ , $4\pi$ , or any other constant that arises from geometry or algebra rather than from a physical measurement. It is also exactly why dimensional analysis alone can never pin down such constants — it can tell you that kinetic energy must have the form $mv^2$ times some dimensionless number, but it cannot tell you that number is $\tfrac12$ . That has to come from the actual derivation of the formula.

How do I find the dimensional formula of a quantity I have never seen before?

Start from its defining equation — the relation that tells you how the quantity is calculated from other, already-known quantities. Write the dimensional formula of every quantity on the right-hand side, then combine them exactly as the equation combines the quantities themselves: multiply dimensional formulae for products, divide for quotients, and raise to a power for an exponent. Add and subtract the powers of $M$ , $L$ , $T$ separately, the way you would with algebraic exponents. This is exactly the method used above for $G$ and for angular momentum, and it works for any mechanical quantity as long as you know one equation that defines it.

Dimensional formulae

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads