Dimensional Formula of All Physical Quantities

What the dimensions of a quantity really are

A unit is the yardstick you measure with. A dimension is the kind of thing being measured. Speed can be given in m/s, km/h or cm/s, but in every case it is a length divided by a time, so its dimensions are [L T⁻¹]. Writing the zero power of mass as well, [M⁰ L¹ T⁻¹], makes it clear that mass plays no part.

Square brackets mean "the dimensions of". So [v] = [M⁰ L¹ T⁻¹] is read as "the dimensions of velocity are length to the first power and time to the minus first". The numbers in the exponents are the only thing that matters. Numbers in front of a formula, such as the ½ in kinetic energy or the $4\pi$ in Coulomb's law, have no dimensions and simply drop out.

Two quantities can only be added, subtracted or set equal if their dimensions match. That single rule, the principle of homogeneity, powers almost every use of dimensional analysis on this page.

The seven SI base quantities and their symbols

Mechanics needs only the first three. Electricity and magnetism add I, heat adds K, and chemistry style questions on gases add N. Luminous intensity almost never appears in school dimensional problems. Everything else is a derived quantity: speed, force, energy, charge, resistance and so on are all built from these seven.

How to work out a dimensional formula in three steps

1. Write a defining formula that contains only quantities whose dimensions you already know.
2. Replace each symbol with its dimensions and drop every pure number.
3. Collect the powers of M, L and T (and I or K if needed).

Table of the dimensional formula of all physical quantities in the syllabus

The three tables group the quantities by chapter. Each linked name opens a page that derives the result line by line.

Motion, force and energy

Rotation, matter, heat and constants

Electricity and magnetism

A quick check on the electrical rows: $\mu_0 \varepsilon_0 = \frac{1}{c^2}$, so multiplying their dimensions must give [L⁻² T²]. It does: [M¹ L¹ T⁻² I⁻²] × [M⁻¹ L⁻³ T⁴ I²] = [M⁰ L⁻² T² I⁰].

Different quantities that share the same dimensions

Examiners love this table. A common one mark question asks which pair has the same dimensions, and the answer is often work and torque or angular momentum and Planck's constant. The flip side matters too: matching dimensions do not make two quantities the same thing. Torque is never measured in joules even though J and N m reduce to the same base units, because torque is a turning effect and work is an energy transfer.

Three uses of dimensional analysis

1. Checking whether an equation can be right

Take $v^2 = u^2 + 2as$. The left side is [L² T⁻²]. On the right, $u^2$ is [L² T⁻²], and $as$ is [L T⁻²][L] = [L² T⁻²]. Every term matches, so the equation is dimensionally consistent. If someone wrote $v^2 = u^2 + 2at$, the last term would be [L T⁻¹], and the equation must be wrong.

2. Converting a unit from one system to another

To change 1 newton into dynes (the CGS unit), use $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$ with a = 1, b = 1, c = −2.

$$n_2 = 1 \times \left(\frac{1000\ \text{g}}{1\ \text{g}}\right)^1 \left(\frac{100\ \text{cm}}{1\ \text{cm}}\right)^1 (1)^{-2} = 10^5$$

So 1 N = 10⁵ dyne. The same method gives 1 J = 10⁷ erg.

3. Guessing how a quantity depends on others

Limits of dimensional analysis and common slips

• No numbers. It cannot give dimensionless constants such as ½, 2π or 4π, as the pendulum showed.
• No sums. It cannot build a formula with two added terms, so it cannot derive $s = ut + \frac{1}{2}at^2$ from scratch.
• No trig, log or exponential relations. Arguments of sin, log and exp must be dimensionless, so their form cannot be found this way.
• Only three unknowns with M, L and T. If a quantity depends on more than three others, there are more unknown powers than equations.
• No direction. It cannot tell a scalar from a vector, and it cannot separate work from torque.

Common slips in tests: writing [M L T⁻²] for pressure instead of [M L⁻¹ T⁻²], forgetting that an angle in radians is dimensionless, and giving $\varepsilon_0$ a positive power of M. When in doubt, go back to one defining formula and rebuild it in three lines.