Dimensional analysis

Behind every physics equation lies a hidden claim of the same rank: both sides describe the same kind of quantity. If one side has the dimensions of length ($L$) and the other has the dimensions of time ($T$), the equation is physically meaningless — even if the numbers happen to match. Dimensional analysis is the tool for extracting and tracking that claim.

The seven fundamental SI dimensions

SI chooses seven base units; every physical quantity is built out of those. For each base unit we write a dimension symbol in square brackets:

Dimension	Symbol	Corresponding SI unit
Length	$[L]$	m (meter)
Mass	$[M]$	kg (kilogram)
Time	$[T]$	s (second)
Electric current	$[I]$	A (ampere)
Thermodynamic temperature	$[\Theta]$	K (kelvin)
Amount of substance	$[N]$	mol (mole)
Luminous intensity	$[J]$	cd (candela)

Note: dimension is not the same as unit. Length can be measured in m, ft, or light-year — those are units; they all share the dimension $[L]$. Dimensional analysis is about dimensions, not any particular choice of unit.

Derived quantities

Any physical quantity can be written as a product of powers of the fundamental dimensions:

$$
[Q] = L^{a}\, M^{b}\, T^{c}\, I^{d}\, \Theta^{e}\, N^{f}\, J^{g}
$$

where $a, b, c, \ldots$ are rational (usually integer) exponents.

Quantity	Symbol	Formula	Dimension
Velocity	$v$	$\Delta x / \Delta t$	$L\, T^{-1}$
Acceleration	$a$	$\Delta v / \Delta t$	$L\, T^{-2}$
Force	$F$	$m\, a$	$M\, L\, T^{-2}$
Work / energy	$W, E$	$F \cdot d$	$M\, L^{2}\, T^{-2}$
Power	$P$	$E / t$	$M\, L^{2}\, T^{-3}$
Pressure	$p$	$F / A$	$M\, L^{-1}\, T^{-2}$
Density	$\rho$	$m / V$	$M\, L^{-3}$
Frequency	$f$	$1 / T$	$T^{-1}$
Electric charge	$q$	$I \cdot t$	$I\, T$
Electric field	$E_{\text{field}}$	$F / q$	$M\, L\, T^{-3}\, I^{-1}$

Dimensionless quantities ($[Q] = 1$): angle (radian), coefficient of friction, particle counts, efficiency, Reynolds number, Poisson’s ratio, any ratio of two like-dimensioned quantities.

The principle of dimensional homogeneity

The core rule:

In a valid physics equation, every additive term and both sides of the equality must share the same dimension.

If you write $\text{energy} = \text{force} + \text{time}$, you don’t need to compute anything — the equation is already wrong, because $[F] = MLT^{-2}$ and $[t] = T$ don’t share a dimension and can’t be summed (any more than 3 meter + 5 second can).

Mental exercise: consider the constant-acceleration kinematic equation:

$$
x(t) = x_0 + v_0\, t + \tfrac{1}{2} a\, t^{2}
$$

• $[x] = L$
• $[x_0] = L$ ✓
• $[v_0\, t] = (L T^{-1}) \cdot T = L$ ✓
• $[a\, t^{2}] = (L T^{-2}) \cdot T^{2} = L$ ✓

Every term carries dimension $L$ — the equation passes the sanity check. Note that this tool is not sufficient: it can’t confirm or reject the numerical coefficient (like 1/2). But it is necessary: any equation that fails this test is definitely wrong.

Practical uses

1. Catching human and machine errors

Suppose you’re working on a pressure-drop problem on paper and end up with:

$$
\Delta p = \rho\, g\, h + \tfrac{1}{2} \rho\, v
$$

• $[\rho\, g\, h] = (M L^{-3})(L T^{-2})(L) = M L^{-1} T^{-2}$ ✓ (pressure)
• $[\rho\, v] = (M L^{-3})(L T^{-1}) = M L^{-2} T^{-1}$ ✗

The second term isn’t a pressure — it was a transcription error. It should have been $\rho\, v^{2}$. Faster than checking the numerical answer.

2. Unit conversion by unit-factor cancellation

The dimension of a quantity can be written as a ratio of units equal to 1. For example:

$$
1 = \frac{1000\,\mathrm{m}}{1\,\mathrm{km}} = \frac{1\,\mathrm{h}}{3600\,\mathrm{s}}
$$

Chain-multiplying such ratios lets us algebraically cancel units:

$$
72\ \frac{\mathrm{km}}{\mathrm{h}} \;\times\; \frac{1000\,\mathrm{m}}{1\,\mathrm{km}} \;\times\; \frac{1\,\mathrm{h}}{3600\,\mathrm{s}} \;=\; 20\ \frac{\mathrm{m}}{\mathrm{s}}
$$

You don’t need to memorize conversion factors — tracking the units alone confirms the answer. Even for unusual conversions:

$$
1\ \frac{\mathrm{atm}\cdot\mathrm{L}}{\mathrm{mol}\cdot \mathrm{K}} \;\times\; \frac{101{,}325\,\mathrm{Pa}}{1\,\mathrm{atm}} \;\times\; \frac{10^{-3}\,\mathrm{m}^{3}}{1\,\mathrm{L}} \;\approx\; 101.3\ \frac{\mathrm{J}}{\mathrm{mol}\cdot \mathrm{K}}
$$

3. Deriving a formula up to a dimensionless constant

The classic example: the period of a simple pendulum. We guess that the period $T$ depends on four ingredients:

• The string length $\ell$, with $[\ell] = L$
• The bob mass $m$, with $[m] = M$
• Gravitational acceleration $g$, with $[g] = L T^{-2}$
• The initial angle $\theta$, dimensionless

Assume:

$$
T \;\sim\; \ell^{a}\, m^{b}\, g^{c}\, \theta^{d}
$$

Left dimension: $T$. Right dimension: $L^{a+c}\, M^{b}\, T^{-2c}$. Matching dimensions gives:

• $L$:   $a + c = 0$
• $M$:   $b = 0$
• $T$:   $-2c = 1 \;\Rightarrow\; c = -\tfrac{1}{2},\ a = \tfrac{1}{2}$

So:

$$
T \;\sim\; \sqrt{\frac{\ell}{g}} \cdot f(\theta)
$$

Without solving any differential equation, we learned:

1. The bob’s mass does not appear in the period (Galileo’s famous result).
2. The period grows as $\sqrt{\ell}$, not linearly with $\ell$.
3. The angle dependence collapses into a single dimensionless function $f(\theta)$ — which dimensional analysis can’t nail down. For small angles, $f(\theta) \to 2\pi$ (giving the full formula $T = 2\pi\sqrt{\ell/g}$).

That’s what dimensional analysis cannot do: it can’t produce the dimensionless coefficients ($2\pi$ above, $\tfrac{1}{2}$ in kinetic energy, $4/3$ in the volume of a sphere). Those require the full physical model or an experiment.

Traps and common mistakes

1. Arguments of transcendental functions must be dimensionless. $\sin$, $\cos$, $\exp$, $\ln$ only accept pure numbers. $\sin(x)$ with $x$ in meters is nonsense. Look at the Taylor series:

$$
\sin(x) = x – \tfrac{x^{3}}{6} + \tfrac{x^{5}}{120} – \cdots
$$

If $[x] = L$, the terms would be $L$, $L^3$, $L^5$ — and dimensional homogeneity would forbid summing them. So $x$ must be dimensionless. In physics you usually see forms like $\sin(\omega t)$ or $\exp(-t/\tau)$ where the argument ($\omega t$ or $t/\tau$) is a ratio of like-dimensioned quantities and is therefore pure.

2. Radians and degrees are both dimensionless, but confusing them is a bug. $\sin(30) \ne \sin(30°)$ — in most math libraries, radians are the default. The correct conversion: $30° = \pi/6 \approx 0.524$ rad.

3. Numerical constants sometimes hide dimensions. For example, G (Newton’s gravitational constant) is not dimensionless:

$$
[G] = M^{-1} L^{3} T^{-2}
$$

If you plug in only the number $6.674 \times 10^{-11}$ without carrying units, the final answer won’t come out in the right units. Always carry the units through, not just the numerical value.

4. Dimensional analysis doesn’t confirm a correct equation — it only rejects a wrong one. The following two equations are dimensionally identical but not equally right:

$$
E = m c^{2} \quad\text{and}\quad E = 42\, m c^{2}
$$

Both pass the dimension test. Which one is correct is a question for physics, not for dimensions.

What you should be able to do

After this section you should be able to:

• Write the dimensions of any classical derived quantity (velocity, acceleration, force, energy, power, pressure, charge, electric field) in terms of $L, M, T, I$
• Check any given equation dimensionally — and if it’s wrong, name the offending term
• Perform chained unit conversions (mile/h → m/s, atm·L → J) using unit factors
• For a simple physics problem, guess the formula up to a dimensionless constant using dimensional analysis
• Know which functions (trig, log, exp) require dimensionless arguments, and why

📚 See also: Halliday Vol 1, Ch 1 — worked examples of unit conversion and end-of-chapter problems solvable by dimensional analysis.
📖 Open reference: NIST SP811 — Guide for the Use of the International System of Units (Chapter 7 on dimensional homogeneity).
📖 Feynman Lectures on Physics — Vol I, Ch 4 (the physical meaning of dimensional consistency).
📖 BIPM — SI Brochure §2 (list of fundamental and derived quantities).