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).

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