Units and Measurements · Lesson 4 of 7

Dimensions and Dimensional Formulae

“Describe physical quantities in terms of base quantities and derive their dimensional formulae.”

Learning Objectives

• Distinguish dimensions from units. • Derive dimensional formulae from definitions. • Interpret zero and negative powers. • Explain dimensionless ratios and unit-system changes.

The physical nature behind a unit

Metres and kilometres are different units of the same kind of quantity: length. Dimensions express that underlying kind without choosing a particular unit. The dimension of length is written [L], mass [M] and time [T]. The remaining base quantities are electric current, thermodynamic temperature, amount of substance and luminous intensity. In mechanics, the quantities needed here can be built from mass, length and time.

Square brackets around a physical quantity mean “the dimensions of”. A dimensional formula lists the powers of base quantities required to represent it. A dimensional equation equates the dimensions of a named quantity to that formula. Neither specifies the quantity’s numerical magnitude. Initial velocity, final velocity and speed all have the same dimensions, even though they describe different aspects of motion.

Building dimensions from relationships

Deriving the dimensions of force

Problem
Force equals mass multiplied by acceleration. Find its dimensional formula.

1.Velocity is displacement divided by time, so $[v] = [L T^{- 1}]$ .
2.Acceleration is change in velocity divided by time, so $[a] = [L T^{- 2}]$ .
3.Therefore $[F] = [m] [a] = [M L T^{- 2}]$ . The exponents are 1 in mass, 1 in length and −2 in time.

Quantity	Defining relationship	Dimensional formula
Area	length × length	$[L^{2}]$
Volume	length cubed	$[L^{3}]$
Speed	distance / time	$[L T^{- 1}]$
Acceleration	velocity change / time	$[L T^{- 2}]$
Density	mass / volume	$[M L^{- 3}]$
Force	mass × acceleration	$[M L T^{- 2}]$
Energy	force × distance	$[M L^{2} T^{- 2}]$

A zero exponent means independence from that base dimension. Volume can be written [M⁰L³T⁰]: it contains three powers of length and no mass or time factor. A negative exponent means division, not a negative physical value. For example, density has three powers of length in its denominator. Dimensions can be multiplied, divided and raised to powers using ordinary exponent rules.

Quantities with no net dimension

A ratio of quantities with the same dimensions is dimensionless. Relative density compares two densities and therefore has no unit. Refractive index compares two speeds. Angle compares arc length with radius. Being dimensionless does not make these ideas interchangeable: an angle is not a refractive index. Named angle units remain useful for identifying meaning.

Changing the base units

The physical energy represented by one calorie does not change when another system of units is adopted. Its numerical value does change. Dimensions tell us the exact powers by which the new base units affect the new energy unit.

Energy in a new unit system

Problem
One calorie is approximately 4.2 J. Let the new mass, length and time units be α kg, β m and γ s. Find its numerical value.

1.Energy has dimensions $[M L^{2} T^{- 2}]$ . One new energy unit therefore equals $α β^{2} γ^{- 2}$ joules.
2.Divide the fixed physical energy by that unit size: $n = 4.2/ (α β^{2} γ^{- 2})$ .
3.The numerical value is $4.2 α^{- 1} β^{- 2} γ^{2}$ . Increasing the size of an energy unit reduces the number needed to represent a fixed energy.

All seven base dimensions

Base quantity	Dimension notation used here	SI unit symbol
Length	L	m
Mass	M	kg
Time	T	s
Electric current	I	A
Thermodynamic temperature	Θ	K
Amount of substance	N	mol
Luminous intensity	J	cd

Symbols for dimensions are conventions. This table distinguishes dimension symbols from unit symbols; the source uses A, K, mol and cd in brackets for the last four base quantities. Here J denotes a luminous-intensity dimension only inside dimensional notation; it is not the joule unit. Likewise N in this table does not denote the newton unit. Mechanics in this chapter uses M, L and T, avoiding those possible symbol ambiguities.

Compound conversions step by step

An acceleration in kilometres per hour squared

Problem
Convert $3.0 m s^{- 2}$ to kilometres per hour squared.

1.Use $1 m = 1 0^{- 3} km$ and $1 s = (1/3600) h$ .
2.Therefore $1 s^{- 2} = 360 0^{2} h^{- 2}$ . The reciprocal time factor is squared.
3.The value becomes $3.0 \times 1 0^{- 3} \times 360 0^{2} = 38880 km h^{- 2}$ .
4.Report $3.9 \times 1 0^{4} km h^{- 2}$ to two significant figures.

Energy and the gravitational constant in CGS units

Problem
Convert one joule and the stated SI value of G to centimetre–gram–second units.

1.For energy, $1 kg m^{2} s^{- 2} = (1 0^{3}) (1 0^{2})^{2} g c m^{2} s^{- 2} = 1 0^{7} g c m^{2} s^{- 2}$ .
2.For G, first replace the newton: $N m^{2} k g^{- 2} = m^{3} k g^{- 1} s^{- 2}$ .
3.Length cubed contributes $1 0^{6}$ and reciprocal kilograms contribute $1 0^{- 3}$ when converting to grams. The net multiplier is $1 0^{3}$ .
4.Thus $G = 6.67 \times 1 0^{- 8} c m^{3} g^{- 1} s^{- 2}$ .

Atomic sizes and volumes

An angstrom, symbol Å, equals 10⁻¹⁰ m. At atomic scales, a spherical model gives an estimate of occupied volume, not a rigid boundary of an atom. A quoted “size” must be interpreted explicitly: taking the same numerical size as a diameter instead of a radius divides the sphere’s volume by eight.

Volume of a mole of spherical particlesLaTeX

V_{particles} = N_{A} \frac{4 π r ^{3}}{3}

Use r in metres to obtain cubic metres per mole of particles. N_A is approximately 6.022 × 10²³ particles per mole.

Estimating the volume of a mole of hydrogen atoms

Problem
Treat the stated atomic size 0.5 Å as a radius. Estimate the total spherical volume.

1.Convert the assumed radius: $r = 0.5 \times 1 0^{- 10} m = 5 \times 1 0^{- 11} m$ .
2.One sphere has volume $4 π r^{3} /3 \approx 5.24 \times 1 0^{- 31} m^{3}$ .
3.Multiply by Avogadro’s number to obtain about $3.15 \times 1 0^{- 7} m^{3}$ per mole before rounding.
4.The input size is approximate and given with one significant figure, so report roughly $3 \times 1 0^{- 7} m^{3}$ . If 0.5 Å is interpreted as diameter, the estimate is eight times smaller.

Gas volume compared with molecular volume

Problem
Using 1 Å as the assumed molecular radius, compare the estimated volume of a mole of spherical hydrogen molecules with 22.4 L of gas.

1.The assumed radius is $1 0^{- 10} m$ . The total sphere volume is approximately $N_{A} (4 π /3) (1 0^{- 10})^{3} = 2.52 \times 1 0^{- 6} m^{3}$ .
2.Convert gas volume: $22.4 L = 22.4 \times 1 0^{- 3} m^{3} = 0.0224 m^{3}$ .
3.The ratio is $0.0224/ (2.52 \times 1 0^{- 6}) \approx 8.9 \times 1 0^{3}$ , of order $1 0^{4}$ . This is a scale estimate rather than a high-precision prediction.
4.Most container volume lies between molecules. If the quoted size were a diameter, the estimated sphere volume would be eight times smaller and the ratio eight times larger. The radius assumption must accompany the numerical result.

Applying measurement to scale

Mean density of the Sun

Problem
Estimate mean density using mass 2.0 × 10³⁰ kg and radius 7.0 × 10⁸ m.

1.For a sphere, $V = 4 π R^{3} /3$ and $ρ = 3 M / (4 π R^{3})$ .
2. $R^{3} = 3.43 \times 1 0^{26} m^{3}$ , giving $V \approx 1.44 \times 1 0^{27} m^{3}$ . Keep these extra digits during calculation.
3.The mean density is about $1.4 \times 1 0^{3} kg m^{- 3}$ to two significant figures.
4.This is comparable to many liquid or solid densities, despite the Sun being plasma. Temperature and state alone do not determine density; pressure and the nonuniform internal structure matter.

Quick check

What are the dimensions of density?

Key Takeaways

• Dimensions describe the kind of quantity independently of the chosen unit. • Derive dimensional formulae from physical definitions. • Zero powers indicate no dependence on a base dimension. • Equal dimensions do not imply identical physical meaning.

Previous · Lesson 3

Calculations with Measured Values

Next · Lesson 5

Checking Equations with Dimensions