Physical quantities, SI units and dimensions

Base and derived quantities, SI units and prefixes, dimensions and checking equations with dimensional analysis.

Base and derived quantities

A physical quantity is anything that can be measured and has a number and a unit, such as $5.0\ \text{kg}$. The SI system builds every unit from seven base units:

Base quantity	SI unit	Symbol	Dimension
Mass	kilogram	kg	M
Length	metre	m	L
Time	second	s	T
Electric current	ampere	A	I
Thermodynamic temperature	kelvin	K	θ
Amount of substance	mole	mol	N
Luminous intensity	candela	cd	J

A derived quantity is defined from base quantities by an equation. Its unit follows from that equation:

velocity $= \dfrac{\text{displacement}}{\text{time}}$, unit $\text{m s}^{-1}$
force $= ma$, unit $\text{kg m s}^{-2}$, named the newton (N)
pressure $= \dfrac{F}{A}$, unit $\text{N m}^{-2} = \text{kg m}^{-1}\text{s}^{-2}$, named the pascal (Pa)
charge $= It$, unit $\text{A s}$, named the coulomb (C)

Prefixes

Prefix	Symbol	Factor	Prefix	Symbol	Factor
tera	T	$10^{12}$	milli	m	$10^{-3}$
giga	G	$10^{9}$	micro	μ	$10^{-6}$
mega	M	$10^{6}$	nano	n	$10^{-9}$
kilo	k	$10^{3}$	pico	p	$10^{-12}$

Careful: $1\ \text{cm}^2 = (10^{-2}\ \text{m})^2 = 10^{-4}\ \text{m}^2$ and $1\ \text{mm}^3 = 10^{-9}\ \text{m}^3$. The prefix is raised to the power too.

Dimensions

The dimensions of a quantity show how it depends on the base quantities. Square brackets mean "the dimensions of":

$[v] = \text{LT}^{-1}$, $[a] = \text{LT}^{-2}$
$[F] = [m][a] = \text{MLT}^{-2}$
$[W] = [F][s] = \text{ML}^2\text{T}^{-2}$ (work, energy, heat)
$[P] = \text{ML}^2\text{T}^{-3}$ (power), $[p] = \text{ML}^{-1}\text{T}^{-2}$ (pressure, stress, Young's modulus)

A pure number, an angle in radians ($\theta = \dfrac{\text{arc}}{\text{radius}}$), strain and refractive index have no dimensions. A quantity can have a unit but no dimensions: the radian is an example. The sound intensity level in decibels is another.

Using dimensional analysis

1. Checking an equation. Every term must have the same dimensions. In $s = ut + \tfrac{1}{2}at^2$: $[s] = \text{L}$, $[ut] = \text{LT}^{-1}\cdot\text{T} = \text{L}$, $[at^2] = \text{LT}^{-2}\cdot\text{T}^2 = \text{L}$. The equation is dimensionally consistent. Dimensions cannot show whether the $\tfrac{1}{2}$ is right.

2. Finding the dimensions of a constant. From $F = 6\pi\eta r v$ (Stokes' law): $[\eta] = \dfrac{[F]}{[r][v]} = \dfrac{\text{MLT}^{-2}}{\text{L}\cdot\text{LT}^{-1}} = \text{ML}^{-1}\text{T}^{-1}$, so the unit of $\eta$ is $\text{kg m}^{-1}\text{s}^{-1} = \text{Pa s}$.

3. Predicting the form of a relation. Suppose the period of a simple pendulum depends on $m$, $l$ and $g$: $T = k,m^x l^y g^z$. $$\text{T} = \text{M}^x,\text{L}^y,(\text{LT}^{-2})^z$$ Matching powers: M gives $x = 0$; T gives $-2z = 1$, so $z = -\tfrac{1}{2}$; L gives $y + z = 0$, so $y = \tfrac{1}{2}$. Hence $T = k\sqrt{l/g}$. Experiment gives $k = 2\pi$.

Example: The speed of a liquid through a tube is written $v = \dfrac{C a^n \Delta p}{\eta l}$, where $C$ has no dimensions. Then $\text{LT}^{-1} = \dfrac{\text{L}^n\cdot\text{ML}^{-1}\text{T}^{-2}}{\text{ML}^{-1}\text{T}^{-1}\cdot \text{L}} = \text{L}^{n-1}\text{T}^{-1}$, so $n - 1 = 1$ and $n = 2$. (A 2023 Paper I question.)

Exam tip: In Paper I, write the dimensions of each symbol on the side first. Most "which equation is correct" questions are then settled in one line.

Scalars and vectors

A scalar has magnitude only (mass, time, energy, temperature). A vector has magnitude and direction (displacement, velocity, force, momentum, field strength). Two vectors of sizes $a$ and $b$ have a resultant between $|a - b|$ and $a + b$: $9\ \text{m}$ and $6\ \text{m}$ can give $4\ \text{m}$ but not $2\ \text{m}$ or $16\ \text{m}$.