Dimensions

Dimension Describes the Type of Quantity

In everyday language, dimension often means spatial size such as length, width, and height. In physics, a dimension shows the base quantities and powers that form a physical quantity.

For example, bolt length, nut diameter, marble radius, and travel distance can use different units, but they are still the same kind of quantity.

Length, diameter, radius, and distance all have the length dimension $[\mathrm{L}]$ .

A dimension does not state the measured value. It identifies the kind of physical quantity being measured.

Reading Powers in Dimensional Formulas

The visual below focuses only on the length dimension. Switch from length to area and then volume. Notice how one length factor becomes $[\mathrm{L}]$ , $[\mathrm{L}]^2$ , and $[\mathrm{L}]^3$ .

From Length to Area and Volume

Choose a shape built from length factors to see its formula, unit, and dimension.

Formula: $s$
SI unit: $\text{m}$
Dimension: $[\mathrm{L}]$

Base Quantities in Dimensional Analysis

The International System of Units or SI uses $7$ base quantities. The dimension of a derived quantity is written as a product of powers of the seven base dimensions.

Base quantity	Example symbol	Dimension
Length	$\ell,\ x,\ r$	$[\mathrm{L}]$
Mass	$m$	$[\mathrm{M}]$
Time	$t$	$[\mathrm{T}]$
Electric current	$I$	$[\mathrm{I}]$
Thermodynamic temperature	$T$	$[\Theta]$
Amount of substance	$n$	$[\mathrm{N}]$
Luminous intensity	$I_v$	$[\mathrm{J}]$

Dimensions are written in square brackets so they are not confused with units. For example, $[\mathrm{L}]$ is the dimension of length, while $\text{m}$ is the unit meter.

Deriving Dimensions from Formulas

Dimensional analysis works by replacing every quantity in a formula with its dimension, then simplifying the powers.

\begin{aligned} [A] &= [l][w] = [\mathrm{L}][\mathrm{L}] = [\mathrm{L}]^2 \\ [V] &= [l][w][h] = [\mathrm{L}]^3 \\ [v] &= \frac{[\Delta s]}{[\Delta t]} = \frac{[\mathrm{L}]}{[\mathrm{T}]} = [\mathrm{L}][\mathrm{T}]^{-1} \\ [a] &= \frac{[\Delta v]}{[\Delta t]} = \frac{[\mathrm{L}][\mathrm{T}]^{-1}}{[\mathrm{T}]} = [\mathrm{L}][\mathrm{T}]^{-2} \end{aligned}

Force, work, and power can also be read as dimension structures.

\begin{aligned} [F] &= [m][a] = [\mathrm{M}][\mathrm{L}][\mathrm{T}]^{-2} \\ [W] &= [F][\Delta s] = [\mathrm{M}][\mathrm{L}]^2[\mathrm{T}]^{-2} \\ [P] &= \frac{[W]}{[t]} = [\mathrm{M}][\mathrm{L}]^2[\mathrm{T}]^{-3} \end{aligned}

Checking Dimensional Consistency in a Formula

Dimensions can help check whether a formula form could be correct. Every term being added must have the same dimension.

Take the displacement formula for uniformly accelerated motion:

s = v_0 t + \frac{1}{2} a t^2

Check the dimensions:

\begin{aligned} [v_0t] &= [\mathrm{L}][\mathrm{T}]^{-1}[\mathrm{T}] = [\mathrm{L}] \\ [at^2] &= [\mathrm{L}][\mathrm{T}]^{-2}[\mathrm{T}]^2 = [\mathrm{L}] \end{aligned}

Both terms on the right have dimension $[\mathrm{L}]$ , so the formula passes the dimensional check. Passing this check cannot prove that a formula is correct. The check can still catch impossible formulas quickly.

$s+t$ is invalid here because the dimensions differ: $[\mathrm L]\ne[\mathrm T]$ .

One Object Can Have Many Quantities

A bolt and a nut look like small objects, but measuring them can involve several different quantities. That is why one object sometimes needs more than one measuring tool.

What is checked	Suitable tool	Dimension
Bolt length	ruler or vernier caliper	$[\mathrm{L}]$
Outer diameter	vernier caliper or micrometer screw gauge	$[\mathrm{L}]$
Cross-sectional area	calculated from diameter	$[\mathrm{L}]^2$
Material volume	calculated from spatial size	$[\mathrm{L}]^3$
Mass	balance	$[\mathrm{M}]$
Density	mass divided by volume	$[\mathrm{M}][\mathrm{L}]^{-3}$

So, two measuring tools can measure quantities with the same dimension, but they serve different contexts. A ruler is enough for a large length that does not need high precision. A vernier caliper or micrometer screw gauge is better when a small diameter must be read more carefully.

Dimensional analysis can test a proposed equation and help derive the unit of an unknown quantity.