MA1 - Dimensions and consistency

MA1 - Dimensions and consistency

Dimensions describe the kind of quantity a formula represents, independently of the units used to measure it. You will build dimensions from mass, length and time, test proposed equations and see why a formula can pass that test yet still be wrong.

Dimensions M L T

Could both s=ut+at2s=ut+at^2 and s=ut+ats=ut+at describe a displacement? Before substituting any numbers, ask whether every term represents a length.

A journey of 2 km2\,\mathrm{km} is also a journey of 2000 m2000\,\mathrm{m}. The numerical value and unit change; the physical quantity remains a length. A dimension records that physical kind. Square brackets mean “the dimensions of”, so [s]=L[s]=\mathrm{L}. They do not mean the numerical value of ss.

Base dimensions

In mechanics, mass has dimension M\mathrm{M}, length has dimension L\mathrm{L}, and time has dimension T\mathrm{T}. A dimensionless quantity has dimension 1=M0L0T01=\mathrm{M}^0\mathrm{L}^0\mathrm{T}^0. These capital letters represent dimensions, not kilograms, metres and seconds.

Area is length times length, so its dimension is L2\mathrm{L}^2. Volume has dimension L3\mathrm{L}^3. Velocity is displacement divided by time, giving [v]=LT−1[v]=\mathrm{LT}^{-1}. Acceleration is velocity change divided by time, giving [a]=LT−2[a]=\mathrm{LT}^{-2}. Gravitational acceleration gg has this same dimension. A negative velocity still has dimension LT−1\mathrm{LT}^{-1}: direction changes its sign, not its physical kind.

QuantityA possible SI unitDimension
Mass mmkg\mathrm{kg}M\mathrm{M}
Displacement ssm\mathrm{m}L\mathrm{L}
Time tts\mathrm{s}T\mathrm{T}
Velocity vvm s−1\mathrm{m\,s}^{-1}LT−1\mathrm{LT}^{-1}
Acceleration aam s−2\mathrm{m\,s}^{-2}LT−2\mathrm{LT}^{-2}

Ratios of quantities with the same dimensions are dimensionless: a length divided by another length has dimension 11, even when you must first convert their units before calculating its value.

Key idea

A unit tells you how a quantity is measured; its dimension tells you what kind of quantity it is. Changing metres to kilometres leaves L\mathrm{L} unchanged.

Use that distinction in this check.

Dimensions of compound quantities

Build an unfamiliar dimension from a defining equation whose quantities you already understand. Multiply, divide and raise dimensions to powers just as you do algebraic factors.

Newton's second law gives [F]=[m][a]=MLT−2[F]=[m][a]=\mathrm{MLT}^{-2}. Thus a newton is kg m s−2\mathrm{kg\,m\,s}^{-2}, but the dimension of force is MLT−2\mathrm{MLT}^{-2}. Density is mass per volume, so [ρ]=ML−3[\rho]=\mathrm{ML}^{-3}, not ML−1\mathrm{ML}^{-1}.

Worked example 1: work from force and distance

Question. Work done by a constant force acting along a displacement is W=FsW=Fs. Find [W][W].

Plan. Replace force and displacement by their dimensions.

[W]=[F][s]use the defining product=(MLT−2)(L)=ML2T−2add length powers\begin{aligned} [W]&=[F][s] &&\text{use the defining product}\\ &=(\mathrm{MLT}^{-2})(\mathrm{L})\\ &=\mathrm{ML}^2\mathrm{T}^{-2} &&\text{add length powers} \end{aligned}

Answer. Work has dimension ML2T−2\mathrm{ML}^2\mathrm{T}^{-2}.

Check. Kinetic energy 12mv2\tfrac12mv^2 has dimension M(LT−1)2=ML2T−2\mathrm{M}(\mathrm{LT}^{-1})^2=\mathrm{ML}^2\mathrm{T}^{-2}, matching work, which transfers energy.

Keep the work quantity and change only to work per unit time. Dividing by time decreases its time exponent by one.

Worked example 2: power from work

Question. Power is P=W/tP=W/t. Find [P][P].

Plan. Divide the dimension of work by the dimension of time.

[P]=[W][t]=ML2T−2T=ML2T−3subtract time exponents\begin{aligned} [P]&=\frac{[W]}{[t]}\\ &=\frac{\mathrm{ML}^2\mathrm{T}^{-2}}{\mathrm{T}}\\ &=\mathrm{ML}^2\mathrm{T}^{-3} &&\text{subtract time exponents} \end{aligned}

Answer. Power has dimension ML2T−3\mathrm{ML}^2\mathrm{T}^{-3}.

Check. The alternative equation P=FvP=Fv gives (MLT−2)(LT−1)=ML2T−3(\mathrm{MLT}^{-2})(\mathrm{LT}^{-1})=\mathrm{ML}^2\mathrm{T}^{-3}.

A pressure, in contrast, is force per area; its length power becomes 1−2=−11-2=-1. Always start from what the quantity means.

Watch out

Wrong: [ρ]=M/L[\rho]=\mathrm{M}/\mathrm{L}. Density is mass per volume, so [ρ]=M/L3=ML−3[\rho]=\mathrm{M}/\mathrm{L}^3=\mathrm{ML}^{-3}. Keep the power on the whole denominator.

Try another rate of transfer.

Checking dimensional consistency

An equation is dimensionally consistent when both sides, and every term being added or subtracted, have the same dimensions. You can add two lengths; adding a length to a velocity cannot produce a physical length.

For a square root, halve all exponents: [as]=L2T−2=LT−1[\sqrt{as}]=\sqrt{\mathrm{L}^2\mathrm{T}^{-2}}=\mathrm{LT}^{-1}. For a sum, check each term separately: do not multiply the dimensions of terms that are added.

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