6.01a–6.01e - Finding and using dimensions

6.01a–6.01e - Finding and using dimensions

Use mass, length and time to uncover the structure of a mechanics formula. You will translate units into dimensions, recognise dimensionless quantities, check equations and find unknown powers in proposed models. You will also see exactly what dimensions can tell you about motion, and what needs another argument or a measurement.

Dimensions from units

If a simple pendulum is modelled using only its length, its bob's mass and the acceleration due to gravity, must a heavier bob change its period? By the end of this lesson you will answer that question without solving the pendulum's equation of motion.

Start with a distinction you already use. A distance of 1 m1\,\mathrm{m} is also 100 cm100\,\mathrm{cm}: the unit and numerical value change, but the kind of quantity is still length. Dimensions describe that kind of quantity, independently of a particular measuring unit.

You already know

Speed is distance divided by time, acceleration is change in velocity divided by time, and Newton's second law is F=maF = ma. In SI units, 1 N=1 kg m s−21\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}. These are assumed mechanics facts; use them to identify dimensions.

The three basic dimensions used here are mass, length and time. Keep their capital letters distinct from the lower-case symbols for units or quantities.

Dimensions

Write [q][q] for the dimensions of a quantity qq. The basic dimensions are MM for mass, LL for length and TT for time. A derived dimension is a product of powers of these, such as [v]=LT−1[v] = LT^{-1} for speed. For example, a speed measured in m s−1\mathrm{m\,s^{-1}} or in km h−1\mathrm{km\,h^{-1}} has the same dimensions LT−1LT^{-1}. The unit m s−1\mathrm{m\,s^{-1}} is not itself a dimension.

Dimensions obey the multiplication and index rules of algebra. Replace each factor by its dimensions before simplifying; a numerical factor such as 22 does not change them.

Multiplying dimensions

For physical quantities pp and qq, with a non-zero denominator where needed,

[pq]=[p][q],[pq]=[p][q],[qr]=[q]r[pq] = [p][q], \qquad \left[\frac{p}{q}\right] = \frac{[p]}{[q]}, \qquad [q^r] = [q]^r

Here rr is a fixed numerical exponent. In particular, [a]=LT−2[a] = LT^{-2} and [F]=MLT−2[F] = MLT^{-2}.

Not in the formula booklet: learn it.

This gives a small translation table. You can derive each row rather than memorising a separate rule for every quantity.

QuantityA possible unitDimensions
Aream2\mathrm{m^2}L2L^2
Volumem3\mathrm{m^3}L3L^3
Density, mass per unit volumekg m−3\mathrm{kg\,m^{-3}}ML−3ML^{-3}
Speedm s−1\mathrm{m\,s^{-1}}LT−1LT^{-1}
Accelerationm s−2\mathrm{m\,s^{-2}}LT−2LT^{-2}
ForceN\mathrm{N}MLT−2MLT^{-2}

The table is not supplied in the formula booklet. If an unfamiliar quantity appears, its units or a defining relationship will let you derive its dimensions; the name alone need not be familiar.

Worked example 1: a quantity measured in newtons per metre

Question. A quantity qq has units N m−1\mathrm{N\,m^{-1}}. Determine its dimensions.

Plan. Replace the newton by the dimensions of force, then divide by length.

[F]=[m][a]=MLT−2Newton’s second law[q]=(MLT−2)L−1force per length=MT−2combine powers of L\begin{aligned} [F] &= [m][a] = MLT^{-2} && \text{Newton's second law} \\ [q] &= (MLT^{-2})L^{-1} && \text{force per length} \\ &= MT^{-2} && \text{combine powers of } L \end{aligned}

Why this step? The symbol N\mathrm{N} hides mass, length and time. Expanding it before cancelling reveals the required powers.

Answer. [q]=MT−2[q] = MT^{-2}.

Check. N m−1=kg m s−2 m−1=kg s−2\mathrm{N\,m^{-1}} = \mathrm{kg\,m\,s^{-2}\,m^{-1}} = \mathrm{kg\,s^{-2}}, which has dimensions MT−2MT^{-2}.

Worked example 2 changes one feature: the unit contains an extra factor of seconds. That adds one to the power of TT, without affecting MM or LL.

Worked example 2: an extra time factor

Question. A quantity cc has units N s m−1\mathrm{N\,s\,m^{-1}}. Determine its dimensions.

Plan. Expand the force dimension and include every unit factor before simplifying.

[c]=(MLT−2)TL−1translate all three factors=ML1−1T−2+1add exponents=MT−1simplify\begin{aligned} [c] &= (MLT^{-2})T L^{-1} && \text{translate all three factors} \\ &= M L^{1-1}T^{-2+1} && \text{add exponents} \\ &= MT^{-1} && \text{simplify} \end{aligned}

Why this step? Keeping the extra TT visible prevents you from treating N s m−1\mathrm{N\,s\,m^{-1}} as though it were N m−1\mathrm{N\,m^{-1}}.

Answer. [c]=MT−1[c] = MT^{-1}.

Check. Multiplying cc by a speed gives dimensions (MT−1)(LT−1)=MLT−2(MT^{-1})(LT^{-1}) = MLT^{-2}, agreeing with the unit calculation (N s m−1)(m s−1)=N(\mathrm{N\,s\,m^{-1}})(\mathrm{m\,s^{-1}}) = \mathrm{N}.

Your final line must show the dimension product, even if units helped you reach it.

Watch out

Wrong: [c]=[MT−1][c] = [MT^{-1}], or [c]=kg s−1[c] = \mathrm{kg\,s^{-1}} when dimensions are requested. Square brackets enclose the quantity whose dimensions you want, not the dimension product. Right: [c]=MT−1[c] = MT^{-1}; kg s−1\mathrm{kg\,s^{-1}} is a possible SI unit.

Now change the time factor once more and track its effect on the exponent.

Dimensionless quantities and unit choices

Dividing one length by another removes the length dimension: a slope, a strain and an angle measured as arc length divided by radius all have this feature.

For instance, a slope with rise 2 m2\,\mathrm{m} and horizontal run 5 m5\,\mathrm{m} is 2/52/5. Measuring both lengths in centimetres gives 200/500=2/5200/500 = 2/5, the same number. The dimensions cancel because the two quantities have the same dimensions.

Dimensionless quantity

A quantity is dimensionless when its dimension product is M0L0T0=1M^0L^0T^0 = 1. For example, if xx and ll are lengths with l≠0l \ne 0, then [x/l]=L/L=1[x/l] = L/L = 1. A speed is not dimensionless: [v]=LT−1[v] = LT^{-1}.

A coefficient of friction is dimensionless because it is a ratio of forces. An angle in radians is dimensionless because it is a ratio of lengths. A dimensionless quantity can still have a named descriptive unit, such as the radian; “dimensionless” describes its powers of MM, LL and TT.

Pure numerical constants, including π\pi, are dimensionless. A letter called a constant need not be: gg may be treated as constant in a model, but [g]=LT−2[g] = LT^{-2}. The dimensionless-ratio rule and these dimension statements are not in the formula booklet: learn them.

Key idea

Dimensions ignore the size of the chosen units. Before calculating a numerical ratio, express both quantities in compatible units.

For example, 36 km h−1=10 m s−136\,\mathrm{km\,h^{-1}} = 10\,\mathrm{m\,s^{-1}}, since 36×1000/3600=1036 \times 1000/3600 = 10. The ratio of these speeds is 11, although dividing the two printed numbers gives 3.63.6. Dimensionless does not mean that unit conversion can be skipped.

Checking every term of an equation

An equation can fail before you substitute a single number: its terms may represent different kinds of quantity.

You can add 3 m3\,\mathrm{m} to 2 m2\,\mathrm{m}, after putting them in the same unit, but you cannot add 3 m3\,\mathrm{m} to 2 s2\,\mathrm{s}. The same restriction applies to every term in a proposed mechanics equation.

Dimensional homogeneity

An equation is dimensionally homogeneous, or dimensionally consistent, when every term being added or equated has the same dimensions. For example, v=u+atv = u + at is homogeneous because [v]=[u]=[at]=LT−1[v] = [u] = [at] = LT^{-1}. The expression u+au + a fails this test.

The constant-acceleration equation v=u+atv = u + at is in the formula booklet. Dimensional homogeneity is not a formula in the booklet: learn the principle. The dimensions of a sum are the common dimensions of its terms; they are not the sum or product of dimension symbols.

For the examples below, power is measured in watts, where 1 W=1 N m s−11\,\mathrm{W} = 1\,\mathrm{N\,m\,s^{-1}}. Hence [P]=ML2T−3[P] = ML^2T^{-3}. This dimension result is not in the formula booklet. It immediately shows that a driving force multiplied by velocity has the dimensions of power, because [Fv]=(MLT−2)(LT−1)=ML2T−3[Fv] = (MLT^{-2})(LT^{-1}) = ML^2T^{-3}.

Worked example 3: checking a sum of power terms

Question. A proposed model is P=αFv+βmvaP = \alpha Fv + \beta mva, where PP is power, FF is force, vv is speed, mm is mass, aa is acceleration and α,β\alpha, \beta are dimensionless constants. Determine whether it is dimensionally consistent.

Plan. Find the dimensions of the left-hand side and each complete right-hand term, including the constants, then consider the sum.

[P]=ML2T−3from the watt[αFv]=1(MLT−2)(LT−1)[α]=1=ML2T−3combine powers[βmva]=1(M)(LT−1)(LT−2)[β]=1=ML2T−3combine powers\begin{aligned} [P] &= ML^2T^{-3} && \text{from the watt} \\ [\alpha Fv] &= 1(MLT^{-2})(LT^{-1}) && [\alpha] = 1 \\ &= ML^2T^{-3} && \text{combine powers} \\ [\beta mva] &= 1(M)(LT^{-1})(LT^{-2}) && [\beta] = 1 \\ &= ML^2T^{-3} && \text{combine powers} \end{aligned}

Why this step? The plus sign demands that the two right-hand terms have the same dimensions. Checking only one term would leave half the argument unfinished.

Answer. Both right-hand terms have dimensions ML2T−3ML^2T^{-3}, so their sum has those dimensions, matching [P][P]. The equation is dimensionally consistent.

Check. If F=maF = ma, then FvFv and mvamva are the same type of quantity, as the dimension calculation predicts.

Worked example 4 changes just one feature: the acceleration in the second term is squared. Its extra factor affects both the length and time powers.

Worked example 4: one incorrect power

Question. With the same meanings of the symbols and dimensionless constants, determine whether P=αFv+βmva2P = \alpha Fv + \beta mva^2 is dimensionally consistent.

Plan. Retain the known dimensions of PP and αFv\alpha Fv, and recalculate the changed term.

[P]=[αFv]=ML2T−3unchanged terms[βmva2]=1(M)(LT−1)(LT−2)2include the square=ML3T−5combine powersML3T−5≠ML2T−3different dimensions\begin{aligned} [P] = [\alpha Fv] &= ML^2T^{-3} && \text{unchanged terms} \\ [\beta mva^2] &= 1(M)(LT^{-1})(LT^{-2})^2 && \text{include the square} \\ &= ML^3T^{-5} && \text{combine powers} \\ ML^3T^{-5} &\ne ML^2T^{-3} && \text{different dimensions} \end{aligned}

Why this step? A dimensionless β\beta cannot supply the missing powers. You must use the stated nature of the constants rather than silently changing their dimensions.

Answer. The equation is not dimensionally consistent: its right-hand terms cannot be added as quantities of a common dimension.

Check. Dividing the faulty second term by the correct second term leaves a factor of aa, which has dimensions LT−2LT^{-2}, not 11.

Passing this test is useful evidence, but it cannot establish the physics or a numerical coefficient.

Watch out

Wrong: “The dimensions agree, so the equation is correct.” For a particle starting from rest under constant acceleration, both s=at2s = at^2 and s=12at2s = \tfrac12 at^2 have dimensions LL, but only the second is the correct displacement formula. Dimensional consistency is necessary for a valid quantity equation; it is not sufficient.

In a full argument, name each term, show the products of dimensions and explicitly say why the sum has the same dimensions as the left-hand side.

Continue this lesson.

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