Dimensional Analysis

Dimensional analysis is the art of reasoning about physics from the units of quantities alone — without solving any equation of motion. It is astonishingly powerful for so simple an idea: it catches algebraic errors instantly (a wrong formula is almost always dimensionally wrong), it predicts the form of physical laws up to a dimensionless constant, and it underlies the scaling arguments that explain why small animals can fall from great heights unhurt and why model aircraft must be tested at adjusted speeds. The whole subject rests on one principle — that you can only add or equate quantities of the same dimensions — and on three fundamental mechanical dimensions: mass $M$M, length $L$L and time $T$T. This lesson develops the method rigorously, applies it to checking and to deriving formulae, and is honest about what it can and cannot do.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M). Dimensional analysis is a self-contained Further-Mechanics topic that rewards careful, systematic algebra: the "derive the form of a law" questions are pure AO2/AO3 (set up and solve simultaneous equations in the exponents, then interpret), while "check this formula" parts are quick AO1 marks. Paper 3 weights these objectives AO1 40% / AO2 25% / AO3 35%. There are no heavy prerequisites beyond solving simultaneous linear equations and confident index laws — but precision with signs of exponents is everything, and the topic ties neatly to the variable-force and SHM material elsewhere in this course.

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2 Core theory — dimensions and homogeneity

Every mechanical quantity has a dimension, written in square brackets, expressible as a product of powers of the three fundamentals:

$$[Q] = M^{\alpha}L^{\beta}T^{\gamma}.$$[Q]=MαLβTγ.

Dimension	Symbol
Mass	$M$M
Length	$L$L
Time	$T$T

From the definitions of the quantities, their dimensions follow by substitution:

Quantity	Definition	Dimensions
Displacement	$s$s	$L$L
Velocity	$s/t$s/t	$LT^{-1}$LT−1
Acceleration	$v/t$v/t	$LT^{-2}$LT−2
Force	$ma$ma	$MLT^{-2}$MLT−2
Momentum	$mv$mv	$MLT^{-1}$MLT−1
Impulse	$Ft$Ft	$MLT^{-1}$MLT−1
Energy / Work	$Fs$Fs	$ML^2T^{-2}$ML2T−2
Power	$W/t$W/t	$ML^2T^{-3}$ML2T−3
Pressure	$F/A$F/A	$ML^{-1}T^{-2}$ML−1T−2
Angular velocity	$\theta/t$θ/t	$T^{-1}$T−1
Moment of inertia	$mr^2$mr2	$ML^2$ML2
Torque	$Fr$Fr	$ML^2T^{-2}$ML2T−2
Spring constant	$F/x$F/x	$MT^{-2}$MT−2
Gravitational constant $G$G	$Fr^2/(m_1 m_2)$Fr2/(m1​m2​)	$M^{-1}L^3T^{-2}$M−1L3T−2

Two observations are worth internalising. First, angles are dimensionless (radian = arc length / radius = $L/L$L/L), which is why $\omega = \theta/t$ω=θ/t has dimension $T^{-1}$T−1. Second, energy and torque share dimensions $ML^2T^{-2}$ML2T−2 yet are physically different — dimensional analysis cannot tell them apart, a limitation we return to.

How does one find the dimensions of an unfamiliar quantity? Always trace it back to a defining equation built from quantities whose dimensions you know. For the gravitational constant, rearrange Newton's law of gravitation $F = Gm_1m_2/r^2$F=Gm1​m2​/r2 to $G = Fr^2/(m_1m_2)$G=Fr2/(m1​m2​), then substitute: $[G] = (MLT^{-2})(L^2)/(M\cdot M) = M^{-1}L^3T^{-2}$[G]=(MLT−2)(L2)/(M⋅M)=M−1L3T−2. For the spring constant, $F = kx$F=kx gives $[k] = MLT^{-2}/L = MT^{-2}$[k]=MLT−2/L=MT−2. For pressure, $P = F/A$P=F/A gives $(MLT^{-2})/L^2 = ML^{-1}T^{-2}$(MLT−2)/L2=ML−1T−2. The method never requires memorising a long table; it requires knowing a handful of definitions and substituting carefully. This is why the table above is best regarded not as something to learn by rote but as the output of the procedure — you should be able to regenerate every row in a few seconds from its defining formula.

A particularly important corollary concerns ratios of like quantities. Any ratio of two quantities with the same dimensions is itself dimensionless: a strain (length/length), an efficiency (energy/energy), a Mach number (speed/speed). These pure numbers are the raw material from which the more advanced theory (the Buckingham $\Pi$Π theorem in the stretch section) builds, and they are invisible to a dimensional check — which is both the strength and the limitation of the method.

The principle of dimensional homogeneity

A physically valid equation must be dimensionally homogeneous: every additive term has the same dimensions. You cannot add a length to a time, any more than you can add metres to seconds. This single rule does almost all the work: it is necessary (though not sufficient) for correctness, so any dimensionally inhomogeneous formula is certainly wrong. It is the mathematician's first line of defence against algebraic blunders: at the end of any derivation, glance at the dimensions of each term, and if they fail to match you know — before checking a single number — that an error has crept in somewhere. Many an examination mark has been saved by a candidate who noticed at the last moment that a "$v$v" should have been a "$v^2$v2" because the dimensions refused to balance.

A corollary: the argument of any transcendental function ($\sin$sin, $\cos$cos, $\ln$ln, $e^x$ex) must be dimensionless. $\sin(\omega t)$sin(ωt) is fine because $[\omega t] = T^{-1}\cdot T = 1$[ωt]=T−1⋅T=1; $\sin(x)$sin(x) for a length $x$x is meaningless.

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3 Worked examples — checking equations

Example 1 — a correct equation

Check that $v^2 = u^2 + 2as$v2=u2+2as is dimensionally consistent.

$$[v^2] = (LT^{-1})^2 = L^2T^{-2}, \quad [u^2] = L^2T^{-2}, \quad [2as] = (LT^{-2})(L) = L^2T^{-2}. \quad (\textbf{M1})$$[v2]=(LT−1)2=L2T−2,[u2]=L2T−2,[2as]=(LT−2)(L)=L2T−2.(M1)

All three terms are $L^2T^{-2}$L2T−2; the equation is homogeneous and so passes the check. (A1) (Note $2$2 is dimensionless and is ignored.)

Example 2 — spotting an error

Is $E = mgh + \tfrac12 mv$E=mgh+21​mv dimensionally correct?

$$[mgh] = M(LT^{-2})(L) = ML^2T^{-2}, \qquad [\tfrac12 mv] = M(LT^{-1}) = MLT^{-1}. \quad (\textbf{M1})$$[mgh]=M(LT−2)(L)=ML2T−2,[21​mv]=M(LT−1)=MLT−1.(M1)

These differ ($ML^2T^{-2}$ML2T−2 vs $MLT^{-1}$MLT−1), so the formula is wrong: the kinetic term should be $\tfrac12 mv^2$21​mv2, giving $ML^2T^{-2}$ML2T−2 to match. (A1)

Example 3 — checking an impulse formula

A student writes the impulse as $J = Ft^2$J=Ft2. Check it.

$$[Ft^2] = (MLT^{-2})(T^2) = ML. \quad (\textbf{M1})$$[Ft2]=(MLT−2)(T2)=ML.(M1)

But impulse $= \Delta(\text{momentum})$=Δ(momentum) has dimensions $MLT^{-1}$MLT−1. Since $ML \ne MLT^{-1}$ML=MLT−1, the formula is wrong; the correct form is $J = Ft$J=Ft (dimensions $MLT^{-1}\,\checkmark$MLT−1✓). (A1)

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4 Deriving a formula by dimensional analysis

If a quantity $Q$Q is believed to depend on variables $x_1, x_2, \ldots$x1​,x2​,…, assume a power-law form $Q = k\,x_1^a x_2^b\cdots$Q=kx1a​x2b​⋯ with $k$k dimensionless, then equate the powers of $M$M, $L$L, $T$T on each side and solve for the exponents.

Example 4 — period of a simple pendulum

The period $T$T is assumed to depend on length $l$l, mass $m$m and gravity $g$g. Find the form.

$$T = k\,l^a m^b g^c, \qquad [l]=L,\ [m]=M,\ [g]=LT^{-2}. \quad (\textbf{M1 assume power law})$$T=klambgc,[l]=L, [m]=M, [g]=LT−2.(M1 assume power law) $$T^1 M^0 L^0 = M^{b}\,L^{a+c}\,T^{-2c}.$$T1M0L0=MbLa+cT−2c.

Equate powers (M1):

$$M:\ b = 0, \qquad T:\ -2c = 1 \Rightarrow c = -\tfrac12, \qquad L:\ a + c = 0 \Rightarrow a = \tfrac12.$$M: b=0,T: −2c=1⇒c=−21​,L: a+c=0⇒a=21​.

Hence $T = k\,l^{1/2}g^{-1/2} = k\sqrt{\dfrac{l}{g}}$T=kl1/2g−1/2=kgl​​. (A1) The mass drops out ($b=0$b=0) — dimensional analysis predicts that the period is independent of mass, before any equation of motion is solved. The constant $k$k is undetermined by this method; theory gives $k = 2\pi$k=2π.

(M1 power-law ansatz; M1 equate $M, L, T$M,L,T powers; A1 $T = k\sqrt{l/g}$T=kl/g​. The key insight — mass cancels — is worth stating explicitly for the AO2 mark.)

Example 5 — drag on a sphere

The drag force $F$F on a sphere is assumed to depend on fluid density $\rho$ρ, speed $v$v and radius $r$r. Find the form.

$$F = k\,\rho^a v^b r^c, \quad [F]=MLT^{-2},\ [\rho]=ML^{-3},\ [v]=LT^{-1},\ [r]=L. \quad (\textbf{M1})$$F=kρavbrc,[F]=MLT−2, [ρ]=ML−3, [v]=LT−1, [r]=L.(M1) $$M^1 L^1 T^{-2} = M^{a}\,L^{-3a+b+c}\,T^{-b}.$$M1L1T−2=MaL−3a+b+cT−b.

Equate (M1): $M:\ a = 1$M: a=1; $T:\ -b = -2 \Rightarrow b = 2$T: −b=−2⇒b=2; $L:\ -3a + b + c = 1 \Rightarrow -3 + 2 + c = 1 \Rightarrow c = 2$L: −3a+b+c=1⇒−3+2+c=1⇒c=2. Therefore

$$F = k\,\rho\, v^2 r^2. \quad (\textbf{A1})$$F=kρv2r2.(A1)

This is exactly the structure of the standard drag law $F = \tfrac12 C_D\,\rho\, v^2(\pi r^2)$F=21​CD​ρv2(πr2), with the dimensionless drag coefficient absorbed into $k = \tfrac12 C_D\pi$k=21​CD​π. Dimensional analysis gets the $\rho v^2 r^2$ρv2r2 dependence for free; only the dimensionless $C_D$CD​ needs experiment.

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5 Dimensionless quantities and the limitations

A dimensionless quantity has dimensions $M^0L^0T^0 = 1$M0L0T0=1: examples are the coefficient of restitution $e$e, the coefficient of friction $\mu$μ, angles in radians, ratios of like quantities, the Reynolds number, and pure numbers like $\pi$π. These are invisible to dimensional analysis and appear only as the undetermined constant $k$k.

The method has clear limitations:

1. It cannot find dimensionless constants ($2\pi$2π, $\tfrac12$21​, $C_D$CD​, …).
2. It cannot distinguish quantities with the same dimensions — energy and torque are both $ML^2T^{-2}$ML2T−2, so it cannot tell $mgh$mgh from $\tfrac12 mv^2$21​mv2 by dimensions alone.
3. It fails when there are more independent variables than fundamental dimensions (here three: $M, L, T$M,L,T) — the exponent equations are then under-determined, and dimensionless groups (à la Buckingham's theorem) appear.
4. It cannot detect additive structure — $a+b$a+b and $a-b$a−b have the same dimensions, so it cannot reveal a sum versus a difference, or a missing additive term.
5. A homogeneous formula can still be physically wrong (right dimensions, wrong coefficient or wrong combination) — homogeneity is necessary, not sufficient.

These limitations are not weaknesses so much as a precise statement of the method's scope. Dimensional analysis is a filter, not an oracle: it rejects the dimensionally impossible and constrains the possible, but it cannot select among dimensionally-equivalent candidates or supply pure numbers. The art lies in choosing the input variables wisely — include too few and you miss a dependence; include too many (especially several with related dimensions) and the exponents become under-determined. A physicist's judgement about which quantities a result can sensibly depend on is therefore an essential, irreducible ingredient. When the pendulum derivation returned $b = 0$b=0 for the mass exponent, that was the method correctly telling us mass is irrelevant — but only because we had the wisdom to offer $l$l, $m$m and $g$g as candidates and let the algebra decide. Offer the wrong list and the method faithfully gives the wrong answer; the discipline is in the modelling, and the algebra merely enforces consistency.

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6 Specimen-style exam question

(specimen-style — not from any past paper)

The speed $v$v of a transverse wave on a stretched string is believed to depend on the tension $T_{\!s}$Ts​ in the string (a force) and on the mass per unit length $\mu$μ of the string. (a) Write down the dimensions of $v$v, $T_{\!s}$Ts​ and $\mu$μ. (b) Use dimensional analysis to find how $v$v depends on $T_{\!s}$Ts​ and $\mu$μ. (c) Explain why the constant of proportionality cannot be found this way, and state one assumption the method requires.

Model solution. (a) $[v] = LT^{-1}$[v]=LT−1; $[T_{\!s}] = MLT^{-2}$[Ts​]=MLT−2 (a force); $[\mu] = ML^{-1}$[μ]=ML−1 (mass per length).

(b) Assume $v = k\,T_{\!s}^{\,a}\mu^{b}$v=kTsa​μb. Then

$$LT^{-1} = (MLT^{-2})^a (ML^{-1})^b = M^{a+b}\,L^{a-b}\,T^{-2a}.$$LT−1=(MLT−2)a(ML−1)b=Ma+bLa−bT−2a.

Equate powers: $M:\ a + b = 0$M: a+b=0; $T:\ -2a = -1 \Rightarrow a = \tfrac12$T: −2a=−1⇒a=21​; $L:\ a - b = 1$L: a−b=1. From $a = \tfrac12$a=21​ and $a + b = 0$a+b=0, $b = -\tfrac12$b=−21​ (consistent with $a - b = 1\,\checkmark$a−b=1✓). So

$$v = k\,T_{\!s}^{1/2}\mu^{-1/2} = k\sqrt{\frac{T_{\!s}}{\mu}}.$$v=kTs1/2​μ−1/2=kμTs​​​.