University Physics I · Mathematical & Physical Foundations · 1.2
Dimensional Analysis & Scaling
Track the powers of mass, length and time. They test every equation you write, hand you the form of simple laws, and tell you what changes when an object changes size.
Build the model
Connect the measurement to the mechanism.
A physical quantity carries more than a number: it carries a dimensional signature, a set of powers of mass, length and time that survives every unit conversion. Three things follow. Every additive term in a correct equation must share one signature, which makes dimensions a free error check.
If you can name the quantities a result depends on, matching signatures often fixes the exponents and leaves only a dimensionless constant unknown — a functional form obtained from bookkeeping alone. And when an object keeps its shape but changes size, areas go as L², volumes as L³, and everything built from them scales by a predictable power. The same habit powers order-of-magnitude estimates: get the powers of ten right and a factor of two rarely matters.
- Simple definition
- Dimensional analysis tracks the powers of mass, length and time inside a quantity, so equations can be tested and simple relationships inferred before a single number is substituted.
- Example
- [v] = L T⁻¹, so v = √(2gh) passes: √(L T⁻² × L) = L T⁻¹. A candidate whose two sides disagree is wrong no matter how neat the algebra looks.
[v] = L T⁻¹ · [F] = M L T⁻² · [E] = M L² T⁻²
base set M, L, T; add Θ, I, N, J when needed
A necessary test: fail it and the equation is already wrong.
sin, exp and ln take dimensionless arguments only
Each base dimension gives one equation for the exponents.
C is dimensionless and the method cannot supply it
Mass drops out: matching M forces its exponent to zero.
L = 1.00 m, g = 9.81 m s⁻² → T = 2.01 s
Double every length: area ×4, volume and weight ×8.
similar shapes only, where one length sets them all
One group fixes the law; two or more leave a free function.
n quantities built from k independent base dimensions
Dimensions outlast units
A unit is a choice; a dimension is not. Report a speed in m s⁻¹, km h⁻¹ or miles per hour and the dimensional signature is the same: [v] = L T⁻¹. Write [Q] for the signature of Q and build it from the base dimensions M, L and T; thermodynamics adds Θ and electromagnetism adds I. Force is M L T⁻² whatever units it is printed in, and energy is M L² T⁻². Pure numbers — angles in radians, refractive index, efficiency, the ratio of two lengths — have signature 1 and are called dimensionless. The signature is the coarser test — 5 m + 3 cm fails a unit check and passes a dimensional one — but it costs one line and survives every conversion later.
Every term in an equation carries the same dimensions
You may add 5 m to 3 m. You may not add 5 m to 3 s. So in a correct physical equation, every term joined by + or − and both sides of the = share one signature. Test v² = u² + 2as: the terms are L² T⁻², L² T⁻², and (L T⁻²)(L) = L² T⁻². Consistent. Test the misremembered v² = u² + 2as²: that last term is L³ T⁻² while the others are L² T⁻², so the equation is dead before any algebra. The rule extends to functions. The series eˣ = 1 + x + x²/2 + … adds different powers of x, and those terms share one signature only if x is dimensionless; sin and ln demand the same. That is why decay is written e(−t/τ) with τ a time, never e(−t).
Assume a power law and let dimensions fix the exponents
Suppose the period T of a small-amplitude pendulum depends only on the bob mass m, the string length L and g. Write T = C mᵃ Lᵇ gᶜ with C dimensionless. In signatures: [T] = T = Mᵃ Lᵇ (L T⁻²)ᶜ = Mᵃ L(b+c) T(−2c). Matching each base dimension gives three equations: a = 0 from M, −2c = 1 from T, and b + c = 0 from L. So a = 0, c = −½, b = ½, and T = C √(L/g). The bookkeeping has produced the whole functional form and predicted that mass does not matter — a claim you can test on a bench. What it cannot give is C. Solving the small-angle equation of motion returns C = 2π, so with L = 1.00 m and g = 9.81 m s⁻² the period is 2π√(1.00/9.81) = 2.01 s.
Know where the method runs out
Counting explains both the power and the limit. With n quantities built from k independent base dimensions, you can form n − k independent dimensionless groups. The pendulum used n = 4 (period, m, L, g) and k = 3, giving exactly one group, T√(g/L), which must therefore equal a constant — that is why the answer came out complete. Allow a finite amplitude θ₀ and you gain a second group, so the best dimensions can say is T√(g/L) = f(θ₀), with f an unknown function. Quantities sharing a signature are also invisible to the method: torque and energy are both M L² T⁻², and no dimensional check will tell you which one belongs in your equation.
Change the size and the physics changes with it
For a family of geometrically similar objects, one length L sets everything: areas go as L², volumes and masses as L³, and the surface-to-volume ratio as L⁻¹ — for a sphere it is exactly 3/r. A bone's strength is set by its cross-section, so it scales as L², while the weight it carries scales as L³; the compressive stress therefore grows as L, and an animal scaled up by two is twice as stressed. Falling bodies obey the same counting. Quadratic drag gives terminal speed vₜ = √(2mg / (ρ Cd A)), for fluid density ρ, drag coefficient Cd and frontal area A. With m ∝ L³ and A ∝ L², vₜ ∝ √L: a hailstone twice as wide falls about 1.41 times as fast.
An estimate is a scaling argument with one anchor number
An order-of-magnitude estimate asks for the power of ten, not the digits. Break the quantity into factors you can each guess within a factor of a few, keep the units attached, and let the errors partly cancel. The air in a 12 m × 8 m × 4 m lecture room occupies 384 m³, and air near sea level has density about 1.2 kg m⁻³, so the room holds roughly 4.6 × 10² kg of air — close to half a tonne. Finish an estimate the way you finish a derivation: check the dimensions of the result, check a limiting case, and ask whether the number is physically plausible. A sound estimate wrong by 30% is worth more than an exact formula applied to the wrong quantity.
Change one variable at a time
Make the relationship visible.
Start from the naive first-powers guess T = C m L and drive all three bars to zero — c settles the time bar, then b settles the length bar. The mass bar only vanishes at a = 0, which is dimensions insisting that the bob's mass cannot enter the period.
M RESIDUAL a1.0
L RESIDUAL b + c1.0
T RESIDUAL −2c − 1-1.0
TOTAL MISMATCH3.0
Live interpretationM RESIDUAL a: 1.0. L RESIDUAL b + c: 1.0. T RESIDUAL −2c − 1: −1.0. TOTAL MISMATCH: 3.0
Catch the common trap
Explain before calculating.
A block of mass m hangs from a spring of stiffness k, where k is measured in N m⁻¹. Which combination has the dimension of time?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA student writes the distance a body falls from rest as s = ½ g t³. Test it with dimensions, then repair the exponent.
- Left side: [s] = L.
- Right side: ½ is dimensionless, so [g t³] = (L T⁻²)(T³) = L T. A length on the left, a length × time on the right — the equation is dead before any algebra.
- Keep g and ask which power of t works: [g tⁿ] = (L T⁻²)(Tⁿ) = L T(n−2), which is L only when n = 2.
- So the form must be s = C g t². Dimensions stop there; integrating a = g twice from rest supplies C = ½, giving s = ½ g t².
Answers = ½ g t³ is dimensionally wrong — L T against L. The correct form is s = ½ g t².
MediumThe speed v of a transverse wave on a stretched string depends only on the tension F and the mass per unit length μ. Find the form of v, then evaluate it for F = 60 N and μ = 4.0 × 10⁻³ kg m⁻¹, given that the dimensionless constant is 1.
- Signatures: [v] = L T⁻¹, [F] = M L T⁻², [μ] = M L⁻¹.
- Ansatz v = C Fᵃ μᵇ gives L T⁻¹ = (M L T⁻²)ᵃ (M L⁻¹)ᵇ = M(a+b) L(a−b) T(−2a).
- Match T: −2a = −1, so a = ½. Match M: a + b = 0, so b = −½. Check L: a − b = ½ + ½ = 1 ✓.
- So v = C √(F/μ). With C = 1: v = √(60 N ÷ 4.0 × 10⁻³ kg m⁻¹) = √(1.5 × 10⁴ m² s⁻²) = 1.2 × 10² m s⁻¹.
Answerv = C √(F/μ); with C = 1, v = 1.2 × 10² m s⁻¹
HardA solid bronze statue 1.8 m tall has mass 480 kg. A geometrically similar maquette in the same alloy stands 0.30 m tall. Find its mass, and compare its surface-to-volume ratio and its ankle stress with the full statue's.
- Geometric similarity means one length ratio sets them all: k = 0.30 m ÷ 1.8 m = 1/6.
- At fixed density, mass follows volume, which goes as L³: m = 480 kg × (1/6)³ = 480/216 = 2.2 kg.
- Area goes as L², so A/V goes as L⁻¹: the maquette's surface-to-volume ratio is (1/6)⁻¹ = 6 times the statue's, which is why the small casting cools far faster.
- Compressive stress is weight over cross-section, L³/L² = L. The statue's ankles carry 6 times the maquette's stress — scaling a shape up is never free.
Answerm ≈ 2.2 kg; the maquette's A/V is 6× the statue's, and its ankle stress is 1/6 of the statue's.