Using Dimensional Analysis — T.1 Tools of Physics — IB Physics Notes

6 Lesson 6 of 9 · T.1 Tools of Physics

Using Dimensional Analysis

Every term in a valid physics equation must have the same dimensions; checking this catches wrong equations instantly, finds the units of constants, and can even predict how one quantity depends on others — though it can never supply dimensionless numbers like ½ or 2π.

🎯 Hook – measuring a nuclear bomb from a photograph

In 1950, physicist G. I. Taylor estimated the explosive yield of the first atomic bomb test using nothing but declassified photographs of the fireball at known times — no classified data at all. He reasoned that the fireball radius R could only depend on the energy released E, the air density ρ, and the time t, and dimensional analysis alone forced the relationship $R \propto (Et^2/\rho)^{1/5}$. Checking which combination of quantities has the right dimensions can reveal a physical law before a single experiment is run.

💡 Diagnostic question — a student writes $v=u+at^2$. Without checking the numbers, how can you tell it's wrong? Dimensionally, $[v]=\text{LT}^{-1}$ but $[at^2]=\text{LT}^{-2}\times\text{T}^2=\text{L}$ — you cannot add a velocity to a length, so the equation must be wrong (it should be $v=u+at$).

📏 Dimensions and the principle of homogeneity

The dimension of a quantity shows how it depends on the base quantities, in square brackets: mass [M], length [L], time [T], current [I]. Working in SI base units is equivalent.

Quantity	Defining equation	Dimensions	SI base units
Velocity	v = s/t	LT⁻¹	m s⁻¹
Force	F = ma	MLT⁻²	kg m s⁻²
Energy	E = Fs	ML²T⁻²	kg m² s⁻²
Pressure	p = F/A	ML⁻¹T⁻²	kg m⁻¹ s⁻²
Frequency, angular speed	f = 1/T	T⁻¹	s⁻¹

Principle of homogeneity$$[\text{every term in a correct equation}] \text{ must match}$$

So s = ut + ½at²: [L] = [LT⁻¹][T] + [LT⁻²][T²] = [L] + [L] ✔ — dimensionally consistent. Pure numbers and angles are dimensionless, and the argument of sin, cos, exp and log must always be dimensionless: in $x=A\sin(\omega t)$, $\omega t$ is dimensionless, so ω has dimensions T⁻¹; in $N=N_0e^{-\lambda t}$, λ has dimensions T⁻¹.

🧮 Units of constants and deriving a relationship

Units of constants: rearrange the equation to make the constant the subject, then substitute dimensions. Example: $G=Fr^2/(m_1m_2) \to \text{M}^{-1}\text{L}^3\text{T}^{-2}$.

Method of dimensions — to derive the form of a relationship:

Guess which quantities the result depends on.
Write $y=kx_1^ax_2^bx_3^c$ (k dimensionless).
Write the dimensions of both sides.
Equate powers of M, L, T; solve for a, b, c.

Fig 1 — a dimension-balance table: equate powers of M, L, T column by column.

🚧 Limitations

Cannot find dimensionless constants — the 2π in $T=2\pi\sqrt{l/g}$ never appears.
Cannot distinguish quantities with the same dimensions (work and torque — both ML²T⁻²).
Fails for sums of different terms or exponentials/logs of dimensional quantities.
Needs a correct guess of the relevant variables; too many unknowns with too few base dimensions makes the solution non-unique.
A dimensionally correct equation may still be wrong; a dimensionally incorrect one is definitely wrong.

📊 Graph mastery — testing a dimensionally derived power law

Graph type	Axes (X, Y)	Shape	What to read
ln v vs ln T (testing v = k√(T/μ))	ln T · ln v	Straight line	Gradient = predicted power (½ here); intercept gives k
v vs √T, v² vs T	√T or T · v or v²	Straight lines through the origin	Gradient related to k and μ

Fig 2 — ln v vs ln T: a straight line of gradient ½ confirms v ∝ √T.

✍️ Worked example 1 — the pendulum and a wave on a string

Pendulum: suppose $T=kl^ag^bm^c$. $[T]=[L]^a[LT^{-2}]^b[M]^c$: M gives c=0; T gives $-2b=1\to b=-\tfrac12$; L gives $a+b=0\to a=\tfrac12$.

Answer: $T=k\sqrt{l/g}$, independent of mass (experiment gives $k=2\pi$, which dimensional analysis cannot supply).

Wave on a string: $v=kT_{\text{tension}}^a\mu^b$: M gives $a+b=0$; T gives $-2a=-1\to a=\tfrac12,\ b=-\tfrac12$; L checks: $a-b=1$✔.

Answer: $v=k\sqrt{T_{\text{tension}}/\mu}$ (k = 1 here).

✍️ Worked example 2 — identifying a quantity from its units

Identify: $x=\sqrt{\gamma p/\rho}$, γ dimensionless.

Data: $p/\rho = (\text{kg m}^{-1}\text{s}^{-2})/(\text{kg m}^{-3}) = \text{m}^2\text{s}^{-2}$.

Answer: $x$ has units $\mathbf{m\,s^{-1}}$ — a speed. This is the speed of sound in a gas.

🌍 Real-world: the Planck length

Dimensional analysis combines the fundamental constants $\hbar$, G and c into the only quantity with dimensions of length: the Planck length, $\sqrt{\hbar G/c^3}\approx1.6\times10^{-35}$ m — the scale at which quantum effects and gravity are expected to become comparably important.

✓ Examiner tip: build a small dimension table (M, L, T columns) for every quantity in the question before writing any equation — it prevents sign and power slips.

🚀 Going further — Buckingham π theorem

With n variables and k independent dimensions, a physical law can always be written in terms of $n-k$ dimensionless groups. Engineers use this to scale wind-tunnel models using the Reynolds number, a dimensionless combination of speed, size, density and viscosity.

Challenge: drag force F depends on ρ, v and A (3 variables, M L T = 3 dimensions) — how many independent dimensionless groups exist? $n-k=3-3=0$: in this simple case F/(ρv²A) must itself be the single dimensionless constant k in F = kρv²A.

📋 Quick practice — answers in italics

(a) Write the dimensions of impulse. Impulse = Δp, so MLT⁻¹.

(b) Check $v=\sqrt{2gh}$ for homogeneity. [LT⁻¹]² should equal [LT⁻²][L] = [L²T⁻²] ✔ — consistent.

(c) Find the SI base units of the Planck constant h from $E=hf$. $h=E/f\to(\text{kg m}^2\text{s}^{-2})/\text{s}^{-1}=\text{kg m}^2\text{s}^{-1}$.

(d) In $x=A\cos(\omega t+\varphi)$, state the dimensions of ω and φ. Both must be dimensionless-when-multiplied-by-their-partner: ω has T⁻¹, φ is dimensionless (an angle).

⚠️ Trap corner – common mistakes

Misconception	Correct view
Dimensions and units are different checks	[MLT⁻²] and kg m s⁻² carry the same information — use whichever notation the question asks for
Numbers and angles carry dimensions	Pure numbers, ratios and angles are always dimensionless — they drop out of a dimensional check entirely
Dimensionally correct means correct	A dimensionally correct equation can still have the wrong constant or even the wrong physics (E = mc² and E = 2mc² are both dimensionally fine)
Terms of different dimensions can be added if the numbers work out	Never add or equate terms of different dimensions — it is meaningless, whatever the numbers say (e.g. ½mv + mgh is invalid)
"Per" doesn't change the sign of a power	"Per" in words means a negative power: kg per m³ = kg m⁻³
rad s⁻¹ has dimension T⁻² like acceleration	The radian is dimensionless, so rad s⁻¹ has dimensions T⁻¹, same as ordinary frequency

⚠️ Most common slip: concluding an equation is "proved correct" because it is dimensionally consistent — homogeneity is a necessary check, never a sufficient one.

[F] = MLT⁻² [E] = ML²T⁻² [p] = ML⁻¹T⁻² sin/cos/exp argument: dimensionless homogeneous ⇏ correct can't find dimensionless constants y = kx₁ᵃx₂ᵇ → match powers of M,L,T

Practice check — multiple choice

1. The dimensions of force are:

A MLT⁻¹

B ML²T⁻²

C MLT⁻²

D ML⁻¹T⁻²

2. Which equation is dimensionally incorrect?

A v = u + at²

B s = ut + ½at²

C v² = u² + 2as

D s = ½(u+v)t

3. Dimensional analysis cannot determine:

A whether an equation is homogeneous

B the units of a constant

C the powers in a product relationship

D the value of a dimensionless constant such as 2π

4. The dimensions of pressure are:

A MLT⁻²

B ML⁻¹T⁻²

C ML⁻²T⁻²

D ML²T⁻²

5. In y = A sin(ωt), the dimensions of ω are:

A T

B L

C T⁻¹

D dimensionless

6. The quantity √(γp/ρ), with γ dimensionless, has the dimensions of:

A acceleration

B energy

C pressure

D speed

7. Which pair of quantities have the same dimensions?

A Work and torque

B Force and momentum

C Power and energy

D Pressure and force

8. If the period of a pendulum is T ∝ lᵃgᵇ, then:

A a = 1, b = −1

B a = ½, b = −½

C a = −½, b = ½

D a = 2, b = −2

9. An equation is found to be dimensionally consistent. This shows that it:

A is definitely correct

B is definitely wrong

C may still be wrong

D contains no constants

10. The dimensions of the gravitational constant G are:

A M⁻¹L³T⁻²

B ML³T⁻²

C M⁻²L²T⁻²

D M⁻¹L²T⁻²

Practice check — fill in the blanks

Complete the statements:

Quantities can only be added or equated if they have the same . The argument of sin, cos, exp and log must always be . A dimensionally correct equation still be physically wrong. In y = A sin(ωt), the angular frequency ω has dimensions .

Exam‑style questions

Exam question · Determine / Suggest

Q1. The density ρ and pressure p of a gas are related to a quantity x by $x=\sqrt{\gamma p/\rho}$, where γ is dimensionless. (a) Determine the unit of x in SI base units. (b) Suggest the physical quantity x represents. [4 marks]

✓ (a) $p/\rho=(\text{kg m}^{-1}\text{s}^{-2})/(\text{kg m}^{-3})=\text{m}^2\text{s}^{-2}$, so $x$ has units $\mathbf{m\,s^{-1}}$.
✓ (b) A speed — specifically the speed of sound in the gas.

Exam question · Show that / State / Suggest

Q2. The drag force F on a sphere is assumed to depend on fluid density ρ, speed v and cross-sectional area A. (a) Use dimensional analysis to show that $F=k\rho v^2A$. (b) State one limitation of this derivation. (c) Suggest why this form fails at very low speeds. [6 marks]

✓ (a) $F=k\rho^av^bA^c$: $\text{MLT}^{-2}=(\text{ML}^{-3})^a(\text{LT}^{-1})^b(\text{L}^2)^c$. M: a=1. T: $-b=-2\to b=2$. L: $-3a+b+2c=1\to c=1$. So $F=k\rho v^2A$.
✓ (b) The dimensionless constant k (≈½C_d) cannot be found by this method.
✓ (c) At very low speeds, viscous (Stokes) drag dominates, which is proportional to v, not v² — a different physical regime the derivation didn't assume.

Exam question · Show that / Suggest / Explain

Q3. A student writes the rotational kinetic energy of a body as $E=\tfrac12I\omega$, with I in kg m² and ω in rad s⁻¹. (a) Show that this is not homogeneous. (b) Suggest the correct power of ω and justify it. (c) Explain why a homogeneous equation is not necessarily correct. [5 marks]

✓ (a) $[I]=\text{ML}^2$, $[\omega]=\text{T}^{-1}$ (rad is dimensionless); $[\tfrac12I\omega]=\text{ML}^2\text{T}^{-1}$, but energy needs $\text{ML}^2\text{T}^{-2}$ — the powers of T don't match, so it is not homogeneous.
✓ (b) $E=\tfrac12I\omega^2$ gives $\text{ML}^2\text{T}^{-2}$✔ — consistent with the translational analogy $\tfrac12mv^2$, with I replacing m and ω replacing v, both squared.
✓ (c) Homogeneity only checks that dimensions balance; it cannot catch a wrong dimensionless constant or a wrong combination of otherwise correctly-dimensioned terms.

Exam question · Paper 1B practical skills — testing a dimensionally derived law

Q4. Dimensional analysis predicts $v=k\sqrt{T/\mu}$ for waves on a string with $\mu=5.0\times10^{-3}\ \text{kg m}^{-1}$:

T/N	v/m s⁻¹
1.00	14.3
2.00	19.8
4.00	28.0
6.00	35.0
8.00	40.3

(a) Explain why ln v vs ln T should be a straight line, and state its predicted gradient. (b) Calculate ln v and ln T. (c) Determine the gradient and comment on the prediction. (d) Use the intercept to determine k. (e) Suggest a different graph giving a straight line through the origin, and what its gradient represents. [10 marks]

✓ (a) Taking logs: $\ln v = \ln(k\mu^{-1/2})+\tfrac12\ln T$ — linear in ln T with predicted gradient ½.
✓ (b) ln T: 0, 0.693, 1.386, 1.792, 2.079. ln v: 2.660, 2.986, 3.332, 3.555, 3.696.
✓ (c) Gradient $\approx(3.696-2.660)/(2.079-0)\approx\mathbf{0.50}$ — matches the predicted ½ well.
✓ (d) Intercept at ln T = 0 gives ln v = 2.660, so $k\mu^{-1/2}=e^{2.660}=14.3\Rightarrow k=14.3\sqrt{0.005}\approx\mathbf{1.0}$.
✓ (e) $v^2$ vs T gives a straight line through the origin with gradient $k^2/\mu$.

Exam question · Explain

Q5. Explain why the principle of homogeneity must hold for any correct physics equation. [2 marks]

✓ Adding or equating two quantities of different dimensions (e.g. a length and a time) is physically meaningless — there is no way to compare them.
✓ A true physical law describes a real relationship between quantities, so every term that is added or set equal must represent the same kind of physical quantity, hence the same dimensions.

Exam question · List / Explain

Q6. List the limitations of dimensional analysis and give an example of each. [4 marks]

✓ Cannot find dimensionless constants — the 2π in $T=2\pi\sqrt{l/g}$ is invisible to the method.
✓ Cannot distinguish same-dimension quantities — work and torque are both ML²T⁻² but are physically different.
✓ Fails for sums of different terms or for exponentials/logs of dimensional quantities — it cannot analyse $E_k+E_p$ term by term in the same way.
✓ Needs a correct guess of every relevant variable — an omitted or extra variable gives a wrong or non-unique answer.

Exam question · Explain

Q7. Explain how G. I. Taylor could estimate the energy of an explosion from photographs using dimensional analysis. What had to be assumed? [3 marks]

✓ Assuming the fireball radius R depends only on the energy E released, the air density ρ, and the time t since detonation, dimensional analysis forces $R\propto(Et^2/\rho)^{1/5}$.
✓ Measuring R and t from photographs at known times, and knowing ρ for air, allowed E to be estimated without any classified information.
✓ This required assuming no other variable (ambient pressure, radiation losses, geometry) significantly affected R — a simplification that happened to work well.

Exam question · Explain

Q8. Explain why the argument of a sine or exponential function must be dimensionless. [2 marks]

✓ Functions like sin and e^x are defined by power series that add together different powers of their argument (x, x², x³, …).
✓ Adding different powers of a dimensional quantity together is not dimensionally valid (x and x² would have different dimensions), so the argument must be a pure, dimensionless number for the series — and the function — to make sense.

Exam question · Discuss

Q9. Work and torque have the same dimensions but are different quantities. Discuss what this shows about the relationship between dimensions and physical meaning. [2 marks]

✓ Work (a scalar, energy transferred) and torque (a vector, a turning effect) are physically distinct quantities that happen to reduce to the same combination of base dimensions, ML²T⁻².
✓ This shows dimensions describe only how a quantity scales with mass, length and time — not its full physical meaning; two quantities can share dimensions without being the same thing, so dimensional analysis alone can never fully identify a quantity.

Score: 0 pts

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