What are dimensions?
Every physical quantity can be written using the 7 base quantities. The dimensions of a quantity are the powers to which the base quantities are raised to show it. We write them in square brackets.
- Length → [L], mass → [M], time → [T], current → [A], temperature → [K]
- Area = length × length → [L²]
- Speed = length ÷ time → [L T⁻¹]
- Force = mass × acceleration → [M L T⁻²]. We say force has dimension 1 in mass, 1 in length and −2 in time.
The expression [M L T⁻²] is called the dimensional formula of force, and "[F] = [M L T⁻²]" is its dimensional equation.
Dimensions do not care about the unit or about number factors. Speed in km/h or m/s, both are [L T⁻¹].
Dimensional formulas of common quantities
| Quantity | Formula | Dimensions |
|---|---|---|
| Velocity | distance/time | [M⁰ L T⁻¹] |
| Acceleration | velocity/time | [M⁰ L T⁻²] |
| Force | m a | [M L T⁻²] |
| Work, energy | F × s | [M L² T⁻²] |
| Power | work/time | [M L² T⁻³] |
| Momentum | m v | [M L T⁻¹] |
| Pressure, stress | F/area | [M L⁻¹ T⁻²] |
| Density | m/V | [M L⁻³] |
| Frequency | 1/time | [T⁻¹] |
| Gravitational constant G | F r²/(m₁m₂) | [M⁻¹ L³ T⁻²] |
| Planck's constant h | E/ν | [M L² T⁻¹] |
| Angle, strain, refractive index | ratio | [M⁰ L⁰ T⁰] (dimensionless) |
Some things have a unit but no dimension (angle in radian). Some constants have dimensions (G, h), called dimensional constants.
Use 1: Checking an equation (principle of homogeneity)
Principle of homogeneity: you can add, subtract or compare only quantities with the same dimensions. So in a correct equation, every term has the same dimensions.
Check s = ut + ½at²: [L] = [L T⁻¹][T] + [L T⁻²][T²] = [L] + [L]. All terms are [L], so it is dimensionally correct.
If the dimensions do not match, the equation is surely wrong. If they match, it may still be wrong (the number ½ could be missing), so dimensional checking is a test that can only catch errors.
Arguments of sin, cos, log and e^x must be dimensionless. So in y = A sin(ωt), ωt has no dimension, and ω must be [T⁻¹].
Use 2: Converting units between systems
A quantity is the same, whatever the unit: n₁u₁ = n₂u₂. If the quantity has dimensions [Mᵃ Lᵇ Tᶜ], then
n₂ = n₁ (M₁/M₂)ᵃ (L₁/L₂)ᵇ (T₁/T₂)ᶜ
Example: 1 J into erg. Energy [M L² T⁻²], a = 1, b = 2, c = −2. n₂ = 1 × (1 kg/1 g)¹ × (1 m/1 cm)² × (1 s/1 s)⁻² = 10³ × (10²)² = 10⁷. So 1 J = 10⁷ erg.
Use 3: Finding how quantities depend on each other
If we know which quantities matter, we can guess the formula. Pendulum time T may depend on length l, mass m and g. Write T = k lᵃ mᵇ gᶜ.
[T] = [L]ᵃ [M]ᵇ [L T⁻²]ᶜ = [Mᵇ Lᵃ⁺ᶜ T⁻²ᶜ]
Match powers: M: b = 0. T: −2c = 1 → c = −½. L: a + c = 0 → a = ½.
So T = k √(l/g). Experiments give k = 2π. Mass does not matter at all.
Try it
Hang a key on a 25 cm thread and a 100 cm thread. Count 10 swings for each. The long one should take about twice the time, just as √4 = 2 says.
Limits of dimensional analysis
- It cannot find number constants like 2π or ½.
- It cannot tell whether a quantity is a vector or a scalar.
- It fails if a quantity depends on more than 3 other quantities (with only M, L, T we get just 3 equations).
- It cannot derive formulas with sums (like v = u + at) or with sin, log, e^x.
- Two different quantities can share dimensions (work and torque are both [M L² T⁻²]), so matching dimensions do not prove a formula.
Key formulas and definitions
- [Force] = [M L T⁻²]; [Energy] = [M L² T⁻²]; [Power] = [M L² T⁻³]
- [Pressure] = [M L⁻¹ T⁻²]; [Momentum] = [M L T⁻¹]
- [G] = [M⁻¹ L³ T⁻²]; [h] = [M L² T⁻¹]
- Conversion: n₂ = n₁ (M₁/M₂)ᵃ (L₁/L₂)ᵇ (T₁/T₂)ᶜ
- 1 N = 10⁵ dyne; 1 J = 10⁷ erg
- Pendulum: T = 2π√(l/g)
Worked examples
1. Find the dimensional formula of work.
Work = force × distance = [M L T⁻²] × [L] = [M L² T⁻²].
2. Find the dimensions of pressure.
Pressure = force ÷ area = [M L T⁻²] ÷ [L²] = [M L⁻¹ T⁻²].
3. Check whether ½mv² = mgh is dimensionally correct.
Left: [M][L T⁻¹]² = [M L² T⁻²]. Right: [M][L T⁻²][L] = [M L² T⁻²]. Same, so it is dimensionally correct (the ½ cannot be checked).
4. Find the dimensions of G in F = G m₁m₂/r².
G = F r²/(m₁m₂) = [M L T⁻²][L²] ÷ [M²] = [M⁻¹ L³ T⁻²].
5. Convert 1 newton into dyne.
Force [M L T⁻²]. n₂ = 1 × (1 kg/1 g)(1 m/1 cm)(1 s/1 s)⁻² = 10³ × 10² × 1 = 10⁵. So 1 N = 10⁵ dyne.
6. The CGS value of g is 980 cm s⁻². Convert it to SI using dimensions.
[L T⁻²]: n₂ = 980 × (1 cm/1 m)¹ × (1 s/1 s)⁻² = 980 × 10⁻² = 9.8 m s⁻².
7. A body moving in a circle needs a force F that depends on mass m, speed v and radius r. Find the formula.
F = k mᵃ vᵇ rᶜ. [M L T⁻²] = [M]ᵃ [L T⁻¹]ᵇ [L]ᶜ. M: a = 1. T: −b = −2 → b = 2. L: b + c = 1 → c = −1. So F = k mv²/r (k = 1).
8. In y = A sin(kx − ωt), find the dimensions of k and ω.
The angle kx − ωt must be dimensionless. kx: k × [L] = 1 → [k] = [L⁻¹]. ωt: ω × [T] = 1 → [ω] = [T⁻¹].
Common mistakes
- Thinking a dimensionally correct equation must be right. The number factor may still be wrong.
- Giving a dimension to angle or strain. They are ratios, so dimensionless.
- Writing the dimensions of speed as [L T]. Dividing by time gives T⁻¹.
- Trying to derive v = u + at by dimensions. It has a sum, so the method fails.